### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# CALCULUS II MATH 1020

RPI

GPA 3.66

### View Full Document

## 9

## 0

## Popular in Course

## Popular in Mathematics (M)

This 295 page Class Notes was uploaded by Jacinthe Brakus on Monday October 19, 2015. The Class Notes belongs to MATH 1020 at Rensselaer Polytechnic Institute taught by Harry McLaughlin in Fall. Since its upload, it has received 9 views. For similar materials see /class/224791/math-1020-rensselaer-polytechnic-institute in Mathematics (M) at Rensselaer Polytechnic Institute.

## Reviews for CALCULUS II

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/19/15

CALCULUS 11 Spring 2011 LECTURE 6 Harry McLaughlin revised 2511 edited by This lecture prouides o a model for areas of surfaces of reuolution o a formula for computing areas of surfaces of reuolution b A 2w y1 yltzgtgt2dz I o examples of surface area calculations yz 1 m 2 yzz7 nggh rgt0 cone m 7r S m g r sphere Mm 7 1g m lt oo Gabriel s Horn 0 if z 0 y1if0ltz 1 Recall We recall from Lecture 5 o a de nition of arc length of a curve modeled by Mm a 3 z 3 b o a theorem asserting that the integral f 1 302 dz can be used for arc length evaluation of smooth real valued functions 0 examples of arc length calculations End of Recall Introduction In a previous lecture we were able to de ne the arc length of a planar curve modeled by Mm a g x g b We approximated the curve by a piecewise linear function computed the length of the piecewise linear function and then used the least upper bound of such lengths as the length of the curve Our success there hinged on the fact that we had a standard namely the length of a planar chord For a surface de ned by a real valued function of two real variables 2 fm there is an analogous process It is possible to triangulate the sur face sum the areas of the triangular facets and then nd the sums least upper bound over all triangulations However that study transcends the mathematics of this treatise This leaves us in the awkward position of wanting to compute surface area without having an underlying de nition of surface area Our work around is to de ne a notion of surface area for a special class of surfaces For surfaces in this class our de nition of surface area is based on a limiting process that in turn leads to an integral We remain mindful of the fact that our surface area calculations are based on our special de nition 7 that a surface in this class may also fall in another class for which another de nition of area is provided In such an event we have a potential for con ict Surface Area In order to get started we mentally draw the curve 1 g x g 2 on an m y plane1 Having done that we imagine the 3 d surface obtained 1The words Curve yc 1 g x g 2 on and Lyplane77 mean the point set modeled by ay R21y571SI 2 by rotating this curve about the z axis We call such a surface a surface of revolution We seek to model our intuitive notion of the area of this surface2 We continue to think behind the scenes in terms of 1 g x g 2 but consider a general non negative real valued continuous function ym a g x g 1 One partitions the interval 11 by a mo lt 1 lt lt MEI lt zn b For each i one focuses on the arc of this curve between the two points i71yi1 and mi y and generates the ith surface by rotating this arc around the z axis We let yi 0 g i g The whole surface is then the composite of the ith surfaces This brings us to the crucial question in our modeling process How can one model the area of the ith surface As a rst approximation one can think of the ith surface as a horizontal cylinder having two circular bases each of radius y or 344 and having height xi 7 MA The lateral surface area excluding the two bases of the Cylinder lS 7 miil or 2717 1 7 i71 To model the complete surface area one could sum over i the areas of the cylinders and then let the size of the partition go to zero This results in an integral of the form f 27rym dm Unfortunately this does not yield gen erally accepted values for areas of classically studied surfaces for example cones and spheres We suspect that the problem lies with the fact that the height ofthe ith cylin der should be something closer to the length ofthe chord between MA 351 and mi Using this intuition we are able to make the following de nition De nition Let denote a continuous real valued non negative function de ned on the interval 11 Let xi0 be an arbitrary partition of 11 and let Ax z 7 MA i 1 71 Let S denote the lateral 3 d surface obtained by rotating about the m axis the function ym a 3 x g b The 2One needs a notion of surface area in order to determine the amount of paint needed to paint a surface One needs a notion of surface area in order to determine the heat radiation characteristics of a surface etc area of the surface S is de ned to be 71 hm Z 2W9ltMgtV i 961202 yi712 i1 provided the limit exists as either a real number or as 00 It is understood that the limit is taken over all partitions a mo lt ml lt lt mn b of 11 as max Ami goes to zero This is a good mathematical de nition of surface area but as in the case of arc length it is not amenable to direct calculation So assuming that y m exists on a b and mimicking the analysis of the arc length de nition one uses the mean value theorem to write 7 I 2 i i712yi yiil2 1lt7yl ylilgtim71 mi 7 miil V1 iiED i 96121 Where 141 lt lt 13 1S i S 77 In this case the radius of the ith cylinder can be taken to be y or 351 giving the ith cylinder a surface area of for example 27WltMW 1 M50 9 i 90H Summing all of these yields TL 2 27ryiv 1 Nil 9 i 90H i1 Whose limit is the total surface area This gives rise to following theorem Theorem Let be a non negative real valued function de ned and con tinuous on 11 and differentiable on 11 Let ml g be an arbitrary partition of 11 and let Ami xi 7 m4 i 1 71 Let S denote the 3 d surface formed by rotating around the z axis The area of S denoted here by A is given by A lim 2 2mm 1 y i2 AM i1 provided the limit exists as either a real number or as 00 It is understood that the limit is taken over all partitions a mo lt ml lt lt mn b of 11 as max Ami goes to zero It is also understood that the 51 s are given by the mean value theorem y izi 7 l 71 yi 7 yiil 1 g i g 71 But the limit given in the theorem looks like an integral Indeed if the Riemann integral 7 2w yltzgtv1 yltzgtgt2dz exists as a real number then the limit in the theorem exists as a limit of Riemann sums and one has 17 A 2w WW 21mm We have taken some care with this statement It says that if the Riemann integral exists then its value is the area of the lateral surface It does not assert that if the area of the lateral surface exists as a nite number then it is given by the value of the integral It may happen that the area of the lateral surface exists as a nite number and yet the Riemann integral does not exist We don t pursue this phenomenon here At the risk of being pedantic we capture these ideas in a formally stated theorem The theorem deals with nite surface areas Theorem The surface area A of the surface generated by rotating about the m axis the graph of a non negative real valued function Mm continuous on 11 and differentiable on 0119 is given by b A2w yltzgt1 ltyltzgtgt2dz provided that the integral exists as a real number that is provided that the function 1 y z2 a 3 x g b is Riemann integrable This is our jumping off point for surface area calculations It is noted that the hypotheses of the theorem do not require that y m exists at z a or at z b This phenomena is encountered below MODELING EXAMPLE 7 SURFACE AREA COMPU TATION Compute the lateral area A of the surface formed by rotating the function ym 1 m 2 about the z axis The surface area does not include the two circular bases Solution We compute the surface area by the integral 2 A 27r yx 1y m2dx 1 In order to evaluate the integral we make a preliminary calcula tion for 1 g x g 27 WE 96 290 9075 1 2 i i ltyltzgtgt 7 4m 1 2 i i 1ltyltzgtgt i 1 i 4m1 7 4m 4m1 2 1ltyltzgtgt 7 435 Using this calculation one computes a value for the surface area as follows 2 A 2w1yx 1y m2dx 2 QW 41dz 1 4m 27139 2 T V4m1dm 1 7 21 g Wg 41gt21i The domain of the function y in this example was chosen With care y z exists on 12 However if the domain is chosen to be 0 1 then one needs to deal With the potential problem that lim 40 y ooi Fortunately the Whole integrand namely yz 1 9 is Well behaved near I 0 and the integral 2 yaw1 ltyltzgtgt2dz exists as a nite number The computational steps above cover this case and one nds that for this case the area A of the lateral surface is MODELING EXAMPLE 7 SURFACE AREA OF A CONE Compute the lateral surface area A of a right circular cone whose base has radius r and Whose slant height is l where r gt 0 and l gt 0 We don t compute the area of the base The length of the chord from any point on the rim of the base to the apex is 1 Solution We model one generator of the cone as a planar chord that in turn is modeled by yx nggh Where12r2h2 The cone is then modeled by rotating this chord about the z axis It is noted that the equation 2 r2 712 de nes the height h of the cone We compute the surface area of the cone by the integral h AAwmw wmrm In order to evaluate the integral we make a preliminary calcula tion WE 05 M90 wm 2 1me V 1wmrig f Using this one computes the surface area as follows A h a ymwwome 27rJh dz In the next example we use the notation R3 to denote the set of triples of real numbers MODELING EXAMPLE 7 SURFACE AREA OF A SPHERE Compute the surface area of a sphere of radius r where r gt 0 Solution We assume that the surface of a sphere S of radius r is the point set in R3 modeled implicitly by m2y222727 73472 6R3 that is S m7y7z 6R3 m2y222 72 We model the surface of the sphere by rotating about the m axis the planer curve modeled by Vrzima 42572 ln turn7 we compute the surface area A of the sphere by the integral A 27r7 yx 1y m2dx Some care is needed here It turns out that y m becomes un bounded near z 7r and z r So we need to ensure that the integral 27139 f 1 y m2 dz exists as a nite number Keeping all of this in mind we proceed with the calculations7 checking each step carefully In order to evaluate the integral we make a preliminary calcula tion for 7r lt z lt r was was 1 m 572 27 2 7 7 7 762352 m Maw 72 7 m2 2 2 1y 1T272 72 7 m2 m2 72 7352 T2 72 7352 2 7 T 1ltyltzgtgt i Using this one is able to compute the value of the integral 7 2w W 1 2de 7 This is done next where we note that 27139 f yz 1 y m2 dz exists as a Riemann integral 2wyltzgtf1ltyltigtgt dm aruggz m T 27W dm 77quot Thus the area of a sphere of radius r is modeled by The integral above7 fT x 7 2 7 12 W dz exists as a Rie mann integral since the integrand is equal to 7 for 77 lt z lt Ti With regard to the value of the integral7 the values of the integrand at z 77 and z 7 are not of concern The next example is fairly well known and oft repeated in calculus lectures The results of the analysis are counter intuitive MODELING EXAMPLErGABRIELls HORN Let i 1 3 z lt 00 and rotate its graph around the z axis Even though the solid region generated called Gabriel s Horn is unbounded one can both model and compute its volume V by the improper integral co 1 2 V 7r lt7 dx 1 m Further one can both model and compute its surface area A by the improper integral CO A27r l 1i4dx 1 xv z Evaluate the rst integral and bound below the second integral then comment on whether or not Gabriel s Horn can be lled with paint Solution On one hand one has On the other hand one has 1 I 1 A 27r 7 1 jdm 1 z z 00 Z 27139 1 dm 1 z 27rln 27rln 7 1 00 Thus one has 12 One concludes that Gabriel s Horn has a nite volume but an in nite surface area Using paint with in ntessimaly small molecules it can be lled with a nite amount of paint but its surface can not be painted with a nite amount of paint Nifty huh Looking Back In an ideal world we could characterize those point sets in R3 that we believe should be called surfaces Having done that we could then de ne a notion of surface area one that is consistent with human intuition But ours is not an ideal world we don7t have such a characterization of a surface and accompanying de nition of surface area So we do the next best thing we de ne notions of surface area for special classes of 3 d point sets special classes of surfaces In this lecture we have focused on 3 d point sets generated by rotating 2 d curves about the z axis We called these point sets surfaces of reuolutz39on and provided a de nition for their areas Having done that we found an integral formula that allows computation of these areas provided that the 2 d curve is smooth enough In the examples we veri ed that the integral formula provides what most people believe to be the areas of cones and spheres The example dealing with Gabriel s Horn shows how fragile human intuition is when dealing with areas and volumes Final Note Arc Length uersus Surface Area It is noted that the de nition of surface area given above depends upon our understanding of arc length When we de ned arc length we did it for all curves independently of smooth ness In fact we observed that every curve has an arc length However when we de ned the notion of area of a surface of revolution we quietly slipped in the word continuous as a modi er of the function being rotated One asks why cannot one just rotate any 017 curve77 about the m axis for ex ample a step function and use the given de nition of surface area to obtain an area ofthe corresponding surface of revolution One can do exactly that However the surface area so obtained is probably not what one wants 13 What goes wrong The following example is illustrative EXAMPLE Let S be the surface obtained by rotating about the z axis7 the function 35 7 0 if m 0 y 1 if 0 lt m g 1 Use the expression n hm Z 27ry90i i 961202 yi712 i1 to model the area of the surface of S Compare the computed value to the lateral surface area of the cylinder of height 1 and radius 1 Let mlMg be a partition of 01 Then 71 lim 2 27ry90i 90139 M702 241quot M702 i1 hm 271794901 951 7 9002 241 y02 1im Z 27ry90i i 901202 M702 i2 lim 27r1 m1 7 02 1 7 02 lim Zn 27r1 7 mil 0 i2 hm 27mm 12 hm Z 27rm 7 mm i2 27139 hm Mm 12 27139 hm 2m 7 mm i2 27r1 27r1 47r The surface S can be thought of as the lateral surface of a cylin der of height 1 and radius 1 In this case7 one would want the area to be 27r121 27139 14 The formula that we have used in de ning the area of a surface of revolution namely hm Z 2W9ltMgtV i 961202 yi712 i1 yields the value 47139 It evidently computes the area of one end of the cylinder in addition to its lateral surface area It is mainly due to the above example that we have insisted when comput ing the area of a surface of revolution that the to be rotated curve Mm be continuous A Loose End Of course most of this lecture has hinged upon the reader s willingness to accept without de nition the notion of a surface of revolu tion For completeness we offer a formal de nition De nition Let denote a non negative real valued function de ned on the interval 11 and let S denote the surface of revolution obtained by rotating about the z axis the function y Then S is a subset of R3 given by S 9571472 6 R3 242 22 WW2 90 6 la7bl CALCULUS H Spring 2011 LECTURE 8 Harry McLaughlin revised 22011 edited by This lecture provides 0 a de nition of a derivative of a vector valved function 0 a de nition of a tangent line to parametric curve 0 models for tangent lines to parametric curves 0 parametric curves that can have more than one tangent line at a given point 0 a parametric model for a cycloid and its tangent lines Recall We recall from Lecture 7 o a vector is a point in R2 or R o p q7 p7 foo lt lt 00 models a line in R2 o r cos trsin 257 0 g t lt 27139 models a circle in R2 o a cos t7 bsin 257 0 g t lt 27139 models an ellipse in R2 End of Recall Introduction Our goal in this lecture is to investigate models of tangent lines to paramet ric curves This raises the question Can a vector valued function have a derivative Having answered that we are able to provide a de nition of a tangent line to a parametric curve Tangent Lines to Parametric Curves Derivatives of VectorValued Functions Working informally we ex plore the notion of a derivative ofthe vector valued function rt m t yt where it is understood that t belongs to some real interval One chooses an arbitrary time value of the parameter say to and computes the vector difference quotient t t0 At 7 rto me At yt0 M mto7yto At At 90750 At 7 90050 me At yto At ltzt0 At 7 we yt0 At 7 MW At At 39 One notes that the two components of this last vector are difference quo tients identical to those used in de ning derivatives of a real valued func tions Thus if z t0 and y t0 exist it can be shown that this last expression has a limit as At approaches zero Speci cally one has me Atgt Wax We rm gtW 1i 1m At AtaO Hence the value of lim rt0 At 7 rt0 AtaO At is exactly what one would hope for7 that is7 ow we 2050 ln words7 one says that the derivative of a vector valved function is the derivatives of its components Based on this observation one makes the following de nition and comment De nition Let the vector valued function rt7 be de ned on some real interval7 I and let to E I One denotes the derivative of 1 t7 at t07 by r t0 and de nes it by rt0 I39t0 7 I39t0 lim AtaO provided the limit exists2 Comment If mt and yt are real valued7 differentiable functions7 such that rt ztyt7 t E I then for each to E I one has 1 750 tot 2050 This is7 in fact7 the working de nition77 of r t0 EXAMPLE For the parametric curve rlttgt mmw tit foo lt t lt oo and for foo lt to lt 07 nd an explicit expression for r t0 Solution By inspection one determines that 1 50 tot 1050 37537 1 Of particular interest is the fact that r 0 07 1 imore on this later 1The equality limmao W W x to yto is appealing it is correct But it needs more attention than we have given We discuss it in more depth below 2It is understood that if the limit exists7 it exists as a Vector not a scalar Since this is our rst use of the notion of a limit applied to a vector valued function we spend some time getting comfortable with it What does the expression r t At 7 r t hm 0 0 Awe At mean We assume that the vectorvalued function rt is de ned on some real interval I and that to E I It is observed that the numerator that is rt0 At 7 rt0 is the difference of two vectors one of which depends upon the scalar At The whole fraction can be thought of as the vector rt0 At 7 rt0 multiplied by the scalar if So changing the scalar At from one value to another changes the dif ference quotient rt0 At 7 rt0 At from one vector to another To say that t At 7 t hm F0 F0 Awe At equals another vector say V think r t0 means that as At gets small the difference between rt0 At 7 rt0 At and V is a small vector This raises the question What does small mean Particularly what does it mean in the context of a vector An answer is a vector is small if its magnitude its length is small This brings us back to our original question We say that lim rt0 At 7 rt0 V Awe At if for e gt 0 there exists 6 gt 0 such that whenever to At 6 I and 0 lt lAtl lt 6 then rt0 At 7 rt0 At 7 V lt 6 If there exists such a vector V we relabel it we call it r t0i De nition of Tangent Line and Examples Geometrically one can think of the difference quotient I39t0 At 7 I39t0 At as a scaled secant vector from rt0 to rt0 At As At a 0 this scaled secant vector approaches what looks like a 77tangent vector77 to the para metric curve rt at rt0 We use this geometric perception in de ning a tangent line De nition Denote by C a parametric curve in R2 that is modeled by the vector valued function rt Mt where t and yt are real valued differentiable functions de ned on some real interval I and let to E I Further assume that r t0 7E 0 Then at rt0 a tangent line to C as modeled by rt is the straight line in R2 that is modeled by t rt0 Mao 700 lt lt 00 It is noted that if r t0 0 then the above expression does not de ne a line in R23 However it is possible that C has a tangent line at rt0 even though r t0 0 We investigate that below For a parametric curve modeled by the vector valued function rt the derivative vector r t0 is frequently called the tangent vector to C at rt0 even when rt0 0 We have been careful to de ne a tangent line to C as modeled by rti That is our de nition of a tangent line is modeldependent However we assert that in practice one nds that at a given point tangent lines determined by two different parametric models are iden tical When speaking of tangent lines to a parametric curve we sometimes omit the words as modeled by rt when the model is clear Further we have the instinct that if C is a point set in R2 that looks like it has tangent lines and smells like it has tangent lines then it 3The bold face zero 0 denotes the zero Vector that is 0 00 ought to have tangent lines We state Without proof that it is possible to de ne7 in a modelindependent manner7 the notion of a tangent line to a point set in a meaningful way The following two examples use the de nition of tangent line MODELING EXAMPLE 7 TANGENT LINE TO A PARA METRIC CURVE Find a vector model for a tangent line to the parametric curve C7 as modeled by rlttgt ltzlttgt7ylttgtgt coslttgt7sinlttgtgt7 o g t 2w at the point corresponding to t Solution We set to g and use the model t rt0 Art07 700 lt lt 00 for a tangent line One computes r g t7y tlg sint7008tlg e 7 Since rltggt lteos em lt gtgt 9 3 one determines that the tangent line can be modeled by the vector valued function t7 foo lt lt 007 where tAlt gtlti gt7 fooltltoo MODELING EXAMPLE TANGENT LINE TO A PARA METRIC CURVE Find a vector model for a tangent line to the parametric curve C as modeled by rt at the point corresponding to t 14 7 M04103 75276 foo lt t lt 00 Solution We set to 1 and use the model W mm M050 foo lt lt 00 for a tangent line One computes r 1x t7y t lt1 2t 225 lt1225 Since r1 15 one determines that the tangent line at r1 can be modeled by the vector valued function t foo lt lt oo given by W 15 M2 25 700 lt lt 001 For this example the tangent line has a more elegant model One can write t 1 e 2 25 15 21e 121e fooltltoo Replacing the scalar 1 2 with the symbol u the Greek letter mu one obtains another model for the tangent line denoted here by T the Greek letter tau it is prescribed by T Mlt1757 00 lt ILL lt 0039 4This is an interesting parametric curve It doesn t have a tangent line at 01 as modeled by rt even though each of the coordinate functions t2 and e2 is Well behaved77 at t O In fact if one assumes that the curve is the trace of a bug starting at t 700 and moving for all time then at t O the bug turns around and retraces its path The bug Visits twice every point on the curve y e 0 g I lt oo lt accidently happens that the tangent line goes through the origin Stating the above de nition of tangent line is somewhat treacherous What if a point set in R2 can be modeled both as a curve and as a parametric curve 7 can it have two different species of tangent lines at the same point The answer is no If a point set is modeled both as a curve and as a para metric curve then the two corresponding tangent lines at a point7 if they exist7 are the same That which we call a tangent line by any other name would appear as a tangent line 7 with apologies to William Shakespear6 For example7 a tangent line to a parabola7 at its vertex7 is a property of the particular pointset7 and should be independent of whether the parabola is modeled by y I2 700 lt z lt 007 or is modeled by rt t7t27 700 lt t lt 007 or by some other model In this regard7 one anticipates a theorem saying that if a pointset in the plane is modeled both explicitly y and parametrically rt I t7 yt7 then at a given point7 the tangent lines prescribed by both models are identical We need this understanding for peace of mind7 however we donlt expand upon it here The next example serves as a launching platform for the discussion following it MODELING EXAMPLE7TANGENT LINE TO A PARA METRIC CURVE Find a vector model for a tangent line to the parametric curve modeled by 3 W 0790 75 7t 00 lt t lt 00 at the point corresponding to t 0 5That is7 if the point set is modeled by y yc a g x g b and also by rt Mt7 yt7 t E I can it have two di erent tangent lines at the same point 6From Romeo and Juliet VVhat s in a name that which we call a rose By any other name would smell as sweet77 Solution We set to 0 and use the model WA 1quot750 Arto7 700 lt A lt 00 for a tangent line One computes 1quot 0 t7y t lto 375271 lto 071 F rom this one determines that a tangent line can be modeled by 0 0 A0 1 foo lt A lt 00 This can be written in terms of the vector valued function tA foo lt A lt oo given by tA A0 1 700 lt A lt 00 The preceding example is of substantial interest The parametric curve of1 this example can also be modeled by the scalar valued function y xi foo lt z lt 00 Using this function one is not able to model a tan gent line to the point set at 00 since y 0 doesn t exist even though the parametric model provides for a tangent line at 00 Here we are in an unfortunate position We have to live with the fact that being able to determine the existence of tangent lines to a point set at a given point may depend upon the model of the point set As we discussed in an earlier lecture 7 with some effort it is possible to de ne a notion of a tangent line that is model independent Having done this one can then check to see which models predict the tangent line and which models don t see it However such an investigation takes us a eld of our study of the calculus we live with the fact that determining the existence of a tangent line may be model dependent In spite of these remarks we are relatively safe As noted above if two different models of the same point set each predict the existence of tangent lines at a point of the set then the tangent lines are the same Multiple Tangent Lines at a Point In the next two modeling examples and the DILEMMA we discuss two more pathologies dealing with tangent lines MODELING EXAMPLE7MORE THAN ONE TAN GENT LINE The symbol 0 that we use to represent in nity is a parametric curve We don t need an explicit parametric model here7 Does it have a tangent line where the parametric curve crosses itself Solution From the picture it looks as if it should have two tan gent lines at the crossing point On the other hand there are those that say that a curve can have at most one tangent line at a point Looking closely at our de nition of a tangent line at a point on a parametric curve we see that indeed it has two tangent lines at the crossing point Our de nition of a tangent line to a parametric curve leaves open the possibility for the existence of multiple tangent lines at the same point on the parametric curve If a crossing point is given by both rt0 and rt1 then according to the de nition the two lines modeled by We r t0 00 lt lt 00 and rt1 r t1 00 lt lt oo are both tangent lines at the point of intersection provided that rt0 and rt1 are non zero vectors MODELING EXAMPLE 7MORE THAN ONE TAN GENT LINE AGAIN Find vector models for two tangent lines at the point where the parametric curve modeled by rt 29253 7 2t 700 lt t lt 00 7The parametric curve modeled by the VectorValued function r cos sin 29 0 g 6 g 27r appears as a gure eight when plotted on an x yCoordinate system 11 intersects itself8 Solution Even though the problem statement asserts implic itly that the curve intersects itself we need to verify that What we verify is that there are two distinct values of the parameter say 3 and z with s lt 2 such that rs The strategy is to assume that s and z exist ii nd values for such 3 and z and then iii verify that rs We then compute rs and r z and use those vectors to model the two tangent lines One de nes the real valued functions t and yt by rt ztyt t2t3 7 2t 700 lt t lt 00 Setting Ms one obtains 52 22 Since 3 7 2 one concludes that z 73 Since ys one concludes that 248 83 i 28 242 83 28 Using the facts that z 7s and s 7 2 hence 3 3A 0 one has the following computation 33 7 23 7s3 7 273 233 7 4s 0 33272 0 s i Thus if rt does indeed intersect itself it must be that r7 We compute r7 and and verify that they are equal We evaluate rst Here 9005 t2ltx 2 t3 2tlt 8We rely on the reader s willingness to interpret appropriately the expression intersects itselfquot xElt L 2 0 One concludes that 20 We evaluate next r7 Here M tzlt7 52 2 t3 7 2tlt ixE exW i 2 0 One concludes that I397 20 Thus 20 is a point where the parametric curve crosses itself In order to model the two tangent lines we compute the two vectors r 72 and r 2 To that end we note that r t 225 3252 7 2 foo lt t lt 00 From this one nds that MixE 224 and r N24 Thus the two tangent lines are modeled by two vector valued functions t1 foo lt lt 00 and t2 foo lt lt oo given by t120 Mix54 700 lt lt 00 and t2 20 224 700 lt lt oo One can simplify these expressions obtaining lt1ltAgtlt2ogtAlt71 igt foo ltA lt col and t2 20A1 700 lt A lt 00 DILEMMA7TANGENT LINE OR NOT We found above that there exists a tangent line at 00 for the parametric curve modeled parametrically by rt 25325 700 lt t lt 00 In fact the tangent line can be modeled by the vector valued function t foo lt lt oo given by t 0 1 700 lt lt 00 However the same parametric curve can be modeled parametri cally by rt t9 23 700 lt t lt 00 Using this alternate model one nds that rO 928352 M 00 Thus the assumption r 0 7 0 needed in the de nition of a tangent line for a parametric curve is not met So using this alternate model it is not possible to determine whether or not the parametric curve has a tangent line at 00 Finally we recall that according to a previous discussion for the point set de ned by rt t3t foo lt t lt 00 one is not able to prescribe a tangent line at 00 using the model yx 7ooltmltoo There is a way around all of the uncertainty introduced in the previous dilemma a model independent de nition of tangent line can be prescribed However as mentioned earlier that takes us a eld of our study of the calcu lus Most of the time we are able to deal with tangent lines for parametric curves using the de nition above We stick with this de nition because of its ease of use The Cycloid The cycloid has been studied since antiquity But shortly after the intro duction of the calculus it was found to solve the tautochrone problem and the brachistochrone problem9 Since then it has played a fundamental role in calculus modeling MODELING EXAMPLE7MODELING A CYCLOID Let r gt 0 Center a circle of radius r at the point 0r The bottom of the circle is at the origin at 00 Label this point P and track the location of P as the circle rolls along the m axis in the positive direction The curve generated by P is called a cycloid After the circle has rolled a bit the point P has coordinates x From the center of the circle draw a line to P and draw another line straight down to the circles resting point on the z axis Label the angle formed at P between these two lines by the symbol t Then the length of the arc of the circle from the point where it rests on the m axis to the point P is rt Thus the center of the circle is at the point rt r From this observation one is able to determine that z rt 7 r sin t and y r 7 r cos Thus a parametric model of a cycloid is mt rt 7 rsin t and yt r7rcost 7oolttlt 00 9The reader may nd an online search of these two problems of some interest 15 One notes that in allowing the parameter t to take on negative values7 we are admitting the possibility that the circles rolls in the negative z direction MODELING EXAMPLE7TANGENT LINES TO A CYCLOID Find a vector model for a tangent line to a cycloid at an arbi trary point on the cycloid Investigate the expression at points where the cycloid has cusps We rely on the reader s willingness to geometrically visualize a cusp Solution One models a cycloid parametrically by Mt rt 7 r sin 25 and yt r776cost7 7oolttlt 00 The vector form of this model is rt rt 7 rsint7 r 7 rcos 257 700 lt t lt 00 From this one obtains rt r 7 rcos trsin 757 700 lt t lt 00 Using this7 one models the tangent line to the cycloid at rt0 as follows t rt0 r t0 rte 7 rsin to7 r 7 r cos 250 r 7 rcos to7 r sin 2507 where 700 lt to lt 00 Thus a model for a tangent line to the cycloid at rto7 is given by t7 700 lt lt 00 where t rte 7 rsin 250 7 rcos 250 r 7 rcos to7 rsin t0 16 This expression serves as a model of a line only when the tangent vector r776 cos to7 r sin 250 is not the zero vector7 that is7 only when to is not of the form 27m 71 0j17 i2 When one graphs the cycloid7 one observes that these are the points where the parametric curve has cusps Looking Back We have been able to de ne and model a tangent line to a parametric curve rt at rt0 by the expression rt0 rto7 foo lt lt 00 provided that r t0 7 0 This leads to two foundational questions i Can a parametric curve that crosses itself have more than one tangent line at the crossing point ii Can a parametric curve have a tangent line at rt0 even though r t0 0 We have found that the answer to both question is yes This analysis suggests one more question iii Is it possible for a smooth parametric curve to not have a tangent line at one or more of its points The answer here is also yes but we don t undertake that investigation Finally we have modeled a cycloid along with its tangent vectors The cycloid has been introduced because of its role in calculus modelingiit solves both the bmchistochrone problem and the tautochrone problem CAUHEUSH Spring 2011 LECTURE 10 Harry McLaughlin revised 3311 edited by This lecture prouides 0 models and computations of areas for regions of the plane that are de ned by some special parametric curues d A ytz tdt 0 models and computations of areas for regions of the plane that are de ned by polar curues b 1 A 77620 d0 a 2 0 areas of 7 inside an ellipse 7 inside one cycle of a cycloid 7 inside of a spiral 7 inside a cardioid Recall We recall from Lecture 9 o the arc length formula 6 f m t2 y t2 dt 0 arc lengths may be model dependent o a de nition of a polar curve 0 the arc length formula 6 f T2t9 r 02 d0 0 a de nition of polar coordinates End of Recall Area Using Parametric Models Early in our study of the calculus we used an integral f x dz for model ing the area of a planar region bounded by the m axis and the non negative real valued function7 m a g x g 1 Our goal now is to explore the form of this integral when the curve m7 a g z 3 b is modeled parametrically by 057 34057 c g t g d For that7 we use the change of variables formula To get started we recall the change of variables formula 7 stated as it was in a previous lecture Theorem ChangeofVariables Let Mt denote a real valued differen tiable function de ned on a non degenerate interval 07d and let a do and b Let 1 denote a real valued function de ned and continuous on an interval I such that zc7 1 Q I If z t is continuous on gel then Ab dz cdfmtx t it It is noted that if t is chosen such that a zd and b zc then the change of variable formula becomes lb dz Ac fxt t it If the curve modeled by the function neglecting smoothness constraints for the moment yf907 19019 is identical to the parametric curve modeled by t7yt7 c S t S d then necessarily fmt yt7 c g t g 1 Thus if do a and zd b7 the change of variables theorem asserts that the area of the planar region bounded by the m axis and the non negative function f is given by b d d a mm fltzlttgtgtz lttgtdt ylttgtz lttgtdt 3 That is using the parametric model one computes the area by the integral 1 ytmt dt Similarly if do b and zd a one has 17 c c mm fltzlttgtgtz lttgt dt mant dt 1 d d and one computes the area by the integral 0 ytmt dt d These are the forms of the integral that we are after Before using them we reset above we assumed known a curve modeled by x and then hypothesized the existence of a parametric model for the same curve However in the problems below we are given only a parametric curve Before using the integral formula fed ytm t dt or ijtz t dt to compute area it is important to determine that in fact there does exist an identical curve de ned by a real valued function fm a 3 x g I even though we don t need its explicit form It is easy to argue that if 05 is strictly monotone then x exists For our area computations below this observation suf ces But monotonicity is not necessary for the existence of This is seen in one of the immediately following examples EXAMPLES7EXPLORING THE EXISTENCE OF f1 i Let zt t2 yt 23 0931 Then the corresponding parametric curve is identical to the curve modeled by the real valued function mg nggl 1We are reminded that we have de ned the notion of area underneath the curve y a g x g b by f dx It is this de nition of area that we want to exploit ii Let Mt sin 25 yt 17 0 g t 3 7r Then the corresponding parametric curve is identical to the curve modeled by the function In this case t is not monotone and every point on the curve de ned by x is Visited twice by the vector valued function sin t7 17 0 g t 3 7139 For example the point 7 1 is Visited by sin 25 1 at bothtandt37 iii Let t 7 1 yt 257 0931 In this case the parametric curve is E R2 z 1 and 0 g y g 1 It is not a curve modeled by a scalar Valued function 1n the examples below the choice of the two formulas for computing an area7 A7 is determined according to if 05 is strictly increasing then d A yont dtt C if t is strictly decreasing then A d ytmtdt We are now armed for area calculations2 In the next example we are able to model and compute the area of an ellipse We nd that the area is 7139 times the product of the lengths of the semi major and the semi minor axes MODELING EXAMPLE AREA OF AN ELLIPSE Let a and b be positive real numbers Model and compute the area A inside of the ellipse as modeled parametrically by Mt a cos t yt 1 sin t7 0 S t S 27139 Since sin 25 7r 7 sin 57 0 S t 3 7139 we model the area inside of the ellipse by twice the area of the region between the m axis and the parametric curve ztyt a cos t7 1 sin 257 0 g t g 7139 That is7 we model and compute the area A of the upper half7 of the ellipse and multiply that area by two It is empha sized that this doubling process is part of our model of the area inside of the ellipse Since Mt7 0 g t 3 7139 is strictly decreasing we model and com pute the area A using the formula f7 ytm t dt Using the facts that z7r 7a and M0 a and noting that z t 7a sin 75 0 S t g 7r7 one writes 2Our focus on increasing and decreasing functions t is for ease of computation The important consideration is whether or not xc a and dd b or ViceVersa A ytztdt b sin t7a sin m dt interchange the limits of integration 7r ab sin2 t dt 0 7r 1 cos 2t cos2t 202 t 75in2 t abO 7 2 dt 1 7 2sin2 t W m2 if cos2t ab 7 O cos 2t dtgt 2 2 7r ab 7 isinet 0 ab iWi 2 Thus7 according to our model7 the area A inside the complete ellipse is given by 7rab One notes that When a b the ellipse becomes a circle and the area just computed becomes the familiar expression for the area of the circle WCLZ In the next example we are able to compute the area under one cycle of a cycloid We nd that the area is three times the area of the rolling circle used to generate the cycloid MODELING EXAMPLE7AREA OF A CYCLOID Let r gt 0 We model parametrically one cycle of a cycloid by Mt rt 7 r sin 75 yt r7rcost7 0 325 27139 In this case t is strictly increasing So the area7 A7 of the region above the m axis and below the cycloid is modeled by 27r ytmt dt 0 Using this one computes A 7 Almaad 07r7quot 7 7 cos tT 7 7 cos dt 7 2 027rl 7 cos t2 dt T227r17 200s t 0052 dt r2 2 7 2sin WE 27r T2 M d7 cos2t cos2 u 7 m2 u 0 2 2 2cos2 t 7 1 1 1 cos t 27r39r2 7 0 7T7 2 7 21 sin 27 6052 t E 27r39r2 7rquot2 0 37m Thus the area under one cycle of the cycloid is given by 37mg Area in Polar Models We imagine a polar curve modeled by the continuous real valued function7 7607 a g 9 g b where 0 lt bia S 27139 We think of r0 as being continuous but later only require that it be integrable The corresponding parametric curve is given by 19 Tt9 cos 9 yt9 Tt9 sin t97 a S 9 S bi The polar curve de nes a pieshaped region point set R in the plane con sisting of all points of the form r0 cos t97 r0 sin where a g 9 g b and 0 g g 1 Our goal is to nd a model in terms of r0 for the area of the region R But we don t know what area is in this context We seek a de nition of area for a pieshaped region Below we are able to nd a de nition in terms of an integral To motivate our de nition we provide an intuition based heuristic model of such an area Using the expression wrz for the area of a circle of radius r one is able to model the area A of a sector of the circle with central angle A0A0 Z 07 using the equation A0 7 A E 7 7T7 2 39 From this one obtains A rw Letting a 00 lt 01 lt 02 lt lt 0 b be a partition of 071 one lets A0k 0k 7 0k1 for 1 g k g n One can model approximately the area of each of the sectors Rk between 0k1 and 0k by r2 kA0k where 5k is arbitrarily chosen such that 0k1 5k 0k We are using a circular arc to approximate the polar curve r0 for 0k1 S 9 0k Summing over all k one nds an approximate model for the area of R to be mama M 1 2 w H 1 We are summing areas of sectors of circles each with a possibly different radius It is emphasized that even though we use the words area of R we don t yet know how to de ne the area of R Recognizing this as a Riemann sum one lets the size of the partition go to zero to obtain an integral model for the area of R namely7 b 1 E7620 d0 It is emphasized that we did not prove that the area of R is given by f 39 t9 dt9i We only offered a plausible argument justifying the integral as a model for the areal Having done the analysis we formalize a corresponding de nition De nition A polar curve is modeled by a real valued7 integrable function7 M0 a S 9 g b where 0 lt b7 a g 27139 Let R denote the point set consisting of all chords each of which is constructed between the origin and a point on the curve Then the area of R as modeled by r097 is de ned to be 17 1 E7620 d0 It is noted that M0 can take on negative values and that M0 is not required to be continuous only integrable At the risk of beating a dead horse77 we remind the reader that eg the area inside of the upper half of the unit circle can be computed two ways using the formula fOW 6209 10 resulting from the polar model and ii using the formula fir ytm t dt resulting from the general parametric model Hopefully they give the same result3 We expand on this idea with one of the examples below We use this de nition below in two examples In the statements of the two examples we use the word insida as in inside a curve We model the inside of a curve using the set R as de ned above 3They do In fact they are both equal to Ill 1 7 zzdz g which is the area resulting from the model of the form y 1 7 2 71 g x g 1 MODELING EXAMPLE 7 AREA IN POLAR MOD ELS Model and compute the area A ofthe region of the plane inside of the spiral as modeled by the polar equation r0 030g 27139 The spiral is modeled parametrically by 19 woos 9 MO sin t97 0 S 9 S 27 Using the integral de nition for area of the region de ned by a polar curve7 one models and computes A by 2W1 A 7r20d0 0 2 1 27r 2 E0 x d0 1 27r 7 0d0 2 One concludes that7 according to our model7 the area of the region inside of the spiral is A7r2 MODELING EXAMPLE AREA OF A CARDIOID Model and compute the area A of the region of the plane inside of the cardioid as modeled by the polar equation r01sin0 0 g 0 g 27139 11 The cardioid is modeled parametrically by 19 1 sin cos 9 yt9 1 sin sin t97 0 g 9 g 27L To graph the cardioid one makes the following table 0 900 240 0 1 0 g 0 2 7r 71 0 3 7quot 0 0 27139 1 0 One models and computes the area by4 27r 1 A 420 d0 0 2 gfhm sin 02 d0 027r12sin0sin20d0 1 27 1 20 WCOS0W j 57 d0 1 wimgigsmm 37139 7 7 0 2 37139 2 One concludes that7 according to our model7 the area A inside the cardioid is 4In the computation We use the identity 2 if sin 972 There is another approach to the problem of the previous example Using the polar model for the cardioid 79 1 sin t97 0 g 0 271397 one models the cardioid parametrically by 09 1 sin cos t97 1 sin sin 97 0 g 9 lt 27139 Using the parametric model we are able to compute the area inside of the cardioid we compute the area inside of the right half and double it To do that one notes that the right half of the cardioid can be modeled parametrically by mom0 ltlt1 sin 0 07 lt1 smlt0gtgt sin lt0 7 s 0 wl 7T 2 Then the area A inside of the right half can be modeled by A May0 d0 1n the next example we verify that such a computation gives exactly the same area inside of the whole cardioid as was just computed using the formula 1 27r 147 ame 2 0 MODELING EXAMPLE7AREA OF A CARDIOID AGAIN Model and compute the area A inside the right half of the car dioid as modeled parametrically by 7139 z0y0 1sin cos t97 1sin sint97 75 S 9 S Use that to model and compute the area inside of the whole cardioid Solution We use the formula A 7 Man0 d0 One rst makes the computations7 for 7 S 0 S m0 1 sin cos 0 cos 0 sin 0 cos 0 cos 0 sin 20 1 sin sin 0 sin 0 sin2 0 cos 0 2 sin 0 cos 0 cos 0 sin 20 cos0 sin 20 cos 0 sin 20 cos2 0 3cos 0 sin 20 sin2 20 cos2 0 gcos 02 sin 0 cos 0 sin2 20 cos2 0 3 sin 0 cos2 0 sin2 20 Using this one computes A cos2 0 3sin 0 cos2 0 sin2 20 10 cos2 cos 0 d0 1 cos20 tT 3 d0 2 1 1 cos40 7 77 d0 2lt2 2 gt 7 I 7 3 g I 7 20 cos 0740 7T 7T 7 070 7 0 2 4 37139 4 One concludes again that the area 2A inside the whole cardioid is 2A Looking Back Prior to this lecture our tool box for computing areas con tained the formula b M 1967 1 where x 2 07 a g x g b Using this formula we have computed the area underneath the curve m a g x g b In this lecture we imagined that the curve x is modeled parametrically by ztyt7 c g t g d We found that the area underneath the curve m7 a 3 x g b can be computed by the formula 1 ytmt dt Shifting gears7 we examined the area inside a curve whose polar model is M0 a g 9 g b where 0 lt b 7 a g 27139 There we found the area formula to be bl 7r20d0 a 2 Examples illustrated these last two formulas CALCULUS 11 SPRING 2011 LECTURE 16 Harry McLaughlin revised 31111 edited by This lecture provides 0 a de nition of a directional derivative Dwfx0y0 0 a de nition of a gradient vector Vfm0y0 o computations of directional derivatives 0 a theorem asserting that Dw mm yo Vflt 07ld0gt W o a theorem asserting that Dw mmyo is maximized in the direction of the gradient vector 0 a de nition of tangent planes for surfaces modeled by z fx7 y validates previous lecture o a theorem asserting that a tangent plane contains tangent vectors to all curves through a given point each found by intersecting surface with planes perpendicular to the 71 plane Recall We recall from Lecture 15 o the de nition of a partial derivative 0 a working de nition of a tangent plane 0 models of tangent planes for several different surfaces End of Recall Introduction The goal of this lecture is to argue that at a point of tangency the tangent plane to a surface contains the tangent line to every curve on the surface containing the point We don7t quite get there but we are able to make the argument for a special class of surface curves In the previous lecture we provided a working de nition of a tangent plane but in the process we swept under the rug some important mathematical considerations In this lecture we lift the corner of the rug Could it be that at a given point on a surface the tangent plane to a sur face doesn t contain the tangent line to some surface curve Could it be that some reasonable surfaces don t even have tangent planes One also wonders whether or not more than one tangent plane can exist at a point on a surface We investigate some of these issues below but need some getting started machinery Topological De nitions The next de nition allows one to distinguish between boundary points and interior points of a subset of R2 Boundary points and interior points are de ned below De nition Let mo yo 6 R2 and 0 lt T An open ball B of radius r and centered at 0 yo is de ned to be the set of all points in the plane whose distance from 0 yo is less than 711 More speci cally one has B 6R2 xi m02 yi y02 lt 72 De nition Let mo yo be an element of a subset D of R2 The point mo yo is an interior point of 7 if there exists an open ball centered at 0 yo 1The word open signals the fact that the ball does not contain its boundary that is it does not contain the Circle modeled by x 7 x02 y 7 yo2 r2 that is contained in D A point of R2 is a boundary point of D if every open ball centered at the point contains at least one point in D and at least one point not in D De nition A subset D of R2 is open if for every x y E D there exists an open ball centered at z y and contained in D Said in another way a subset of R2 is open if it is empty or if it con tains only interior points Example We show that an open ball D is an open subset of R2 Speci cally we make the argument for the unit ball One de nes the open unit ball D by D 6R2 m2y2 lt 1 One lets mo7 yo 6 D and r M Then one de nes the open ball B of radius r and centered at 07 yo by Bmy 6R2 m2y2 ltlirz Since B is contained in D it follows that D is an open subset of R2 In the analysis of this lecture the notion of open appears repeatedly What is its role A partial answer is provided in the parenthetical remark following the de nition of a directional derivative Continuity of f 393a y Analogous to the idea that a real valued function of one real variable being continuous at a point is the notion of a real valued function of two real vari ables being continuous at a point We introduce the notion now because it is needed in the hypothesis of one of the theorems of the next section Here is the needed de nition De nition Let 7 denote a subset of R2 and let mo7 yo 6 D A real valued function m y de ned on D is continuous at 07 yo if 1 y ee yo lim gtyquot0yo The expression 179 f107y0 lim gty 10gtyo means that when 17y E D and when 17y is close to 107 then fz7 y is close to fzoy0i This can be said more precisely the function is continuous at 107 if for every 6 gt 0 there exists a corresponding 6 gt 0 such that WWW 1079M lt 5 whenever I y E D and I 7 zo2 y 7 yo2 lt 6 There are interesting phenomena associated with the de nition of conti nuity of fmy the de nition warrants careful study However7 our goal is to investigate directional derivatives and their consequences Theorems dealing with directional derivatives may need the word continuity in their hypotheses7 but7 in some sense directional derivatives exist independently of continuity So we have chosen to give lip service77 to continuity and forge ahead with directional derivatives Directional Derivative The next de nition builds on the notion of a partial derivative of m It de nes the notion of rate of change of f in an arbitrary direction not just the z direction or the y direction De nition Let mo7 yo be an element of an open subset D in R2 Let m y be a real valued function de ned on 7 Let W 1471 be a unit vector in R2 The directional derivative of f at mayo in the direction of w is denoted here by Dwfzo yo and de ned by Dw mwo i135 provided that the limit exists it is noted that fzo tu yo tv is de ned for t With sufficiently small magnitude so the expression Eng 10 my 90 it 10790 makes senselli This is Where the fact that D is open is used From where does the de nition of directional derivative come If mo yo 6 R2 and 1412 is a unit vector in R2 the line in R2 containing mo yo and with direction vector 1412 is modeled by 930790 t 75017107 00 lt t lt 00 or more speci cally by ztyt x0 tu yo tv 700 lt t lt 00 The real valued function fz0 tuy0 2512 of the one real variable t is de ned on this line lts derivative at t 0 is given by 91 950 W7 240 ttv 109607 yo provided that the limit exists So Dwfm0y0 is the value at t 0 of the derivative of the real valued function fz0 tu yo 2512 One can write d Dwf07y0 10900 twyo 7511 t0 ln words the directional derivative of fzy at 0343 in the direction w records the rate of change at 0343 of the height of the surface 2 fmy lying directly above below the line modeled by 0340 tw foo lt t lt oo There is a subtle idea here that merits a careful look If W u v is replaced by 7W iu 71 then even though the corresponding lines modeled by 10tuy0tv 700 lt t lt 00 and 10 t7uy0 tv7 700 lt t lt 00 are identical the corresponding directional derivatives are not the same In particular Diwfmyo hmf10tu7yotvfonyo tgt0 711m 10 Whyo 45 10790 EHO 7 7 lim zao Dwfx07y0 0 tu790 75 xo o 75 That is the directional derivative off at zoy0 in the direction W is the negative of the directional derivative of f at zoy0 in the direc tion iwi So for example 101950790 Dwaoyyo When W 170 fw107yo Dwf107yo When W 710 To get a feeling for this phenomenon we examine the case ay 17 any 6 R2 With 10790 070 and W Here Dwfltzoyogt m f10 0i However Dwfzoy0 77570 hm Ho 7 mm 71 Ammo The following examples explore directional derivatives EXAMPLE 7 DIRECTIONAL DERIVATIVE OF AN ELLIPTIC PARABOLOID Let 1 1 2 2 2 z z m ER andw77 rm 1mg Compute the value of the directional derivative of f at 07 0 in the direction of w Solution In order to ease the notation one writes the vector W f f as w 7471 and then writes 0 t 0 t 7 0 0 hm w v 1 taO t i tzuz t2 2 7 0 hm taO t lim tuz 25122 taO 0 Dwf07 0 Thus7 Dwf07 0 0 One notes that the above computation is independent of Wi One concludes that the directional derivative of f at the origin is 0 for all directions The next example changes the point of evaluation of the previous example EXAMPLE7DIRECTIONAL DERIVATIVE OF AN ELLIPTIC PARABOLOID Let 2 2 R2 d w i i WM 90 y7m7y an Compute the value of the directional derivative of f at 17 1 in the direction of w Solution As above7 in order to ease the notation one writes w 7471 and then writes m flt1tu71tvif171 Dwflt171gt a t 2 27 hm 1tu 1tv 2 taO t i 1 12tut2u212tvt2v272 7 3 t lir2utu22vtvz 2u2v 2 2 77 V5 V5 NE Thus Dwf11 NE The above computation can be made perhaps slightly more easily by writing 1 Dwf171 3K1 75102 1 tv2llt0 21tuu 21tvvlt0 Qu 21 WE 2V5 The next theorem is a powerful one It says that most of the real valued functions of two real variables that we deal with have directional derivatives in all directions Further it provides a formula for the evaluation of direc tional derivatives Theorem Let mo7 yo be an element of an open subset D in R2 Let m y be a real valued function7 with continuous partial derivatives7 de ned on 7 Let w 1471 be a unit vector in R2 Then the directional derivative of f at 07 yo in the direction w exists and is computed by Dwf9007 yo fz9007y0u leg07 yov The writer nds this result quite amazing It says that if the directional derivatives in the z and y directions are known7 then the directional derivatives in all directions are known Before proving the theorem we make the following two remarks 1 Because the partial derivatives fmz0y0 and fyz0y0 ap pear so frequently in this context it is useful to seek a more compact notation The gradient vector denoted by Vfzy is de ned by V y fm7y7fym7y7 any 6 domain 1 Using this notation one nds the appealing expression Dw mowo VJWEOWO W In words the directional derivative of f in the direction w is the dot product of the gradient of f and w 2 For emphasis it is remarked that the gradient operator con verts a real valued function of two real variables into a vector valued function of two real variables The term operator re ects the idea that the gradient vector results from an operation on the scalar valued function m Proof of the theoremi As long as 10 tu7 yo tv lies in the open set D one uses the mean value theorem to write fzo tu yo tv f107y0 lo W790 W 16on yo tvl f107y0 7511 16on y0l fx 7yotvtufyIo 7tv for some 5 in the open interval determined by 10 tu and 10 and for some 77 in the open interval determined by yo tv and go Thus since 5 is squeezed into 10 and 77 is squeezed into yo as t A 07 one has t t 7 7 Dw zoyyo tl io uyo t 0 90 f1 5 yo tvu 11m fylt10777v EHO f1lt107 90 fylt107y0v where we have used the fact that both f7 and fy are continuous in Di This completes the proof In the following example we are able to use the gradient in a directional derivative computation EXAMPLE iDIRECTIONAL DERIVATIVE ON A HY PERBOLIC PARABOLOID Let 1 2 m m z ER2 and w 77 f y y y lt 6 6 Compute the value of the directional derivative of f at 17 1 in the direction of w Solution We use the formula To compute the gradient vector one writes fz7y y fm171 1 leg9079 90 fit171 1 Vfuv 1 f11717fy171 11 Thus Dwf11 Vf11w 1 2 11 77 l VS v5 i 7 One concludes that Dwf171 An interesting question related to directional derivatives is posed and an swered in the following example EXAMPLEfDIRECTIONAL DERIVATIVES ON A HYPERBOLIC PARABOLOID Let rm y2 7 9027 907 y E R2 The surface 2 m7 y is a hyperbolic paraboloid Find the direction in which 1 is increasing most rapidly at z y 17 1 Solution For m y y2 7 2 one nds the unit vector w in R2 that maximizes The unknown here is the direction vector w We use the formula Dwf171 Vf171 w and write Dwf11 lVf11l lwlcos0 where 9 is the angle between the vector Vf11 and the un known vector w Thus Dwf17 1 is maximized when w is cho sen so that cos 0 17 that is when w is parallel to the gradient vector Vf11 Since w must have length one7 one concludes that V l 1 w 7 lVf171l as long as lVf11l gt 0 It remains to evaluate w To that end one writes fmmy 72m M171 2 filmy 2y M171 2 Vf11 722 may NE From this one determines that 7 Vf171 lWlt171gtl 72 2 7 71 1 a e Finally7 the direction w in which the hyperbolic paraboloid is increasing most rapidly7 at 171 is 13 Having found the direction of maximum rate of increase one can compute the value of the directional derivative in that direction7 that is7 compute the magnitude of the maximum rate of increase It is given by NE Thus the maximum value of the directional derivative at 11 is 2f It is noted that this value is exactly lVf11l this is not an accident The ideas of the last example result in the following theorem Theorem Let mo7 yo be an element of an open subset D in R2 Let m7 y be a real valued function7 with continuous partial derivatives7 de ned on 7 Further assume that the gradient vector is not the zero vector at 0310 Then the directional derivative of f at 07 yo is a maximum in the direction of the gradient vector at 07 yo The maximum value is lV mO7 y0l The directional derivative is a minimum in the direction of the negative gra dient vector at 07 yo The minimum value is ilV zmyo The proof of this theorem is just a rehash of the previous example we don t provide a formal proof Next we give a formal de nition of tangent plane lt formalizes the work ing de nition77 of tangent plane used in the last lecture Knowing about directional derivatives allows us to then argue that the tangent plane at a point on a surface contains the tangent lines7 at that point7 to all curves on the surface resulting from intersecting the surface with vertical fences De nition Let 040 be an element of an open subset D in R2 Let m7 y be a real valued function7 de ned along with its partial derivatives7 on 7 Then the tangent plane to the surface 2 m7 y7 7 y 6 D7 at 073 z0g07 is the plane modeled by M9007 yew 0 Mme WW 7 yo i Z 7 20 07 907 y Z 6 R3 where 20 mo yo It is emphasized that our de nition only applies to surfaces modeled by realvalued functions of the form 2 fz7 For example7 it does not apply to all implicit surfaces7 that is7 surfaces modeled by an ex pression of the form fz7 y 2 0 Discussion Leading to a Theorem Let mayo be an element of an open subset D in R2 Let m7 y be a real valued function7 with continuous par tial derivatives7 de ned on 7 Let 5 denote the surface modeled by z m7 y7 7 y E 7 Let 141270 be a unit vector in R3 and let 1 denote the line in R3 modeled parametrically by 07 34070 25u7 127 07 foo lt t lt 00 The line is con ned to the m7 y plane One constructs a plane containing 6 and perpendicular to the m7y plane This plane intersects the surface S in a curve call it C The curve C is modeled parametrically by 0 tu yo w mo tu yo tv where it is understood that the values of the parameter t are constrained so that the points 0 tu yo 2512 lie in D We have seen above that if w 1412 then Dw mm yo fz9007y0u fylt9007 240 and that d Dw mowo 10000 75147 yo tvlt0 One concludes from the de nition of directional derivative that f950 75147 yo tvlt0 fz9507 240W fylt07 y0U Thus at 0 yo mo 340 a tangent vector t to the curve C is modeled by t 147117 fz9507y0u leg95079011 Now one writes t in a fancy way as the sum of two vectors namely t M17 07 fwlt0790 00717fy07yo Since the cross product 11 1707fz07yo X 0717fy9007yo is orthogonal to both of these vectors it follows that n t 0 One notes that n 1707101950790 X 071716410790 1 j k 1 0 f zo yo 0 1 fylt10790 fx107 90 fy107 yo 1 and hence n is normal to the tangent planed to S at 0 yo fm0 y0 Recall the de nition of tangent plane given immediately above This says that n is orthogonal to the tangent vector to C at 0 yo mo 340 But since n is the normal vector for the tangent plane to S at 0 yo mo 30 one concludes that the tangent plane to S at 0310 mo 340 contains the 16 tangent line to C at 07 yo mo 340 We have just proved the following theorem Theorem Let mo7 yo be an element of an open subset D in R2 Let m y be a real valued function7 with continuous partial derivatives7 de ned on 7 Let 5 denote the surface modeled by z fmy my 6 7 Let C be the curve of intersection of S and any plane that both contains mo7 300 and is perpendicular to the z y plane Then the tangent plane to S at 07 yo mo 340 contains the tangent line to C at 9007140 f9007y0 All of this leaves open one more question If C is any smooth parametric curve on S that contains the point 071407 mo 307 does the tangent line to C at mayo Km 30 lie in the tangent plane to S at 07 yo mo 340 The answer is yes We leave the analysis to the next lecture Finally we recall that at the outset of this lecture we wondered whether or not there might exist more than one tangent plane to a surface at a given point According to our de nition of tangent plane this is not possible However it is possible to model a self intersecting surface parametrically and from a reasonable de nition of tangent plane conclude that more than one tangent plane can exist at a point on a surface This is not unlike our experience with multiple tangent lines at a point where a parametric curve intersects itself Looking Back We have been able to de ne the notion of a directional derivative 7514 751 7 7 DWWMO HEW Using this we were able to compute directional derivatives for several differ ent surfaces We provided a formal de nition of a tangent plane to a surface modeled by z x y fzmo7yo 0 M9007 WW 7 yo i Z 7 20 0 17 This de nition validates the tangent plane computations of the previous lec ture We were able to argue that tangent planes contain tangent lines to a speci c class of surface curves 7 namely curves of intersection of a surface and planes perpendicular to the m y plane CALCULUS 11 Spring 2011 LECTURE 1 Harry McLaughlin revised 12211 edited by This lecture provides examples of trigonometric logarithmic and exponential integrals It introduces o fcos2 dm fcos3 Mala etc o ftanz dm ftan3 dm etc o fsin2 mcos2 dm fsin3 mcos3 dm etc o fsin2 mcos3 dm etc o the notion of integration by parts 0 f1nzdz o fmcosdz o fxem dm 0 fxzem dm 0 f cos l dm Recall We recall from previous calculus experience 0 the integration technique called u substitution A M du d futu t dt 0 the integral formulas fsinzdzicosz07 foo lt mlt oo fcoszdz sinmC7 foo lt z lt oo ftanmdm ilnc0sz C7 7 lt z lt fsecz dz tanm C7 71 lt z lt 1 femdmem07 fooltmltoo End of Recall Introduction To date we have investigated integrals of the six main trigonometric func tions sin z cos x etc and the exponential function em Of interest now are integrals of functions of the form eg sin3 and sin3 cos2 Further the integral of the natural logarithmic function ln z remains to be investigated Functions of the form zsin z mew and ln can be integrated using the method of integration by parts We at tend to these matters in this lectures Lectures 2 and 3 deal with additional functions not mentioned here One might think that there are an in nite number of function types whose integrals are interesting So why are these notes focused on a select few One answer is that with regard to pencil and paper calculations the integrations selected for these notes comprise most of what one is likely to encounter It is true that in practice most occurring integrals are suf ciently involved that there is no hope for a pencil and paper evaluation lnstead one routinely uses numerical schemes However pencil and paper integrations are possible frequently enough that they are worth studying independently of numerical schemes To help anchor all of this one notes that there is no known closed form expression for the inde nite integral fsin 2 dm for example the de nite integral fol sin 2 dz must be computed numerically Before launching our study of further integration techniques we review the fundamental theorem of calculus 7 all pencil and paper integrations use this theorem Fundamental Theorem 0f Calculus The fundamental theorem of calculus says that the de nite integral of the derivative of a real valued function is completely determined by the func tion s values at the end points of the interval We give a precise statement Theorem Fundamental Theorem 0f Calculus Let a and 1 denote real numbers With a lt I Let 1 denote a real valued function de ned and continuous on 11 and differentiable 0171 If f is Riemann integrable on 11 then1 b a fmdx 1017 m That s it7 in all of its elegancel It says that if one needs to integrate7 from a to b7 a function gm7 ie fgm dm then 1 think ofgm as the derivative of some other function gm f z and to evaluate that other function at two points fa and fb then sub tract This is sometimes written 17 a f dfZfbifa Proof of the Fundamental Theorem ofCalculusi Let n denote a positive integer and a 10 lt 11 lt lt In bi By the mean value theorem for derivatives7 there exists E zinlwi l g i S n7 such that n N u Elf509 M71 i1 Since the left hand side of this equation is independent of both n7 and the choice of the zis7 so too is the right hand side Since f is 1liklen though f a and fb may not be de ned7 one can arbitrarily assign values to f a and f b and note that the integrability of f is independent of the assignment Clearly the the Value of the integral is independent of the assignment Riemnann integrable it must be that the Riemann sums on the right hand side are all be equal to the integral of fquot That is n b Extmm 7 7 mm i1 for all positive integers n and all choices of the I si Thus 12 fax 7 W 7 new This completes the proof Before going on we try to capture the statement and proof of the fundamen tal theorem of calculus in a thumb nail sketch n n b M 7 M 7 Z W 7 few 7 Zf cnmi 7 Nudes i1 R Here the rst equality is just an identity the second equality follows from the mean value theorem and the last equality follows from the de nition of the Riemann integral Since this theorem has been used extensively in our study of the integral we don t offer more examples at this point Integration of Trigonometric Functions The reader is advised that in the following we are able to lean heavily on the computational tricks77 of our predecessors They have reduced our problem solving from days to minutes Powers of Sines and Cosines Rather than trying to identify a com plicated77 general approach to integrating powers of trigonometric functions we illustrate key ideas through examples The rst example is familiar EXAMPLE Evaluate the integral Solution Thus 7r A cos Using the fundamental theorem of calucus one writes 7r 03 cos dz The next example transforms the problem of computing the integral of cos2 into that of cos EXAMPLE Evaluate the integral Solution 2 cos2 dm 0 One writes 2 cos dz cows y cos as cos y 0 75inxsiny 1 COS cos2ac 205200 75in2w 7 7 dz 7 x 7 A lt2 2 7 2 1 w 1 202 an 3 M I E 2 cos 21 d 2 2 7 7 z 2 0 0 2 7r sin 21 g i y 4 4 0 7r sin 7r sin 0 r lt gt m 4 4 4 7r 4 Thus fog cos2 dz Having successfully integrated cos and cos2 one asks about the inte grals of cos3 z7 cos4 z7 etc The next two examples deal with this question in the guise of sin3 and sin4 They uncover what is needed to inte grate higher powers of the sine and cosine functions However we don t dig deeper than the two examples EXAMPLE Evaluate the integral 7r 3 sin3 dm 0 Solution Using the fact that sin2 cos2 1 one writes 3 g 2 s1n dz s1n s1n dz 0 0 g 17 cos2 35 sin x dz 0 7r 5 sin m 7 cos2 m sin 35 dz 07f 2 i 2 d s1ndm cos micosdm 0 0 dm Thus EXAMPLE Evaluate the integral 7r 2 sin4 dz 0 Solution Using the identity sin2 cos2 1 one writes sin4 1 dz cos z y cos z cos y 0 isinz siny cos2z cos2 z 7 sm2 z sin2 sin2 dz 17 2mg 76 0 2 1 cos 2x 4 sm z E 2 1 7 cos2 sin2 dz 0 sinzy sinzcosy sin2 7 cos2 sin2 dz cos zsin y 0 sin2z 21nzcosz 1 2 4 1 lt2zgt lt2zgt dz m71lt2wgt 7 o o 0 2 2 4 7r sin 2z 1 1 cos 4z d 7 7 7 7 7 7 7 7 I 8 4 0 4 0 2 2 7r 1 7r 1 7lt170gt7 A cos4zdz wmow 0 3 1 1 W ltsm m i 1 mW 4 Thus Powers of Tangents and Cotangents We investigate the integrals of tan z7 tan2 z7 tan3 and tan4 z7 leaVing powers of the cotangent func tion to the worksheets The next example is familiar EXAMPLE Evaluate the integral 1 tan dm 0 OZ tandz 0 dz Solution One writes Thus f0 tan dz ln This can also be written f0 tan z dz ln 2 The next example explores the integral of tan2 The technique differs from that of the integral of cosZ EXAMPLE Evaluate the integral 2 tan dm 0 Solution Using the identity tan2 secz 7 1 one writes 7r 0Z tan2 z dz fade z 7 1 dz 7 tanltzgt1 l 7r 4 1771 171 Thus fog tan2 dz 17 g The next two examples deal with integrals of tan3 and tan4 EXAMPLE Evaluate the integral tan3 dm 0 Solution Using the identity tan2 seCZ 7 1 one writes Ma 0tan3xdz 0 tantan2d e1 tan zsec2 7 1 dz tan 1 0 K zl tan 90 d9 7 0X tan QB dag mi lmwsw 0 1 1 5701nlt gt4nlt1gt 7ln 70 7ln 39 Thus This can also be written f0 tang m dz 1 71112 EXAMPLE Evaluate the integral 7r Z tan4 dm 0 Solution Using twice the identity tan2 sec2 7 1 one writes 7r I tan4 dz 0 3 4 I tan 74sec271d 3 0 1 i a g o tanl0lo 1 7r 37170170 L2 7 4 339 Thus l f0 tan4 zdz g 7 This completes our investigation of integrals of powers of sin z7 cos z7 tan and cot Powers of sec and csc are left to a later lecture Products of Powers of Sines and Cosines In this section we deal with integrals of products of powers of sine and cosine functions In the middle of one of the computations above we were able to write 7r 7 I 1 2 4 sin2 cos2 d1 1 Mn 0 0 4 ll 1 cos 41 5 T 19 7r sin41 E 32 0 7 7T a This leads naturally to the next example where we increase the powers of sin and cos from two to three EXAMPLE Evaluate the integral 2 cos3 sin3 d1 0 Solution One writes 2 cos3 sin3 d1 0 sin1 y sin 1 cos y 1 cos1siny g sin3 21 d1 sin21 2in1cos1 1 0quot cos3 15in3 1 sing 21 7 sin2 21 sin 21 d1 0 j 17 cos2 sin 21 d1 1 l g 2 d cos 21 s1n 21 d1 0 cos d1 12 gO l 7 cos 21 1 cos3 21 0 16 0 1 1 s WHAHEHA 7 1 1 7 8 24 i 7 12 Thus fog cos3 sin3 dz There is an alternate computation for the previous example We introduce it next since it helps getting started with the example following it EXAMPLE Evaluate again the integral cos3 sin3 dz Solution One writes 7r 03 cos3 sin3 dz o O m a A H V A H l o O m m A H V V 9 I A H V amp 8 Thus f0 cos3 sin3 dz Fortunately we obtained the same answer in both computations The above examples don t treat integrals of the form fsinm mcos dz where the non negative integers m and n are distinct We include the next example to provide light on that case EXAMPLE Evaluate the integral 2 sin2 cos3 dm 0 Solution One writes 2 sin2 mcos3dm 2 sin2 cos2 cosdm 0 0 g sin2 m1 7 sin2 35 cos m dz 0 E 2 E 4 s1n zcoszdzi s1n mcosmdm 0 0 I d A2 sin2 sinmdm I d 7A2 sin4sindm i sin3 sin5 7 3 0 i 5 0 i 1 1 7 3 7 3 i 2 7 E Thus This ends one phase of the investigation of integrals of powers of trigono metric functions We still have to worry about powers of sec and csc We pause to look back and ask What makes the previous computations successful For answers here is a rst cut 14 o For the example f0 cos2 dz we used the identity 1 cos 2z 2 7 7 7 cos 2 2 o For the example f0 sin3 dz we wrote sin3 sinz sin 17 cos2 sin 717 cos2 cos o For the example f0 sin4 dz we wrote sin4 sinz 81112 1 7 cos2 sinZ m 7 ltzgt ltzgt LMM 7 1 cos2z 1 1 cos4z 7 2 2 4 2 2 39 o For the example f0 tanz dz we wrote tanz secz 7 1 o For the example f0 tan3 dz we wrote tan3 tan tanz tanzse02 7 1 tan tan 7 tan 0 For the example f0 tan4 dz we wrote tan4 tan2 tanz tanz zsec2 7 1 tanz tan 7 tan2 tanz tan 7 sec2 7 1 o For the example fog cos3 sin3 dz we wrote 1 cos3 sin3 g sin3 2z sin2 2zsin 2z 11 7 cos2 235 sin Qz 8mm coszwdd OOlHOOl 75 2 and in the alternate version we wrote cos3 sin3 cos3 sin2 sin cos3 7 cos2 sin 7cos3 7 cos5 cos c For the example f0 sin2 cos3 dz we wrote sin2 cos3 sin2 cos2 cos sin2 7 sin2 cos sin2 cos 7 sin4 cos sin2 sin 7 sin4 sin In the next lecture7 we investigate7 still further7 trigonometric integrals In particular we investigate sec3 z dz which seems to occur more than it share of the time in modeling problems We also calculate a value of 1 sin 27rz cos 37rz dz 0 lntegrals of this type play a central role in the study of Fourier series But before going to the next lecture we introduce the idea of integration by parts Integration by Parts One recalls that the product of two differentiable real valued functions each de ned on the same interval is differentiable Further the accompanying product rule is ltultzgtvltzgtgt ultzgtultxgt unias Computing the inde nite integral on each side and rearranging gives umvm dm 7 dm where we have replaced dz by This is the so called integration by parts formula It is stated carefully in the next theorem Theorem Integration by Parts Let u and 1 denote real valued differen tiable functions each de ned on a non degenerate real interval I Further assume that there exists an antiderivative of the function vmu z that is fumu zdz exists Then dz exists and umvx dz mom 7 Uzux dz z e 1 This is one form of the integration by parts formula 7 the inde nite integral form It is sometimes written Ud U LL U7 Udu which is good for recall Later we have a look at the inherent notational ambiguity The n Proof Using the fact that the product of two differentiable realvalued functions is differentiable and using the product rule one Writes Thus uzvz vzuz z E L uzvz 7 vzuz z E L Since the inde nite integral f dz exists by de nition and since dz exists by hypothesis the inde nite integral of the righthand side of the equation existsi Hence the inde nite integral of the lefthand side existsi Thus uzv z dz 7 dz z 6 I ext example illustrates the power of the integration by parts formula EXAMPLE Express the anti derivatives ofthe function z cos z 700 lt z lt oo in terms of elementary functions Solution Making the identi cation z 700 lt z lt 00 and vz sin z 700 lt z lt oo in the integration by parts formula one obtains zcos dz Thus zsin 7 zsin 7 zsinzcoszC 7ooltzlt oo fzcoszdzzsinzcoszC 7ooltzlt oo The integration by parts formula has a de nite integral form It is given in the next theorem Theorem Integration by Parts Let u and 1 denote real valued differen tiable functions each de ned on a non degenerate real interval7 11 Fur ther assume that two of the three integrals f vzu m dm fuzv m dm f dz exist Then all three exist and b b uzvm d1 7 U Ltd This is another form of the integration by parts formula 7 the de nite integral form It is sometimes written 17 b udvuvlzi 39udu I 1 Proof Using the fact that the product of two differentiable realvalued functions is differentiable and using the product rule one writes uzvz vzuz z E a7bli Since the de nite integrals of two of these three functions exist7 it fol lows that all three de nite integrals existi Using the fact that f dz one obtains b b uzvz dz 7 vzuz dzi The next example deals with the de nite integral form of the previous ex ample EXAMPLE Evaluate the integral 7r 2 z cos dm 0 Solution Making the identi cation m 0 g z 3 g and vz sin z7 0 g x g g in the integration by parts formula7 one obtains d g d d 0 mcosz z 7 0 81n H A ml 3 l O l lt5 NH 9 E A R V H amp R gcosltzgti 7139 7 071 2lt gt 7139 771 2 Thus As noted above the de nite integral form of the integration by parts formula is sometimes written as b b udv uvlzi Udu 04 04 This is a good way to remember the formula but it needs to be read not too literally In most contexts the expression fa udv is taken to mean the de nite integral of a function 141 a g u g b However in the context ofin tegration by parts the expression fju dv is taken to mean fumvz dm ln words one says the integral ofu dee v is uv minus the inte gml ofv dee u The next example lls a hole in our so far generated list of anti derivatives of transcendental functions Recall that we routinely use the fact that an anti derivative of e1 is em However in these notes we have not yet found an explicit form for an antiderivative of ln The computation of the example uses a trick namely writing lnzdz lnm 1dz lnzzdz Luckily we are agian able to draw on the insights of earlier mathematicians 20 EXAMPLE Evaluate the inde nite integral lnIo lI7 0 lt Ilt 00 Solution Using the integration by parts formula Ud U LL U7 Udu and making the identi cation ln I7 0 lt I and UI I 0 lt I one writes lnIdI lnI1dI lnIIdI armyz mam IlnI7dI IlnI7IC Thus an antiderivative of ln I7 I gt 0 is given by flndln707 OltIltoo ln words7 the integral of ellen I is I ellen I minus I Armed with the integration by parts formula we are able to tackle the in tegral7 erm dI It is known how to integrate I and how to integrate em separately this example shows how to integrate their product EXAMPLE Evaluate the integral 1 I5m dz 0 21 Solution Using the integration by parts formula 17 b udv 1Lle 7 Udu I I one writes 1 1 d mew dm m fem dm 0 0 dm 1 meml 70 emdim d1 1 5707em1dm 0 7 e 7 em 1 e 7 e 7 1 1 Thus Above we were able to evaluate f mew dm Having done that one wonders if there is a pencil and paper approach to evaluating f mnem dm n 27 37 r The next example uncovers the core idea in such calculations EXAMPLE Evaluate the integral 1 mzem dx 0 Solution Using the integration by parts formula 17 b udv uvlzi Udu 1 1 twice7 one writes I l d mzem d1 m2 fem dm 0 0 dm l mzeml i eQOdm 0 l d 1 7 570720 mac dm 1 572xeml iemldz 0 57257075ml5 572ei571 572i 572 Thus fol 1325 all e 7 2 Finally the next example shows the role of integration by parts in comput ing integrals of inverse trig functions Although it is just one example it tells the whole story Practice integrating other inverse trig functions is left to the worksheets EXAMPLE Evaluate the inde nite integral cos 1ydy7 foo lt y lt 00 Solution Using the integration by parts formula and recalling that 1 V1in icov1 y 71 1 dy ltylt one obtains7 for 71 lt y lt 17 cos 1ydy cos 1 1dy d 71 i 1 cos wdyy y 7 29 chs 1y7fy2dy 23 242 dy 1 1 1 d 24008 y W1 yc0s4ltygte 2 o geos me w Thus fcos 1ydyycos lg7 1ig2C7 71ltylt1 The de nite integral fil cos 1 dy is dealt with in one of the worksheets The fundamental theorem of calculus can be employed here even though the deriva tive of cos 1 does not exist at y i1 Some care with the reasoning is needed Experience shows that integrals of the form 0 fmnsindz o fzncosdz O fznem d1 0 fxnlndx 0 fm cos 1 zdz o fznsin ldz o fzntan 1 dz 0 etc scream for integration by parts Looking Back We have successfully evaluated the following integrals o fooszdz o fcos2xdz 0 sin3 dz 24 0 sin4 dz 0 ftan dz 0 tan2 dz 0 tan3 dz 0 tan4 dz 0 f sin2 0052 dz 0 f sin3 cos3 dz 0 f sin2 cos3 dz 0 f z cos dz 0 f In dz 0 fze dz 0 fz2e dz 0 feos l dz 25 CALCULUS 11 Spring 2011 LECTURE 2 Harry McLaughlin revised 12211 edited by This lecture provides 0 fsecsdz o fsin mm cos dz 0 fcosmzcosdz o a reissue of the change of uariables formula 17 d MW fltzlttgtgtz lttgtdt I 0 cos t 1sin t dt 0 an39 t J t39 to 39 t 39 bytrig 7 It 39 andother substitutions ft3 sint2dt o fcosEdz o f 1 7 m2 dm 1 39 l W 19 17212 d l m 9 o a remark that u substitution and trig substitution are two di erent ways of looking at the change of uariables formula Recall We recall from Lecture 1 0 evaluation of the following integrals f cos dz f 0052 dz f sin3 dz f sin4 dz f tan dz f tan2 dz f tan3 dz f tan4 dz f sin2 0052 dz f sin3 cos3 dz f sin2 cos3 dz f z cos dz f ln dz f ze dz f zQEE dz f cos 1 dz 0 the integration by parts formula Lbd U LL U7 Udu End of Recall Introduction The previous lecture dealt with integrals involving powers of sin z cos x tan and cot It tiptoed around integrals of powers of sec and csc This lecture begins with a look at integrals of higher powers of sec and csc After that integrals of the product functions sin mm cos and cos mm cos are considered Finally the method of trig substitu tion is introduced The Integrals of sec3 and sec4 One recalls that we know how to integrate sec z sec2 z csc and csc2 Our goal in this section is to investigate the integrals of sec3 and sec4 Before going there we review in the next two examples the integrals of sec and sec2 M EXAMPLE Evaluate the integral sec dm 0 Solution One writes1 A sec dx 111SeC l tan 1n 17ln10 Inlt 1 1Recall that x sec tan W sec 70 tan I W wv I mm Thus f0 sec dz ln 1 EXAMPLE Evaluate the integral 7r Z sec2 0 4sec2zdz tanml 0 Solution One writes Thus f0 sec2 dz 1 These two examples set up the question How about f0 sec3 dz The integral of sec3 rears its head in a substantial number of calculus mod eling problems2 So we make a special effort on its behalf Here it is EXAMPLE Evaluate the integral 7r Z 3 sec 0 Solution Using the integration by parts formula 17 b udvuvlzi Udu 04 1 tan2 9 sec2 9 7 17 and the identity 2For example it occurs in the computation of the arc length of the parabolic arc modeled byyxx27 ngg 1 one writes 1 W 7 ft W 7 Z tan sec tan dz f7O7Ozsecztan2zdz 7 7 gsecz sec2z7 z 7 ltgtlt Ww 7 7 Zsecilz z Zsecz z 7 ltgtd J ltgtd Since the integral f0 sec3 dz appears on both sides of the equation7 one continues with the computation by adding f0 sec3 dz to both sides of the equation This yields 2Zsec3zdz lnsecztanz ln 17ln10 1n 1 Thus f0 sec3zdz g ln 1 Working with the corresponding inde nite integral7 one obtains sec3 dz ln l sec tan sec ztanzC on all intervals on which both sec and tan are de ned The constant C may be interval dependent The integral 1 csc3 dz 4 can be treated in a similar manner The next example is provided in order to complete the picture EXAMPLE Evaluate the inde nite integral sec4wdw7 0 w lt Solution Using the integration by parts formula 17 Ud U LL U7 Udu 1 one writes ltwgtdw sec2ltwgtsec2ltwgtdw se02wtanwdw sec2 w tan w 7 tan w2 sec w sec w tan w dw se02wtanw72tan2wseczwdw sec2 w tan w 7 2 tan2 m tan w dw M secz w tan w 7 2 C Thus 7 Ztam3 fsec4wdwse02wtanw 3 0 O wltg The integral of csc4 w can be treated in a similar manner Although it is tempting to derive general formulas for sec w and cs0 w7 n 1727 we leave these for another treatise Fourier Examples The next two examples are typical of integrations that need to be made in the study of Fourier Series EXAMPLE Evaluate the integral 1 sin 27m cos 37m dm 0 Solution One writes 1 sin 27m cos 37m dz 0 O sin 57m sin iwz dz 1 sin 57m 7 sin dz 1 700s 57m 1 cos 7m 1 7 7 2 57r 0 7r 0 1 11 1 1 1 2 57r 57r 2 7r 7r i 57r 4 577 Thus fol sin 27m cos 37m dm add sin accos y sin as y 1 ism as e y The following example yields to a similar technique EXAMPLE Evaluate the integral 1 cos 37m cos 27m dm 0 7 Solution One writes 1 cos 37m cos 27m dz cos 1 y cos wees y 0 7sinx sin y 1 1 cosac 7 y cos accos y 7 cos 571 cos dz sin acsin y 2 0 add 1 sin 57m 1 sin 7m 1 COSwcosy 7 cosac y 7 7 4 2 2 5 0 7r 0 cos an 7 y 1 1 7 7 0 7 0 7 0 7 0 2lt gt 2lt gt 0 Thus fol cos 37m cos 27m dz 0 We reluctantly leave these Fourier type integrals7 resisting the temptation to dig deeper Change of Variables The changeof Variables formula plays a dominant role in the computation of integrals We recall7 once more7 an exact statement Theorem ChangeofVariables Let Mt denote a real valued differen tiable function de ned on an interval 07d and let a do and b Let 1 denote a real valued function de ned and continuous on an interval I such that zc7 dl Q I If z t is continuous on 0 1 then f ddz d awt dt 1f t is chosen such that a zd and b do then the changeof Variable formula becomes abfdz dcfztggt dt39 ln stating the changeof variables theorem7 we have used the symbol 1 to represent two different mathematical constructsi In the expression Ab 161 dz the symbol 1 represents an arbitrary point in the domain of the func tion The expression dz indicates that the integration takes place with respect to the variable In the expression 01 frtr t dt the symbol 1 represents a realvalued function de ned on the interval 0dli We acknowledge the potential for confusion but believe that when dealing with a change of variable7 this notation is to be preferred When we used this formula earlier we started with an integral of the form 1 A fltzlttgtgtz lttgt dt in which we could identify 1 and Having done that we evaluated that integral by evaluating the integral gym day We called this the method of u substitution The following two examples help to recall the method of u substz39tutz39on EXAMPLE Evaluate the integral 7r 3 cos t dt39 0 1 s1nt One lets mt1sint0gt gandfz1 z 2 So 0d 0 g I 1 2 and a 11 2 Using the change of Variable formula fmdz d fltzlttgtgtz lttgt dt one makes the calculation cost i g 1 d 0 Wait 7 0 Hs mw 1smtdt 0 fztz tdt b 7 MW ln2iln1 7 ln2 One concludes that I t H 1133 dt ln 2 One can put all of this in the language of usubstz39tutz39ons the formal expression 1132 dt is of the form of the for 1 mal expression idui Hence the integral fog 105 dt can be evaluated by evaluating the integral f du between ap propriate limitsi 3For many problems using usubstitutions the Writer makes computations in a less formal Way For example7 one can write 75 cost 7 v 1 d 4 O mdt 7 O ma JrsmODdt d O Eh1s1ntdt lt1nlt1gtsinlttgtgtl ln 2 7 ln1 ln But this is mainly a matter of taste EXAMPLE Evaluate the integral 2 t3 sin t2 dt 0 One lets Mt 2527 0 S t 0 Thusc07 d 7 I g g and x zsin zt7 foo7 gt07 a 01 Noting that z t 2257 0 g t 3 one uses the changeof Variables formula to make the following computation 1 f 2 t3 sin t2dt 2 t2 sin t2tdt 0 0 1 f 5 2 t2 sin t2 2tdt 0 g manwrit gjmdm 1 f 7 zsindm 2 0 Thus the problem has been reduced to evaluating the integral 1 g 7 zsin 2 0 Using the integration by parts formula 17 b udvuvlzi Udu I I this is done as follows 1 1 d 50 zs1nmdz 7 50 micosmdz ltz7coszl 7A 7cosz1dzgt 11 whalele le One concludes that folE t3 sin t2 dz One can put all of this in the language of usubstitutions the formal expression t3 sin t2 dt is ofthe form of the for mal expression lu sin dui Hence the integral fog t3 sin t2 dt 2 can be evaluated by evaluating the integral u sin du be tween appropriate limitsi In a footnote to the previous example we proposed a less formal way to write the solution For this problem there is a corresponding method It starts out as I i 2t3sint2dt 2t2sint22tdt 0 0 lt2zsinzdzgt dt However7 one may not be initially attracted to this method Having become familiar with the change of variables formula in the context of u sulostitution7 one notes that the change of variables formula can be used in another way In seeking to evaluate an integral of the form abfmdm 12 one can choose distinct real numbers 0 and d and choose any real valued function Mt de ned on 0 1 such that 1 z t is continuous on cd i ii C a and Md b EmmWJDQMM It is noted that iii can be replaced by zc dl Q domain 1 Then as long as t is so chosen it follows that b d mfmmwww I C Theoretically this is a powerful result Practically one needs considerable experience and a bit of luck in choosing 05 so that d wmwow can be evaluated with pencil and paper In general this is no small task We offer next four examples where one is able to choose Mt successfully In the rst example an appropriate choice of t is not such a stretch EXAMPLE Evaluate the integral A4cos dm Solution One uses the changeof Variables formula 17 d MMf W 1 C and searches for an appropriate function One notes that if were replaced by x then the integration would be immediate One way to effect this change is to let mt t2 0 g t g 2 Then 9505 2t 0 g t g 2 350 0 and M2 4 Using the changeof Variables formula and the integration by parts formula7 one obtains 4 2 Amman Ocos xtxtdt 2 cost2tdt 0 2 d 20 t sintdt 2 2tsintlgi2 sintgt1dt 0 4sin2702costlg 4sin22cos272cos0 4sin22cos272 Thus f04 cos dz 4sin 2 2 cos 2 7 2 It is interesting to note that fcoszdm and cosE dz can be evalu ated in closed form however there is no known closed form expression for f cos 2 dm In the next example t is chosen with the bene t of experience EXAMPLE Evaluate the integral 1 V 1 7 m2 dx Solution4 One uses the changeof Variables formula 117mm d fltzlttgtgtz lttgt dt 4One thinks Is it possible to replace x by xt and dx by if dt so that the resulting integrand is immediately integrable and searches for an appropriate function A Experience leads one to choose Mt sin t7 7r 6 7 Thuscgdgamc andbzd 90 1 and z t cost g g g g Letting fz t v17 m2 g x g 17 the change of variable formula yields 1 6 1 V1712d1 1 i d frtr tdt c g 2 l 7 sin t cos t dt 7 COS2 t dt cosacy cosxcosy 75inxsiny 1 cos2ac coszw7sin2 an 2 1M d7 2c052m71 1 2 2 2 1 6 cos an E1cos2w 7 l sin2t 6 4 i 6 17r xi lto gt w w 7 6 8 Thus fixlim2dx7 For the next example7 experience again aids in the choice of 5 Hindsight in action 7r lisin2 t cos2 t cost7 lt t lt EXAMPLE Evaluate the integral 2 1 27 dx 7 1 Solution One uses the changeof Variables formula 17 d mm manwrit 1 C and searches for an appropriate function Experience leads one to choose Mt sec 25 Thusc16rd7amc l andbm 22and z t sec ttant7 t g Letting f W E 3 x g 27 the changeof Variables formula yields 2 1 g 1 W dm 4 W sec t tan t dt tanlt sec ttan tdt 7r 3 sectdt l l E G l E i ln2iln 3 1nlt1 2 1 7 2 f mdanilnlt 1gt l 3 Thus 6 Hindsight in action vsec2 t 7 tar t tant lt t lt The next integral looks impossible It is somewhat amazing that letting 05 sin 25 reduces it to the integral of a cosine function EXAMPLE Evaluate the integral 1 7 2m2 71 72 dm W V17 m Solution One lets 17 2952 71 1 7 7 lt lt 7 NE W 2 m 2 and chooses Mt sint 1 lt t lt I 7 4 7 7 439 Having done this one notes that z t2 is contlinuous on 7 10247 E and me W in was g 7 jg So am 7 1 and 07d em Using the change of Variables formula one determines that 1212 7 dz L W t d fsin sin t dt dt cos an y 20570 cosy i f51ncostdt 75in as sin y if 2 cos2ac cos2x7sin2 an I 172sin t 1725m2m cos t dt f 417 sin2 t 1 L cos 2t cos 7 cos t dt 77 cos 2t dt sin 2t 2 pi M N N law M One concludes that The evaluations of the above four examples were successful be cause we were able to choose mt appropriately This raises the question In general7 how does one know how to choose Mt One answer is that considerable experience with radicals in the integrand leads us to look for t in the form of a trigonometric function After that7 choosing mt is an art Choosing mt as a trigonometric function is known in the liter ature as integration by trig substitution Sometimes it is possible to change variables in more than one way and still be able to evaluate in terms of simple functions the resulting integral We 18 revisit one of the above examples7 proposing two other changes of variables one of which is fruitful and the other7 not so We examine rst another changeof variables that works EXAMPLE Evaluate the integral 1 V 1 7 m2 dx Solution One uses the changeof variables formula 17 d mm fltzlttgtgtz lttdt 1 C and searches for an appropriate function Instead of the choice Mt sin t7 g t g g as aloove7 one chooses Mt cos t7 0 g t S n ase7 c 07 d g a zd a and b do z0 1 and z t isint 0 g t g g Letting m 1 7352 3 x g 1 the changeof variables formula yields 1 1 Wm fztz t dt d my sin t dt A sin2 75 7 sin dt 7 sin2 dt cos as y cos accos y g sin2 t dt isinxsiny 0 cos2ac 202 an isinzw 1 2t 125in2x 7700M dt 2 71 0 2 2 sm an 7 317cos2x 7 Tr gt sin 2t 7 3 4 0 mha mama LOP i A O wis ogt ii h cab H 3 C U This is consistent with the answer that we obtained above Next we examine another changeof Varialoles7 one that doesn t work EXAMPLE Evaluate the integral 1 1 V172d E 20 Solution One uses the changeof variables formula 17 d mm manwrit 1 C and searches for an appropriate function Instead of choosing zt sin 25 or zt cos t we try zt at ln g t g 0 Using this change of variables one obtains l 0 lizzdz xlieZtetdt 1116 Even though this equation is correct mathematically it is not clear how to evaluate directly the right hand side with pencil and paper Although we have computed f 1 7 z2 dz by trig substitution we have not considered f dz as a candidate for trig substitution lndeed7 it can be 1 11712 V liz2 evaluated by trig substitution7 but we recognize it of sin 1 as the derivative Looking Back We have successfully evaluated the following integrals o f sec dz 0 f sec2 dz 0 f sec3 dz 0 f sec4 dz 0 f sin 27rz cos 37rz dz 0 f cos 27rz cos 37rz dz 39 f 15180 dt 0 f t3 sin t2 dt 0 f cos dz 0 f H dz 0 f dz o 21 CALCULUS 11 Spring 2011 LECTURE 3 Harry McLaughlin revised 13011 edited by This lecture provides several examples of integral evaluations us ing the method of partial fractions The examples include 1 o f 274 dm 5z2m3 o fima dm ml 39 fm37m274m4 d7 39 f milm1 d9 32 0 fi1 dz Recall We recall from Lecture 27 o integrals of the form f sec m dag f sec2 m dag f secs m dag f sec4 m dag f sin 2 cos 3 day f cos 27m cos 37m dz 0 an introduction to integration by trigonometric substitution 0 the changeof Variables formula A M dz d fltzlttgtgtz lttgt dt End of Recall Introduction The goal of this lecture is to offer a glimpse of the integration technique called partial fraction decomposition We have chosen not to dwell on the underlying algebraic theory but instead look only at examples Partial Fraction Decomposition The integration technique called partial fraction decomposition is appli cable when the integrand is a rational function a polynomial divided by a polynomial We look at integrals of several generic types of rational functions distinguished one from the other by the forms of their denomina tors EXAMPLE Evaluate the integral Solution If the denominator of the integrand were 2 4 we would be able to compute the integral in terms of the arctangent function But it is not so we decompose the integrand into the sum of two fractions each of whose denominators is linear This is done by writing 1 i 1 i A B 2747x721727m72 m2 and calculating the corresponding values of A and B To do this one writes 1 i A2Bx72 72m2 i 72m2 equate numerator 0z1 ABx2A7B equate like powers of a iolentical polynomials have iolentical coef cients A B 0 2A 7 2B 7 1 solve for A and B A 7B 2713 7 2B 1 74B 1 1 TX 1 Z This allows the one to evaluate the integral as follows 1 1 1 i 1 7 Am A dxtO m 1 iinlze2loeilnlz2ll 1 1 Zln1 71112 7 Ztlln 3 e In W 7i1n21fl3 ln2 1 iZln3 Thus We allow a digression here in order to focus on the difference between the two integrands 2174 and m The rst was treated above in the next example we examine the second EXAMPLE Evaluate the integral 1 1 0 m 4 Solution The integral looks something like delmtan l mdm but not exactly1 We adjust it as follows 1 1 1 1 1 om24d Zo 21d 2 1 1 1 7 7 7 d 4 g21lt2gt 35 1 1 1 d m 50 g 1amp6 dag 1 Thus fd lm W114 6 lt is noted that in this example there was no need to decompose the integrand into partial fractions We continue with partial fraction decompositions When the denominator of the integrand is the product of a linear factor and an irreducible quadratic factor instead of the product of two linear factors the procedure is slightly different The next example illustrates the process EXAMPLE Evaluate the integral 25m2m3 3 dm 1 m Fm 1Recall di tan 1 g 700 lt I lt oo Solution We factor the denominator into the product of a linear function and a quadratic function This yields 25x2m3 25x2m3 3 dm 2 dx 1 m m 1 Mm 1 Next one decomposes the integrand into the sum oftwo fractions as follows 5m2m3 7 A BmC x3m 7 m21 i Am2ABz2Cz 7 mz21 7 ABz20zA new equate numerator 5z2z3 ABz20zA equate like powers of an AB 5 C 1 A 3 solve for B B l l E0 This allows one to evaluate the integral as follows 2 2 2 2 5 Wm dm21a 1 m m 1 m 1 m 1 2 2m 31 2 7d nlxll11 12H x 2 1 1 Wm 3ln27ln tan 1m 31n21n 57ln 2 tan 12 itan 11 ln20tan 12i 1l 110302 1W l VAVA I 4 Thus ff dz ln20tan 127 In the next example the denominator of the integrand is the product ofthree distinct linear functions EXAMPLE Evaluate the integral 1 5 m 1 d 0x37m274x4 Hint The denominator vanishes at z 12 Solution Since the linear polynomial m 7 1 is a factor of the denominator one uses long division to divide the denominator by m 7 1 to obtain 3717274m4 71m72x2 From this one writes L1 i 1 i 1 L 37m274z47m71 72 7172 and calculates the corresponding values of A7 B and C This can be done as follows m 1 A B C 37x274z4 x71m72z2 Az274Bm2x72Cm273m2 x 7 1x7 2m 2 z2A B 0 zB 7 30 74A7 2B 20 7 1x7 2m 2 equate numerator 2Finding roots of polynomials is an industry We Work With text book examples in order to make Clear the methods of partial frantion decomposition 0x2x1 z2ABCzB73C74A72B2C equate like powers of an ABC 0 13730 1 74A72B2C 1 solve for A and B 2 A 3 3 B Z 1 o E This allows the one to evaluate the integral as follows I11 e12 e E 1 d 73d 4 d 12 d A1371274z4 I 3171 I0 172 I0 12 I 1 1 igln 7 ll 5 3 0 lln zin i 0 1 e EInoltac2gt1gt 7 1n 71111 2 1n 71112 7113 1n 71112 eguna 71112 71111 ln3 7 1112 7 1112 71172 n57ln2 71112 1112 e g ln3 1n5 2 3 1 igln2 lln3 7 Eln5l A 2 0 Thus 1 5 11 f0 137m274m4 d9 In the next example the denominator of the integrand is the product ofthree linear functions7 two of which are the same EXAMPLE Evaluate the integral 3 1 dz 2 90 12 1 Solution One decomposes the integrand into the sum of three functions as follows 4 i 1 i L x712x1 7 71 71 Az71x 1 Bz1 Cxi 12 90 12 1 equate numerator 1 Az271Bz1Cm272m1 0352 035 1 A 1352 B e 2Cz 41 13 0 equate like powers of an A C 0 B e 20 0 7A B C 1 solve for AB and C 14 B HgtlelH C This allows one to evaluate the integral as follows 3 1 2 ltz71gt2ltz1gt dz 3 if 3 1 7 4 2 7 Axildz2 1712dz 3 dz NH 2 11 1 3 1 1 11n1xil112 E 1 11nlt1z11gt13 3 2 1 1 1 71ln271n1 1 1 11n 4 7 1n in ln 1 1 1 1 11 4 7 711n21 1n2711n3 152 1 1 1 3quot 7 11n2711n31 1 1 7 11n271n31 11 2 1 7 4n 3 4 Thus 3 12 m71m1d 1111 Looking back one wonders why we decomposed the inte grand according to i1i1i 1712z1 7171 1712 117 that is7 why this decomposition and not something else One answer is that this form is prescribed by a theorem not given here Investigating the theorem 1eads away from our focus7 so we let it rest The important thing is our decomposition worksi 1n the following example a new twist77 is needed EXAMPLE Evaluate the integral 1 3 2 dx 0 x 1 Solution Since the degree of the numerator is not less than the degree of the denominator one can use the long division algo rithm to write m32 m1 1 27 1 7 m x 1 Then one computes the integral as follows 1m32 1 2 1 1 Om1dm Ox 7m1dxO mdm 133 132 1 1 377zgt0mlt1z11gt10 Ll11 271 1 7 3 2 H H 2ln2 Thus fol L312 dz 3 ln 2 m In order to not get lost in the forest we back away and ask how does one integrate a fraction whose denominator consists of exactly one irreducible quadratic factor Even though such an integral has been treated earlier in the study of integration we review the technique here in the next example EXAMPLE Evaluate the integral 1 1 11m22m5dm 11 Solution One notes that the discriminant of the quadratic factor is 4 7 20 lt 0 Thus the quadratic factor is irreducible In order to evaluate the integral one completes the square in the denominator and writes 1 1 1 1 dz dz add andsubtract 7112215 1 1124 ch f1 1 e square 0 5 1 2dx 4 710 1 2 1 gtan 1 l 7 tan 1 1 7r i lt1 0 7 I 7 8 Thus 1 1 L1 z22z5 d7 39 Looking Back We have successfully evaluated the following integrals f 21 41 f 124 dx 1 dx mil f 1371274144 dx 1 fm x3i2 f 11 dx o f These examples involve integrands whose denominators are a product of linear factors ii a product of a linear and a quadratic factor and iii a product of linear factors one of which is repeated However there are other naturally occurring cases For example we did not treat the case of an integrand whose denominator has a repeated quadratic factor Our goal is to expose some of the phenomena associated with partial fraction decompo sition but not to dwell on it In the examples above the assumed partial fraction decomposition are pro totypical For example the decomposed form for W was given by 4 i r i r L x712m17x71 71 m139 This form as well as the others used seems intuitively correct their appro priateness is guaranteed by a theorem that is not discussed here Additionally most of the examples above depend upon the assumed but not formally stated theorem that asserts that every polynomial can be written as the product of linear and irreducible quadratic factors We didn t prove that theorem here CALCULUS 11 Spring 2011 LECTURE 4 Harry McLaughlin revised 2111 edited by This lecture prouides o seueral examples of improper integrals 1 1 f0 W dm 1 1 f0 2 d7 1 1 L1 2 d7 00 1 f1 Ed 00 1 f1 Edd 00 1 700 1z2 dm 1 f02 sec zdz f0 e m dm fooo zrfm dm 0 discussion of the terms conuergent and diuergent o a fundamental theorem for improper integrals Recall We recall from Lecture 3 0 every polynomial can theoretically be written as a product of linear and quadratic factors 0 integration of certain rational functions can be effected by partial frac tion decomposition of the integrand End of Recall Introduction Looking back one notes that a Riemann integrable function is necessarily bounded and ii the Riemann integral is de ned for real valued functions de ned on bounded intervals These observations help crystallize the questions ls it possible to extend the de nition of the Riemann integral so that7 eg7 i fol dz makes sense and ii f6 dm makes sense This lecture addresses these two questions the answers are yes We call the extensions improper integrals Our approach to improper integrals is phenomenological We nesse careful de nitions and rely on examples to support understandings1 Improper Integrals We answer the question What is an improper integralf2 with two examples both of which are examined in some detail later 1 For 0 lt t lt 1 one computes ftl dz 2 7 25 and notes that 1 1 l 7d 2 tiIngt m 1Careful de nition rests upon the notion of the union of nested intervals In this case the area of the region under the curve 0 lt z 3 1 is assigned the value 2 However7 fol dz does not exist as a Riemann integral The integrand is not bounded So7 strictly speaking7 the expression fol dz has no meaning it needs a de nition This is explored in the rst example below 1 m2 to For M gt 0 one computes flM dz 1 7 and notes that M 1 lim 7 dz 1 M4100 1 In this case case the area of the region underneath the curve 2 1 g z lt 00 is assigned the value 1 However pdz does not exist as a Riemann integral The domain of the integrand is not a bounded interval So7 strictly speaking7 the expression fl dz has no mean ing it too needs a de nition This is also explored in an example below The examples suggest a need for de ning an integral that is slightly more general than the Riemann integral We do that in this section by example But7 in order to get our feet on the ground we recall brie y the de nition of the Riemann lntegral De nition Let 1 denote a real valued function de ned on the closed and bounded interval 11 Let L denote a real number such that for each 6 gt 0 there exists 6 gt 0 such that if zi0 is a partition of db with maxlglSn lt 67 13118112 lt6 2 Ham 7 L i1 for every choice of the 51 s constrained by zFl g 51 lt zi 1 lt 2 lt n The number L is called the integral of f or the Riemann integral of 1 Instead of writing the symbol L7 one writes the integral of f by gym dz 2As usual Azi zi iziil 1g 139 g n If a real valued function de ned on a7 1 has a Riemann integral then it is said to be integrable or Riemann integrable Most functions that we deal with are Riemann integrable7 but not all The function 1 i 0 lt m lt 1 a 0 M is not Riemann integrable3 But it has an integral in another sense7 called an improper integral This is our rst example below De nition For a real valued function7 1 de ned on a7 b7 a Riemann sum is an expression of the form 71 Zflt5 gtAi i1 where n is a positive integer7 mlMg is a partition of ab7 the points 51 where i1 3 51 g m 1 g i g n are arbitrarily chosen and Axl mi7i717 Roughly speaking7 one says that the integral of an integrable function f is the limit7 as n a 007 of its Riemann sums Abusing our de nition of a limit we write 1 n l fltgtdwmganogfltmm We use this thinking in the analysis of the next example The next example is not that of a Riemann integral but of an improper Rie mann integral The integrand7 ix is not Riemann integrable on 07 17 since it is not bounded there So an additional limiting process is imposed We work through the example and then look back to see what happened EXAMPLE Evaluate the improper integral 1 1 7 dx 0 E In making the evaluation we implicitly de ne the notion of an improper integral 3One recalls that a Riemann integrable function is necessarily bounded It is understood that the function i is not de ned at z 0 But as is seen below7 the expression fol dz is still meaningful Solution One writes I 1 i 1 1 idz lim x 2dx 0 taOt t 1 i A lim 2x2 taOt t 3151 lt2 7 2 272 lim 25 ta0 Thus7 the value of the improper integral is given by This example is prototypical of a whole class of improper integrals7 so we take it apart First we verify that x i is not Riemann integrable on 014 To do that we note that if 0 mo lt 1 lt 2 lt lt mn 1 is a partition of 01 and if 5139 E mnhxi 2 S i S n are arbitrarily chosen7 then the Riemann sum ELI f imi 7 zinl is determined except for the term f 1m1 7 me But since x ix is unbounded in the interval zeal one can make the sum ELI f imiimi1 as large as one wishes by choosing 51 closer and closer to zero Hence m 113 Hema does not exist as a nite number and thus the real valued func tion7 m 0 g x g 17 de ned by 4We can assign arbitrarily For completeness we assume O is not Riemann integrable Having established that fol dz does not exist as a Riemann integral we are able to look back at the computation and note that ft dz does exist as a Riemann integral for positive if no matter how small the value of t We further note that the limit limtn0 ftl dz exists Guided by our intuition we use that limit as an integral of im 0 lt z 3 1 and write it as fol dz 2 In this context the integral fol dz is an im proper Riemann integral Finally apropos of this example the readers attention is di rected to the Fundamental Theorem of lmproper Integrals given at the end of this lecture In the next example we replace i by i and run the experiment EXAMPLE Evaluate the improper integral 1 1 7dz 0 z Again it is noted that the function i is not de ned at z 0 Solution Using the ideas of the previous example one writes 1 1 m 213 351 In ad 351011 lt1 71nlttgtgt 7315111103 Thus One says that the improper integral fol dz is convergent it assumes the value 2 and the improper integral fol idm is divergent it does not assume a value in R If the improper intergral f x dz has a nite value it is said to converge otherwise it is said to diverge In the case f0 idz the integral diverges but in the extended real numbers it has a value namely 00 Before going on we comment on the equation that we wrote above 1 ldzooi 0 I There is not complete agreement in the literature on the suitability of this expression It says that 11 lim idzooi 240 t I This in turn says that by Choosing t positive and Close enough to zero the value of the integral f dz can be made arbitrarily large All of this leaves unanswered the question Does fol dz exist Our answer is yes it exists as an extended real number and it equals 00 The next example extends the thinking of the previous example The re mark at the end of the next example is of particular interest EXAMPLE Evaluate the improper integral 1 1 m Solution One notes that near z 0 the integrand approaches both foo and 00 depending upon the sign of m With that in mind one writes l t 1 3 dm lim 1 all lim 1 dm 1 x taO 1 z taOt t z 1 1 t 1 1 1 1351 EMMA 35 BMW l ltilll llilt 112111 nltgt1t3511nltgt nltgt1 700 00 Thus7 since the expression foo 00 is not de ned in our ex tended number system7 on concludes that the improper integral fil idm does not exist In this case one concludes that f idm is divergent Remark It is tempting to calculate f idm by the following 1 1 1 d96 N 1Hllli1 1Z1 ln1iln1 0 7 thus obtaining an erroneous nite value for the improper inte 5 gral Looking back one notes that we have uncovered two types of divergent in tegrals The two improper integrals foli and are both divergent However the rst has a value in the extended real number system it equals 00 and the second does not have a value in the extended real number system 5We have just introduced another way of possibly de ning an improper integral This is studied elsewhere under the subject heading Cauchy pm39ncz39ple value But we don t admit Cauchy principle values in these notes Before exploring the next example one notes that the expression 00 1 7dm 1x does not represent a Riemann integral since the upper limit is not a real number However7 it is possible to think of flee dz as an improper Rie mann integral and assign to it a real number That is done in the next example EXAMPLE Evaluate the improper integral Solution One writes 00 1 jdz 1 z One concludes that the by CO 2 dx 1 x M 1 lim 7 dz Mace 1 1 M lim lt77 Mace 1 1 1 Tm lim lt7igt lim 1 M4100 M M4100 0 1 1 value of the improper integral is given In this case one says that 00 1 1 f1 p dz 1s convergent The following example is similar to the previous one7 but has a different ending EXAMPLE Evaluate the improper integral 00 1 idx 1 1 Solution One writes 00 1 M 1 idm lim 7dm 1 m M4100 1 m M 1 1 M13100 ML Mug ln M 7 In 11 lim ln 0 M4100 00 Thus fooo idm is divergent One concludes that Making a comparison of the two integrands and 27 one nds that 00 1 f1 T2 dz 1s convergent it assumes the value 1 and f1 idm is divergent but it assumes the value 00 It is interesting to note that both of the functions i and converge to zero as x a 00 However the rst converges so fast that the area of the region underneath the curve 712 1 3 z lt 00 is nite it is 1 And the area of the region underneath the curve 2 3 z lt 00 is not nite it is 00 In one of the worksheets there is a chance to study the corresponding integral of the in loetween function7 g 1 g z lt 006 There it is seen that 1 idmoo 2 mlnz The next example crystallizes still another phenomena associated with im proper integrals EXAMPLE Evaluate the improper integral 1 72011 7001m Solution One chooses any real numloer7 call it 017 and writes 1 a 1 1 7d 7d 7d 1001m2 1001m2 ma 1m2 In our calculation we choose a 07 for ease of computation One has 1 0 1 00 1 7d 7d 7d W1m2 W1m2 3 1m2 1 0 1 1 M 1 NEIEmdem l nm 0 17 0 M lim arctan N M4100 lim arctanm Naioo 0 lim arctan 0 7 arctan Naioo lim arctanMiarctan M4100 7 lim arctanN lim arctanM Naioo M4100 lt gt 2 2 7139 6One notes that i lt lt l 1 lt x lt 00 x2 xlnx x7 i Thus7 co 1 700 112 dm One says that 1 72 dz is convergent 00 1 m With regard to this example it is tempting to equate If g dx to M 1 lim 7 dx Mace 7 M 1 x2 and then compute M 1 M lim 7 dx lim arctan Mace 7M1x2 Mace M Mh moJarctan 7 arctan lim 2 arctan M NX 7r This is consistent With the answer obtained above but is a dangerous proce dure Using this procedure one could obtain the erroneous result 0 M x dx N lim x dx 7w Mace 7M x2 M hm 7 Mace 2 7M M2 M2 Alli T 7 new 0 When7 in fact 00 o M xdx hm xdx lim xdx 7w N ioo N Mace O 2 2 NEE l MILE 7 700 00 Which is not de ned in our extended number system In this example If x dx diverges but If xdx doesn t equal either a real number or ioo 13 One says that If z dz is divergent We insert two more examples EXAMPLE Evaluate the improper integral 2 sec dz 0 Solution Noting that see is not de ned7 one writes t sec dz lim sec dz 0 ta7 0 lim ln sec tan my lim ln sec 25 tan 7 ln sec 0 tan my oo 7 0 00 Thus f0 sec dz 00 One says that f0 sec dz is divergent The next example is worthy of a good look EXAMPLE Evaluate the improper integral CO ze m dz 0 This is a particularly interesting example On one hand one nds that 000 z dz 00 and on the other hand few 6 dz 1 So what happens to the integral of the product of these two functions one of which has a convergent integral and 14 the other not In our analysis below we nd that the de Cay of 6 dominates the growth of z as z approaches 007 that is fem ze dz is nite Although we clonlt make the argument here one can show that few zne dz is nite for every nonnegative integer n Solution Using integration by parts and l Hospital7s Rule one writes 00 zei dz 0 M lim zeim dz Mace 0 M d MliinooA Ig 71 dI udu uuiUdu M Mlignmre l3 Mlignm e 1dr 7 7M 7 7 7x M MAW e gt 0 Allin 0 Maw 7M Mace 07 lim 6 71 M Mew M wM 0701 1 7 1 Thus f6 ze m dz 1 One says that 00 1 i f0 ze dz 1s convergent One observes the values of the two improper integrals CO ze m dz 1 0 CO e mdz 1 0 15 and One concludes that the multiplier x in the integrand xe m does not effect the value of the improper integral in this case In teresting huh Looking Back o It is tempting to nail down an explicit de nition of improper integral Making such a de nition is possible7 but we have resisted the tempta tion to do so we are content with examples of improper integrals With regard to an improper integral we have used the word exists to mean that the value ofthe improper integral exists as an extended real number7 that is7 it is either a nite number or ioo We have used the words convergent and divergent in several contexts 7 An improper integral is convergent if it assumes a nite value 7 An improper integral is divergent if either i it assume the value 00 or foo or ii it assumes no value at all Additionally7 we have introduced a potential notational problem by using the expression fa x dx in two different ways 1 1t denotes a Riemann integral when x is Riemann integrable and ii it denotes an improper Riemann integral when x is not Riemann integrable but the improper Riemann integral of x exists We take some com fort in knowing that this ambiguity does not generally hinder under standing In the analysis of the rst example of this lecture we could have taken some liberties and written 11d M1 7x m2 0 0 270 2 This is of course the correct answer but arrived at by devious methods Prior to the introduction of the notion of an improper inte gral the starting expression namely fol dm has no meaning since the function i 0 lt z 3 1 is not Riemann integrable But these steps leading to the solution are so appealing that we cast about for mathematical justi cation The following theorem is inserted here for that reason It is used extensively later in these notes Theorem Fundamental Theorem for Improper Integrals Let x be a real valued function de ned and continuous on 11 Further assume that f z exists on ab Then if the improper integral fz dz exists as a real number it follows that b Mm fltbgt ma 8 W O l a m H V In the example above m 2 Proof Let 0 lt h lt b 7 at Then using the fundamental theorem of calculus and the fact that is continuous at z a one has I b dz ling h dz 334M 7 fah f b i f a This completes the proof CALCULUS 11 Spring 2011 LECTURE 5 Harry McLaughlin revised 2311 edited by This lecture provides 0 a de nition of arc length every curve has an arc length 0 an integral model of arc length 17 v1ltyltmgtgt2dx 0 several examples of arc length calculations ym m 0 m 1 W 0 g m S 1 ymx7m71 z 4 cosh z7 0 g z 3 1 Recall We recall from Lecture 4 o Improper integrals arise When integrals are unbounded and when in tervals are unbounded o lmproper integrals are evaluated by computing limits of Riemann in tegrals End of Recall Introduction Our goal is to i to make a precise mathematical de nition of the term arc length of a planar curve ii then nd a computable mathematical expression that yields arc length a number for a planar curve Arc Length We assume that a curve C is modeled by a real valued function de ned on an interval 11 that is CmyER2yyx a ng1 One can imagine a continuous curve but continuity is not necessary for our de nition One chooses points 133 0 S i S n on the curve so that i71 lt m 1 g i g 71 These points partition the curve into n arcs Then one approximates the curve C by a piecewise linear curve consisting of n chords each chord is determined by two partition points MA 344 and mi yi 1 S i S n We write y for Since the length of a chord in the plane has already been de ned one can model the length of the piecewise linear curve by summing the lengths of the chords that is by the expression n 2 V M M702 M 7 gm i1 1The question What is a curve77 does not have an easy answer One can look for characteristics of special point sets in R2 or R3 that permit the point sets to be called curves Alternatively a point set can be called a curve if it has a particular model One such model is xy E R2 y yx a g x g b where y is a realvalued function de ned on the interval 11 In these notes we will investigate this and two other models of curves If the length of each ofthese chords is small enough the sum of chord lengths is an approximation for what one intuitively believes the length of C should be It is noted that this approximation depends upon both the number and placement of the points yi0 One also notes that if one more point is adjoined to a given partition then the sum of the lengths of the chords does not decrease and almost always it will increase This suggests considering partitions with any nite number of points and de ning the length of the curve as the least upper bound of all possible nite sums of the chord lengths If one admits the symbol 00 to represent a nonfinite least upper bound then it follows that every curve de ned on ab independently of smoothness has an arc length In what follows we partition the interval 1 1 instead of the curve itself This is done without loss since it is assumed the the curve is modeled by a real valued function de ned on 1 1 De nition Let be a real valued function de ned on the interval 1 1 whose graph is a curve denoted by C The arc length of the curve C as modeled by ym is de ned to be the least upper bound of M H 90139 i 901202 24121 where the least upper bound is computed over all nite partitions of 1 b and where yi 0 g 1 S n Sometimes the term arc length is shortened to length2 It is noted that this least upper bound could be 00 In fact there exist continuous real valued functions de ned on 1 1 whose 2 ractice one frequently attaches units to arc lengths for example centimeters inches miles etc However the mathematics behind our study of arc lengths demands no units Consequently speci c units don t appear in our calculations graphs have in nite length3 One observes that if the curve C is translated or re ected through one of the coordinate axes then the arc length doesn t change4 So for exam ple the curves x m2 0 g z 3 1 and gm zz 0 g x g 1 and hz 2 1 0 g x g 1 all have the same arc length Before going on we comment that up to this point we have made no as sumptions about how smooth is it could be wildly discontinuous In particular we have been able to de ne the notion of arc length of a curve independently of whether its function model is differentiable or even con tinuous For example according to our de nition the arc length of the step function 0 if 71 g m g 0 f9 1if0ltggg1 is 3 Having de ned the notion of arc length we argue that the arc length of C denoted here by C or just 1 is given by the integral 1 10 b 1 ywzwdz provided that is continuous on 1 b and ii y m exists on 1 b and iii 1 2 is integrable on 1 1 Assuming these three hypotheses one partitions the interval 1 b by 1 mo lt 1 lt lt mnnl lt m b and uses the mean value theorem to write 3One such function is M 3mm 23 1 4In fact using a more general de nition of arc length one can argue that the arc length of a curve is invariant under every rigid motion However it is not our intent to investigate such a theory here n E Ii712 M702 2391 397 I 2 1 lt91 yzilgt gt 7 7amp2 Ii 1121 M M W H H Jr A 2 ll SZ S LL V m A E l E l H V a 2 5 Ar 0 a m m a lt m E 0 1y 5i2zrzi71 yweywki ycxx wki i where it is understood that zFl lt 51 lt m 1 g i g n In using the mean value theorem we have used the facts that each yi is of the form and that satis es the hypotheses and ii above This last sum is a Riemann sum for the function 1 gz27 a g x g 1 Finding the least upper bound7 over all partitions of 11 of the sums 90139 i 901202 yi712 M H is accomplished by computing the least upper bound over all partitions of 11 of the sums 1 yEi2i 95121 M H But this least upper bound is the Riemann integral of 1 y m2 pro vided f 1y z2 dm a g x b7 exists as a nite number Thus 17 z 10 1 yltzgtgt2dz 1 provided f H1 y m2 dm a g z 3 1 exists and is a nite number Summarizing7 one has the following theorem Theorem Let denote a continuous real valued function de ned on 11 and differentiable on 017 1 Then the arc length of its graph7 C7 as modeled by Mm is given by the Riemann integral 1 i 1 yltzgtgt2dz provided that the integral exists as a nite number We pause to crystallize a subtlety that is partially camou aged by the theoremi The theorem asserts that if the Riemann integra b 1 ltyltzgtgt2dz exists then it equals arc length The theorem does not address the question ls is possible for C to have an arc length which is not com putable as a Riemann integral The answer to this question is yes An example is seen below where we deal with the curve modeled by zgyli We explore the use of the theorem in the following section Arc Length Calculations In the next four examples the functions all satisfy the hypotheses of the previous theorem In the fth example there is need to augment the theoremThe sixth example provides an opportunity to review the role of trig substitutions in integral evaluations MODELING EXAMPLE iARC LENGTH COMPUTA TION Compute the arc length l of the curve C as modeled by 2 3x3 03 m S 1 Solution We compute the arc length l by lOl 1y m2dx 7 In order to evaluate the integral we make the following compu tations For 0 g x g 17 MM g5 JW 5 Hz 7 z 1ym2 1x Using this7 the value ofl is computed by 1 z M1y m2dz 0 1 V 1 m dm 0 1 gg 0 Thus The value ofl can also be expressed in the equivalent forms ll NM ltM71gt L MODELING EXAMPLE7ARC LENGTH COMPUTA TION Compute the arc length l of the curve C as modeled by ymv27m2 nggl Solution We compute the arc length using the formula 1 l x1y2dx 0 8 In order to evaluate the integral we make the following compu tations For 0 g x g 17 W W ew 1 24 From this one determines the arc length by the computation Alma Al zdz WW 1 x arcsin 0 ltarcsin 7 arcsin 2 i 0 g One determines that the arc length l of the curve modeled by V2 imz 0 g x g 1 is given by Another Approach If my satis es the equation x2iz27 0 g x g 1 then y2z2 27 so the point my lies on a circle with radius x2 and centered at the origin Conversely7 if y2 2 2 and 0 g x g 1 then y V2 7 m2 Thus the curve modeled by 2 7m27 0 3 x g 1 is the arc of the circle 1th between 07 and 11 So its length is g of the circumference of the circle That is7 its length is g 27r2 g MODELING EXAMPLE ARC LENGTH COMPUTA TION Compute the arc length l of the curve C as modeled by 1 ym z7z1 m 4 Solution We compute the arc length using the formula 1 14 1 y m2dm In order to evaluate the integral we make the following compu tations For 1 g x g 4 W ym 7 m 1 1 gt 7 m 7 2 m 1171 2 m5 1 x272m1 2 7 was 4 m 4z1 272x1 12902 4m x22z1 4m 7 x 12 7 4m 1 1 2 a y ltzgtgt M From this one determines the arc length by the computation 14 1 yltzgtgt2dz H N 1 98 3 RH amp R 1 8 1 ltg2gtlt1gt 14 4 3 3 10 3 E 2 7 Thus7 the arc length l of the curve modeled by z 1 xi 1 m 4is given by MODELING EXAMPLE iARC LENGTH COMPUTA TION Compute the arc length l of the curve C as modeled by cosh z7 0 g x g 1 Solution One recalls that cosh We compute the arc length using the formula 1 z 1y m2 dz 0 Preliminary computations yield7 for foo lt z lt oo 62172em57m672m 4 52z572z72 4 452m572m72 4 7 1 2 21 72m 1 5 5 l WY 1 was 3 ex 64 2 cosh We note in passing that we have just found that if cosh then 1 y m2 Mm z E R5 In order to compute the arc length one recalls that sinh exit2 2 z E R and writes I l 0 1y z2dz Ocoshmdz 5The computation could have been shortened a bit by using the identity 1sinh2 cosh2 x 700 lt I lt 00 and then Writing yc cosh yx sinh 1y2 1Sillh2I cosh2 1y x2 coshx sinh l5 sinh 1 7 sinh 0 sinh From this one determines that the arc length l of the curve mod eled by7 cosh z7 0 g z 3 17 is given by h1 It is noted that this arc length can also be written as l le7e l MODELING EXAMPLE 7 ARC LENGTH COMPUTA TION Compute the arc length l of the curve modeled implicitly by It is noted that this curve can not be modeled by Mm 71 g x g 1 Thus we cannot use our de nition of arc length directly Instead we use human intuition and model the arc length of the implicit curve by computing the arc length of the curve modeled by 1 7 0 g x g 1 and multiplying the result by four We are computing the arc length of the implicit curve restricted to the rst quadrant and multiplying that length by four Additionally7 we encounter another obstacle It is seen below that the arc length integral needs to be interpreted as an im proper integral since 1 y z2 0 lt x g 1 is not bounded Fortunately we are able to realize that the value of the improper integral yields the arc length in the rst quadrant as de ned at the outset A careful argument is needed but such an argument transcends the focus of these notes The calculation of y m is facilitated by differentiating implicitly the equation mg 34 1 This yields 2 3 we W 7 0ltz1 My 3 7Oltm 1 2 1 2 0ltlt1 m 1 7271 0ltzlt1 z W e oltz1 1 y m2 m0ltz 1 Using an improper integral of this last obtained formula7 one can calculate the arc length in the rst quadrant by l 1 1y x2dm m7dm 0 0 3 21 7x3 2 0 3 7170 2lt i 3 7 2 Thus the arc length in the rst quadrant is found to be multi plying this by four one nds that according to our multiply by four model7 the arc length l of the whole curve modeled by is given by We offer one more example of arc length calculations The computation revisits integration using trig substitutions MODELING EXAMPLE ARC LENGTH COMPUTA TION Compute the length l of the parabolic arc modeled by Solution We compute the arc length 1 using the formula 1 01 1 y m2dm A preliminary computation7 for 0 S x g 17 yields m2 290 290 y 2 4962 WW H4252 2 V 1 4x2 From this one obtains l01 1y z2dm01 V14m2dm At this point one looks for a trigonometric substitution 1t to be used in the change of variable formula 1 d 01412dz 14It21tdt 15 Where 15 0 and 1d 1 One Choice of 1t that works is 1 1t E tan t7 0 g t3 tan 1 In this case one has7 for 0 S 75 312811710 1 t 1 E sec2 t 14zt2 H1 4 tan2 t 14ltzlttgtgt2z lttgt 1 tan2 t S C2 t S C2 sec2 t sec3 With the change of variables Mt tan 0 S t tan 1 2 one is able to write 01 1 yltzgtgt2dz 7 1 V1 4x2 dm 0 d 14zt2x tdt tan 1 2 g sec3 t dt 0 j sec3 t dt secttant 1nectta t C 1 an Zsec ttan 25MB 12 1 axr 11nsecttant3 2 1 EM 0 Inlt 3 2 7 1nlt1 0W ilt2 51n52gt Thus the length l of the parabolic are is given by l251nx52 It is worthwhile remarking that there are relatively few curves in R2 whose arc lengths can be computed by pencil and paper In practice one almost al ways resorts to numerical evaluations of the integrals The examples above are text book examples they are representative of the handful of arc lengths that can be evaluated by pencil and paper Looking Back At the outset we provided a de nition of arc length of a curve C There we insisted that the curve be modeled by a real valued func tion This means that the arc length of C depends upon the way it is modeled If the point set C is modeled in another way7 say implicitly or parametrically7 we are not yet able to determine its arc length without rst nding a model in the form of a real valued function Later in these notes we are able to de ne arc length for parametrically mod eled curves and argue that this de nition is compatible with the de nition given here But this is only one step in being able to de ne the notion of arc length for arbitrary planar curves More generally7 trying to characterize point sets in R27 that look like curves and look as thought they have nite lengths7 opens a Pandora s box We don t go there Later we argue that there doesn t seem to be much hope for nding a model independent de nition of arc length However we are able to live successfully without such CALCULUS 11 Spring 2011 LECTURE 6 Harry McLaughlin revised 2511 edited by This lecture prouides o a model for areas of surfaces of reuolution o a formula for computing areas of surfaces of reuolution b A 2w y1 yltzgtgt2dz I o examples of surface area calculations yz 1 m 2 yzz7 nggh rgt0 cone m 7r S m g r sphere Mm 7 1g m lt oo Gabriel s Horn 0 if z 0 y1if0ltz 1 Recall We recall from Lecture 5 o a de nition of arc length of a curve modeled by Mm a 3 z 3 b o a theorem asserting that the integral f 1 302 dz can be used for arc length evaluation of smooth real valued functions 0 examples of arc length calculations End of Recall Introduction In a previous lecture we were able to de ne the arc length of a planar curve modeled by Mm a g x g b We approximated the curve by a piecewise linear function computed the length of the piecewise linear function and then used the least upper bound of such lengths as the length of the curve Our success there hinged on the fact that we had a standard namely the length of a planar chord For a surface de ned by a real valued function of two real variables 2 fm there is an analogous process It is possible to triangulate the sur face sum the areas of the triangular facets and then nd the sums least upper bound over all triangulations However that study transcends the mathematics of this treatise This leaves us in the awkward position of wanting to compute surface area without having an underlying de nition of surface area Our work around is to de ne a notion of surface area for a special class of surfaces For surfaces in this class our de nition of surface area is based on a limiting process that in turn leads to an integral We remain mindful of the fact that our surface area calculations are based on our special de nition 7 that a surface in this class may also fall in another class for which another de nition of area is provided In such an event we have a potential for con ict Surface Area In order to get started we mentally draw the curve 1 g x g 2 on an m y plane1 Having done that we imagine the 3 d surface obtained 1The words Curve yc 1 g x g 2 on and Lyplane77 mean the point set modeled by ay R21y571SI 2 by rotating this curve about the z axis We call such a surface a surface of revolution We seek to model our intuitive notion of the area of this surface2 We continue to think behind the scenes in terms of 1 g x g 2 but consider a general non negative real valued continuous function ym a g x g 1 One partitions the interval 11 by a mo lt 1 lt lt MEI lt zn b For each i one focuses on the arc of this curve between the two points i71yi1 and mi y and generates the ith surface by rotating this arc around the z axis We let yi 0 g i g The whole surface is then the composite of the ith surfaces This brings us to the crucial question in our modeling process How can one model the area of the ith surface As a rst approximation one can think of the ith surface as a horizontal cylinder having two circular bases each of radius y or 344 and having height xi 7 MA The lateral surface area excluding the two bases of the Cylinder lS 7 miil or 2717 1 7 i71 To model the complete surface area one could sum over i the areas of the cylinders and then let the size of the partition go to zero This results in an integral of the form f 27rym dm Unfortunately this does not yield gen erally accepted values for areas of classically studied surfaces for example cones and spheres We suspect that the problem lies with the fact that the height ofthe ith cylin der should be something closer to the length ofthe chord between MA 351 and mi Using this intuition we are able to make the following de nition De nition Let denote a continuous real valued non negative function de ned on the interval 11 Let xi0 be an arbitrary partition of 11 and let Ax z 7 MA i 1 71 Let S denote the lateral 3 d surface obtained by rotating about the m axis the function ym a 3 x g b The 2One needs a notion of surface area in order to determine the amount of paint needed to paint a surface One needs a notion of surface area in order to determine the heat radiation characteristics of a surface etc area of the surface S is de ned to be 71 hm Z 2W9ltMgtV i 961202 yi712 i1 provided the limit exists as either a real number or as 00 It is understood that the limit is taken over all partitions a mo lt ml lt lt mn b of 11 as max Ami goes to zero This is a good mathematical de nition of surface area but as in the case of arc length it is not amenable to direct calculation So assuming that y m exists on a b and mimicking the analysis of the arc length de nition one uses the mean value theorem to write 7 I 2 i i712yi yiil2 1lt7yl ylilgtim71 mi 7 miil V1 iiED i 96121 Where 141 lt lt 13 1S i S 77 In this case the radius of the ith cylinder can be taken to be y or 351 giving the ith cylinder a surface area of for example 27WltMW 1 M50 9 i 90H Summing all of these yields TL 2 27ryiv 1 Nil 9 i 90H i1 Whose limit is the total surface area This gives rise to following theorem Theorem Let be a non negative real valued function de ned and con tinuous on 11 and differentiable on 11 Let ml g be an arbitrary partition of 11 and let Ami xi 7 m4 i 1 71 Let S denote the 3 d surface formed by rotating around the z axis The area of S denoted here by A is given by A lim 2 2mm 1 y i2 AM i1 provided the limit exists as either a real number or as 00 It is understood that the limit is taken over all partitions a mo lt ml lt lt mn b of 11 as max Ami goes to zero It is also understood that the 51 s are given by the mean value theorem y izi 7 l 71 yi 7 yiil 1 g i g 71 But the limit given in the theorem looks like an integral Indeed if the Riemann integral 7 2w yltzgtv1 yltzgtgt2dz exists as a real number then the limit in the theorem exists as a limit of Riemann sums and one has 17 A 2w WW 21mm We have taken some care with this statement It says that if the Riemann integral exists then its value is the area of the lateral surface It does not assert that if the area of the lateral surface exists as a nite number then it is given by the value of the integral It may happen that the area of the lateral surface exists as a nite number and yet the Riemann integral does not exist We don t pursue this phenomenon here At the risk of being pedantic we capture these ideas in a formally stated theorem The theorem deals with nite surface areas Theorem The surface area A of the surface generated by rotating about the m axis the graph of a non negative real valued function Mm continuous on 11 and differentiable on 0119 is given by b A2w yltzgt1 ltyltzgtgt2dz provided that the integral exists as a real number that is provided that the function 1 y z2 a 3 x g b is Riemann integrable This is our jumping off point for surface area calculations It is noted that the hypotheses of the theorem do not require that y m exists at z a or at z b This phenomena is encountered below MODELING EXAMPLE 7 SURFACE AREA COMPU TATION Compute the lateral area A of the surface formed by rotating the function ym 1 m 2 about the z axis The surface area does not include the two circular bases Solution We compute the surface area by the integral 2 A 27r yx 1y m2dx 1 In order to evaluate the integral we make a preliminary calcula tion for 1 g x g 27 WE 96 290 9075 1 2 i i ltyltzgtgt 7 4m 1 2 i i 1ltyltzgtgt i 1 i 4m1 7 4m 4m1 2 1ltyltzgtgt 7 435 Using this calculation one computes a value for the surface area as follows 2 A 2w1yx 1y m2dx 2 QW 41dz 1 4m 27139 2 T V4m1dm 1 7 21 g Wg 41gt21i The domain of the function y in this example was chosen With care y z exists on 12 However if the domain is chosen to be 0 1 then one needs to deal With the potential problem that lim 40 y ooi Fortunately the Whole integrand namely yz 1 9 is Well behaved near I 0 and the integral 2 yaw1 ltyltzgtgt2dz exists as a nite number The computational steps above cover this case and one nds that for this case the area A of the lateral surface is MODELING EXAMPLE 7 SURFACE AREA OF A CONE Compute the lateral surface area A of a right circular cone whose base has radius r and Whose slant height is l where r gt 0 and l gt 0 We don t compute the area of the base The length of the chord from any point on the rim of the base to the apex is 1 Solution We model one generator of the cone as a planar chord that in turn is modeled by yx nggh Where12r2h2 The cone is then modeled by rotating this chord about the z axis It is noted that the equation 2 r2 712 de nes the height h of the cone We compute the surface area of the cone by the integral h AAwmw wmrm In order to evaluate the integral we make a preliminary calcula tion WE 05 M90 wm 2 1me V 1wmrig f Using this one computes the surface area as follows A h a ymwwome 27rJh dz In the next example we use the notation R3 to denote the set of triples of real numbers MODELING EXAMPLE 7 SURFACE AREA OF A SPHERE Compute the surface area of a sphere of radius r where r gt 0 Solution We assume that the surface of a sphere S of radius r is the point set in R3 modeled implicitly by m2y222727 73472 6R3 that is S m7y7z 6R3 m2y222 72 We model the surface of the sphere by rotating about the m axis the planer curve modeled by Vrzima 42572 ln turn7 we compute the surface area A of the sphere by the integral A 27r7 yx 1y m2dx Some care is needed here It turns out that y m becomes un bounded near z 7r and z r So we need to ensure that the integral 27139 f 1 y m2 dz exists as a nite number Keeping all of this in mind we proceed with the calculations7 checking each step carefully In order to evaluate the integral we make a preliminary calcula tion for 7r lt z lt r was was 1 m 572 27 2 7 7 7 762352 m Maw 72 7 m2 2 2 1y 1T272 72 7 m2 m2 72 7352 T2 72 7352 2 7 T 1ltyltzgtgt i Using this one is able to compute the value of the integral 7 2w W 1 2de 7 This is done next where we note that 27139 f yz 1 y m2 dz exists as a Riemann integral 2wyltzgtf1ltyltigtgt dm aruggz m T 27W dm 77quot Thus the area of a sphere of radius r is modeled by The integral above7 fT x 7 2 7 12 W dz exists as a Rie mann integral since the integrand is equal to 7 for 77 lt z lt Ti With regard to the value of the integral7 the values of the integrand at z 77 and z 7 are not of concern The next example is fairly well known and oft repeated in calculus lectures The results of the analysis are counter intuitive MODELING EXAMPLErGABRIELls HORN Let i 1 3 z lt 00 and rotate its graph around the z axis Even though the solid region generated called Gabriel s Horn is unbounded one can both model and compute its volume V by the improper integral co 1 2 V 7r lt7 dx 1 m Further one can both model and compute its surface area A by the improper integral CO A27r l 1i4dx 1 xv z Evaluate the rst integral and bound below the second integral then comment on whether or not Gabriel s Horn can be lled with paint Solution On one hand one has On the other hand one has 1 I 1 A 27r 7 1 jdm 1 z z 00 Z 27139 1 dm 1 z 27rln 27rln 7 1 00 Thus one has 12 One concludes that Gabriel s Horn has a nite volume but an in nite surface area Using paint with in ntessimaly small molecules it can be lled with a nite amount of paint but its surface can not be painted with a nite amount of paint Nifty huh Looking Back In an ideal world we could characterize those point sets in R3 that we believe should be called surfaces Having done that we could then de ne a notion of surface area one that is consistent with human intuition But ours is not an ideal world we don7t have such a characterization of a surface and accompanying de nition of surface area So we do the next best thing we de ne notions of surface area for special classes of 3 d point sets special classes of surfaces In this lecture we have focused on 3 d point sets generated by rotating 2 d curves about the z axis We called these point sets surfaces of reuolutz39on and provided a de nition for their areas Having done that we found an integral formula that allows computation of these areas provided that the 2 d curve is smooth enough In the examples we veri ed that the integral formula provides what most people believe to be the areas of cones and spheres The example dealing with Gabriel s Horn shows how fragile human intuition is when dealing with areas and volumes Final Note Arc Length uersus Surface Area It is noted that the de nition of surface area given above depends upon our understanding of arc length When we de ned arc length we did it for all curves independently of smooth ness In fact we observed that every curve has an arc length However when we de ned the notion of area of a surface of revolution we quietly slipped in the word continuous as a modi er of the function being rotated One asks why cannot one just rotate any 017 curve77 about the m axis for ex ample a step function and use the given de nition of surface area to obtain an area ofthe corresponding surface of revolution One can do exactly that However the surface area so obtained is probably not what one wants 13 What goes wrong The following example is illustrative EXAMPLE Let S be the surface obtained by rotating about the z axis7 the function 35 7 0 if m 0 y 1 if 0 lt m g 1 Use the expression n hm Z 27ry90i i 961202 yi712 i1 to model the area of the surface of S Compare the computed value to the lateral surface area of the cylinder of height 1 and radius 1 Let mlMg be a partition of 01 Then 71 lim 2 27ry90i 90139 M702 241quot M702 i1 hm 271794901 951 7 9002 241 y02 1im Z 27ry90i i 901202 M702 i2 lim 27r1 m1 7 02 1 7 02 lim Zn 27r1 7 mil 0 i2 hm 27mm 12 hm Z 27rm 7 mm i2 27139 hm Mm 12 27139 hm 2m 7 mm i2 27r1 27r1 47r The surface S can be thought of as the lateral surface of a cylin der of height 1 and radius 1 In this case7 one would want the area to be 27r121 27139 14 The formula that we have used in de ning the area of a surface of revolution namely hm Z 2W9ltMgtV i 961202 yi712 i1 yields the value 47139 It evidently computes the area of one end of the cylinder in addition to its lateral surface area It is mainly due to the above example that we have insisted when comput ing the area of a surface of revolution that the to be rotated curve Mm be continuous A Loose End Of course most of this lecture has hinged upon the reader s willingness to accept without de nition the notion of a surface of revolu tion For completeness we offer a formal de nition De nition Let denote a non negative real valued function de ned on the interval 11 and let S denote the surface of revolution obtained by rotating about the z axis the function y Then S is a subset of R3 given by S 9571472 6 R3 242 22 WW2 90 6 la7bl CALCULUS 11 Spring 2011 DRAFT LECTURE 7 Harry McLaughlin revised 21311 edited by This lecture provides a de nition of a vector in R2 a de nition of a line in R2 a parametric model of a line containing two given points conversion of a parametric model of a line to a scalar model a de nition of a vector valved function a de nition of a parametric curve a parametric model of a circle a parametric model of an ellipse a model of the path of an ant moving to and fro on a line two di erent parametric models of the same point set 1 a parametric model ofy mi 700 lt m lt oo Recall We recall from Lecture 6 o a de nition of areas of surfaces of revolution 0 several examples of surface area computations o Gabriel7s Horn End of Recall Introduction In this lecture we begin our study of vector valued functions that is func tions de ned on real intervals and taking values in the Euclidean plane in 2 We nd vector models parametric models for straight lines Then we move on to more general planar curves Here we nd vector models parametric models for circles and ellipses which cannot be mod eled with single scalar valued functions eg y But the foundational idea is that of a vector We have a look Vectors The study of vector valued functions rests on the basic mathematical con struct called a vector space and its elements called vectors The abstract notion of a vector space is perhaps more general than what is needed for the study of curves in the plane and later in R3 So initially we focus on vectors in R2 so called 2 11 vectors Vectors in the Plane One recalls that the Euclidean plane denoted here by R2 is de ned by R2 z ERandy ER In words the Euclidean plane is the set of all ordered pairs of real numbers It is understood that z y u v if and only if both z u and y v For example the ordered pair 2377139 is an element of R2 One writes gm 6 R2 lts z coordinate is 23 and its y coordinate is 7139 The Eu clidean plane exists as a mathematical construct independently of any geo metric notions However7 one frequently models the Euclidean plane on a piece of paper or a computer screen using two calibrated orthogonal lines 7 the m axis and the y axis One can think of each ordered pair z y in R2 as a point7 on the piece of paper or on the computer screen7 located so that its projection on the m axis has the value x and so that its projection on the y axis has the value y It is emphasized that the medium of the paper and the medium of the computer screen offer only models of the Euclidean plane De nition The elements of R2 are called 2d vectors or7 in the context of this lecture7 just vectors We emphasize that a vector is an element of R2 and not an arrow drawn on a piece of paper 7 more about that later Vectors can be added 2 71gt N31 21gt lt2 V3 20 This is called a vector sum Vectors can be multiplied by a scalar by a real number 1 21 7 V321 1 7 A gt v3 This product is called the scalar product The scalar 1 multiplies the vector V3721 It is common to denote a vector in the plane by a single bold face symbol For example if a 27 71 and b 217 then ab2 20 It is this seemingly innocuous notational preference that gives vectors some of their power a lot of information can be encoded in a single symboli In our case7 two coordinates are encoded simultaneously in a single symboli Further7 vectors have lengths If a 271 and b 03721 then the lengths of a and b are denoted by lal and lbl respectively They are given by lal 22712 and lbl72212 pig3 and lbl2111 The length of a vector is sometimes called its magnitude1 That is It is noted that for a vector a and for a scalar it follows that Mal W W where it is understood that vertical bars have been used to denote both the magnitude of a vector and the absolute value of a scalar of a real number For example7 if a 27 71 then Maiixmeuiiwxyeai 4v4A2Mx CF AHm In particular for a non zero vector a one has 21 lal 1 lal In words dividing a vector by its length yields a vector of length 1 fre quently called a unit vector If u 27 3 and V 717 72 the vector di erence or just di erence between u and V is a vector it is given by uev tekLeD 2 17 3 2 3 5 1One notes that we have used the expression lal to denote the length of a vector 3 In the expression lal we use the same vertical bars to denote the absolute value of the real number a It is almost always clear from the context whether or not the vertical bars represent length of a vector or the absolute value of a real number Thus u 7 V 35 We have interpreted the expression 77172 to mean 7171 72 1 The magnitude of the difference between u and V is written lu 7 V and is given by lu7Vl V3252 The magnitude of the difference is called the distance between u and V The vector u 7 V is frequently referred to as the vector from V to u For example one can write u V u 7 V and think of u as the vector V plus the vector from V to u This has special appeal later in these notes when we model points on a line plane closest to an external point One can model geometrically a vector at b on an x y coordinate system by a darkening a point at the location a 1 Additionally one can model the vector at b by drawing a chord from the origin to the point whose co ordinates are a b In addition to these two models one frequently sees the vector ab modeled by putting an arrow on the cord from the origin to the point ab at the ab end Thus the vector ab can be modeled by the point the chord and the arrow Finally one also sees the vector at b modeled by translating the arrow above to an arbitrary position in the zy plane The geometric length of the chordarrow models the length of the vector at b We count 4 different geometric models of the same vector abl In the preceding paragraphs we referred to ab as both a point and as a vector We remark that the terms point and vector are not used consistently in the literature Some authors argue strongly that the two terms name different mathematical constructs other authors argue that p0intuect07 In these lectures we use both terms when speaking of an element of the INSERT URES HERE FIG7 Euclidean plane or Euclidean threespace2 Our choice of vector instead of point is frequently governed by common usage For example we speak of adding two vectors but not adding two points On the ip side we speak of a line through a point but not a line through a vector3 There exist quite elegant geometric models of addition of two vectors and of scalar multiplication of vectors These ideas are shown in the examples of the accompanying two gures Finally it is worthwhile commenting on the geometric signi cance of the difference of two vectors One frequently sees eg the difference vector 51 7 23 referred to as the vector from the point 2 3 to the point 5 1 and repre sented geometrically by an arrow drawn from 2 3 to 51 and pointed at 51 Below we think of such vectors as direction vectors Some authors reserve the term direction vectors77 for vectors of length one the so called unit vectors but we allow direction vectors to have any positive length We have not mentioned the notion of multiplication of vectors There is a commonly used notion of the product of two vectors called the dot product Here the product is not another vector but instead a real number More about that later During our investigation of vectors in R3 we introduce the notion of cross product However for the time being we limit our notions of product to the scalar product de ned above 2Euclidean threespace is denoted by R3 At an abstract level it looks just like the Euclidean plane except its elements are ordered triples of real numbers One usually writes an arbitrary element in R3 in the form x y z where 1 y z are real numbers 3We could speak of adding two points and a line through a vector without saying some thing incorrectly However most working mathematicians would consider these phrases a bit awkward INSERT URES HERE FIG The above introduction to vectors and the accompanying terminology is a bit tersei One can use the examples below to get comfortable77 with most of the ideas Lines in R2 Our goal in this section is to de ne the notion of a line in the plane and then to investigate various models of lines De nition A point set L in R2 containing two distinct points p and q and having the property that every point in L can be expressed as 17pq7 forsome 7 7ooltltoo is called a line in R2 Such a line is said to be determined by p and q For example7 if p 23 and q G 7 then every point 17y on the line determined by p and q has the form my 17 A2 A 1 7 A3 A 3 7 27 for some real number A De nition Let a and b denote two distinct points on a line in R2 Then the vector a 7 b is said to be a direction vector of the line Similarly7 the vector b 7 a is said to be a direction vector of the line One notes that if v say v a 7 b is a direction vector for a line then v7 E R2 7 0 is also a direction vector for the same line De nition Two lines are parallel is they share a common direction vector De nition Let p and q denote distinct points vectors in the plane Each of the expressions 17pq7 fooltltoo and pqip7 fooltltoo is called a vector model of the line containing p and q or the line deter mined by p and q4 The same expressions are sometimes called paramet ric models of a line The symbol is called a parameter One notes that the two expressions 17pq7 foo lt lt 00 and pqip7 foo lt lt 00 model the same line L in R2 Further7 if the two distinct points u and V belong to the line determined by p and q then both p and q belong to the line determined by u and V This requires a proof MODELING EXAMPLE iVECTOR MODEL OF A LINE THROUGH THE ORIGIN Let p 00 and q 21 Then the line determined by p and q is the set of all z y satisfying any 070 271 271 for some E R We make two observations 4For the expression pq7p one thinks Go out to p7 look in the direction of q multiply the Vector from p to q by A and add the result to p i The point z y lies on the line if and only if z y 2A A for some real number A For A 7S 0 the ratio of the y component to the the z component is always this sup ports our previous experience with modeling lines in the plane through the origin that is by y mm foo lt z lt 00 Here m 12 ii If A lies between 0 and 1 then x y lies between 00 and 21 if A gt 1 then 21 lies between 00 and zy if A lt 0 then 00 lies between z y and 215 Notation The expression p Aq 7 p foo lt A lt 00 used to model a line is frequently denoted by the single expression rA that is rA pAq7p foolt A lt 00 This is not unlike replacing the expression 2x 3 foo lt z lt 00 by the single expression fm in which case fm2m3 fooltmltoo MODELING EXAMPLE7 PARAMETRIC MODEL OF A LINE CONTAINING TWO POINTS Let p 12 and q 35 Letting rA represent an ar bitrary point on the line determined by p and q one writes a vector model for the line by rA pAqip7oolt Alt oo Fixing A and setting rA z y one has m lt172gtAltlt375gt7lt172gt 7 1 2A 2 3A One concludes that zy lies on the line if and only if z y 1 2A 23A for some some value of the parameter A One says that the line is modeled parametrically by zA where 5Here we are pulling the wool over the reader s eyes When one plots the line determined by 00 and 2 1 we are able to interpret the word between in a familiar way However at an abstract level the word between is de ned in terms of the Value of A In spite of this we hope the intent is Clear 10 V V V V 1 2 2 37 fooltltoo In the literature one frequently sees the expressions vector model and pam metric model used interchangeably We do that here We have provided a vector approach to de ning and modeling lines One may ask here7 how do the two sclar models perhaps more familiar ambyc and y 0 9w 6R2 771z l b7 fooltxltoo relate to the parametric vector model The next two examples deal with this question MODELING EXAMPLE 7 CONVERTING A PARA METRIC MODEL TO A SCALAR MODEL To do this one assumes that z y lies on the line modeled para metrically by mxyx 1 2 2 3A7 700 lt lt 00 and makes the following computations for foo lt z lt 00 m y 12 23 1 595 1 23m71 One concludes that if my lies on the line modeled by z 12andy2 l 37 fooltltoothen NlH At this point one asks Whether or not every point on the line modeled by y z foo lt z lt 007 lies on the line modeled parametrically by z 12 y 237 fooltltoo7 We argue that if z y E R2 and if y z then there exists a such that z 12 y 23 No matter how z is chosen there exists a value of 7 call it A0 such that z 1 2A0 In fact one can choose 0 mgl Since z and y are related by the equation y z and z 1 2A07 one can make the calculation 31 4 y 2 2 3 1 712A0 2 3 1 n 3 7 2 02 230 Thus z 1 1 2A0 and y 2 3A0 One concludes that if z y E R2 and y z then there exists a such that z12 and y23 The above example provides an algorithm for converting a parametric model of a line to a scalar model of the form y mm b it does not provide an algorithm for converting a scalar model of the form y mm b to a parametric model6 We handle this in the following example MODELING EXAMPLE 7 CONVERTING A SCALAR MODEL TO A PARAMETRIC MODEL Find a parametric model for the line modeled by in lt lt y72z 200zoo Solution We step through a general algorithm If x 1 then y 2 so 1 2 lies on the line If x 3 then y 5 so 3 5 lies on the line Letting p 1 2 and q 3 5 one models the line containing p and q by the expression pqp7 00lt lt 00 Thus pqip 12 371572 172 lt2A73Agt 12A23 Hence one suspects that a parametric model of the line is given by m 12 and y 23 foolt ltoo We haven7t argued explicitly that the two sets z y E 2 y and E R2 both z 1 2 and y 2 3 for some are the same However the needed ideas are visible in the previous 6In going from the scalar model above to the parametric model we didn t have to discover the form of the parametric model we said at the outset that it has the form ac 12A y23 fooltltoo 13 exam ple Using these ideas one can assert that our suspected model MA 12 and y 23 foolt ltoo is indeed the parametric model sought One might ask Does the parametric model found above de pend upon the arbitrarily selected points 1 2 and 3 5 If one selected two other points on the line would one nd a different parametric model The answer is yesl We choose two other points If I 0 then y so 0 lies on the line If I 2 then y g so 2 lies on the liner Letting p 0 and q 2 g one models the line containing p and q by the expression pAq7p foo lt lt 00 Thus lto gtAlt2ogi gt a lt2A13gt700ltlt00i Hence a parametric model of the line is given by z2 and y3 700ltAltooi It is emphasized that both parametric models model the same line the same pointset in R This ends our introduction of linear parametric models VectorValued Functions and Parametric Curves Our discussion above introduces the notion of modeling point sets in the plane by vector valued functions Our focus was on straight lines In this section we consider other non linear point sets We motivate our study with a story of an ant on the patio table Anticipating that an ant is going to appear on our patio table we construct an m y coordinate system on the table top When the ant does appear we are able to project its position at each instant of time onto each of the two axes the x axis and the y axis This provides a way for recording the mo tion of the ant its m projection as a function of time t and its y projection as a function of time t The ant leaves us with two functions x as a function of time zt and y as a function of time We record the motion of the ant by the ordered pair Mt Here the time variable t is constrained to some nite interval that coincides with the time the ant remains on the table At each instant of time to the ants location is modeled by a point in the my plane namely the point zt0yt0 But this is just a vector in R2 As noted earlier it is customary to denote the vector by a single symbol say rt0 So for every if one has rt Thus the vector valued function rt models the ants location as a function of time De nition Let I denote a real interval A vectorvalued function de ned on I is a function from I into R2 The de nition just given is a special case of a substantially more gen eral notion of vectorvalued functions The generalization transcends the mathematical constructs of this treatise So we are not able to include it here This is a succinct de nition but it doesn t provide the structure handles one needs for modeling paths such as the ants path7 So we take it apart 7We use the term path rather than curve since the term curve has already been de ned in terms of a scalarValued function and the path of the ant may not be a curve 15 Let I denote a real interval and let rt be a vector valued function de ned on I Then F05 t7yt7 t6 1 where mt and yt are both real valued functions de ned on I In words a uector ualued function is an ordered pair of real valued functions These real valued functions provide the needed handles they are called the com ponents of the vector valued function One imagines that as time increases7 the vector valued function generates a path in R2 Sometimes one calls this path7 a parametric curve The term parametric curue is de ned formally in the next de nition De nition Let I denote a real interval A parametric curve in the plane in R27 as modeled by the vector valued function7 rt Mt7 yt7 t E I is the subset of the plane given by z7y E R2 my rt for some t E I One frequently speaks of a parametric curue without the phrase as modeled by the vector valued function when the underlying vector valued function is clear or its exact form is not relevant The vector valued function rt7 t 6 I7 that corresponds to a parametric curve is said to be a parametric model or a vector model of the para metric curve Sometimes the parametric model is written mt7yt7 t6 1 and sometimes rt7 t E I Splitting Hairs The term curve as modeled by Without the modi er parametric has been used to name the graph of the realValued function For example if f R gt R is de ned by x2 700 lt I lt 00 then the corresponding curve the function is a subset of R X R de ned by x y E R2 y x2 However7 a parametric curve is not analogously de ned For example if r R gt R2 is de ned by rt t2 t37 700 lt t lt 00 then the parametric curve7 as modeled by rt is not the graph of the function r it is not t 41 6 R X R2 rt t2t3 Instead it is x y E R2 Ly rt for some t E R Appreciating this subtlety is not necessary for Working effectively With parar metric curves This discussion has been added for completeness 16 MODELING EXAMPLE PARAMETRIC MODEL OF A CIRCLE We model a circle parametrically Let r gt 0 and de ne the parametric curve 09 by8 09 r cos 9 340 rsin t97 0 S 0 lt 27r From the computation 209 3420 r2 cos2 9 r2 sin2 9 762cos2 9 sin2 2 r one determines that every point on the parametric curve belongs to the circle of radius r centered at the origin To argue that every point on the circle is a point on the para metric curve is straightforward7 but it needs a case analysis We do not do that here We note in passing that this circle can also be modeled by 19 TCOS 79 yt9 Tsin 797 0 g 9 lt 2 and by 19 7 cos 29 yt9 Tsin 29 0 g 9 lt 7L In fact there are an in nite number of distinct parametric models for the same circlei MODELING EXAMPLE7 PARAMETRIC MODEL OF AN ELLIPSE We model an ellipse parametrically Let a 3A 0 and b 7 O For ease of geometric understanding we assume a gt 0 and b gt 0 De ne a parametric curve 09 by 8Here 6 is a parameter 09 acos0 W bsin0 0 0lt2m From the computation cos 0 sin 0 1 one concludes that for every value of 0 0 S 9 lt 27139 the point 09 lies on the ellipse modeled by 2 2 95 y 2 Eb71 xy ER To argue that every point on the ellipse is a point on the para metric curve is straightforward but it needs a case analysis We do not do that here The next example is particularly interesting we are able to model in a sim ple way an ant as it leaves the origin travels in a straight line to the point 1 1 turns around and travels back to the origin retracing its original path MODELING EXAMPLE7 PARAMETRIC MODEL OF A CORD TRAVERSED ONCE IN EACH DIRECTION We model the path traced by an ant moving in a straight line from 00 to 11 and then back to 00 retracing its path The model is simple it needs no explanation it is rt sin t sin 25 0 g t 7r MODELING EXAMPLE iANOTHER PARAMETRIC CURVE WITH DIFFERENT MODELS 18 The parametric curves corresponding to the vector valued func tions rt 25252 700 lt t lt 00 and rt 253756 700 lt t lt 00 are the same But their underlying vector valued functions are different The parametric curve7 modeled by these two vector valued func tions7 can also be modeled by the function y 27 foo lt z lt oo Veri cation of these assertions is left to the reader EXAMPLE CURVE SKETCHING Sketch the parametric curve modeled by the vector valued func tion rt 25325 700 lt t lt 00 Since rt ztyt 25325 foo lt t lt 007 it follows that for every 257 one has yt That is7 every point on the parametric curve belongs to the curve y z foo lt z lt 00 Next we argue that every point z g on the curve y z foo lt z lt oo belongs to the parametric curve Setting t y one obtains t3 y z Thus every point on the curve y z foo lt z lt oo belongs to the parametric curve modeled by rt mtyt 2537 foo lt t lt 00 Thus the para metric curve is identical to the function y z foo lt z lt 00 The technique of the previous example does not always work In fact7 for most parametric curves7 one is not able to nd an explicit form for y as a function of m We encounter an example below However it is true that every curve in the plane7 modeled by a real valued function f is a parametric curve has a parametric model For suppose that 19 I is a real interval and f I a R Then the curve modeled by the function f is exactly the parametric curve modeled by rt t ft t E I That is if rt t ft 256 I one has zy R2yfmandzEI rt tEI This is a lot of symbols to say something fairly simple every point of the form x fz z E I is a point of the form t ft t E I9 The next example provides a parametric curve that cannot be modeled as a real valued function of the form y fz10 It involves some thought since the vector model of the parametric curve has a singularity it is not de ned at t 71 The behavior of the parametric curve as t approaches in nity is also of some interest EXAMPLE For the parametric curve t2 til 25 7 7 7 t t 71 ro 2231 oolt ltoo y compute the eight missing vectors in the following table11 Determine the behavior of the parametric curve as 25 goes to in nity 9This looks like gobbledygook It is but it is necessary to state since not every parametric curve is a function Later in these notes one nds the parametric curve rt t2t3 7 2t 700 lt t lt 00 which is not a curve modeled by a scalarvalued function it crosses itself 10It is a getyourhandsdirty example 11We are taking a small liberty here The de nition of a parametric curve requires that the de ning vectorvalued function be de ned on a real interval In this example rt is de ned on an interval With a hole in it t 71 However Without affecting the phenomena uncovered in the analysis We could de ne r71 to be any arbitrarily chosen vector eg r71 00 But in order to not muddy the Water We don t do that 20 Determine the behavior of the parametric curve at t approaches Solution By substitution for t in the formula for rt one is able to ll in part of the table as follows 1172 07 g 1472 1quot0 Lil r 0 1quot1 17 0 r 1 1quot2 l 5 1 To obtain values for r one makes the following computation for t in the domain of r7 25 2 t7 1 t rm t22 t31gt 252 21 7 t 22t t3 11 7 t 713t2gt r 5 752 22 7 ta Dz t2272t274t t3173t33t2 lt 92 7 ml l 7t274t2 72t33t21 lt W2 7 ml By substitution for t in the formula for r t one is able to ll in the rest of the table as follows 114 07 Fi2 7 1quot0 17 1 FO 671 1quot1 170 Fl i 1quot2 t 1 fsre gl To determine the behavior of rt as t approaches in nity we introduce the notation 05 522 7 3111 and make the following computations i i 75 t1i gt10t 7 21 1 Z lim 2 taootf O and i i til 330340 7 l i thrgot2 0 Since both t and yt approach zero as t approaches in nity it follows that limtH00 rt 00 That is the parametric curve rt approaches the origin as t ap proaches in nity To determine the behavior of rt as t approaches 1 one notes that t 7 1 2 t 1 777 W taginlt31 0 l 7 Hilly y 00 1 t 1 t 1 2 Hilly 7 Hilly t3 1 7 0 We have abused the mathematical notation in the above ex pressions For example we use the expression 72 to mean that limta1 yt foo Although such expressions are not per mitted in our mathematics they are commonly used we suc cumb to common usage To complete the analysis one notes that limtnnl Mt Thus hmta71 r05 17 00 and limt klf I39t One says that as t approaches 71 the parametric curve rt is asymptotic to the line modeled by z g 22 Looking Back We were able to give a de nition of a 2 d vector and note that vectors can be added and multiplied by scalars real numbers We provided a de nition of a vector model of a straight line in R2 and con verted vector models to scalar models and scalar models to vector models We provided a de nition of a vector valued function and noted that a vector valued function is just an ordered pair of real valued functions7 each de ned on the same domain We provided a de nition of a parametric curve and found vector models for circles and ellipses 23 CALCULUS H Spring 2011 LECTURE 8 Harry McLaughlin revised 22011 edited by This lecture provides 0 a de nition of a derivative of a vector valved function 0 a de nition of a tangent line to parametric curve 0 models for tangent lines to parametric curves 0 parametric curves that can have more than one tangent line at a given point 0 a parametric model for a cycloid and its tangent lines Recall We recall from Lecture 7 o a vector is a point in R2 or R o p q7 p7 foo lt lt 00 models a line in R2 o r cos trsin 257 0 g t lt 27139 models a circle in R2 o a cos t7 bsin 257 0 g t lt 27139 models an ellipse in R2 End of Recall Introduction Our goal in this lecture is to investigate models of tangent lines to paramet ric curves This raises the question Can a vector valued function have a derivative Having answered that we are able to provide a de nition of a tangent line to a parametric curve Tangent Lines to Parametric Curves Derivatives of VectorValued Functions Working informally we ex plore the notion of a derivative ofthe vector valued function rt m t yt where it is understood that t belongs to some real interval One chooses an arbitrary time value of the parameter say to and computes the vector difference quotient t t0 At 7 rto me At yt0 M mto7yto At At 90750 At 7 90050 me At yto At ltzt0 At 7 we yt0 At 7 MW At At 39 One notes that the two components of this last vector are difference quo tients identical to those used in de ning derivatives of a real valued func tions Thus if z t0 and y t0 exist it can be shown that this last expression has a limit as At approaches zero Speci cally one has me Atgt Wax We rm gtW 1i 1m At AtaO Hence the value of lim rt0 At 7 rt0 AtaO At is exactly what one would hope for7 that is7 ow we 2050 ln words7 one says that the derivative of a vector valved function is the derivatives of its components Based on this observation one makes the following de nition and comment De nition Let the vector valued function rt7 be de ned on some real interval7 I and let to E I One denotes the derivative of 1 t7 at t07 by r t0 and de nes it by rt0 I39t0 7 I39t0 lim AtaO provided the limit exists2 Comment If mt and yt are real valued7 differentiable functions7 such that rt ztyt7 t E I then for each to E I one has 1 750 tot 2050 This is7 in fact7 the working de nition77 of r t0 EXAMPLE For the parametric curve rlttgt mmw tit foo lt t lt oo and for foo lt to lt 07 nd an explicit expression for r t0 Solution By inspection one determines that 1 50 tot 1050 37537 1 Of particular interest is the fact that r 0 07 1 imore on this later 1The equality limmao W W x to yto is appealing it is correct But it needs more attention than we have given We discuss it in more depth below 2It is understood that if the limit exists7 it exists as a Vector not a scalar Since this is our rst use of the notion of a limit applied to a vector valued function we spend some time getting comfortable with it What does the expression r t At 7 r t hm 0 0 Awe At mean We assume that the vectorvalued function rt is de ned on some real interval I and that to E I It is observed that the numerator that is rt0 At 7 rt0 is the difference of two vectors one of which depends upon the scalar At The whole fraction can be thought of as the vector rt0 At 7 rt0 multiplied by the scalar if So changing the scalar At from one value to another changes the dif ference quotient rt0 At 7 rt0 At from one vector to another To say that t At 7 t hm F0 F0 Awe At equals another vector say V think r t0 means that as At gets small the difference between rt0 At 7 rt0 At and V is a small vector This raises the question What does small mean Particularly what does it mean in the context of a vector An answer is a vector is small if its magnitude its length is small This brings us back to our original question We say that lim rt0 At 7 rt0 V Awe At if for e gt 0 there exists 6 gt 0 such that whenever to At 6 I and 0 lt lAtl lt 6 then rt0 At 7 rt0 At 7 V lt 6 If there exists such a vector V we relabel it we call it r t0i De nition of Tangent Line and Examples Geometrically one can think of the difference quotient I39t0 At 7 I39t0 At as a scaled secant vector from rt0 to rt0 At As At a 0 this scaled secant vector approaches what looks like a 77tangent vector77 to the para metric curve rt at rt0 We use this geometric perception in de ning a tangent line De nition Denote by C a parametric curve in R2 that is modeled by the vector valued function rt Mt where t and yt are real valued differentiable functions de ned on some real interval I and let to E I Further assume that r t0 7E 0 Then at rt0 a tangent line to C as modeled by rt is the straight line in R2 that is modeled by t rt0 Mao 700 lt lt 00 It is noted that if r t0 0 then the above expression does not de ne a line in R23 However it is possible that C has a tangent line at rt0 even though r t0 0 We investigate that below For a parametric curve modeled by the vector valued function rt the derivative vector r t0 is frequently called the tangent vector to C at rt0 even when rt0 0 We have been careful to de ne a tangent line to C as modeled by rti That is our de nition of a tangent line is modeldependent However we assert that in practice one nds that at a given point tangent lines determined by two different parametric models are iden tical When speaking of tangent lines to a parametric curve we sometimes omit the words as modeled by rt when the model is clear Further we have the instinct that if C is a point set in R2 that looks like it has tangent lines and smells like it has tangent lines then it 3The bold face zero 0 denotes the zero Vector that is 0 00 ought to have tangent lines We state Without proof that it is possible to de ne7 in a modelindependent manner7 the notion of a tangent line to a point set in a meaningful way The following two examples use the de nition of tangent line MODELING EXAMPLE 7 TANGENT LINE TO A PARA METRIC CURVE Find a vector model for a tangent line to the parametric curve C7 as modeled by rlttgt ltzlttgt7ylttgtgt coslttgt7sinlttgtgt7 o g t 2w at the point corresponding to t Solution We set to g and use the model t rt0 Art07 700 lt lt 00 for a tangent line One computes r g t7y tlg sint7008tlg e 7 Since rltggt lteos em lt gtgt 9 3 one determines that the tangent line can be modeled by the vector valued function t7 foo lt lt 007 where tAlt gtlti gt7 fooltltoo MODELING EXAMPLE TANGENT LINE TO A PARA METRIC CURVE Find a vector model for a tangent line to the parametric curve C as modeled by rt at the point corresponding to t 14 7 M04103 75276 foo lt t lt 00 Solution We set to 1 and use the model W mm M050 foo lt lt 00 for a tangent line One computes r 1x t7y t lt1 2t 225 lt1225 Since r1 15 one determines that the tangent line at r1 can be modeled by the vector valued function t foo lt lt oo given by W 15 M2 25 700 lt lt 001 For this example the tangent line has a more elegant model One can write t 1 e 2 25 15 21e 121e fooltltoo Replacing the scalar 1 2 with the symbol u the Greek letter mu one obtains another model for the tangent line denoted here by T the Greek letter tau it is prescribed by T Mlt1757 00 lt ILL lt 0039 4This is an interesting parametric curve It doesn t have a tangent line at 01 as modeled by rt even though each of the coordinate functions t2 and e2 is Well behaved77 at t O In fact if one assumes that the curve is the trace of a bug starting at t 700 and moving for all time then at t O the bug turns around and retraces its path The bug Visits twice every point on the curve y e 0 g I lt oo lt accidently happens that the tangent line goes through the origin Stating the above de nition of tangent line is somewhat treacherous What if a point set in R2 can be modeled both as a curve and as a parametric curve 7 can it have two different species of tangent lines at the same point The answer is no If a point set is modeled both as a curve and as a para metric curve then the two corresponding tangent lines at a point7 if they exist7 are the same That which we call a tangent line by any other name would appear as a tangent line 7 with apologies to William Shakespear6 For example7 a tangent line to a parabola7 at its vertex7 is a property of the particular pointset7 and should be independent of whether the parabola is modeled by y I2 700 lt z lt 007 or is modeled by rt t7t27 700 lt t lt 007 or by some other model In this regard7 one anticipates a theorem saying that if a pointset in the plane is modeled both explicitly y and parametrically rt I t7 yt7 then at a given point7 the tangent lines prescribed by both models are identical We need this understanding for peace of mind7 however we donlt expand upon it here The next example serves as a launching platform for the discussion following it MODELING EXAMPLE7TANGENT LINE TO A PARA METRIC CURVE Find a vector model for a tangent line to the parametric curve modeled by 3 W 0790 75 7t 00 lt t lt 00 at the point corresponding to t 0 5That is7 if the point set is modeled by y yc a g x g b and also by rt Mt7 yt7 t E I can it have two di erent tangent lines at the same point 6From Romeo and Juliet VVhat s in a name that which we call a rose By any other name would smell as sweet77 Solution We set to 0 and use the model WA 1quot750 Arto7 700 lt A lt 00 for a tangent line One computes 1quot 0 t7y t lto 375271 lto 071 F rom this one determines that a tangent line can be modeled by 0 0 A0 1 foo lt A lt 00 This can be written in terms of the vector valued function tA foo lt A lt oo given by tA A0 1 700 lt A lt 00 The preceding example is of substantial interest The parametric curve of1 this example can also be modeled by the scalar valued function y xi foo lt z lt 00 Using this function one is not able to model a tan gent line to the point set at 00 since y 0 doesn t exist even though the parametric model provides for a tangent line at 00 Here we are in an unfortunate position We have to live with the fact that being able to determine the existence of tangent lines to a point set at a given point may depend upon the model of the point set As we discussed in an earlier lecture 7 with some effort it is possible to de ne a notion of a tangent line that is model independent Having done this one can then check to see which models predict the tangent line and which models don t see it However such an investigation takes us a eld of our study of the calculus we live with the fact that determining the existence of a tangent line may be model dependent In spite of these remarks we are relatively safe As noted above if two different models of the same point set each predict the existence of tangent lines at a point of the set then the tangent lines are the same Multiple Tangent Lines at a Point In the next two modeling examples and the DILEMMA we discuss two more pathologies dealing with tangent lines MODELING EXAMPLE7MORE THAN ONE TAN GENT LINE The symbol 0 that we use to represent in nity is a parametric curve We don t need an explicit parametric model here7 Does it have a tangent line where the parametric curve crosses itself Solution From the picture it looks as if it should have two tan gent lines at the crossing point On the other hand there are those that say that a curve can have at most one tangent line at a point Looking closely at our de nition of a tangent line at a point on a parametric curve we see that indeed it has two tangent lines at the crossing point Our de nition of a tangent line to a parametric curve leaves open the possibility for the existence of multiple tangent lines at the same point on the parametric curve If a crossing point is given by both rt0 and rt1 then according to the de nition the two lines modeled by We r t0 00 lt lt 00 and rt1 r t1 00 lt lt oo are both tangent lines at the point of intersection provided that rt0 and rt1 are non zero vectors MODELING EXAMPLE 7MORE THAN ONE TAN GENT LINE AGAIN Find vector models for two tangent lines at the point where the parametric curve modeled by rt 29253 7 2t 700 lt t lt 00 7The parametric curve modeled by the VectorValued function r cos sin 29 0 g 6 g 27r appears as a gure eight when plotted on an x yCoordinate system 11 intersects itself8 Solution Even though the problem statement asserts implic itly that the curve intersects itself we need to verify that What we verify is that there are two distinct values of the parameter say 3 and z with s lt 2 such that rs The strategy is to assume that s and z exist ii nd values for such 3 and z and then iii verify that rs We then compute rs and r z and use those vectors to model the two tangent lines One de nes the real valued functions t and yt by rt ztyt t2t3 7 2t 700 lt t lt 00 Setting Ms one obtains 52 22 Since 3 7 2 one concludes that z 73 Since ys one concludes that 248 83 i 28 242 83 28 Using the facts that z 7s and s 7 2 hence 3 3A 0 one has the following computation 33 7 23 7s3 7 273 233 7 4s 0 33272 0 s i Thus if rt does indeed intersect itself it must be that r7 We compute r7 and and verify that they are equal We evaluate rst Here 9005 t2ltx 2 t3 2tlt 8We rely on the reader s willingness to interpret appropriately the expression intersects itselfquot xElt L 2 0 One concludes that 20 We evaluate next r7 Here M tzlt7 52 2 t3 7 2tlt ixE exW i 2 0 One concludes that I397 20 Thus 20 is a point where the parametric curve crosses itself In order to model the two tangent lines we compute the two vectors r 72 and r 2 To that end we note that r t 225 3252 7 2 foo lt t lt 00 From this one nds that MixE 224 and r N24 Thus the two tangent lines are modeled by two vector valued functions t1 foo lt lt 00 and t2 foo lt lt oo given by t120 Mix54 700 lt lt 00 and t2 20 224 700 lt lt oo One can simplify these expressions obtaining lt1ltAgtlt2ogtAlt71 igt foo ltA lt col and t2 20A1 700 lt A lt 00 DILEMMA7TANGENT LINE OR NOT We found above that there exists a tangent line at 00 for the parametric curve modeled parametrically by rt 25325 700 lt t lt 00 In fact the tangent line can be modeled by the vector valued function t foo lt lt oo given by t 0 1 700 lt lt 00 However the same parametric curve can be modeled parametri cally by rt t9 23 700 lt t lt 00 Using this alternate model one nds that rO 928352 M 00 Thus the assumption r 0 7 0 needed in the de nition of a tangent line for a parametric curve is not met So using this alternate model it is not possible to determine whether or not the parametric curve has a tangent line at 00 Finally we recall that according to a previous discussion for the point set de ned by rt t3t foo lt t lt 00 one is not able to prescribe a tangent line at 00 using the model yx 7ooltmltoo There is a way around all of the uncertainty introduced in the previous dilemma a model independent de nition of tangent line can be prescribed However as mentioned earlier that takes us a eld of our study of the calcu lus Most of the time we are able to deal with tangent lines for parametric curves using the de nition above We stick with this de nition because of its ease of use The Cycloid The cycloid has been studied since antiquity But shortly after the intro duction of the calculus it was found to solve the tautochrone problem and the brachistochrone problem9 Since then it has played a fundamental role in calculus modeling MODELING EXAMPLE7MODELING A CYCLOID Let r gt 0 Center a circle of radius r at the point 0r The bottom of the circle is at the origin at 00 Label this point P and track the location of P as the circle rolls along the m axis in the positive direction The curve generated by P is called a cycloid After the circle has rolled a bit the point P has coordinates x From the center of the circle draw a line to P and draw another line straight down to the circles resting point on the z axis Label the angle formed at P between these two lines by the symbol t Then the length of the arc of the circle from the point where it rests on the m axis to the point P is rt Thus the center of the circle is at the point rt r From this observation one is able to determine that z rt 7 r sin t and y r 7 r cos Thus a parametric model of a cycloid is mt rt 7 rsin t and yt r7rcost 7oolttlt 00 9The reader may nd an online search of these two problems of some interest 15 One notes that in allowing the parameter t to take on negative values7 we are admitting the possibility that the circles rolls in the negative z direction MODELING EXAMPLE7TANGENT LINES TO A CYCLOID Find a vector model for a tangent line to a cycloid at an arbi trary point on the cycloid Investigate the expression at points where the cycloid has cusps We rely on the reader s willingness to geometrically visualize a cusp Solution One models a cycloid parametrically by Mt rt 7 r sin 25 and yt r776cost7 7oolttlt 00 The vector form of this model is rt rt 7 rsint7 r 7 rcos 257 700 lt t lt 00 From this one obtains rt r 7 rcos trsin 757 700 lt t lt 00 Using this7 one models the tangent line to the cycloid at rt0 as follows t rt0 r t0 rte 7 rsin to7 r 7 r cos 250 r 7 rcos to7 r sin 2507 where 700 lt to lt 00 Thus a model for a tangent line to the cycloid at rto7 is given by t7 700 lt lt 00 where t rte 7 rsin 250 7 rcos 250 r 7 rcos to7 rsin t0 16 This expression serves as a model of a line only when the tangent vector r776 cos to7 r sin 250 is not the zero vector7 that is7 only when to is not of the form 27m 71 0j17 i2 When one graphs the cycloid7 one observes that these are the points where the parametric curve has cusps Looking Back We have been able to de ne and model a tangent line to a parametric curve rt at rt0 by the expression rt0 rto7 foo lt lt 00 provided that r t0 7 0 This leads to two foundational questions i Can a parametric curve that crosses itself have more than one tangent line at the crossing point ii Can a parametric curve have a tangent line at rt0 even though r t0 0 We have found that the answer to both question is yes This analysis suggests one more question iii Is it possible for a smooth parametric curve to not have a tangent line at one or more of its points The answer here is also yes but we don t undertake that investigation Finally we have modeled a cycloid along with its tangent vectors The cycloid has been introduced because of its role in calculus modelingiit solves both the bmchistochrone problem and the tautochrone problem CALCULUS H Spring 2011 LECTURE 9 Harry McLaughlin revised 22711 edited by This lecture prouides a de nition of arc length of a curue as modeled by rt an arc length formula 6 Walt the arc lengths of a circle and of a cycloid a note on how arc length of a curue may be model dependent a de nition of a polar curue an arc length formula 6 Wold calculation of two spiral arc lengths a de nition of polar coordinates Recall We recall from Lecture 8 r t limmH computes a derivative of a vector valued function r A 7r 0 H At t t rt0 r to7 foo lt lt 007 models a tangent line to a parametric curve multiple tangent lines at the same point on a parametric curve are possible rt rt 7 rsintr 7 rcost7 00 lt t lt 00 models a cycloid generated by a circle of radius r it is possible to model explicitly a tangent line to a cycloid End of Recall Introduction In a previous lecture we found a model for the arc length Z of a curve modeled by y Mm a 3 x g b There we wrote 17 Z V 1 y m2dx In this lecture we tackle the same problem for parametric curves We uncover a corresponding integral formula 17 a zmu y t2dt Additionally we introduce the notion of polar curves and nd a formula for their arc lengths it is z b Mme Howe Arc Length of a Parametric Curve We sense that a parametric curve when plotted on an m y coordinate sys tem has a length We don t have a mathematical model for this our goal is to develop such Roughly the idea is to partition the curve by a nite number of its points and connect each two consecutive points by a chord after that sum the lengths of the chords If the points are close enough together the sum of the chord lengths looks as if it should be a good approximation for our intuitive idea of arc length We proceed with this idea Let C denote a 2 d parametric curve modeled by the vector valued function rt a g t g 1 Rather than partitioning C by pointing to the points directly we partition C by pointing to parameter values1 In that regard we let n denote a positive integer and choose 71 1 values ti 6 11 such that a to lt 151 lt 152 lt lt tn 1 Below we denote such a partition of 11 by or by 1532 when n is needed in the discussion Similarly we write or rti0 to denote the corresponding partition points on C If the z and y components of rt are given by rt zlttgt7ylttgtgt7 a 1 b7 then for each i the length of the chord between rti a 1 d rti is given by 111751 11751701 V 95051 i 900514 24051 7 2405171 The sum of the chord lengths denoted by ELI lrti 7 rti1l is an ap proximation for the still to be de ned arc length of C It is understood that V L 2 1111507 rltmilgt11rltt1gte rose 1rltt2gterlttlgt11rlttngte roam i1 MODELING EXAMPLE ARC LENGTH COMPUTA TION We approximate crudely the arc length ofthe parametric curve modeled by rlttgt ltzlttgt7ylttgtgt 221712 71 1 1 Let n 4 and choose the partition as given in the following table 1 ti 0 71 71 0 1 1 3 1 5 5 1 2 0 0 1 1 1 3 3 E 5 1 4 1 1 0 1This seemingly innocuous act hides an important understanding It is after all the parameter Values that order the points on the parametric curve The ordering enables us to partition the curve and connect each two consecutive points by a chord Using this data one obtains l9113i19 4 16 4 16 4 1 4 16 13 5 7 2 162 E 2 wm This evaluates to approximately 29208 Using integration tech niques developed later one nds that a better approximation to the arc length is 29579 If the points rti0 on C are close enough together the sum of the chord lengths is a good approximation to our intuitive notion of arc length2 The approximation depends upon two things i the way the points are chosen for a xed 71 and ii the value of 71 itself Getting Comfortable We pause to think about the idea of sum ming chord lengths In our discussion here we assume that 2It is not cricket to call the sum of the chord lengths an approximation to arc length since we are still in the process of de ning arc length but we do it anyway rt a g t g b is continuous For a xed value of n the number of chords one can choose the partition points ti0 in an in nite number of ways It turns out that for a xed 71 and for all reasonable parametric curves there is at least one way to choose the partition points so that the sum of the chord lengths is a maximum value that is for a xed 71 the sum of the chord lengths for a xed curve is as large as possible We imagine that we have this set of n 1 partition points 71 chords in hand then we adjoin one more point to the set creat ing a set of n 2 partition points The sum of the chord lengths for this new set of n 2 partition points is not less than that for the set of n 1 partition points If C is nowhere linear then for n 2 points the new sum of chord lengths is strictly greater than the sum for n 1 points This thought experiment reveals that in general there is no way to partition C partition 0 b with a nite but variable num ber of points so that the sum of the chord lengths is a maximum value But resting on our intuition about arc lengths we suspect that there must be some number that acts as an upper bound for the sum of the chord lengths That is no matter how 71 is chosen and how the partition points are chosen the corresponding sum of chord lengths will not exceed this upper bound If there is such an upper bound then there is a least such upper bound3 Below we use this least upper bound to de ne arc length De nition Let C denote the parametric curve modeled by the vector valued function rt a S t S 1 Assume that 21lrtii rti1l is bounded above independently of the choice of n and independently of the choice of the partition points 25 where a to lt 251 lt lt 25 b The arc length of C as modeled by rt is the least upper bound of all possible values of 2 MW WHN i1 3One can think of the existence of this least upper bound as an axiom of our mathe matics where the integer n and corresponding partition points7 t1 0 are arbi trarily chosen We have been careful to use the words arc length of C as modeled by rt instead ofjust arc length Below we see that it is possible to assign many values to the arc length of a circle of xed radius7 depending on how the circle is modeled We rephrase the de nition of arc length Alternate De nition Let C denote the parametric curve modeled by the vector valued function rt7 a g t S b The real number L is the arc length ofC as modeled by rt if i for every positive integer n and every partition ti7 where at0ltt1ltlttnbonehas 71 2 Mil 1quotti71l S L i1 and ii If L lt L then there exists a partition of 11 such that 71 2 mm 7 rt1l gt L i1 It is generally impossible7 using the above de nitions7 to determine whether or not a number L represents the arc length of a given parametric curve The above de nitions are valuable since they provide a precise mathematical model for the notion of length of a parametric curve However7 they don t provide a closed form formula for computing arc length That is done below using the notion of the Riemann integral In practice one frequently computes numerical approximations for arc lengths by a brute force77 algorithm That is7 one selects n and corresponding par tition points7 a to lt 251 lt lt tn b7 then computes the sum 2 Mil 1quotti71l Z 9575139 900271 2475139 240271 il i1 lf7 in increasing n a little bit7 the value of the sum doesn t change much say in the 6th or 7th decimal place then the sum is declared to be the arc length4 4The number of unchanged decimal places usually depends upon the application In the above example in order to approximate the arc length of the para metric curve we computed numerically a value for n Z 950 9005171 905139 7 240514 i1 for a particular 71 n 4 and a particular choice of the partition points We could have improved the approximation by choosing n to be larger In fact one can imagine letting 71 go to in nity and in the process generating a sequence of better and better approximations to the arc length However it is sometimes possible to avoid the brute force algorithm and represent arc length as an integral We explore this idea next An Integral Formula for Arc Length We focus on the sum displayed immediately above and assume that rt is continuous on 11 and that r t exists and is continuous on 11 Using the mean value theorem each term of the form 7 zt12 can be written in the form i2Ati2 where 5139 is a particular point in the interval 221 ti and Ati ti 7 221 Similarly using the mean value theorem each term of the form 7 yti12 can be written in the form ym2Ati2 where m is a particular point in the interval 221 ti and Ati ti 7 tiil Then the sum of lengths of chords can be written 2 9075139 90051712 905139 240514 i1 H H i2Ati2 y m2Ati2 H H M50 2010 Ati For each 71 one can identify the largest At and call that the norm of the the partition ti0 Letting this norm get very small one can replace the m s by the 51 s without a catastrophic result5 What results is the Riemann sum n 2 V QB502 MED Ni i1 Letting the partition norms go to zero one obtains the Riemann integral b zw ylttgtgt2dt 1 provided that the integral exists The above discussion provides a pseudo proof7 for the following theorem Theorem Let C denote the parametric curve modeled by the vector valued function rt a g t g 1 Assume that rt Mt yt a S t g b where 05 and yt are continuous on 11 Further assume that r t exists and is continuous on a 1 Then the arc length 6 of C as modeled by rt is given by a b zw iitwat provided that the integral exists as a nite number6 In alternate notation one writes 17 z lr tl dt Before investigating more thoroughly this formula we note that if C is mod eled by the real valued function yt a g t g b then C can be modeled by the vector valued function rt t yt a g t g 1 Thus under the right conditions the arc length of C is computed by z b 1y t2dt 5This requires a proofinot given here It analyzes the replacement of m by on each of the intervals except the rst and last intervals 61f x t and y t are continuous on the closed interval 11 then the integral must exist This is the formula used earlier when we computed arc lengths of curves modeled by Mm a 3 x g b MODELING EXAMPLE7ARC LENGTH OF A CIR CLE Compute the circumference7 Z of a circle of radius r r gt 0 Solution One models such a circle parametrically by rt rcos trsin 157 0 g t g 27139 We interpret the words circumference of a circle77 to mean the arc length of the circle modeled by rt One makes the compu tations for 0 g t g 27139 Mt 7 cos t zt fr sin 15 3105 7 sin 15 yt 7 cos From this one obtains 27r 0 z112 2711111 27r 0 1 fr sin 152 7 cos 152 dt 27r 0 r sint2cost2dt 27r rdt 0 27W N l l Thus the circumference Z of a circle7 that is7 the arc length of a circle as modeled by rt 7 cos trsin 157 0 g t g 27139 is given by At this point one is tempted to enter an example dealing with the arc length of an ellipse Unfortunately such arc length calculations are not easy The resulting integrals are called elliptic integrals and are widely studied in the mathematical literature7 MODELING EXAMPLE7ARC LENGTH OF A CY CLE OF A CYCLOID Compute the length 6 of one cycle77 of a cycloid generated by a circle of radius r r gt 0 Solution One models such a cycle77 parametrically by rt rt 7 rsin 25 7 rcos 757 0 g t S 27139 We interpret the words length of one cycle of a cycloid77 to mean the arc length of the cycle as modeled by rt One makes the computations Mt rt 7 r sin 75 zt r 7 rcos t yt r 7 rcos t yt r sin 25 with some algebra 05 34 2T2 7 272 cos 7It is interesting to note that determining the area of an ellipse ab is straightforward yet determining the arc length of an ellipse is less straightforward From this one obtains 27r z 7 zlt1 71 d1 0 27r V272 7 272 cos t dt 0 27r 7 x2l7costdt 0 27r t cos2t 202 t 7st t 7 4sin2 7 dt 172518 0 2 2 2m t 17cos2t 27r t t 7 27 sin lt7gt dt 45mg 3 7 217 WSW 0 2 27r t 27 sin lt7 dt 0 2 27f 74Tcoslt3gt 2 0 7 4T7 cos 7r 7 7 cos 4m 1 87x Thus the length 6 of one cycle of a cycloid as modeled by rt r157 rsin 157 r 7 rcos 157 0 315 27139 is given by It is curious that the length of a cycle of a cycloid doesn t ex plicitly show the number 7139 Discussion In de ning arc length we have been careful to use the words arc length ofC as modeled by rt These words hint at the possibility that the arc length of C may depend upon its vector model This is indeed the case Example The parametric curve C modeled by the vector valued function rt cos tsin 157 0 g t g 27139 is also modeled by the vector valued function st cos tsin 157 0 g t g 47139 The arc length ofC as modeled by rt is computed to be 27139 and the arc length of C as modeled by st is computed to be 47139 Choosing an appropriate length for C depends upon the applica tion If one wants to model the length of string needed to wrap once around the unit circle7 then the most appropriate length may be 27139 If one wants to model the length of string needed to wrap twice around the unit circle7 then the most appropriate length may be 47139 Polar Curves and Their Lengths In this section we investigate arc lengths of a special class of parametric curves the so called polar curves These are parametric curves whose models have the form z0 r0 cos 0 y0 r0 sin 07 a g 0 b7 where r0 is a real valued function de ned on an interval 1718 This parametric model is sometimes called a polar model The correspond ing curve is called a polar curve Before going on we emphasize that the polar model z0 r0 cos 0 y0 r0 sin 07 a g 0 b7 where r0 is a real valued function de ned on an interval 11 uses the symbols 70 and 0 even when r0 lt 0 and 0 07 27139 It is noted that a polar model of a parametric curve is completely determined once the function r0 is speci ed We use this idea in the next example EXAMPLE7POLAR MODEL A polar curve is modeled by r000 0 27139 Thus it is a parametric curve modeled by 8One can think of the point M07 710 in vector form r0 cos 0 r0 sin r0cos 0 sin In this form one sees that the vector cos 0 sin which lies on the unit circle is scaled by the real number r0 Thus a polar curve can be thought of as a scaling of each point on an arc of the unit circle 09 0cos0 M0 0sin0 00g2w The curve is a spiral Whose initial point is my 00 It unwinds as 9 increases lts terminal point is zy 27r0 How to think of the plot of this polar curve Wrap the t9 interval 0 2 around the unit circle in a counterclockwise direction With 9 0 and t9 27f both placed at zy l 0 Each point on the unit circle has coordinates cos 9 sin for some 9 E 0 270 To nd the corresponding point on the polar curve namely 76 cos t9sint9 scale the unit vec tor from the origin to cos 9 sin by the real number Tt9 Which in this case is just 9 We say it another way Since an arbitrary point on a polar curve can be Written as Tt9 cos 9 sin one can think of the point as lying on a halfline emanating from the origin and containing the point cos t9sin The distance from the point to the origin is Recall Tt9 9 in our example As mentioned above a parametric curve with a polar model is sometimes called a polar curve9 Even though a polar model of a curve is a vector model as noted above a polar curve can be described by the scalar valued function 79 This is a powerful idea it reduces a vector model to the speci cation of a scalar valued function 9It is noted that sometimes the same curve a pointset in the plane can be called i a scalar curve or ii a parametric curve or iii a polar curve The name of the curve depends upon the context EXAMPLE 7POLAR MODEL A polar curve is modeled by M 0lt0 In this case the corresponding parametric model is given by 950 M The polar curve is again just a spiral lts initial point is z y 17 0 it unwinds in a counter clockwise direction as 9 increases until reaching its terminal point at my 54720 529 cos 0 mwyogog For polar curves one can use the formula for arc length7 Z of a parametric curve lf 9 is used instead of t it becomes 17 a mim fwwm da In terms of M0 this formula has a particularly elegant form We assume that T09 is differentiable on 11 and derive the formula below One has M0 M0 cos 9 we rlt0gtsinlt0gt m0 r0 cos 9 7 M0 sin 9 309 r0 sin 9 M0 cos 9 m 2 7 0 Adding these last two expressions by inspection one is able to write 17 b a z0 W de WW wlt0gtgt2da This analysis leads to a proof of the following theorem Theorem The arc length 6 of a polar curve as modeled by M0 a g 9 g b7 Where r is continuous on 071 and 09 exists and is continuous on 11 is given by b 6 76209 r 02 d0 1 provided that the integral exists The next two examples explore the use of this formula MODELING EXAMPLE7ARC LENGTH OF A PO LAR CURVE Calculate the length7 Z of the curve Whose polar model is do 02 0 0 g 27139 Solution The arc length of the polar curve as modeled by do 02 0g 0g 2w is given by 27r Z 76209 T02 d0 0 To evaluate this integral one makes the following calculations M0 04 W0 2W 402 Thus 27r z i720w02da 0 27r 04402d0 0 27r 0x024d0 0 27r x02420d0 0 1 2 d 50 x024024d0 16 use ursubstitutio n G 9 WW l 2 32 7 347139 4 38 7r2171 Thus the length 6 of the polar curve as modeled by r0027 0027r is given by 7r2171 MODELING EXAMPLE7ARC LENGTH OF A PO LAR CURVE Calculate the length7 Z of the curve Whose polar model is do 529 0 g 0 g 27139 Solution The arc length of the polar curve as modeled by do 5 0 g 0 g 27139 is given by 27r Z V7309 7609 10 0 To evaluate this integral one makes the following calculations M0 e49 W0 2529 4549 Thus a 2W Mme amaze 0 27r e49 4549 d0 0 27r V5529 d0 0 27r 529 Hence7 the arc length 6 of the polar curve as modeled by do 529 0 g 0 g 27139 is given by e4quot71 Polar Coordinates Closely related7 to the notion of a polar form7 is the notion of polar coordi nates We introduce the idea here and then expand upon it in a following lecture lf7 for some 0 0 g 9 lt 271397 and for some r gt 07 the point my has the form r cos t97 r sin then 9 can be interpreted geometrically as the angle measured counterclockwise from the positive m axis to the chord from 07 0 to z And T can be interpreted as the length of the chord De nition If my 6 R2 and if z rcos 0 y r sin t97 for some r gt 0 and for some 0 0 g 9 lt 271397 then r and 9 are said to be polar coordinates of the point The polar coordinates of my 07 0 arer0and00 Examples of polar coordinates are given by M M 07 0 07 0 171 W57 i 071 17 g 717 0 177139 i 71 27 7 One notes that according to our de nition7 the Tvalue in polar coor dinates is always nonnegative Our de nition of polar coordinates leaves us with two questions Does ev ery z y have polar coordinates and ii Are polar coordinates for a given my unique We leave the existence question to a later lecture where it is answered in the a irmative The uniqueness question answered in the a irmative is settled by the next theorem Theorem Uniqueness of polar coordinates Let my 6 R2 have polar coordinates r 0 and p gt Then T p and t9 1 Proof To prove that 7 p it suf ce to note that z Tcost9 19 y T sin 9 12 y2 7 20082 9 sin2 l recall that 7 gt O T x12y2i Since this expression for T is independent of the value of 9 it follows that T pi To argue that 9 45 one notes that if 7 76 T7 then T cos 9 T cos and T sin 9 T sin and hence cos 9 cos and sin 9 sin From this one is able to write cos 9 7 cos 9 cos sin 9 sin cos2 9 sin2 9 1i Relabeling7 if necessary7 we assume that 9 2 45 and note that 0 S 9 7 45 lt 27L Since cos 9 7 17 one concludes that 9 i This completes the proof Looking Back For a parametric curve7 modeled by rt7 a S t b7 we have been able to provide a clean de nition of arc length of the curve as modeled by rt namely the least upper bound of 2 W750 1quotti71l i1 computed over all partitions ti0 of 0 b and over all n We are attracted to this de nition because it does not depend upon continuity or differentia bility of the vector valued function rt 20 Further for parametric curves modeled by rt ztyt a S t S b we have found an elegant integral formula for arc lengths namely 17 zwawwmrw In doing this we had to restict rt slightly we required rt to be continuous on 11 and r t to be continuous on a 1 And the integral must exist Using the formula we were able to compute the circumference of a circle and the length of one cycle of a cycloid We were able to crystallize the relationship between the length of a para metric curve and its parametric model We found that arc length may be model dependent The notion of polar curves M ww UWNmWLM M La0b came up naturally as a special class of parametric curves Assuming that r0 and r 0 have certain properties we found an integral formula for arc length namely 17 Z V7209 7092 10 1 which we used successfully on several examples Finally we provided a formal de nition of polar coordinates and proved that they are unique 21 CAUHEUSH Spring 2011 LECTURE 10 Harry McLaughlin revised 3311 edited by This lecture prouides 0 models and computations of areas for regions of the plane that are de ned by some special parametric curues d A ytz tdt 0 models and computations of areas for regions of the plane that are de ned by polar curues b 1 A 77620 d0 a 2 0 areas of 7 inside an ellipse 7 inside one cycle of a cycloid 7 inside of a spiral 7 inside a cardioid Recall We recall from Lecture 9 o the arc length formula 6 f m t2 y t2 dt 0 arc lengths may be model dependent o a de nition of a polar curve 0 the arc length formula 6 f T2t9 r 02 d0 0 a de nition of polar coordinates End of Recall Area Using Parametric Models Early in our study of the calculus we used an integral f x dz for model ing the area of a planar region bounded by the m axis and the non negative real valued function7 m a g x g 1 Our goal now is to explore the form of this integral when the curve m7 a g z 3 b is modeled parametrically by 057 34057 c g t g d For that7 we use the change of variables formula To get started we recall the change of variables formula 7 stated as it was in a previous lecture Theorem ChangeofVariables Let Mt denote a real valued differen tiable function de ned on a non degenerate interval 07d and let a do and b Let 1 denote a real valued function de ned and continuous on an interval I such that zc7 1 Q I If z t is continuous on gel then Ab dz cdfmtx t it It is noted that if t is chosen such that a zd and b zc then the change of variable formula becomes lb dz Ac fxt t it If the curve modeled by the function neglecting smoothness constraints for the moment yf907 19019 is identical to the parametric curve modeled by t7yt7 c S t S d then necessarily fmt yt7 c g t g 1 Thus if do a and zd b7 the change of variables theorem asserts that the area of the planar region bounded by the m axis and the non negative function f is given by b d d a mm fltzlttgtgtz lttgtdt ylttgtz lttgtdt 3 That is using the parametric model one computes the area by the integral 1 ytmt dt Similarly if do b and zd a one has 17 c c mm fltzlttgtgtz lttgt dt mant dt 1 d d and one computes the area by the integral 0 ytmt dt d These are the forms of the integral that we are after Before using them we reset above we assumed known a curve modeled by x and then hypothesized the existence of a parametric model for the same curve However in the problems below we are given only a parametric curve Before using the integral formula fed ytm t dt or ijtz t dt to compute area it is important to determine that in fact there does exist an identical curve de ned by a real valued function fm a 3 x g I even though we don t need its explicit form It is easy to argue that if 05 is strictly monotone then x exists For our area computations below this observation suf ces But monotonicity is not necessary for the existence of This is seen in one of the immediately following examples EXAMPLES7EXPLORING THE EXISTENCE OF f1 i Let zt t2 yt 23 0931 Then the corresponding parametric curve is identical to the curve modeled by the real valued function mg nggl 1We are reminded that we have de ned the notion of area underneath the curve y a g x g b by f dx It is this de nition of area that we want to exploit ii Let Mt sin 25 yt 17 0 g t 3 7r Then the corresponding parametric curve is identical to the curve modeled by the function In this case t is not monotone and every point on the curve de ned by x is Visited twice by the vector valued function sin t7 17 0 g t 3 7139 For example the point 7 1 is Visited by sin 25 1 at bothtandt37 iii Let t 7 1 yt 257 0931 In this case the parametric curve is E R2 z 1 and 0 g y g 1 It is not a curve modeled by a scalar Valued function 1n the examples below the choice of the two formulas for computing an area7 A7 is determined according to if 05 is strictly increasing then d A yont dtt C if t is strictly decreasing then A d ytmtdt We are now armed for area calculations2 In the next example we are able to model and compute the area of an ellipse We nd that the area is 7139 times the product of the lengths of the semi major and the semi minor axes MODELING EXAMPLE AREA OF AN ELLIPSE Let a and b be positive real numbers Model and compute the area A inside of the ellipse as modeled parametrically by Mt a cos t yt 1 sin t7 0 S t S 27139 Since sin 25 7r 7 sin 57 0 S t 3 7139 we model the area inside of the ellipse by twice the area of the region between the m axis and the parametric curve ztyt a cos t7 1 sin 257 0 g t g 7139 That is7 we model and compute the area A of the upper half7 of the ellipse and multiply that area by two It is empha sized that this doubling process is part of our model of the area inside of the ellipse Since Mt7 0 g t 3 7139 is strictly decreasing we model and com pute the area A using the formula f7 ytm t dt Using the facts that z7r 7a and M0 a and noting that z t 7a sin 75 0 S t g 7r7 one writes 2Our focus on increasing and decreasing functions t is for ease of computation The important consideration is whether or not xc a and dd b or ViceVersa A ytztdt b sin t7a sin m dt interchange the limits of integration 7r ab sin2 t dt 0 7r 1 cos 2t cos2t 202 t 75in2 t abO 7 2 dt 1 7 2sin2 t W m2 if cos2t ab 7 O cos 2t dtgt 2 2 7r ab 7 isinet 0 ab iWi 2 Thus7 according to our model7 the area A inside the complete ellipse is given by 7rab One notes that When a b the ellipse becomes a circle and the area just computed becomes the familiar expression for the area of the circle WCLZ In the next example we are able to compute the area under one cycle of a cycloid We nd that the area is three times the area of the rolling circle used to generate the cycloid MODELING EXAMPLE7AREA OF A CYCLOID Let r gt 0 We model parametrically one cycle of a cycloid by Mt rt 7 r sin 75 yt r7rcost7 0 325 27139 In this case t is strictly increasing So the area7 A7 of the region above the m axis and below the cycloid is modeled by 27r ytmt dt 0 Using this one computes A 7 Almaad 07r7quot 7 7 cos tT 7 7 cos dt 7 2 027rl 7 cos t2 dt T227r17 200s t 0052 dt r2 2 7 2sin WE 27r T2 M d7 cos2t cos2 u 7 m2 u 0 2 2 2cos2 t 7 1 1 1 cos t 27r39r2 7 0 7T7 2 7 21 sin 27 6052 t E 27r39r2 7rquot2 0 37m Thus the area under one cycle of the cycloid is given by 37mg Area in Polar Models We imagine a polar curve modeled by the continuous real valued function7 7607 a g 9 g b where 0 lt bia S 27139 We think of r0 as being continuous but later only require that it be integrable The corresponding parametric curve is given by 19 Tt9 cos 9 yt9 Tt9 sin t97 a S 9 S bi The polar curve de nes a pieshaped region point set R in the plane con sisting of all points of the form r0 cos t97 r0 sin where a g 9 g b and 0 g g 1 Our goal is to nd a model in terms of r0 for the area of the region R But we don t know what area is in this context We seek a de nition of area for a pieshaped region Below we are able to nd a de nition in terms of an integral To motivate our de nition we provide an intuition based heuristic model of such an area Using the expression wrz for the area of a circle of radius r one is able to model the area A of a sector of the circle with central angle A0A0 Z 07 using the equation A0 7 A E 7 7T7 2 39 From this one obtains A rw Letting a 00 lt 01 lt 02 lt lt 0 b be a partition of 071 one lets A0k 0k 7 0k1 for 1 g k g n One can model approximately the area of each of the sectors Rk between 0k1 and 0k by r2 kA0k where 5k is arbitrarily chosen such that 0k1 5k 0k We are using a circular arc to approximate the polar curve r0 for 0k1 S 9 0k Summing over all k one nds an approximate model for the area of R to be mama M 1 2 w H 1 We are summing areas of sectors of circles each with a possibly different radius It is emphasized that even though we use the words area of R we don t yet know how to de ne the area of R Recognizing this as a Riemann sum one lets the size of the partition go to zero to obtain an integral model for the area of R namely7 b 1 E7620 d0 It is emphasized that we did not prove that the area of R is given by f 39 t9 dt9i We only offered a plausible argument justifying the integral as a model for the areal Having done the analysis we formalize a corresponding de nition De nition A polar curve is modeled by a real valued7 integrable function7 M0 a S 9 g b where 0 lt b7 a g 27139 Let R denote the point set consisting of all chords each of which is constructed between the origin and a point on the curve Then the area of R as modeled by r097 is de ned to be 17 1 E7620 d0 It is noted that M0 can take on negative values and that M0 is not required to be continuous only integrable At the risk of beating a dead horse77 we remind the reader that eg the area inside of the upper half of the unit circle can be computed two ways using the formula fOW 6209 10 resulting from the polar model and ii using the formula fir ytm t dt resulting from the general parametric model Hopefully they give the same result3 We expand on this idea with one of the examples below We use this de nition below in two examples In the statements of the two examples we use the word insida as in inside a curve We model the inside of a curve using the set R as de ned above 3They do In fact they are both equal to Ill 1 7 zzdz g which is the area resulting from the model of the form y 1 7 2 71 g x g 1 MODELING EXAMPLE 7 AREA IN POLAR MOD ELS Model and compute the area A ofthe region of the plane inside of the spiral as modeled by the polar equation r0 030g 27139 The spiral is modeled parametrically by 19 woos 9 MO sin t97 0 S 9 S 27 Using the integral de nition for area of the region de ned by a polar curve7 one models and computes A by 2W1 A 7r20d0 0 2 1 27r 2 E0 x d0 1 27r 7 0d0 2 One concludes that7 according to our model7 the area of the region inside of the spiral is A7r2 MODELING EXAMPLE AREA OF A CARDIOID Model and compute the area A of the region of the plane inside of the cardioid as modeled by the polar equation r01sin0 0 g 0 g 27139 11 The cardioid is modeled parametrically by 19 1 sin cos 9 yt9 1 sin sin t97 0 g 9 g 27L To graph the cardioid one makes the following table 0 900 240 0 1 0 g 0 2 7r 71 0 3 7quot 0 0 27139 1 0 One models and computes the area by4 27r 1 A 420 d0 0 2 gfhm sin 02 d0 027r12sin0sin20d0 1 27 1 20 WCOS0W j 57 d0 1 wimgigsmm 37139 7 7 0 2 37139 2 One concludes that7 according to our model7 the area A inside the cardioid is 4In the computation We use the identity 2 if sin 972 There is another approach to the problem of the previous example Using the polar model for the cardioid 79 1 sin t97 0 g 0 271397 one models the cardioid parametrically by 09 1 sin cos t97 1 sin sin 97 0 g 9 lt 27139 Using the parametric model we are able to compute the area inside of the cardioid we compute the area inside of the right half and double it To do that one notes that the right half of the cardioid can be modeled parametrically by mom0 ltlt1 sin 0 07 lt1 smlt0gtgt sin lt0 7 s 0 wl 7T 2 Then the area A inside of the right half can be modeled by A May0 d0 1n the next example we verify that such a computation gives exactly the same area inside of the whole cardioid as was just computed using the formula 1 27r 147 ame 2 0 MODELING EXAMPLE7AREA OF A CARDIOID AGAIN Model and compute the area A inside the right half of the car dioid as modeled parametrically by 7139 z0y0 1sin cos t97 1sin sint97 75 S 9 S Use that to model and compute the area inside of the whole cardioid Solution We use the formula A 7 Man0 d0 One rst makes the computations7 for 7 S 0 S m0 1 sin cos 0 cos 0 sin 0 cos 0 cos 0 sin 20 1 sin sin 0 sin 0 sin2 0 cos 0 2 sin 0 cos 0 cos 0 sin 20 cos0 sin 20 cos 0 sin 20 cos2 0 3cos 0 sin 20 sin2 20 cos2 0 gcos 02 sin 0 cos 0 sin2 20 cos2 0 3 sin 0 cos2 0 sin2 20 Using this one computes A cos2 0 3sin 0 cos2 0 sin2 20 10 cos2 cos 0 d0 1 cos20 tT 3 d0 2 1 1 cos40 7 77 d0 2lt2 2 gt 7 I 7 3 g I 7 20 cos 0740 7T 7T 7 070 7 0 2 4 37139 4 One concludes again that the area 2A inside the whole cardioid is 2A Looking Back Prior to this lecture our tool box for computing areas con tained the formula b M 1967 1 where x 2 07 a g x g b Using this formula we have computed the area underneath the curve m a g x g b In this lecture we imagined that the curve x is modeled parametrically by ztyt7 c g t g d We found that the area underneath the curve m7 a 3 x g b can be computed by the formula 1 ytmt dt Shifting gears7 we examined the area inside a curve whose polar model is M0 a g 9 g b where 0 lt b 7 a g 27139 There we found the area formula to be bl 7r20d0 a 2 Examples illustrated these last two formulas CALCULUS H Spring 2011 LECTURE 11 Harry McLaughlin revised 3311 edited by This lecture provides 0 a de nition of a 3 d vector 0 a de nition of the dot product of two vectors 0 a de nition of the length of a vector 0 a de nition of the distance between two vectors 0 a de nition of the angle between two vectors 0 the Cauchy Schwartz inequality 0 a de nition of orthogonal vectors 0 a de nition of the vector projection of one vector on an other a de nition of the scalar component of one vector on an other projba compba 0 a calculation of the point on a 3 d line closest to an external point 0 a best approximation theorem Recall We recall from Lecture 10 o the area underneath a parametric curve can be computed by f ytm t dt 0 the area inside a polar curve can be computed by f r20 10 End of Recall Introduction Our goal in this lecture is to compute a point on a 3 d line that is closest to an external point Our analysis takes place in the setting of R3 Even though R3 has previously been introduced as a vector space we reintroduce it here in a bit more detail Vectors in 3Space A point in R3 is called a vector or a 3d vector recall that a point in R2 is called a 2 d vector One frequently writes a vector X E R3 in terms of its components X 17m27m3 m17m27x3 E R The zero vector in R3 is denoted here by 0 and de ned by 0 07 07 0 As in R2 vectors in R3 can be added and multiplied by a scalar If a and b are 3 d vectors then a b is the 3 d vector whose components are the sum of the respective components of a and b If is a real number then a is the 3 d vector whose components are the respective components of a each multiplied by That is if a a17a27a3 and b 131192123 and E R then ab a1 51412 527a3 53 an a a1a27 a3 Having de ned addition and scalar multiplication one wonders whether it is possible to de ne7 in a meaningful way7 a notion of multiplication of two vectors There are two such commonly used multiplications One is called here the dot product and the second is called the cross product We investigate next the dot product and delay introduction of the cross product until a subsequent lecture The dot product provides a way of modeling the notion of length of a vector and a way of modeling a notion of angle between two vectors We take a look De nition Let a 11012 13 and b 131192123 be 3 d vectors The dot product of a and b is written a b and de ned by a b albl agbg agbg If a 11012 and b b1 b2 are 2 d vectors then a b min agbg It is noted that the dot product is a scalar a real number it is not a vector Example If a 1 075 and b 37241 then ab lt1gtlt3gt lt0gtlt72gt lt75gtlt4gt 717 If a 10 and b 3 72 then ab lt1gtlt3gt lt0gtlt72gt 3 De nition Let a denote a 2 d vector or a 3 d vector The length of a is denoted here by lal and de ned by lalaa The vector a is said to be a unit vector if lal 1 The length of a vector is sometimes called its magnitude Note One notes that lal 0 if and only if a 0 0 or a 000 Example If a 1 2741 then lal Fa a 12 22 if m If a 1 2 then lal W 12 22 From the computation xAa Aa 2aa lAlxap W lal E R one obtains Mal W lal ER ln particular7 if a 3A 0 one lets to obtain 1 i ilal 1 lal lal Thus every non zero vector can be multiplied by a scalar to obtain a unit vector a vector of length one It is noted that in the above expression W W the rst set of double bars denotes the absolute value of a scalar and the second set of double bars denotes the length of a vector In the next de nition we model our intuitive notion of distance in R2 and R3 When we plot a and b we are able to intuitively sense the length of the chord between them The mathematical expression lbial assigns a number to our intuition De nition Let a and b be two 2 d vectors or two 3 d vectors The distance between a and b is de ned to be the length of b 7 a7 that is lb 7 al Example If a 17 274 and b 27 713 then Max071 lt71gt7237lt74gtgtlt1737gt The distance between a and b is given by lbi al 12 32 7 v59 Since laibl lilllaibllabl lbial one concludes that the distance between a and b and the distance between b and a are identical Angles and Projections De nition Let a and b be non zero 2 d vectors or 3 d vectors The angle between a and b denoted here by 9 is de ned by 9 7 arccos al N 39 It is noted that the angle between two vectors always lies in the interval 0n391gt2 One can write a b 9 arccos lt7 la W In words the cosine of the angle between two unit vectors is the dot product of the two unit vectors Although this de nition appears a bit abstract it can be seen to be consis tent with previously encountered notions of angles and their measurements3 We denote a good portion of a subsequent lecture to an in depth look at the de nition But for now we accept it at face value and use it EXAMPLE7ANGLE BETWEEN TWO VECTORS Find the angle 9 between the two 3 d vectors a V310 1 and b lt1 0 21gt Solution One computes ab xi70 2V3 lal N3 02 12 1One needs to Verify that the magnitude of 53 is equal to or less than 1 in order for 6 to be de ned We do that later 2Recall that the range of the arccosine function is 0 7r 3For example if a i and if b 10 then 3 b l and a b 7 71 7 71 1 7 Henceeicos ab7cos 7 lbl 12 02 V3 The angle 9 between a and b is found by arccos W W lt2 gt arccos 7 4 arccos 7 6 One concludes that the angle 9 between a and b is give by One can argue that if a and b are unit vectors in R2 then on the unit circle7 the length of the shortest arc determined by them is exactly arccos a b One says that two non zero vectors a and b have the same direction ifthe an gle between them is zero They have opposite directions if the angle between them is 7139 and they are perpendicular or orthogonalifthe angle between them is The de nitions of same direction and opposite directions leave us with the realization that we have not de ned the term direction of a vector It is tempting to leave the term unde ned7 but because of its use in the litera ture we offer the following The direction of a non zero vector a is the vector a itself This allows one to speak of the direction vector a Whoops The above discussion is based on the de nition 9 7 arccos W W 39 One can ask Is one permitted to make such a de nition That is does the real number lt a 39 b gt lal lbl lie in the domain of the arccosine function More speci cally is it true that ab 71lt 7 lt1 lallblgt That the dot product of two unit vectors lies between 71 and 1 follows from the Cauchy Schwarz inequality The Cauchy Schwarz inequality is ubiqui tous in mathematics so we take time out to state and prove it The proof below uses some yet unproven but easily veri able identities involving dot products Even though we donlt prove them we list them here If a b C are all Sd vectors or all 2d vectors and if i E R then i Mallllal ii abba iii Aababab iv abcacbCl Theorem CauchySchwarz Inequality Let a and b be two 2 d vectors or two 3 d vectors Then la bl S lal M with equality if and only if either b 0 or a b for some real number The pair of vertical bars on the left hand side ofthe inequality denote the absolute value of the real number a b And each pair of veritical bars on the right hand side of the inequality denotes the norm of a vector Proof We treat rst the case b a 0 Since the length of every vector is nonnegative one has a b b 2 lbl2 ab ab 77b 77b a lbw l lbw l a b a b 0 a aa72 3 2 3 4 bb Jig is 47 a 27071393 l322 Thus a39b2 S lalZlbl2 andhence labl laHbl Ifb 0 then clearly labl g lal Recall that ao 0 and lol 0 This completes the proof of the inequality We examine now the equality condition If b o the inequality becomes an equality7 since both sides of the inequality assume the value zeroi Further7 if a Ab for some real number A then la 39 bl W lbl2 W lbl lbl lAbl lbl lal lblA Thus if either b 0 or a Ab equality holds Next we assume that equality holds If b y 0 then the rst inequality in the sixline computation above7 must become an equality Hence Setting A le one sees that a Abi This completes the proof of the equality condition4 It is possible to rest easy77 now The function 9 7 arccos lal lbl is de ned as long as a and b are non zero vectors Another Look The proof of the Cauchy Schwarz inequality depends upon the vector b a 7 7 b a W One might ask From where did this vector come If one rst asks the question What vector of the form a 7 b is orthogonal to b one nds7 from a straightforward calculation that ab 7 lblz o aiabyb abikbb abialblg 7 ab 7 W2 Thus the special vector abb a 7 M2 is orthogonal to b This question of orthogonality is naturally asked when one is trying to deter mine the point on a line in 3 space or 2 space that is closest to an external point We investigate these ideas later in this lecture We have de ned the notion of orthogonality above and used it Because of its importance we enter the following note on the relationship between the angle between two vectors being and the vectors being orthogonal 4We have employed a subtle logic argument In order to prove that if equality holds then either b 0 or a Ab it suf ces to argue that if b 7 0 then a Ab 10 One notes that two nonzero vectors are orthogonal if the angle between them is 55 One also notes that the 3d zero vector is orthogonal to all other 3d vectors similarly for the 2d zero vectori However we have not de ned the angle between two vectors one of which is the zero vec tor We donlt make such a de nition here So we are not able to say I that the angle between two vectors one of which is the zero vector is 2 The Law of Cosines Armed with the fact that the cosine of the angle 9 between two non zero vectors a and b can be written a b cos 9 7 lal lbl one is able to make the following computation laiblz aebgtltaibgt aa72abbb W lblz Qlal lbl0080 Theidentity ae blz W lblz e 2w lblcos0 is a vector form of the familiar law of cosines When a and b are orthogonal it becomes a vector form of the Pythagorean Theorem that is lai blz W W In the following example we are able to compute the distance between two unit vectors knowing only the angle between them In particular their com ponents are not known EXAMPLE iDISTANCE BETWEEN TWO VECTORS Let a and b be unit vectors in R2 or R3 such that the angle 9 5H a b O and a and b are nonzero vectors then b 7r 6 arccos L arccos O 7 lal W 2 between them is Compute the distance between a and b Solution One uses the law of cosines and writes la 7 W lalz lblz Qlal lbl COS t9 117 211 2 7 xi One concludes that the distance between a and b is given by laiblv273 De nition Let a and b be two 2 d vectors or two 3 d vectors and assume that b 7 0 Then the projection of a on b is denoted here by projba and is de ned by projba 7 b a b lblz From where does this de nition come We see below that the point on the line determined by b that is closest to a is exactly projba This observation motivates our study of projections Further a short calculation shows that the length of the projection of a on b is given by 3l lbl 39 EXAMPLE iPROJECTION OF ONE VECTOR ON ANOTHER Let a 3 71 1 and b 20 2 Find the projection of a on b and the projection of b on a Solution We make the following calculations ab 321012 8 lalz 32t71212 11 b Z 220222 8 Using these one obtains b projba 7b 27072 projab 7a Eu 7 One concludes that projba 2707 2 Prolab 377171 Quite by accident it happens that projba 27 O7 2 b and Example Using the data of the above example one determines that the length of the projection of a on b is given by la bl 8 4 b 217 772 lbll lbl N5 bllbl 217 7 lbl lbl This is consistent with l20 2M N5 13 One notes that the projection of a on b is a vector it is a scalar multiple of b It is noted that this use of the term projection is appropriate since i a b aiprOjbababiWbbabia b0 ln words the vector from the projection of a on b back to a is orthogonal to b We think of the projection of a on b as the shadow of a on b A Crystallization Having investigated the notion of projection of one vector on another7 one is able to reinterpret the Cauchy Schwarz inequality The proof shows that the Cauchy Schwarz inequality is just a restatement of the obvious inequality 0 ab afi b 2 b In words the Cauchy Schwarz inequality asserts that the length of the vec tor from the projection of a on b back to a is non negative Said another way the distance from a to its projection on b is non negative De nition Let a and b be two 3 d vectors or two 2 d vectors and assume that b 7 0 Then the component of a on b is denoted here by compba and de ned by a b compba 7 lbl One notes that the component of a on b is a scalar It can be negative The magnitude of compba is exactly the length of the projection of a on b That is6 lcompbal lprojbal EXAMPLE Let a 37171 and b 202 Find the component of a on b and the component of b on a 6The pairs of Vertical bars in this expression have two distinct meanings one is the length of a Vector and the other is the absolute value of a real number Solution Using the computations of one of the above examples one sees that ltabgtb 8 b 8 b 8 b 2 b prOJa 7 7777777 7 b W W xglbl 2 lbl 4 W W and hence compba2 Further ibi ba if 8 i W lal lal r11 W and hence It is worthwhile taking a careful look at the two notions i projba b and H 7 b 11 compba 7 7H proj a a b b b 7 7 W W i b prOJba compba If one writes then it is clear that In words the projection of a on b is the unit vector in the direction of b scaled by the component of a on b This brings us to the focal point of this lecture Closest Point on a 3d Line We imagine a line in 3 space and an external point Our questions are c How does one locate the point on the line that is closest to the external point 0 How does one determine the distance between the line and the exter nal point Let a and b be vector representations of distinct points on the line Let c be a vector representation of an additional point not necessarily on the line We nd the point on the line that minimizes the distance from c to the line Then we compute the distance from c to that point One can model the line7 containing a and b7 by the vector expression aAltb7a7 foolt lt 00 The points on the line are exactly those points in 3 space that can be mod eled by a b 7 a for some value of A The square of the distance D from c to an arbitrary point on the line is given by D laba Clz that is D abiaicabiaic Our goal is to nd the value of 7 say 0 that minimizes D We are using the fact that the distance from c to the line is minimized by the value of that minimizes the square of the distance from c to the line To this end we make the following computations D abacabac aCbaaCba lai d2 2b 7 a a7 c Azlbi alz 16 All of the quantities in this last expression are known except A It is a quadratic expression in A Since the coefficient of A2 is positive7 the minimum value of DA can be found by setting to zero the derivative of the quadratic expression and solving for the corresponding value of A This provides a value A0 such that DA0 DA7 foo lt A lt 00 x AA2BAC iooltgtxltltgto 2AAB 700ltAltltgto O B 7 Thus DA0 DA7 foo lt A lt 00 when A0 2biaaic A0 2lb7al2 biaHC a T lbialz Thus the point p on the line determined by a and b that is closest to c is given by ltb1lagtltbiagt p aA0bia a This can be written in the form p a prOiIHAC a Finally7 the distance from c to the line is given by lcipl The following example illustrates how one computes the distance from c to the line MODELING EXAMPLE 7CLOSEST POINT ON A 3 D LINE Let a 1717 2 and b 307 71 and c 4721 Find 17 the 3 d coordinates of the point p on the line determined by a and b that is closest to c ii Find the distance from c to the line Solution We use the formula for 0 found in the above discussion to locate the point on the line closest to c In particular we let b 7 31 c 7 a lb7al2 bia and p a proibia i a a 0b a Then we compute the distance from c to the line using the ex pression lp 7 cl We make the following calculations 7a 7 421gt7lt1712gt 3351 b7a 307171712 2153 b7ac7a 2153 3351 lt2gtlt3gt 13 lt73gtlt71gt 12 lb7al2 21732173 2212732 14 0 12 6 y Now we use the fact that a vector model for the point p on the line that is closest to c is given by pa0b7a lts coordinates are computed by p 31 0b7a 18 6 17 71 2 271773 12 6 718 lt1777172Tgt Eli 7 7 7 39 The square of the distance from the point c to the line is given by 19 1 4 7 2 7 iii if 16 pl 47271 lt77 7 7 19 1 4 14 77lt2gt7lt1gt 7 9 2 15 2 112 t 7 t 7 1amp1 T 49 49 49 a 49 761 77 61 7 Thus the 3 d coordinates of the point p on the line that is closest to c are given by 2 2 p 417977179 The distance from the point c to the line is W We abstract the above analysis The point p on the line determined by a and b that is closest to c is given by ciab7a PaProlbiaC aa bia z b al 19 Further the square of the distance D from c to the line is given by DC7pC7p and the distance itself is given by i D2 67p 67p This provides an explicit formula for p and an explicit formula for the dis tance from c to the line Examples are found in the Worksheets Discussion It is noted that c7p is perpendicular to the line determined by a and b7 that is c 7 a projbac 7 a b 7 a 07 or written another way c 7 a b 7 a c7a7Wb7a b7a 0 This can be seen by inspection This observation plays a central role in the study of Fourier series Because of that we restate it in its abstract form Theorem Let V and w be vectors in R3 or R If W is not the zero vector then V minus its projection on w is orthogonal to w That is Vw w V777 w0 M W Proof The proof follows immediately from the following calculation lt VW Wgt VWWW V777 w vw7 2 W W W I 2 Wm WM lWl2 0 20 The best approximation theorem given next follows immediately from the previous theorem Theorem Best Approximation Let V and w be vectors in R3 or R If w is not the zero vector then the projection of V on w offers the best approximation to V from the line through the origin determined by w That is if p is the projection ofV on w then for every u that is on the line through the origin determined by w one has lviul Z lvipl with equality if and only if u p that is if and only if u mw lWlZ 39 Proof We write V711 Vipp7ui The proof uses the fact that p 7 u V 7 p 0 Recall that V713 is orthogonal to W and hence to every vector on the line through the origin determined by W The proof follows from the following computation lViulZ VPPu39VPPu lvipl22piuV7Plpul2 lvipl20lpiul2 2 lvipl2 with equality if and only if u p m The proof of the best approximation theorem is of interest It doesn t use calculus constructsl This means that our analysis of the closest point on a line to an external point can be rewritten without using the notion of a derivative We used the derivative above in our analysis of closest points because we wanted a familiar approach to minimization Later using these abstract ideas we are able to nd the point on a plane that is closest to an external point without using calculus constructs 21 Looking Back We have had another look at R3 and in particular the notion of an angle between two non zero vectors in R3 We de ned the angle between a and b by the formula 1 a b 9 cos and then realized that one needs to be assured that a b m always lies in the domain of the inverse cosine function This issue was resolved by introducing and then applying the Cauchy Schwarz inequality We introduced the idea of a projection of a point onto a line Using that we were able to nd the point on a given line that is closest to an external point 7 that was the goal of the lecture At the end we remarked that closest points can be found without recourse to calculus constructs 22 This CALCULUS 11 Spring 2011 LECTURE 12 Harry McLaughlin revised 3311 edited by lecture provides a de nition of a plane in R3 vector and scalar models of planes a de nition of the cross product of two vectors in R3 the theorem a x b 131 1b1sin0 a way to model the plane containing three non colinearpoints in R a way to compute areas of triangles in R3 A a X b a way to compute areas of parallelograms in R3 A a X b a way to compute volumes of parallelepipeds in R3 V 1c a X b a way to nd the point on a plane that is closest to an external point a way to model the line of intersection of two given planes Recall We recall from Lecture 11 o a 3 d vector is an element of R3 o the angle between two non zero vectors is given by 9 cos 1 MM 0 the Cauchy Schwarz inequality is given by la bl S lal lb 0 the de nition of projection of one vector on another 0 the calculation of a point on a line that is closest to an external point End of Recall Introduction The goals of this lecture are to provide a de nition of a plane in R3 and to investigate its consequences In particular we are able to model a plane containing three non colinear points7 ii locate a point on a plane that is closest to an external point and iii model the line of intersection of two planes Along the way we iv compute the areas of 3 space triangles and parallelograms Additionally we v compute the volumes of parallelepipeds Planes One typically uses the term plane to name a at surface For mathematical modeling purposes one needs an explicit mathematical de nition of a at surface How to do it De nition Let n denote a non zero vector in R3 and let p E R3 The plane 73 in R3 with normal vector 11 and containing the point p is the set of all vectors x in R3 such that the vector x 7 p is orthogonal to n That is PXER32HX7p 0 One calls the equation 11 x 7 p 0 a vector model of the plane 73 It is noted that we have not de ned the word plane7 but rather the phrase plane with normal vector n and containing the point p Sometimes when the context permits we use just the word plane Example The plane in R3 with normal vector 11 17273 that contains the point p 107 71 has a vector model given by 1 723 zyz 7 1971 0 my 2 6 R3 Making the indicated multiplications in the vector model yields a scalar model for the plane name y 172y3220 zyz ER This idea is investigated more carefully in one of the exam ples belowi De nition Let 731 and 732 be planes in R3 with respective normal vectors H1 and n2 The planes are parallel if n1 and n2 are parallel that is if n1 n2 for some E R The two planes are perpendicular if their normal vectors are orthogonal that is if n1 n2 0 Properties of Planes We explore some of the properties of planes We as sume that 73 denotes a plane in R3 with normal vector n and containing the point p The rst two properties are technical i If n is a normal vector for a plane then so too is n for every non zero foo lt lt 00 That is nxip Ofor XERS ifand only if nxip0 for every non zero E R The proof is straightforward and omittedi ii If X and q belong to the plane with normal vector n and containing p then X belongs to the plane with normal vector 11 and containing q Proof Since nxip0 and nqip0 onehas nXiq nXip aeoo nXipHnpeq nXipnqep 070 0 From this one concludes that x belongs to the plane With normal vector n and containing qt This completes the proof iii This property is one way of saying that a plane is at If the two distinct points X1 and X2 belong to 73 then every point on the line determined by X1 and X2 belongs to P Proof One recalls that the line determined by x1 and X2 is exactly the set of points p such that p l 7 Ax1 sz for A E K To argue that such a point belongs to 73 one notes that if for some A e R p 1 7 Mp Ap then W AX1 sz P 39 n 1 AX1 P sz 13 39 n 1 AX1P 39nxz P39n 0 0 0 This completes the proof Of course R3 itself satis es this property and R3 may or may not be thought of as at However7 the next property ensures that R3 is not a plane iv This property says that a plane is thin in some sense7 it is two dimensional Let a and b be distinct vectors in a plane 73 with normal vector n and containing the point p Further assume that a7 p and b 7 p are not colinear Then a point X E R3 belongs to 73 if and only if xp1ap2bp for some real numbers 1 and 2 Proof Assume rst that xp11aPI2bP Where 11 and 12 are real numbers Then nXip IlnaipHmnbip 00 0 One concludes that x belongs to 7 Assume next that a7 b and x belong to 73 and that aip and hip are not colinear vectors We argue that X pziaPI2bP for some real numbers 11 and 12 It can be argued7 using elementary arguments that every vector x in R3 can be Written in the form xzon p11a 1312b pl for some real numbers 1011 and 121 Using the fact that ab and x belong to 73 one obtains 0 nXip nronzla7pzgbip Iolnl200 Iollei Since lnl 0 one concludes that 10 0 Thus xPIla PI2b P for some real numbers 11 and 12 This completes the proof The next example is a valuable one for computation purposes It allows one to convert a vector model of a plane given in the de nition to a scalar model 1The argument is not provided here MODELING EXAMPLE7 CONVERT A VECTOR MODEL OF A PLANE TO A SCALAR MODEL Let n 233 and p 100 Find a scalar model in the form ambyczd 0 abcd ER for the plane 73 in R3 whose vector model is nxip 0 Solution One lets X belong to the plane and writes X z y The scalar model emerges from the following computation 0 n X i p 27373 9071472 7 17070 2737 3 95 7172472 2x713y32 2m3y3272 Thus 2m3y32720 Conversely one lets X z y z belong to R3 and 2m3y 327 2 0 It is argued that n x7 p 0 To this end one writes 0 2x3y3272 7 233 myz 7 2 2737 3 9072472 1707 0 H Kip One concludes that a scalar model for 73 is 2m3y32720 It is seen that the values of 011 and c as they appear in the scalar model of a plane are the coordinates of the vector n as they appear in the vector model of the plane For example a normal vector n for a plane modeled by 2m3y 32 7 2 0 is n 2 3 3 So to convert a scalar model of a plane to a vector model one needs only to pick olfl 7 n and nd one point p on the plane Having done that that it suf ces to write n x 7 p 0 And the plane with normal vector n 2 33 has a scalar model of the form 2x 3y 32 1 0 for some 1 E R As seen in the previous example if p 0340213 and n a b c then the plane with normal vector n and containing the point p is modeled by az 7 0 by 7 yo Cm 7 20 O In this case 1 7am0 byo 620 All of this leaves open the question How in general does one nd n The normal vector can be found easily once a scalar model of the plane is known But it turns out that the normal vector can be computed directly once three non colinear points on the plane are known All of this involves a vector construct that we have yet to encounter The next section deals with de ning the notion of a cross product of two vectors Cross Products The idea of a cross product of two 3 d vectors a and b plays a central role in our study it produces a vector that is orthogonal to both a and b The cross product operation assigns to two 3 d vectors a and b another 3 d vector denoted by ax b There are two commonly used algebraic de nitions of the cross product of two vectors a a1a2a3 and b 13112 b3 The rst is a throw at the user formula it is a gtlt b agbg 7 agbg 7a1b3 7 agbl ale 7 12121 So for example the second component of the vector a X b is the number 7a1b3 7 a3b1i This formula is not particularly easy to remember but quite useful in prov ing theorems The second and equivalent de nition is easier to remember it centers on a procedure But it involves the computation of a determinant which is written 1 j k 04 a2 a3 1 b2 3 Because this de nition is frequently used we rst investigate it and its perhaps unfamiliar symbols The throw at the user formula follows imme diately from the determinant de nition The notion of a determinant hinges on the notion of a matrix so we begin by introducing the notion of a matrix through example The notion of a ma trix can be carefully de ned but it is easier and for our purposes probably better to introduce 2x2 matrices two by two matrices through examples Matrices Matrices are usually denoted by square or rectangular arrays of numbers Here the 2 x2 matrices A and B are de ned by AltC1 7 andBlt 1 gt and the 2nd column of B The 1St column of A is the column vector lt 0 is the column vector lt 1 Columns are vertical think columns of a building The 2nd row of A is the row vector 01 and the 1St row of B is the row vector 7312 So 2 x 2 matrices have rows and columns which are numbered according to the scheme of the example Determinants The determinant of the 2x2 matrix Hi is a real number7 denoted by lAl and computed by W 13 7 M72 11 ln words the determinant of A is the product of the elements on the main diagonal minus the product of the elements on the other diagonal Again llt112 21 122 a 13 2 One can well imagine What 3x3 matrices are We deal With them implicitly in the de nition of determinants of 3x3 matrices The determinant of the 3 x3 matrix 714 2 A 1012 72 1 3 written lAl is computed by expanding across the 1St row We do it w ml 13 74 i 13 lt2gt 2 1 the signs in front of each 2x2 determinant alternate each element of 10 71l03 7 112l 7 4l13 7 7212l 2l11 7 720l 7171274421 12 7 16 2 7272 Thus the determinant of the matrix A is given by lAl 7272 Coordinate Vectors Of some use in computing the yet to be de ned cross product are three coordinate vectors in R3 they are given by i100 and j010 and k001 From these de nitions one concludes that every x y z E R3 can be written as any 95 951 24 2k Examplei One writes 371M 3100 71010 001 31 7j Viki Now that we have explored matrices determinants and coordinate vectors we have the ingredients needed for our work with cross products De nition The cross product7 of the two 3 d vectors a a17a27a3 and b b17 b2 b3 is written a x b and de ned by i j k axb a1 a2 a3 b1 b2 b3 Computing this one nds i j k axb a1 a2 a3 b1 52 53 ia2asia1a3 a1a2 1 b2 bgl Jlbl bgl b1 b2 7 1a2b3 7 L352 7 ja1b3 7 03191 ka1b2 7 0251 or 7 agbg 7 agbg7 7a1b3 7 03121 a1b2 7 12121 This results in the throw at theuser formula a X b a2b3 7 agbg7 7a1b3 7 agbl7 0ile 7 gain ln computing cross products7 with pencil and paper7 the writer almost al ways uses the determinant calculation EXAMPLE7COMPUTATION OF A CROSS PROD UCT Compute the cross product of the two vectors a 17 71 2 and b 107 073 Solution We use the determinant de nition rst To that end onewrites 1 J k axb 1 71 2 10 0 73 71 2 1 2 1 71 1 0 73 75 l10 3lkl10 0 H37 0 493 7 20 k0 10 3 23 10 On the other hand7 one can use the formula de nition7 obtaining a gtlt b a2b3 7 03122 7a1b3 7 a3b17a1b2 7 a2b1 17173 7 M0 71173 7 21017 WW 7 7110l 37 237 10 Fortunately7 in both cases one nds that a x b 32310 If one is Willing to use only the formula one doesn7t need to introduce the notion of a determinant The cross product exists simply as a 3 d vector Whose elements are prescribed by the formula However7 the determinant version of the cross product is of value because it frees one from having 12 to remember the order of subscripts in six products and ii it has some value in proving theorems about cross products One might wonder why history has ltered out the not so good looking cross product further what prompted its de nition in the rst place The writer suspects that in the modeling of physics phenomena there was a need for identifying a single vector that is orthogonal to two given vectors In search ing for a solution to this problem7 the cross product emerged We investigate this orthogonality property below Properties of Cross Products One theorem that is easily analyzed by the formula alone is the following Theorem The cross product of two vectors in R3 is orthogonal to both of the vectors That is if a and b belong to R3 then axba0 and axbb0 Proof Let a a17a27a3 and b 121122123 Then7 using the formula for the cross product one has by inspection a X b a agbg 7 a3b2a1 7 albg 7 a3b1a2 a1b2 7 a2b1a3 0 and a X b b agbg 7 aging zl 7a1b3 7 agbl zg a1b2 7 aglzl zg 0 This completes the proof This result is probably the miszm d tre of cross products Another theorem that is easily analyzed by the formula alone is the following Theorem For two vectors a and b in R3 the cross product a x b is the negative of the cross product b x a That is axb7bxa Proof The cross product b X a is found by interchanging the symbols involving the als and bls on both sides of the equation a X b 2123 7 agbg 7a1b3 7 a3b17a1b2 7 win to obtain b X a bgag 7 bgag 7b1a3 7 b3a17b1a2 7 b2a1 and then noting that the expressions on the right hand sides are the negatives of each other This completes the proof The next theorem says that the magnitude of the cross product of two unit vectors is the sine of the angle between them The proof uses the fact that the magnitude of the dot product of the two unit vectors is the magnitude of the cosine of the angle between them Before stating the theorem we recall that the angle 9 between two vectors always lies in the range 077139 and hence the sine of the angle between them is always non negative Theorem Let a and b be two unit vectors in R3 Then la x bl sin 0 where 9 cos 1 a b Proof In the proof we nd an expression for sin2 9 and then show that la gtlt bl2 equals that expression Let a a17a27a3 and b 1217 122123 be two unit vectors First one nds that sin2 9 1 7 cos2 9 1 7 a b2 1 1111 2112 asbs2 1 ail 5173 a 72a1a2b1b2 7 2a1a3b1b3 7 Zagagbgbgi Now it is argued that la gtlt bl2 equals this expression We use the fact that a a3 a 17 etc To this end one writes la gtlt bl2 lia2b3 7 aging 7ja1b3 7 a3b1 ka1b2 7 a2b1l2 7 a b a b aib aibi aib 1512 72a2a3b2b3 7 2ala3b1b3 7 2a1a2b1b2 baa a 011 a Wag a 14 72agagb2b3 7 2a1a3b1b3 7 2a1agb1b2 531 i a 551 i a 17W i a 72agagb2b3 7 2a1a3b1b3 7 2a1agb1b2 123 bi b 7 Inga 7 Inga 7 Inga 72agagb2b3 7 2a1a3b1b3 7 2a1agb1b2 1 all ugh a 72agagb2b3 7 2a1a3b1b3 7 2a1agb1b2 sin2 One concludes that la gtlt bl sin 9 This completes the proof Next one asks how to interpret ax b if the two non zero vectors have lengths that are not necessarily one If a and b are not necessarily unit vectors but are both non zero vectors one observes that according to the de nition of cross product and hence for all non zero vectors a and b one has la x bl lal lblsin0 where 9 cos 1 a b In the formula for a X b just replace a3 3 by W a a an by 1 a2 by g etc Pothole Although one can write b t9 cos71ltaigt W W sin 1 a X b W W since the range of the arcsine function is 77 one cannot equate 9 to Modeling Planes One recalls that a plane 73 in R3 with normal vector n and containing the point p7 can be modeled by nx7p07 XERS Thatis7 PXER32HXpO In practice one frequently needs a preprocessing step to compute the normal vector n The next example shows how one can nd both a vector model and a scalar model of the plane containing three non colinear points in R3 MODELING EXAMPLE A PLANE THRU THREE POINTS Let p 0737717 q 177271 and r 724773 Find a vector model and a scalar model of the plane that contains p7 q and 1quot2 Solution We nd a normal vector n for the plane and then convert its vector model7 namely nlxipQ to a scalar model Since qip and rip are both orthogonal to qip gtlt rip one can de ne n by nmemxem We are using the fact tha p7 q and r are not colinear 2Although We don t provide a proof it is true that there exists exactly one plane containing three given noncolinear points in To compute n one writes 177271 7 073771 1707727311 177572 Clip rip 7274773 7 073771 7270473731 1 J k 1 75 2 72 1 72 1 2 175 172l 5 l72 72lkl72 1 110 7 2 7j72 4 k17 10 877279 Thus the plane is modeled by nltxepgtlt877279gtAltz707y7321gt 0 and hence also by 8x72y7926790 8m72y792730 Modeling With Planes Areas of Triangles and Parallelograms For two non zero 3 d vectors it is possible to use the magnitude of a X b to compute the area of the triangle that has a and b as sides The triangle is of course planar but it lies in R3 Let distinct non zero 3 d vectors a and b model two sides of a triangle Using the notion of triangular area as modeled by one half the length of the base times the heigh 7 one nds that the area of the triangle is gm lbl sin lt0 where 9 is the angle between a and b that is 9 cos 1 33 From this one obtains two useful formulas As just noted the area A of the triangle with two non zero 3 d vectors a and b as sides is given by 1 A 7 b 21a x 1 where 9 is the angle between a and b One notes that the area ofthe parallelogram determined by a and b is twice the area of the triangle determined by a and b ii Thus the area A of the parallelogram determined by the two vectors a and b is given by A axb MODELING EXAMPLE7 AREA OF A TRIANGLE A triangle in 3 space has the three points p 100 q 010 and r 001 as vertices Model the area A of the triangle and within the model compute its area Solution The area A is modeled by 1 A 51qip X FPl 18 To nd a value for A one writes a i q 0717017070 71170 b rip 0707117070 717071 ThenAlaxbland i j k 10 710 711 axb 71 0 1 7 71 0 1 0 1 71 1 71 0 From this one obtains axb Thus laxblx1212123 Using the formula A la x bl one concludes that the area of the triangle is given by It can be argued that if a7 b7 C model sides of a non degenerate triangle in 3 space7 then the area can be modeled by all three expressions 1 1 1 Elaxbl and Elaxcl and Elbxcl and found to be independent of the model One can gain con dence in this assertions by comput ing for this example lpiqxriql7 For parallelograms the story is much the same There are three parallelograms that contain three non colinear points p q r as vertices Since the areas of each of the parallelograms is twice the area of the triangle having p q r as vertices the areas of all three parallelograms are equal In particular one models the area A of such a parallelogram by A lq7p X PPM Volumes of Parallelepipeds Let ab and c be non colinear points in R3 One of the parallelepipeds generated by ab and c has base area given by la gtlt bl and height h given by h lcl l cos where 9 is the angle between the two vectors a x b and c that is 0COS1 M 39 lCl la X bl lts volume V modeled by area of base times height is given by la x bl lcl lcos0l that is by V lcax These areavolume formulas are quite powerful Given threefour points in R3 it is possible to easily determine the areas of the corresponding triangles and parallelograms and the volumes of the corresponding paral lelepipeds The ease of these computations is supported by the underlying vector space structure The computations of areas of parallelograms and volumes of parallelepipeds are left to the Worksheets 20 Best Approximations The next theorem validates one s instincts about the point on a plane that is closest to an external point It says that if e is a point external to a plane 73 and if p E 73 is such that e 7 p is normal to the plane7 then p is the point on the plane closest to e Actually this statement is true even if e belongs to the plane Theorem Best Approximation Let 73 denote a plane with normal vector n Let e E R3 and p E 73 be such that e 7p n for some E R Then p is the unique point of 73 the is closest to e That is7 leiplg leixl7 VXEP with equality if and only if x p Proof Let x E 73 Using the fact that eipx7p An x 7 p 0 one makes the following computation leixl2 leil3l3ixl2 97PXP39ePXP leipl22eip39X7PlXPl2 leipl270lxipl2 leipl2lxipl2 2 leipl2 with equality if and only if X pi One concludes that leipl g leixl7 VXEP with equality if and only if x p This completes the proof The best approximation theorem is quite useful We use it in the following example However7 it tells only one half of the story It doesn t deal with the question of existence of a closest point One wonders7 is it possible that 21 there is no point on a plane that is closest to a given external point In the example we are able to display a closest point so the existence question is resolved Later we resolve the general existence question For now we sweep existence under the rug MODELING EXAMPLE 7 CLOSEST POINT ON A PLANE A plane 73 in 3 space contains the point r 210 and has n 10 71 as a normal vector Find the coordinates of the point p on 73 that is closest to the external point 0 1 0 Also nd the distance from the point 0 1 0 to 73 Solution Let x zy 2 denote an arbitrary point in R3 and r 210 Then 73 is modeled by n x 7 r 0 That is 73 consists of all x E R3 such that n x 7 r 0 We model the point p on the 73 that is closest to 010 by insisting that for some E R p 010 1 n That is the vector from 0 10 to p is normal to the plane 73 Here we are applying the theorem asserting that p is the point on 73 that is closest to 010 if and only if the vector from 0 1 0 to p is perpendicular to 733 The line in 3 space containing 0 1 0 that is perpendicular to the plane is modeled by 010 n foolt lt 00 that is by 010 10 71 700 lt lt 00 Thus every point on this line has the form 1 7A 3It is worth noting that we are Changing the original problem one involving distance to a geometry problem one involving the intersection of a line and a plane 22 for some A E R In particular there is a value of A such that the point lies on 73 We nd that value of A by insisting that the coordinates of the point A 1 7A satisfy the vector model of73 That is 0 n A1 7A 7 210 1071A17A7210 1071A7207A A 7 2 A 2A7 2 One concludes that A 1 Thus p 07170117071 171771 One concludes that the point on 73 that is closest to the point 0 10 is given by p 171771 To compute the distance from 0 1 0 to 73 one writes 071707171771l 71707 71gt 02 12 xE One concludes that the distance from 0 10 to 73 is The next example deals with a closest point However the normal vector to the plane is not given it needs to be computed 23

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.