### Create a StudySoup account

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

Already have a StudySoup account? Login here

# Calculus from an Adv Viewpoint MATH 6102

UNCC

GPA 3.73

### View Full Document

## 18

## 0

## Popular in Course

## Popular in Mathematics (M)

This 38 page Class Notes was uploaded by Mrs. Dangelo Fahey on Sunday October 25, 2015. The Class Notes belongs to MATH 6102 at University of North Carolina - Charlotte taught by David Royster in Fall. Since its upload, it has received 18 views. For similar materials see /class/228906/math-6102-university-of-north-carolina-charlotte in Mathematics (M) at University of North Carolina - Charlotte.

## Reviews for Calculus from an Adv Viewpoint

### 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: 10/25/15

Chapter 12 Integrals 121 Path Intgrals Let 04 a7 b a R2 or R3 be a Cl path The function 15 0d a a7 b is called bijective if and only if it is one to one and onto The composition 6 a o t 07d a R2 or R3 is called a repararnetrization of Lt We saw earlier that Mt a t t Thus7 HB WH l tllla tll so the reparametrization is a change of speed andor direction Now7 o 07 d a a7b must map the endpoints to endpoints Thus7 either 150 a and d b or 150 b and d a In the former case7 60 a 0 aa and 6d a d 0vb and the reparametrization is called orientation preserving The other case is called orientation reversing Theorem 121 A repararnetrization is orientation preserving if and only if 15 gt 0 and orientation reversing if and only if 15 lt 0 De nition 121 The image of a one to one piecewise Cl path 04 ab a R2 or R3 is called a simple curve A simple curve means that no two points of the interval a7b are mapped onto the same point on the curve7 a simple curve cannot intersect itself De nition 122 The image of a piecewise Cl path 04 ab a R2 or R3 that is one to one on ab and is such that aa 0vb is called a simple closed curve 149 150 CHAPTER 12 INTEGRALS 122 Path Integrals of RealValued Functions We used the Riemann sum to de ne the de nite integral of a real valued function on an interval a7 b We will use this same technique to de ne the integral of a real valued function over a path in R Let f R a R be a continuous function and let 0 ab a R be a path in R The composition fot represents the values of the function f along the curve 0 Let P be a partition of a7b7 say P to a7t17tn b We can approximate the path 0 by the polygonal path go 7 whose vertices are 0a ot0ot17 7otn ob We can approximate the length of ol connecting otZ and oti1 as HUti HAti where Atl tZJrL 7 ti Now we form the sums n71 An Z f0ti HU tiHAti i0 If f were positive7 then An could represent the approximate area of the fence77 built along 0 from 0a to ob whose height is determined by f De nition 123 Let 0 ab a R be a Cl path and let f R a R be a function such that the composition fot is continuous on a7b The path integral fafds off along 0 is given by b n71 fds mamatum 39 Zfltalttigtgtldltmlml 7 a i0 if this limit eccists Recall that Example 121 Let fxyz ayz and ot SlHL COS tsint7 t E 07T2 Find fafds First7 f7sint cos tsint ixEsinztcost o t 7 cos t7 ixisin tcos t HIT0H 2 W 2 quot 2 fds ixi sinZtcost dt igsinst g 7 0 0 f 0 t A MATH 6102 090 Spring 2007 123 PATH INTEGRALS OF VECTOR FIELDS 151 Example 122 Let fy 2x y Consider the path integral 7de along the following paths in R2 joining the points 1 0 and 01 H ONT2 Along the straight line segment 02t 10 t71 1 1 7 t t t 6 01 1 LO Counterclockwise along the quarter circle 01t cos t sin t t 6 9quot Along the piecewise Cl path 03 that consists of the path 04t 6 01 followed by 05t 0t t 6 01 H We have ot 7 sintcost and 1 and 1 fds2costsintdt3 71 0 E0 Here we have o t 711 and 02t x2 and 1 fds 27t2dt 72 0 2 Here U4t U5t 1 and 1 1 fds fds fds2i2tdt tdt C73 04 05 0 0 Theorem 122 Let U be a Cl path in R let f be a real valued function continuous on the image of o and let 739 o o t be a reparametm39zatz39on ofo Then desdes Note that this says that the integral is independent of any reparametrization 9 123 Path Integrals of Vector Fields Let F be a vector eld de ned on the area that contains the curve 17 Again let P be a partition of ab and let AU be the vector that runs from ot to ot1 Then A17 is approximately equal to o tAt Then for each 239 0 n 7 1 we can form the dot product Fot A0 We can then form the Riemann sum 71 H o i If this limit exists then it is called the path integral or line integral of F along a and is denoted by 0 F ds MATH 6102 090 Spring 2007 152 CHAPTER 12 INTEGRALS De nition 124 Let a a7 b 7 R be a Cl path and let F R 7 R be a uector eld such that the composition Fot is continuous on db The path integral fUF ds ofF along 0 is given by indwbmga otdt a In the case where F represents a force7 the path integral gives the work done by F along 0 Theorem 123 Let F be a continuous vector eld on R let a db 7 R be a 01 curve and let 7t o t be a reparametrization of o where t c7d 7 db Then F d L F ds if t is orientation preseruing s T 7 L F ds if t is orientation reuersing De nition 125 The line integral fUF ds of a continuous vector eld F around a oriented simple closed curve 0 is called the circulation of F around a 1 24 Double Integrals As we did for a function from R 7 R we can also talk about the integral of a function R2 7 R Here we have two different dimensions along which to partition Divide the intervals a7b and c7 d into n subintervals Pa172771 b andQCy17y277yn1d and form the rectangles RH i7i1 gtlt ya741 ij 17 7n This divides the rectangle R db gtlt cd into n2 subrectangles PW The sides of RH are Adl n1 7 xi and ij yHl 7 pi and its area is AA A iij Choose a point in each RH and build a parallelpiped over RH whose height is equal to the value of f at The volume of the parallelpiped7 fzjyAAlj approximates the volume of the three dimensional region under the surface 2 fdy and above PW The sum of the volumes of all n62 parallepipeds is called the double Riemann sum and is M n le DAAij 1 n ilj approximates the volume ofthe region under the surface 2 x7 y over the rectangle db gtlt cd MATH 6102 090 Spring 2007 124 DOUBLE INTEGRALS 153 A function og de ned on a rectangle R Q R2 is called integrable on R if the limit of the sequence of Riemann sums n n Fin Z Z f95l7y AAij i1 j1 as n a 00 exists and does not depend on the way that points are chosen in each subrectangle PW If f is integrable then the double integral ffRfdA off over the rectangle R is given by le DAAij 1 fdA lim nlimngtooz R Hm i1 7 Theorem 124 Properties of Double Integrals Letf andg be integrable func tions de ned on a rectangle R and let c be a constant Then 1 The function f g is integrable and RfgdARfdARgdA39 2 The function cf is integrable and RcfdAcRfdA 3 IfR is divided into n rectangles 1 i 1n that are mutually disjoint then f is integrable over each R and RfdAlRifdA 4 S gzy on R then MadeAWA 5 The absolute value of the double integral satis es amtrm Theorem 125 Let hy be afunction of two variables let R ab gtlt cd and let fg RaR be so that Way f969y RhzydAlb cw f wdy MATH 6102 090 Spring 2007 Th en 154 CHAPTER 12 INTEGRALS Theorem 126 Fubinils Theorem ff ag is a continuous function de ned on a rectangle R a7b gtlt ed in R2 then RfdAabldf9cydydzldlbf7yddy39 1 25 Triple Integrals We de ne the integral of a function f R3 a R similarly lf W ahbl gtlt a27b2 gtlt a37 b3 then we can form the triple Riemann sum n Z Z 2 1 9 2A iijA2k i j k The triple integral de off over W is de ned by de lim n W Hm Theorem 127 Fubinils Theorem ff fyz is a continuous function de ned on a rectangular boa W ahbl gtlt a27b2 gtlt a37b3 in R then m i ltjlt we cw dz i l lt w lt fzyzdy ax dz etc There are sicc interated integrals that are all equal when this limit exists 126 Change of Variables The change of variables is much like our Chain Rule for functions of one variable This section is mostly de nitions7 just so we can see how things work Assume that f R a R is a function of n variables We will consider it 2 and n 3 for the most part7 but everything is the same for arbitrary n A change of variables is a function T R a R We think of it as setting xi iu1 7un7 i 17 7n7 where m is a differentiable function of the u For n 2 this looks like Tui xuoyui MATH 6102 090 Spring 2007 126 CHANGE OF VARIABLES 155 Standard examples are polar coordinates for n 2 and cylindrical and spherical coordinates for n 3 Let Tui zuvyuv be a 01 function The determinant of the derivative DT of T is called the Jaeobian of T and is denoted by Jxy ui or ew3mm Hence any i hdu dzdv 6ui T dydu 61162 Thus for T R3 a R3 6 7y72 dsdu dzdv dsdw dydu dydv ail3w Jz7y72u7v7w 5u7v7w azau 6261 bzbw and for T R a R Ml3m Ml3W Mldun 6z17 7g gig6111 a zauz J1xnu1unm 6U I 3 7 7 n 7 Mn3m Mn3W Mndun Example 123 For polar coordinates x rcost9 and y rsin0 we have My 7 6r 9 cost irsin sint9 rcost9 For cylindrical coordinates x rcos 0 y rsin0 and z 2 so 2 cos0 7r sin0 0 sin0 rcost9 0 r 6rt9 z 0 0 1 For spherical coordinates x pcos0sin b y psin0sin b and z pcos b and 6W 2 cost9sin 7p sin6sin pcost9cos sin6sin pcos6sin psin6cos pzsina p7 Mb cos 0 ipsinab Theorem 128 Change of Variables Formula n 2 LetD and D be elemen tary regions in R2 and let T D a D be a Cl 0ne t0 0ne map such that TD D For any integrable function f D a R D mam D fltzltuwgtyltuwgtgt mg is the Jaeobian ofT dA My 6ui MATH 6102 090 Spring 2007 where 156 CHAPTER 12 INTEGRALS Thus7 for a change to polar coordinates D WWW Df7 6rdrd6 We can see what the Change of Variables is for general 71 Theorem 129 General Change of Variables Formula Let R and P be el ementary regions in R and let T D a D be a Cl me to one map such that TD D For my integrable function f D a R quot39Df1xndw 6z1 z 7 7 dW Adm l W llama M 3 n where M is the Jaeobz39an of T 6u177un This intimates that for a change to cylindrical coordinates7 fzyzdzdydz frbzrdrdbdz and for a change to spherical coordinates7 fzyzdzdydz fpb p2sm dpdbd MATH 6102 090 Spring 2007 MATH 6102 Spnng 2009 The Deriva ve 3152005 Hlstory Tang his 7 mmm Gmkgmme39ms mg mm 3m BCEMLhma 1 L mm m andApn nunls anm gaL 252mm m 39USEnfm m 1m mmmm Hummus anymva rhyahiammr 5w mammama s m smythemnmunme man u enfm mtrsmakmmmmmmanfdmga 7 dewhpa lhy h sllam 141135 Biashalarmssaldmmanylczymmmnf mamaam m mmm wm ll am Rnlh smenmn Hlstory Mdmd39 xi and rim45w aim mum pm mum 2mm wpaammmm madmaimrhsmmhamm Mmymbnym unmmm mm m w mmWWMm M W m amimmw m m m m azdnzdmhndydngmwf Mum nmmgmnm Mama mmu m mmmm suingth 3162009 Tangent Line to 1 Curve The question of nding the tangent line to a graph or the tangent line problem was one ofthe central questions leading to the development of calculus in the 17th century In the second book of his Geometry Ren q r u m r the tangent to a curve And 39 39 tsefuz and must genernlprublem in geometry that Iknow but even that Ihnve ever desired to knowquot Tangent Line n nnthwnnawnitrmcmnr thmghmd Definition The Difference Quotient fxf01 ea x orby1ettingx u h f11hf11 h These measure the slope ofthe secantline De mtlon 3152005 semnthlwasxgetsdnsetnam M ftxJiimJ H Wm mga Thesemem wslnpenf wmugmthlw De mtlon S mmemcmt mw 0mm whyam M 41 dmmqmwauwnspmhu nniuslmt Example an L 11 Lm7 1 gm 7 MHJru z bani I h v Wifwh 1 h m MW1 h Example Medmrl 2 MAIN 5 X5 3g 5 are w x7 quot39 3152005 Example Methan 3 Maw 1 31th MlY739 15 15 15 z Example mmmd was haw WW wanmluawhemw 3152005 hwyu f0 W Example Mmdma f0 miu M7ka m H7 mw 39 h m 7 Baslc lens wymwm um masumasqux vsw 39lslxasuuqas u n 4x A q vy4XJ7uquot4UJ4Wquotm quotV uomugeq QJEUJQJI39V39 1W warm 39ssma mama W q WWm yxwazqmlmgtp sljwlwiwsannap m npx XV uouxugaq oxseg mm x z x 0st ms 1 z z z Kmgtgtxjsmxns gtXgtXsmx1xs EGGZQIE XP 439 4 XP 57546 X XJXJ XJtxxsm elnalonpmd XJ gtxwmmAD Wm gt1 xJAmW WWW x6 1 Semedogd quot txyteraX W Mnullsnuum Hg MW mmwg nunsmgpamap w JOJEJedO EGGZQIE m Xgtes 1 gm W Xxsm Xsmgxmsgxms xm P P mo 4394 439 m 4 P uouqudv F 1 E X xlxXJ6 1 am JUQIJOHO wsz Mpg thmkt M sagmwsmpxmuaduw 39xmuaduwsqu Mpg mm l fj l w vmwmm sagmsmxmmduw pus uoumoN EGGZQIE 3152005 Chem Rule 4 74 d Wagmjgmwn QXXJ XJ exmnrlam lmahgwd a mum me dam m rule Inverse Functlons km um f1 mm and mm and hasan mm mmmlv 9 mm 15 the yaph nf 91 mmpmd n a yaphnmxy Mame whammnmsw m mudaunssthelmeyd7 Inverse Functlons XXJJ XJEXJ 1 ti Hews 1 2 31 9w Smffalmhngbaan l b if 9 J m p manor had a dermme nf w msnw mmmu xmsmsx 1 4Wmsx anmsx 1 dx Smaxltmsx 3152005 Apphcauo This 15mm Mysausfamnry s msmmmm Ar J3 gmmtxJH 5mm dx 39 mmstx 1 Exponenual Functlons Thlsxsbasedn the 39lnwmglnmt X 11mm a warm Mm a H h MAX 1 r xx r L zquot m w we 1341 mi Exponenual Functlons 11mm muse rmmmnm we have Lw a Smallth mammals Alejuslmulhpbsnf mammallngaldexpnmma 11 1 aiding 1m x 1 gym 2 Ehgllcm 3152005 Fermat s Theorem Lelfbede nei an an open mzmzmmmg c 1 m1 ffm ffmmbleau mm mm Rolle s Theorem anabmdmhy jf1af Lafbemnnnuauson m andbedgymmze b Mammy aba anecsabwhaI39 r Mean Value Theorem Let fbe a mmmm mm an W m 3 Manama an 126 M have u a pain 256113 uh mm cfbJa bra 3152005 Comllaryl Lelfbea d mmbk mm on 91 m m foam for 1116mm 71m f u 11 mm nmmww lma 1H 2 z afc cbJa Com ary2 Lelfami g be a waywabzefmm on 91 m m f g 39 umzzxsmm my mg m mm L39thaI x gxo an my Corollary 2 Lelfami g be a waywabzefmm on 91 m m f g 39 umzzxsmm my mg m mm L39thaI x gxo an my 3152005 L Hosplml s Rule LelM yn lz z z39 nwrnwhmauaml d mmble mmon Wm Ihefollmwng x bmzms L U PM aw mftxJgrggtxj a 0H m L Hosplml s Rule EWW1 P EquotS xh hmx km 2 1 Chapter 11 Implicit Function Theorem Consider the function f R2 a R given by ay x2 y2 7 1 Choose a point zoy0 so that f0y0 0 but no 31 1 71 In this case there is an open interval A in R containing x0 and an open interval B in R containing yo with the property that if z E A then there is a unique y E B satisfying ay 0 We can then de ne a function g A a R by g is that unique element of B for which fzgx 0 In other words there is a function g A a B so that y gzior we have solved for y in terms of the variable x We say that the equation ay 0 has implicitly de ned y as a function of x In this case we are able to explicitly solve for y for if y gt 0 then y g v1 7 x2 Note that this function is differentiable On the other hand if we were to have chosen x0 1 then we would not be able to nd such a function 9 de ned on an open interval containing 1 for some value of z would of necessity be sent to two different values of y In the calculus of one variable you learn the importance of the inversion process We de ne the logarithm y lnx as the inverse to the exponential function The inverse trigonometric functions are de ned as the inverse process to the trigonometric functions The inversion process is also important for functions of several variables For example the process of switching between Cartesian and polar coordinates in the plane involves inverting two functions of two variables Recall from one variable calculus that if y f is a continuously differentiable function and f x0 31 0 then locally near x0 we can solve for z in terms of y that is nd the inverse function x f 1y We learn that f 1 y 1f We need to be able to deal with similar situations in the case of multiple variables Theorem 111 Special Implicit Function Theorem Suppose that F R a R has continuous partial derivatives Denoting points in R 1 by xz where x E R and z E R assume that X020 satis es 9F Fx020 0 and X020 31 0 Then there is an open ball U C R containing x0 and an interval V C R containing 20 such that there is a unique function 2 gx de ned for x E U and z E V that 145 146 CHAPTER 11 lMPLlClT FUNCTION THEOREM satis es Fxgbx 0 Moreover ifx E U and z E V satisfy FXZ 0 then 2 gx Finally 2 gx is continuously di erentiable with the derivative given by 1 Dgx WDmFXZ 7 x2 92 291 where DmF dFdzl dFdxn in other words 3g fidFdz 71 oz BF62 2 You will nd a proof of this an a standard Advanced Calculus book We will not prove it here The general theorem gives us a system of equations in several variables that we must solve What are the criteria for deciding when we can solve for some of the variables in terms of the others or when such an implicit function can be found Assume that we are given a system of n equations f1z1mzmy17mym 0 111 f2z1mzmy17mym 0 112 113 fnd1dny1ym0 114 115 in nm variables x1quot xny1 ym We want suf cient conditions so that we can solve for 1 dn in terms of y1 ym Theorem 112 Implicit Function Theorem Iffl fn are di erentiable func tions on a neighborhood of the point x0y0 d9 x2y 49 in R gtlt R if f1z0y0 f2z0y0 fnx0y0 0 and if the n gtlt n matrico 3x1 3x2 39 quot 3 3x1 3x2 39 H 3 afn afn afn 3x1 3x2 39 H 3 is nonsingular at x0y0 then there is a neighborhood U of the point yo 11 49 in R there is a neighborhood V of the point 0 zd2 in R and there is a unique mapping p U a V such that ltpy0 0 and f1ltpyy fnltpyy 0 for all y in U Furthermore p is di erentiable MATH 6102 090 Spring 2007 147 This says that if we write y My 79421 where g17 gn are differentiable functions on U7 then 191yi77ym 116 zn gny17 7ym is the unique solution to the above system of equations near go If we restrict to a special case7 namely n 3 and m 17 the lmplicit Function Theorem gives us the following corollary Corollary 111 Let f R3 a R be a given function having continuous partial deriva tives Suppose that p07y0720 is a point satisfying 3 fWo o o 0 and 6 o7yo720 7 0 Then there is an open dish U C R2 containing pogo and an open interval V in R containing 20 such that there is a unique function 2 gpy de ned for my 6 U and z E V that satis es f7y799 7y 0 Moreover z gpy is di erentiable with the derivative given by 69761 6f a a a 119 59151 if 6y ay 62 1110 We shall nd this useful when it comes to parametrizing surfaces and doing patch work A special case of the lmplicit Function Theorem is the lnverse Function Theorem In this case we are trying to solve the n equations f1x1zn y1 1111 f2177n 112 fnlt17wngt yn for 1 an as functions of y1 7y that is7 we are trying to invert the equations in this system The question of when we can solve these equations is answered by the MATH 6102 090 Spring 2007 148 CHAPTER 11 lMPLlClT FUNCTION THEOREM general implicit function theorem applied to the functions yi 7 fix1 7x The condition for solvability in a neighborhood of a point x0 is that the determinant of the matrix Dfx0 is nonzero This determinant is called the Jacobian determinant of f and is 3f1 5f f a 3 17quot7 n J ammo f Wu 311 31 Theorem 113 Inverse Function Theorem LetU C R be open andlet fi U a R have continuous partial derivatives Consider the equations 1111 near a solution x07y0 If the Jacobian determinant Jf x0 31 0 then 1111 can be solved uniquely as x gy for x near x0 and y near yo Moreover the function g has continuous partial derivatives MATH 6102 090 Spring 2007 MATH 6102 090 Hyperbolic Trigonometric Functions Hyperbolic Trigonometric Functions Let us quickly recall the analytic definitions of the trigonometric functions of a real variable that is the definition that did not use the right triangles You have seen the definition of the six trigonometric functions in terms of a right triangle sinA opposrte adjacent opposite cosA tanA hypotenuse hypotenuse adjacent We can also define these functions in terms of the unit circle the circle of radius one centered at the origin This does not necessarily help us calculate the sine and cosine of an angle because it takes us no further than the right triangle definitions indeed it relies on right triangles for most angles The unit circle definition does however allow us to de ne the trigonometric functions for all positive and negative real numbers not just for values between 0 and rr 2 radians From the Pythagorean theorem the equation for the unit circle is x2 y2 1 In the figure below some common angles measured in radians are given Measurements in the counter clockwise direction are positive angles and measurements in the clockwise direction are negative angles Let a line through the origin making an angle of 6 with the positive half of the x aXis intersect the unit circle Then the point of intersection of this line with the unit circle is cos t9sint9 In the triangle below the radius is equal to the hypotenuse and has length Spring 2009 Page l 1 MATH 6102 090 Hyperbolic Trigonometric Functions 1 so we have sint9i and cost9 The unit circle can be thought of as a way of 1 1 looking at an infinite number of triangles by varying the lengths of their legs but keeping the lengths of their hypotenuses equal to 1 cos9 sin9 Now I want us to think of this just a little differently Instead of the angle between the x aXis and the segment what is the area of that sector of the circle This means that we have to remember the area of a sector of a circle If an angle 6 measured in radians sweeps out a sector in the circle of radius r the area of the sector is Aiar2 2 So in the unit circle since r 1 the area is A 62 or 19 2A Thus the coordinates of the point on the unit circle swept out by an area of A radians are cos2Asin2A Now instead of a circle of radius 1 think about a different conic cross section the hyperbola We are interested in the unit hyperbola the hyperbola given by the equation x2 y2 1 Note that there is very little different in the equation of the hyperbola and the circle just subtraction instead of addition Can we do a similar construction for the hyperbola that we did for the circle First we need to recall that the asymptotes for the hyperbola are the lines y x and y x If a line passes through the origin with slope greater than 1 or less than 1 that line will not intersect the hyperbola In the figure below consider a ray emanating at the origin and having slope m with 1 lt m lt 1 This ray will intersect the hyperbola at a point P Consider the area of the region bounded by the ray the hyperbola and the Spring 2009 Page l 2 MATH 6102 090 Hyperbolic Trigonometric Functions positive x axis and call this area A Then the coordinates of the point P are cosh2Asinh2A Again this does not help us compute the values of the hyperbolic sine and cosine To do this we appeal to some mathematics that we might cover later Mathematicians have found that there is a relationship between the exponential function and the trigonometric functions This then leads to considering the following two functions 2quot equot sx CX 2 What do the graphs of these functions look like We could plot them on the calculator but because of their pattern we want to try some algebra first Note that 2 ex erx 2 82x 2er2x s X 2 J 4 and 2 equote 2 e2x2e 2quot c x 2 J 4 so that CX2sx2e 2e e 2e 2 4 4 That is we get that these functions satisfy the equation u2 v2 1 In fact each point on the hyperbola has coordinates cxsx for some value of x 2 1 4 Not surprisingly these functions are the hyperbolic cosine cx and the hyperbolic sine sx Thus we often define these functions by sinhx e 37 COSMX i We also have seen the basic hyperbolic trigonometric identity cosh2x sinh2x 1 Once we have the hyperbolic sine and hyperbolic cosine de ned then we de ne the other four functions as Spring 2009 Page l 3 MATH 6102 090 Hyperbolic Trigonometric Functions tanhx i coshx eX e X coshx eX e X cothx 7 x 0 s1nhx eX e X sechx coshx eX e X cschx 1 2 7X x i 0 sinhx 2 2X 2 Based on these definitions and the basic hyperbolic trigonometric identity we find a large number of hyperbolic trigonometric identities that are analogous to the usual trigonometric identities 1tan2xsec2x 1 tanh2xsech2x M tanhx y 1 tanxy 1 x 1 cosx x s1n2 s1nh2 x 1cosx x cos2 cosh2 x 1 cosx x tan2 tanh2 tani tanhi X1 COSX 1 tan tanhi x x x x s1nx 2s1n cos s1nhx 2s1nh cosh sinxi siny y y cosxcosy y y cosx cosy x y y What other properties do they seem to have in common We should consider at least three more things inverse functions and derivatives and graphs Spring 2009 Page l 4 MATH 6102 090 Hyperbolic Trigonometric Functions Inverse Hyperbolic Trigonometric Functions Since the hyperbolic trigonometric functions are defined in terms of exponentials we might expect that the inverse hyperbolic functions might involve logarithms Let us first consider the inverse function to the hyperbolic sine arcsinhx By the de nition of an inverse function y arcsinhx means that x sinhy Thus ey eiy 2 ey e39y 2x ey e yey 2xey X e2y 2xey 1 0 Let u ey then this equation becomes u2 2xu 1 O u 2Xi X24 xixlx21 eyxxlx21 y lnxx2 1 and it must be the positive square root because ey gt O arcsinhx lnx 4x2 1 We should not find this too surprising We would expect the others to be similar Doing similar work we find that arccoshx lnx lx2 1 arctanhx ln lln1 x lln1 x 1 x 2 2 arccothx Int llnx 1 llnx 1 x 1 2 2 1 arcsechx ln arccschx ln 1 1 x2 x Wow No you don t have to memorize all of this Derivatives of Hyperbolic Trigonometric Functions Just like everything else we would expect the derivatives of the hyperbolic trigonometric functions to take an analogous route to those of the regular trigonometric functions Spring 2009 Page l 5 MATH 6102 090 Hyperbolic Trigonometric Functions We can easily find the derivatives since they are defined in terms of the exponential function sinhx Xe 7 j XeX e equot e exe xcoshx d coshx 41 1 11w e4 lex e lex e sinhx dx 2 2 dx 2 2 Thus we have found that these functions have a nice derivative periodicity and we do not have to take negative signs into account figsinhbr coshx coshbr sinhx From these we can compute the other derivatives expecting analogous results dtanhx d s1nhx coshxcoshx sinhxsinhx sech2x dx dx coshx coshx2 coshx2 dc0thx d coshx s1nhxs1nhgr coshxcoshx 1 csch2x dx dx s1nhx s1nhx2 s1nhx2 d sechx i 1 Slnhx sechx tanhx dx dx coshx coshx2 d cschx d 1 cschx cothx dx dx s1nhx s1nhx2 These are pretty close to what we expected Graphs of Hyperbolic Trigonometric Functions We wait until now to look at the graphs so that we can use the derivative to help us First let us look at coshx equot equot coshx Tgt 0 since the numerator is always positive Since asinhbr coshx and coshx gt 0 2 i for all x we see that the hyperbolic sine function is always increasing We also note that it has no critical points since its derivative is always defined and is never 0 Now looking at the graph it is not too surprising to find that it looks like the figure to the right Figure 1 y sinhx Spring 2009 Page l 6 MATH 6102 090 Figure 2 y coshx Hyperbolic Trigonometric Functions Now that we know what the hyperbolic sine looks like we can analyze the hyperbolic cosine Since its derivative is 0 at x 0 we know that it has a critical point there Since the second derivative is always positive this critical point must be a local minimum It is not hard to show that it is a global minimum The graph looks like the gure on the left This may look somewhat like a parabola to you but it actually grows faster than any parabola The interesting thing is that it does describe a physical setting If you put two i pegs at the same height some distance apart and let a rope hang between the two pegs so that the ends just hang over the pegs not tied to them then this rope takes on a shape called a catenary The catenary is a hyperbolic cosine shape Examples are electrical wires hanging between power poles y tanhx y cothx y sechx y cschx Spring 2009 Page l 7 MATH 6102 090 Hyperbolic Trigonometric Functions Figure 3 A Web of Hyperboic Trigonometric Functions Spring 2009 Page i 8 MATH 6102 Spring 2009 Parametric Equations Implicit Function Theorem Related Rates 0 ar2009 MATIV Rep Curves in R2 6102 Parametric Equations 39 cost x cost sint yz sint 1 2 3 cost E L y cost 5 6 7 MATH 6102 4 x sint 8 Tangent Line to a Curve Q dx dt dt 1 gym dx2dx dx dt dt

### 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 used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "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."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### 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.