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# Elementary Functions MTH 112

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This 23 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 112 at Portland Community College taught by Staff in Fall. Since its upload, it has received 33 views. For similar materials see /class/224645/mth-112-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15

Haberman MTH 112 Section IV Parametric and Implicit Equations Module 3 More Implicit Equations In this module we will study the parametric and implicit forms of ellipses and hyperbolas ELLIPSES As we observed in the previous module if we start with the system of parametric equations x cost y sint that defines a circle of radius 1 centered at the origin and multiply both the x and y coordinate by a factor of r and add constants h and k to the x and ycoordinates respectively we obtain the system x h rcost y k rsint that defines a circle of radius r centered at the point h k What would happen if we didn t multiply the x and ycoordinate by the same factor r but instead multiply the x coordinate by the factor a and the ycoordinate by the factor b where a at b If a gt b then the x coordinates will be stretched more than the ycoordinates so we should expect a warped circle or an oval that is longer the horizontally than vertically Similarly if b gt a we should expect an oval that is longer vertically than horizontally The mathematical term for an oval is ellipse If h k a b e R then the system of parametric equations below defines an ellipse centered at the point h k x h acost y k bsint We usually take a b gt 0 The horizontal axis of the ellipse is 2a units and the vertical axis is 2b units 4 EXAMPLE 1 Sketch a graph of the ellipse defined by the system of parametric equations x 2 20050 y 3 65in0 SOLUTION Based on what we observed above we see that the center is 2 3 and the horizontal axis is 22 4 units and the vertical axis is 26 12 units see Figure 1 Figure by x 2 20050 y 3 65in0 You should sketch this system on your graphing calculator and make sure you obtain the same graph EXAMPLE 2 Eliminate the parameter r from the system of parametric equations x 2 20050 y 3 65in0 to obtain an implicit equation that describes this ellipse SOLUTION We can use the Pythagorean Identity to eliminate the parameter The Pythagorean ldentity involves 5in0 and 0050 so we need to first we need to isolate 5in0 and 0050 in the equations in our system x 2 20050 y 3 6sint 2 2cost x 2 and 2 6sint y 3 3 cost 2 3 sint y g3 3 Now we can substitute the expressions x 2 and y6 2 for cost and sint in the Pythagorean Identity and obtain an implicit equation for the ellipse coszt sin2t 1 x22 y732 2 T ltTgt 1 2 2 3 x 2 y 7 3 1 22 62 2 2 x 2 y i 3 3 T 36 1 2622 3732 Thus the ImplIcIt equation 4 36 1 represents the same ellipse defIned by the given system of parametric equations EXAMPLE 3 The system of parametric equations x h acost y k bsint defines an ellipse centered at the point h k with horizontal axis 2a units and the vertical axis 2b units Find an implicit equation that describes the same ellipse SOLUTION We can use the Pythagorean Identity to eliminate the parameter just as we did in Example 2 First we need to isolate sint and cost in the equations in our system x h acost y k bsint 3 acost x h and 3 bsint y k 3 cost xh 3 Sin yik b Now we can substitute the expressions x h and ygk for 0050 and sint in the Pythagorean Identity and obtain an implicit equation for the ellipse coszt sin2t 1 2 2 gt l 1 a b 2 2 H H 1 a2 b2 If h k a b e R then the implicit equation Hr we a2 b2 1 defines an ellipse centered at the point h k with horizontal axis 2a units and the vertical axis 2b units HYPERBOLAS EXAMPLE 4 Sketch the graph of the system of parametric equations x 2tant y 3sect and eliminate the parameter r to obtain an implicit equation that describes the same curve SOLUTION We can use our graphing calculator or any other graphing utility to graph the system see Figure 2 below 3seez Graphs We the ewe m thure 2 are named hyperbolas Hyperbetas have magenat asymatotes that ean be fuund usm the 2 and 3 m the ME fur the x7 and ya cuurdmates respemve y te draw a rectangte vth henzentat Ength 2 2 4 and vemeat Ength 2 3 6 and the asymptutes furthe hypemeta are the magma at the rectangte See gure 3 betew SmEE the parametenzattun x 2tanz y 3seez tnvutves tangent and secant We can use the tdenttty 53521 e m2t 1 tn Ehmmate the parameter 1 a Identity in Section I Module 6 In order to utilize this identity we need to solve the equations involved in our parameterization for tant and sect respectively x 2tant y 3sect 3 WW and 3 sect 3 Now we can substitute 7 and for tant and sect in the identity seczt tan2t 1 seczt tan2t 1 2 2 all 2 2 Thus the implicit equation y JCT 1 describes the same hyperbola as the system of parametric equations x 2tant y 3sect EXAMPLE 5 Sketch the graph of the system of parametric equations x 2 4sect y 3 3tant and eliminate the parameter r to obtain an implicit equation that represents the same curve SOLUTION We should expect the center of the hyperbola to be shifted to the right 2 units and down 3 units 2 and 73quot in the rules for the x and ycoordinates respectively Also we should expect the asymptotes for the hyperbola to the diagonals of a rectangle with horizontal length 24 8 and vertical length 2 3 6 Below is the graph of this system Obvmus y ms hyperbu a upens Inward the hunzunta mremen WW2 the hyperbu a we smmeu m EXamp E 4 ripened m the venma mremen The reasun that these hyperman upen m Merem mreeuens s that m Examp e 4 tangent s mvu ved m the ME furthe x7 cuurdmate and seeam s mvu ved m the ME fur the yrcuurdmate but m m EXamp E we we yrcuurdmate mphmt Equatmns fur the hyperman that upen hunzuntaHy uwer mm these that upen vemcaHy As we mu m Examp e 4 we can use the newly 53521 4mm 1 m USL NEW 1 mvu ved m uurparametenzatmnfur Lana and 5321 respem x 2 4sect y3 31mm and 2 5351 X Z Lana y Nuvv we can subs utute 12 and V fur 5351 and Lana m the newly 53521 7 1mm 53521 7 WM 1 e 1244411 M2 MK T T 1 2 2 Thus the mphmt Equater 139 rm1 desmbes the same hyperbma as the gwen sys LEm uf parameme Equatmns Name that myth the mphmt Equatmn we fuund m EXamp E 4 m mm the has he ew an expression mvowng y mmus an expression mvowng xx ms Equatmn has the farm an expression mvowng x mmus an expression mvowng y x n s m uwerenee that makes the Me hyperman m upen m uwerent dwrectmns If h k a b e R then the system of parametric equations x h atant y k bsect defines a hyperbola that is centered at the point h k and opens in the vertical direction the system x h asect y k btant defines a hyperbola that is centered at the point h k and opens in the horizontal direction The asymptote of the hyperbola can be found by drawing the diagonals of a rectangle centered at the point h k with horizontal length 2a units and vertical length 2b units If h k a b e R then the implicit equation 2 2 we H 1 b2 a2 defines a hyperbola that is centered at the point h k and opens in the verticaldirection while implicit equation 2 2 H we 1 a2 b2 defines a hyperbola that is centered at the point h k and opens in the horizontal direction The asymptotes of the hyperbola can be found by drawing the diagonals of a rectangle centered at the point h k with horizontal length 2a units and vertical length 2b units Math 112 Damped Oscillation Model each of the following situations with an exponentially decaying sinusoidal function of the form dt Ae sinjBtC or dt Ae cosBtC where dt is the distance from the equilibrium position with respect to time Note It s always easier to begin with a diagram of the system or a graph of the function 39I N SpringMass System A mass attached to a spring is suspended from the ceiling The mass is pulled down 10cm and released at time t O The time required to complete one cycle is 12 second You note that at 1 second the mass is 3cm below its equilibrium position Shock Absorbers a When the road ahead gets rough Smooth Move shocllts oscillate completing 4 cycles per minute Write a function that models this behavior Note You don t have any info on amplitude yet so just call it A 939 Just after going over a bump the amplitude of the oscillations is 2cm The amplitude decreases so that after 2 minutes the amplitude is l8cm Assuming the amplitude decreases exponentially and continuously write a function At that models this behavior c Combine the functions from parts a and b to write and graph function that fits the form given in the instructions Note There is some room for interpretation here so you should always be ready to justify whether you used a sine or cosine function Bungee Jumping Bungee Jumping Biff jumped from a platform that was 98ft above the ground Not realizing he was afraid of free fall Biff barfed on the way down causing all his friends to run away and leave him hanging Eugene however was fascinated and noted that once Biff stopped bouncing his equilibrium position was 64ft above the ground He also noted that each bounce toollt Biff 6 seconds to complete and after the first bounce at time t 6 he only reached a max height of 15ft above equilibrium HINT Set this up initially so that y O is the midline Then achieve the desired heights by treating the midline of 64ft as a vertical shift Haberman MTH 112 Section II Trigonometric Identities Module 3 DoubleAngle Identities In this module we will find identities that will allow us to calculate 00526 and sin2t9 if we know the values of 0056 and sint9 We call these doubleangle identitiesquot Let s start by finding the doubleangle identity for sine Recall that the definition of the cosine and sine functions tell us that 0056 and sint9 represent the horizontal and vertical coordinates respectively of the point specified by the angle 0 on the unit circle See Figure 1 below Figure 1 The unit circle with a point P specified by the angle 0 As we ve studied in the previously we can construct a right triangle using the terminal side of angle 0 This triangle has hypotenuse of length 1 unit and sides of length 0056 and sint9 See Figure 2 below Figure 2 Right triangle with angle 0 Armerugh we wu urmnue m fucus un angre a Et s abe the angre Whuse vertex rs pmnt P we eau r a SEE Frgure 3 beruw Figure 3 NEIUEE that sma mfg 2055 We H use thrsfam ater Nuvv Et s cunstrum the mwummage err mrs mangre beruw the xraxrs rn Quadrant m see Frgure 4 Figured r we re r measure 25 We ve emphasrzeg mrs rrrangre rn Frgure 5 by mgrng the umt mrde and the uurdmate ne er me thatthe srue uppusrre 29 rs Hength 2mg srnee rr eunsrsrs uftvvu segments Each u ength sm5 Figure5 We ean use th1str1ang etu ndthe duub eranghe 1dentmesfur cusme and sme Fm 1e1 s 25 0 smee sma cos frum abuveLwe can 2mg 1 The Law uf Smes EH us that subs utute 2055 fur mm 2 sm2 25m5cos 1e ubtam the vame elf sm2 1fvve knuvvthe va1ues 1 2055 and 51115 Nuvv 1e1 s me the duub eranghe Mammy fur cusme We ean use the same mange we cuns trumed abuve We ve named m1 mange bemw m gure B but apmy the Law of osines msiead ufthe Law 1 Smes 25m52 12 12 2 2 11 20525 2 45m2 11 2cos25 2 45m2 2 e 2cos25 2 2cos2 2 e Asmzw 2 50525 12 25m25 Figure 6 The last equation above is the doubleangle identity for cosine Notice that we can use this identity to obtain the value of cos219 if we know the value of sin19 We can use the Pythagorean identity to obtain two other forms of the doubleangle identity for cosine Recall that the Pythagorean identity tells us that 2 2 sm 19 cos 19 1 3 sin219 1 00526 and we can now substitute 1 00526 for sin219 in the doubleangle identity to obtain another form of the identity 005219 1 251n219 2 cos2191 21 00526 2 005219 1 2 2005209 2 005219 2005209 1 This last equation is another doubleangle identity for cosine We can obtain a third double angle identity for cosine by substituting sin219 00526 for 1 005219 Zeosz19 1 3 005219 Zeosz19 sin219 00526 3 005219 Zeosz19 sin219 00526 3 005219 00526 sin219 Below we ve summarized the doubleangle identities for sine and cosine DOUBLEANGLE IDENTITIES sine sin219 251n19cos19 cos2191 251n219 cosine 005219 2005209 1 005219 00526 sin219 We could easily find doubleangle identities for tangent by using the fact that tan2t9 But the resulting identities aren t easy to remember so it makes more sin29 sense to learn the IdentItIes for sIne and cosIne and use the fact that tan2t9 00809 If you ever need to calculate tan2t9 EXAMPLE Suppose that sina and that a is in Quadrant 11 Find cos2a sin2a and tan2a SOLUTION First let s find cosa since we need this value to use the doubleangle identity for sine To find cosa let s use the Pythagorean identity sin2a cosza 1 3 cosza 1 3 cosza 1 3 cosza1 3 cosza g 3 cosa Note that we take the negative square root of since a is in Quadrant 11 so that cosa must be negative Now we can use the doubleangle identities to find cos2a and sin2a Let s start with sin2a sin2a 25ina cosa 2 Li 9 To find cos2a we can use any one of the three doubleangle identities for cosine Let s use cos2a 1 25in2a cos2a 1 25in2a 1 22 L 129 A 19 1 9 You should verify that the other doubleangle identities for cosine give the same value for cos2a To find tan2a we can use the fact that tan2a sin20 tang cos20 M 9 Haberman MTH112 Section 1 Periodic Functions and Trigonometry Module 1 The Unit Circle Radians and ArcLength In this module we will study a few definitions and concepts that we ll use throughout the quarter DEFINITION A unit circle is a circle with a radius r of 1 unit See Figure 1 Figure 1 Now let s take note of some conventions and terminology that we will use when discussing angles within circles like angle 9 in Figure 2 below Figure2 I The angle 9 is measured counterclockwise from the positive xaxis I The segment between the origin 0 0 and the point P is the terminal side ofangle 0 I Two angles with the same terminal side are said to be coterminal angles I The point P on the circumference of the circle is said to be specified by the angle 9 I Angle 9 corresponds with a portion of the circumference of the circle called the arc spanned by 9 See Figure 3 below A are spanned by 639 A x Figure3 Thus far in your mathematics careers you have probably measured angles in degrees Three hundred and sixty degrees 360 represents a complete trip around a circle Le a full rotation so 1 corresponds to 1360th ofa full rotation As noted above angles are measured counterclockwise from the positive xaxis consequently negative angles are measured clockwise from the positive xaxis See Figure elow Figure4 We mermuned abuve that metermma angles share the same terrmna sme Smce 360 represents a mu ratater abuutthe m e fvve add any mteger mump e 6 360 m an ang e amen ubtam an ang e mtermma tu a m uthervvurds the makes a and 92 9 360 k Where keZ are mtermma Fur EXamp E the ang es 45 and 45 360 405 are curlerrmnak see gure 5 be uvv Figure5 The ang es 43 and 405 are cmtermma TradmunaHy me cuurdmate mane 5 deed mm four quadrant see gure B be uvv We Wm u en use the names ufthese quadrants m desmbe the ucatmn ufthe termma see 6 uweremangxes 3f Quadrant z Quadrant II I 3 Figures Fur EXamp E cunswder me makes gwen m ngure 4 the ang e 60 5 m Quadrant 1 vaE 450 5er QuadrantHI Instead of using degrees to measure angles we can use radians DEFINITION The radian measure of an angle is the ratio of the length of the arc on the circumference of the circle spanned by the angle and the radius of the circle See Figure 7 below Figure 7 The angle 9 measures radians Since a radian is a ratio of two lengths the lengthunits cancel Thus radians are a unitless measure NOTE An alternative yet equivalent definition is that an angle that measures 1 radian is defined to be an angle at the center of a unit circle measured counterclockwise which spans an arc of length 1 unit on the circumference of the circle Since on a unit circle the radian measure of an angle and the arclength spanned by an angle are the same value in order to find the radian measure of a complete rotation around a circle ie 360 we need to find the arclength of an entire unit circle Of course the arc length of an entire circle is the circumference of the circle recall that that circumference c of a circle is given by the formula c 2m where r is the radius of the circle Thus the circumference of the unit circle ie arclength of the complete unit circle is c 27r1 2n units Therefore the radian measure of a complete rotation about a circle ie 360 is equivalent to Zn radians We can state this symbolically as follows 360 27239 radians The equation above implies that the following two ratios equal 1 we can use these ratios to convert from degrees to radians and vise versa 27239 rad 360 360 27239 rad EXAMPLE a How many degrees are 8 radians b How many radians are 8 degrees SOLUTION 360 27 rad 39 a In order to convert 8 radians into degrees we can multiply 8 radians by Since this equals 1 it won t change the value of our anglemeasure 3600 83600 8 Jed anad 27139 II w 458370 Thus 8 radians is about 45837 b In order to convert 8 degrees into radians we can multiply 8 by 2 rad Since this 360 equals 1 it won t change the value of our anglemeasure 27rrad167r 8 360 360 rad 2 rad m 014 rad Thus 8 is about 014 radians EXAMPLE a Convert 1 radian into degrees b Convert 90 into radians SOLUTION a In orderto convert 1 radian into degrees we can multiply 1 radian by 3600 3600 1 mad 3 E 7139 m 5730 Thus 1 radian is about 573 27 rad 360 39 b In orderto convert 90 into radians we can multiply 90 by 90 27239 rad 1807r rad 360 360 L 2rad Thus 90 is equivalent to radians EXAMPLE Complete the table below 9 degrees 0 0 30 45 60 90 180 270 360 t9 radians SOLUTION 6 degrees 37r 2 27239 b w M i 6 radians 0 Recall the definition of radian the radian measure of an angle is the ratio of the length of the arc on the circumference of the circle spanned by the angle and the radius of the circle Applying this fact to the circle in Figure 8 below if 6 is measured in radians then 6 arclength i radius r Circle of radius r with an angle 6 spanning an arclength s Figure 8 By solving the equation 6 for s we obtain the following definition DEFINITION The arclength s spanned in a circle of radius r by an angle 6 radians is given by s r6 Note we need the absolute value of 6 so that we obtain a positive arclength if 6 is negative Lengths are always positive Also note that this formula only works if 6 is measured in radians EXAMPLE a What is the arclength spanned by an angle of 2 radians on a circle of radius 5 inches 339 What is the arclength spanned by an angle of 30 on a circle of radius 20 meters SOLUTION a To find the arclength we can use the formula s rt9 srt9 52 10 Thus the arclength spanned by an angle of 2 radians on a circle of radius 5 inches is 10 inches 57 Before we can use the formula s rl l we need to convert the angle into radians ln orderto convert 30 into radians we can multiply 30 by 265 which equals 1 Of course we could use the table we created earlier in this module but we will go ahead and show the computation here 3027239 rad 607239 rad 360 360 l 6 rad Thus 30 is equivalent to radians Now we can find the desired arclength s rt9 l 20 6 107r 3 Thus the arclength spanned by an angle of 30 on a circle of radius 20 meters is IOT meters

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