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## Calculus 1

by: Vernie McCullough

8

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20

# Calculus 1 MATH 161

Vernie McCullough

GPA 3.58

Bruce Ikenaga

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COURSE
PROF.
Bruce Ikenaga
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Class Notes
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20
WORDS
KARMA
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## Popular in Mathematics (M)

This 20 page Class Notes was uploaded by Vernie McCullough on Thursday October 15, 2015. The Class Notes belongs to MATH 161 at Millersville University of Pennsylvania taught by Bruce Ikenaga in Fall. Since its upload, it has received 8 views. For similar materials see /class/223529/math-161-millersville-university-of-pennsylvania in Mathematics (M) at Millersville University of Pennsylvania.

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Date Created: 10/15/15
Math 1610306 127372007 Review Problems for the Final These problems are intended to help you study for the nal However you shouldnlt assume that each problem on this handout corresponds to a problem on the nall Nor should you assume that if a topic doesn t appear here it won t appear on the nal 1 Compute the following derivatives d I d x lnx1 a E 011 b g 1ne 1gte ltgt d1 11 lt1 1 gt1 1 dc c dI n n nz dI d 12 2 d cos I 2 6E81nlt123gt DEV sinz4 g diz5 413 7 lnzsinz37 h did 13 51 17 z z z i 3 Where I sing 7 cosy cosgy 0 j 3 Where y em 3022 k y When I 1 and y 17Where I2y3 7 3 I y 7 2A 71 d 2 Compute sin 1 1 dz 3 Compute 1 tan 1 6 1 i dx i 4 Prove that e gt 1 1 for I f 0 53 3 83 5 Graph y 31 7 gr i 6 In the following problems compute the limit or show that it is unde ned sin 31 7 sin 51 a lim x70 sin 21 4x V12 1 b llm l 1700 31 7 I h50 7 150 C 1133 h 12 7 31 7 4 d 011314 I27 6 i e llIilJr 16 7 12 1273174 1 f 1 i 131412781H6 2 I QUE1 Tf 5 13 7 Compute the following integrals a 21 312 dzi 21 b 114 dzi 62x d c 621 If a lt31ngt cosz e sinz2 2sinz ld1 12z f di lf273127213 I seczl3 2 gI2 3dzi 1 z ln12 1 h W i 12 dz iff1 1 and f2 e j I1412215d mgm m l n m4 8 a A population of amingo lawn ornaments grows exponentially in Calvinls yardi There are 20 after 1 day and 60 after 4 days How many are there after 6 days b A hot pastrami sandwich with a temperature of 150 is placed in a 70 room to cool After 10 minutes7 the temperature of the sandwich is 900 When will the temperature be 80 9 a Find the area of the region in the rst quadrant bounded on the left by y 127 on the right by z y 2 and below by the zaxis b Find the area of the region between y 12 7 12 and y z from I 0 to z 4 10 Approximate the area under y z 7 sin z2 from I 3 to z 5 using 20 rectangles of equal width7 and using the midpoints of each subinterval to obtain the rectangles7 heights 2 11 Approximate the sum 3sin2 3sin3 3sin4 3sin4l 121 222 32 40240 to 5 decimal places 12 Calvin runs south toward Phoebels house at 2 feet per second Bonzo runs east away from Phoebe s house at 5 feet per second At what rate is the distance between Calvin and Bonzo changing when Calvin is 50 feet from the house and Bonzo is 120 feet from the house 13 Find the dimensions of the rectangle with the largest possible perimeter that can be inscribed in a semicircle of radius 1 14 A window is made in the shape of a rectangle with an isosceles right triangle on top 5V2 5V2 a Write down an expression for the total area of the windowi b Write down an expression for the perimeter of the window that is7 the length of the outside edge c eper1me er1sg1ven o e w a v ueo sma es e o a areaamaximumi lfth 39tquot tb47htal f kthttl 397 15 a Find the absolute max and the absolute min of y 13 7 121 5 on the interval 0 S I g 5 b Find the absolute max and the absolute min of 2143 7 15113 on the interval 71 S I S 8 16 Use a limit of a rectangle sum to nd the exact area under y 12 31 from I 0 to z ll 17 Given that I5 413 177 what is f 1 22 18 Find the largest interval containing 1 1 on which the function 314 7 1613 1 has an inverse 71 d is 19 a Compute 2 t2 dt 1 4 b Compute 3 12 1 dzi z 20 a Use the de nition of the derivative as a limit to prove that 7i dz 1 7 4 I 7 42 1 xh b Compute flbini vt4 1dti 21 Let 3 3 ifz lt 1 7 7 I 0 0 9 ifz1 312722 ifz gt 1 Is f continuous at z 1 Why or Why not 2 z 7 7 22 a Suppose that 7 5 and f 7 12 16 Use dlfferentlals to approx1mate f2i99 to 5 places b A differentiable function satis es j y 6 cos 31 and y0 Oili Use differentials to approximate y0i02i z 23 Use 3 iterations of Newton7s method starting at z 2 to approximate a solution to 4 7 12 ea 24 Prove that the function 13 21 7 cosz 5 has exactly one rooti Solutions to the Review Problems for the Final 1 Compute the following derivatives 1 a mlt2z1gt2lt1gtgt2lt2z1gtlt2x D b ltlnltex1gtemlgt76161lt gte mtll U C glnu1nalnmltmgt 1 g 3 d gmannasmxcosz u f d cosz2il cosz2 712 sinz47sinz7cosz2cosz D dz sinz4 2 sinz4 sinz42 g diz5 413 7 lnzsinz 37 15 413 7 lnzcosz sinz 37 lt514 1212 7 i El 1 I h 13 5117z 135117z712 7zg511531251 i I d i d y Where I siny 7 cosy cos 2y 0 z Differentiate implicitly7 then solve for y sin y sinyzycosyysiny72ysin2y 07 y j where y 7 ex 2 Use logarithmic differentiation lny lne 223 I2 3 1116 27 I2 3 e 2 7 y 2zlnex 2 9 61 1 2 y 6 2ch3 lt21 lne 2 L i El k y When I 1 and y 17Where 12y3 7 13 z y 7 2i Differentiate implicitly 312y2y 21y3 7 312 1 yi Pluginzl andyl 3y 2731y 7 2y 27 1 NOW differentiate implicitly 312y2y 612yy 2 6zy2y 6zy2y 2y3 7 6r y 5 Note that the term 312y2y produces three terms when the Product Rule is applied Now set I 1 y1 and y 1 3y 666276 y 2y 140 y 77i El d 2 Compute sin 1 I 1 dz isinilzzl dz 2 d 3 Compute E tan 1e 1i d E tan 1e 1 Er 1 e 12 4i Prove that e gt 1 z for z y 0 mwbmmw 1 05 05 1 15 2 The result is true as the picture shows However a picture is not a proof Let 617 11 f measures the distance between the two curves The derivative is f 61711 f 0 at z 0 since f z ex and f 0 1 gt 0 the critical point is a local mini Since it is the only critical point it is an absolute mini Thus the minimum vertical distance between the curves occurs at z 0 when it is 0 Since this is an absolute min it follows that gt 0 for z y 0 7 that is e 7 1 1 gt 0 for z y 0 This is equivalent to what I wanted to showi El 3 5 Graph y 3153 7 gzgSi The function is de ned for all I you can take the cube root of any number Since y 3153 1 7 gr the zintercepts are I 0 and z 8 The y intercept is y 0 The derivatives are 10 5 5 271 23 53 23 713 23 y7517z7z57z y731 7 z7 131 In working with these kinds of expressions it is better to have a sum of terms when you7re differentiating On the other hand it is better to have everything together in factored form when you set up the sign chartsi Notice how 1 used these two forms in the derivatives above y 0 for z 0 and at z 5 y is de ned for all xi was x70 r1117 x5 181712 The function increases for I S 5 and decreases for z 2 5 There is a local max at z 57 y m 1644760 y 0 at z 2 y is unde ned at z 0i 171175 H f l153 H 18175 W W m The graph is concave down for z lt 0 and for z gt 2 The graph is concave up for 0 lt z lt 2 z 0 and z 2 are inflection pointsi Since the function is de ned for all I there are no vertical asymptotesi lim 3153 7 E18 700 and lim 3153 7 EzgS 700 1700 8 7700 8 The graph goes downward on the far left and far right 15 6 In the following problems7 compute the limit7 or show that it is unde ned sin 31 7 sin51 1 a 0213 sinQI sin 31 sin 51 3 sin 31 5 sin 51 lim Slng i smsx lim I I lim 3x 5x 3i 71 070 s1n21 70 s1n21 170 2 s1n21 2 z 21 b lim 4I 21 1700 31 7 i 4 1 1 1 41x1217 1 7 75 El 113 7 mam 33 5 1 I h50 7 150 C 50 50 1 h 7 1 11 for 1501 Hence7 by the Power Rule7 507 50 lim m 501491 I h7gt0 h 12 7 31 7 4 d 1131 I2 16 1273174 11174 11 5 l39 l39 39 1 31 12716 1317414 x x4 8 D e lim 16 7 121 4 17 For 1 Close to 4 but larger than 4 7 eg 1 4101 7 16 7 12 is negative Since the square root of a negative number is unde ned7 the limit is unde ned El 12 7 31 7 4 911314 I2 7 8116 1 127317471 1741171 11 11311278116 1131 35742 1511174 5 Plugging in gives Moreover7 1 1 11m 00 and lim 1 7001 144 1 7 4 144 1 7 Hence7 the limit is unde ned El 12 7 4 g 11312 1 1 5 1 3 1274 1274 1 74 1274 17212 01121217 1 ung 13 7 5 ag 35 ung 12 ag 12 5 13 513 513 513 513 513 8 5 3 lin12z 7 2z 2 1 12 111112595 3z 2 100 n EH 1 7 ma 7 Compute the following integrals a 2E 6352 dzl 1 2 1 6216312dz6412651661gtdz164x36516661c D 21 b 114dzl 21 7 2u713 7 2u1 7 2 1 7 z14dI7u 4dui u4 dui u3u4 dui 11z17 dudz zu71 1 1 u2 3113 1 1 C W WC e22 Cmdzl e27 627 du 1 du 1 1 d 1 C 1 2 1 Cl 6211I u 262 2 u 2nlulJr 2nle l ll 7211d7221d did I uie 7 ui e z 1726 2 CD3mm 1dI z 2 2 Wdz zdu3u21duu3u0lnz3lnzCl El 1 z u lnz7 du d I dz zdu z cosz d l esinz225inz1 I cosz d 7 cosz du 7 du 7 sinz22sinz1 If u22u1 coszi u22u17 11sinz7 ducoszdz7 dz du COS I 9 du 1 1 u12 7u107sinz1o El 11 12z f d 3273z27213 12z 12z du 1 713 3273127213dzi W 397612z77Eu dui 11273127213 du76zi612dz dzd u 7 7 7612z 1 3 23 7 1 2 323 76211 C774273z 721 0 El 2 sec 113 g seczl32 secu2 2 3 2 1 3 2 3dz2 331du3secudu3tanuC3tanz 0 I z u 113 du z wg dz dz 3123 du El lzln121 00 W lzlnz21d 7211nu dui 0 121 If 1 u 217 j u 10u1 11u2 z 1 21 1 1n2 1 1n2 1 1 1n2 1 du gudw wdw w2 1n2201120111 1 u 0 0 0 4 111217 du2zdz7 dz 2 2 u 2 2 2 d 101nu7 dw u7 duudw u1w0 u27w1n2 El u i 1 1 417 if f1 1 and f2 a x Qfz 7 elfI du 7 edii nue 1 facdggi1 u fI 1 u 7111111 1 11fz7 dufzdz7 dz 11uf11 12uf2e El j ltr 1gt4lt12215gtdm aws u d 1 u 3 u 1 aws z14 dz zl4 2I1 2 4 du 2 1n44 C 21n44 0 du 7 2 7 7 7 uiz l 21 l 57 du7212d172z1dz d172z1 l dzi 3 3 7 75 1 ac dziliS drilnl slf ci El mdz eixlie h dziie du7dufisiniluCiisinilei C eixlie h emxliu2 xliuQ 116717 du7e dz dz7e du El 1 d i l mm I du2tan 1uC2tanilECi demmdu21luz uE dudz dz2Edu El 8 a A population of amingo laWn ornaments grows exponentially in Calvin s yardi There are 20 after 1 day and 60 after 4 days How many are there after 6 days Let F be the number of amingos after t days Then F Foek i When t l7 F20 20 Foe When t 4 F 60 60 Foal Divide 60 Foe by 20 Foek and solve for k 60 Foe 3k 3k ln3 im 3767 ln37lne 73k kiTi Plug this back into 20 Foek 20 ms 3 20F0elt V F0 mi 11 Hence7 0 2 ln3 3 eln 33 6 V F When t 67 F 2 124180503 amingos El 20 60quot 3mg b A hot pastrami sandwich with a temperature of 150 is placed in a 70 room to cool After 10 minutes7 the temperature of the sandwich is 90 i When will the temperature be 80 Let T be the temperature at time t The initial temperature is To 1507 and the room s temperature is 707 so T 70 150 7 70Mt or T 70 seems When t 107 T 90 90 70 80a 20 806 i em 111i ln em 1 1 1 an 10167 Is 0an1 Hence7 T 70 806t110ln14 Set T 80 and solve for t 80 70 806010 ln147 10 806t110ln147 etl10lnl47 In lnetl10ln147 1 1 1 ln t ln t 8 15minutesi El 91 a Find the area of the region in the rst quadrant bounded on the left by y 12 on the right by z y 2 and below by the zaxis The curves intersect at the point z 17 y 11 You can see this by solving y 12 and z y 2 simultaneously iVide the region up into horizontal rectanglesi A typical rectangle has width dyi The right end of a rectangle is on I 2 7 y the left end of a rectangle is on I iiei y 12 Therefore7 the length of a rectangle is 2 7 y 7 V37 and the area of a rectangle is 2 7 y 7 dyi 12 The area is 1 A70127y7mdy 2y72 2 1 5 2 7 32 7 301833331 11 y 3y 0 6 b Find the area of the region between y 12 7 12 and y z from I 0 to z 41 Find the intersection point 12712 1 127171207 14z7307 174 or 131 I 3 is the intersection point between 0 and 41 Use vertical rectangles Between 1 0 and z 37 the top curve is y 12 7 12 and the bottom curve is y 11 Between 1 3 an I 4 the top curve is y z and the bottom curve is y 12 7 121 The area is 3 4 1 1 3 1 1 4 79 1271271 dz 1712712d1 7 121 7 z 7 2 12 7121 z 7 m 261333331 11 0 3 3 2 0 2 3 3 3 101 Approximate the area under y z 7 sin z2 from I 3 to z 5 using 20 rectangles of equal width7 and using the midpoints of each subinterval to obtain the rectangles7 heightsi 573 The w1dth of each rectangle 1s AI T 0111 305315 325 435495 1 l 1 l 1 l 1 quotn39 1 l 1 l 1 l l l l l l l 30 31 32 33 43 49 50 The midpoints start at 3105 and go to 4195 in steps of size 0111 If I 7 sin z27 the rectangle sum 19 011 f3105 011k m 44171071 11 60 111 Approximate the sum 3sin2Jr3sin33sin4 3sin41 121 222 32 40240 to 5 decimal places 40 3s1n2 3s1n3 3s1n4 3s1n41 3s1nn1 m 31311921 12471Jr 222 T 32 T 40240 7 n2n D 13 12 Calvin runs south toward Phoebels house at 2 feet per second Bonzo runs east away from Phoebe s house at 5 feet per second At What rate is the distance between Calvin and Bonzo changing When Calvin is 50 feet from the house and Bonzo is 120 feet from the house Let z be the distance from Bonzo to the house7 let y be the distance from Calvin to the house7 and let 8 be the distance between Calvin and Bonzoi CNWn Bonzo Phoebes house By Pythagoras7 s2 z2 y2i Differentiate With respect to t d3 7 dz dy d3 7 dz dy 28E 7 ZzE 2y7 SE 7 zE yai dz dy E75andai y507s130i So 72 negative7 because his distance from the house is decreasing When z 120 and d3 d3 50 130 1205 50727 a E N 3534615 feet per second El 13 Find the dimensions of the rectangle With the largest possible perimeter that can be inscribed in a semicircle of radius 1 X X The height of the rectangle is y the Width is 2zi The perimeter of the rectangle is p4z2yi 14 By Pythagoras7 12 y2 17 so y 1 7 12 Therefore7 p4z21izQi z 1 gives a at rectangle lying along the diameter of the semicircle z 0 gives a thin rectangle lying along the vertical radiusi dp z 2 7 so p0forz E M17127 dz 3 i The negative root does not lie in the interval 0 S I g 1 1 V3 p 2 2V3 4 max 2 1 4 z ives i The dimensions of the rectan le With the lar est erimeter are 21 and 3 g y E g g p x3 39 the maximum perimeter is p 2V3 m 41472141 El 1 9 7 14 A Window is made in the shape of a rectangle With an isosceles right triangle on top 5V2 a Write down an expression for the total area of the Windowi 71 A E s2 28hi I b Write down an expression for the perimeter of the Window that is7 the length of the outside edge p 2h2s V231 El c If the perimeter is given to be 4 What value of 8 makes the total area a maximum Since 41172h2s2s7 2 h27877si Hence7 1 2 1 1 A s228 2737 gs 5822 87 82782 2 87 827 582 L 2x5 The extreme cases are 8 0 and h 07 Which gives 8 15 The derivative is dA 2 7 2amp3 7 81 d3 Find the critical point N5 02 72 2 7 7 1 f 8 87 S 2 1 Plug the critical point and the endpoints into A 0 Ni 4 S N5 1 2 A 0 1104482 0168629 Ni Wh 7 118 the area is a maximumi El 151 a Find the absolute max and the absolute min of y 13 7 121 5 on the interval 0 S I S 51 The derivative is y 7 312 712 7 312 7 4 3z 7 2z 2 y 0 for z 2 and z 72 however7 only I 2 is in the interval 0 S I S 51 y is de ned for all 11 The absolute max is at z 5 the absolute min is at z 21 El 3 b Find the absolute max and the absolute min of 1143 7 15113 on the interval 71 S I S 81 I75 fz 113 7 51723 I2 f 0 for z 5 and f is unde ned for z 01 Both points are in the interval z 71 8 0 5 fz 15175 718 0 719123723 The absolute max is at z 71 and the absolute min is at z 51 El 161 Use a limit of a rectangle sum to nd the exact area under y 12 31 from I 0 to z 1i 1 Divide the interval 0 S I S 1 up into n equal subintervalsi Each subinterval has length AI 71 1711 n evaluate the function at the righthand endpoints of the subintervals7 Which are 12 3 n71 75757 H7T7 3 The function values are 1249vltzgt23ltzgtlt2gt23lt2gt1 16 The sum of the rectangle area is k 2 k 1 1 7 3 7 1 1 2 1 3 1 Z 3 32k22 17w17mA n n n n n n 6 n 2 161 161 161 The exact area is 3 El 2 1 1 nn12n1 3 nn1 71700 n3 6 n2 2 17 Given that I5 413 17 What is f 1 22 First 514 12121 Then f 1 22 f f 122 1 need to nd f 1221 Suppose f 122 11 Then 22 so I54I31722 and z54zts01 1 can7t solve this equation algebraically but 1 can guess a solution by drawing the graph of y I5 413 7 5 1 05 M 15 2 It looks like I 1 is a solution Check 15 4 13 7 5 01 Thus 22 so f 122 11 Hence 1 1 1 122 1 n f 1 lt 1 1067102 MD 17 181 Find the largest interval containing 1 1 on Which the function 314 7 1613 1 has an inverse The derivative is fz 1213 7 48x2 121 7 4 f z 0 for z 0 and z 41 is de ned for all I Here is the sign chart for y 11711760 x0 11117313 x4 11513011 didil6 x2t2dt6z5 21526x5 2z12i u z 4 z z 4 b Computedigz21gt dzi z mew n0 u I 20 a Use the de nition of the derivative as a limit to prove that 7amp1 1 1 174 zh74 f1fltzhzifltzgt ml mmgm 1747zh74 lie I 7 4W h 7 4 53 1 TNT 7 4 13 ha 7 4Ezhh 7 4 1 1 53 174h74 1174 D xh b Compute fling vt4 1 dt 1 Thel1m1t as h goes to 07 the E So 1 make a guess that the limit in the problem is actually the derivative of a function The problem is to gure out What function is being differentiated i i i i Let t41dti Then 0 and the limits I and z h remind me of the de nition of the derivative xh ac xh 0 xh fzhifz xt41dt7 xt41dt xt41dt xt41dt xt41dti 0 0 0 E E Therefore7 hgt0 xh limE t41dtlbinbfzh71611 gwfz di xt41dtz41i El 1 0 21 Let 3 3 ifzlt1 7 7x 0 0 9 ifz1 3127212 ifz gt 1 1s f continuous at z 1 Why or Why not 3 4 m 1 7 7081 11in f 11117 6 7r 5 7 2 7 7 7 7 1113r 7 0511711117r 3x 212 3 2 2 018 The left and righthand limits agree Therefore lim1 0181 However 019 so lim1 Hence f is not continuous at z 1 El 12 1216 221 a Suppose that 5 and Use differentials to approximate f2199 to 5 places Use fltz dz M 1 dz 9 dz 2 99 7 3 70101 and f 3 g 0136 Therefore f2199 m f3 H3 dz 7 5 013670101 7 499640 El b A differentiable function satis es y 6 cos 31 and y0 011 Use differentials to approximate y0102i 1 dz 0 02 7 0 0102 When I 0 y 60 cos0 1 Therefore I dy dy 7 a dz 7 10102 7 0102 Hence y0102 m y0 dy 01 0102 0112 El 231 Use 3 iterations of Newton7s method starting at z 2 to approximate a solution to 4 7 12 emf Write the equation as 4 7 12 7 e 0 Set 4 7 12 7 6 so 721 7 ea z 1 HI 2 7738906 7113891 135121 7168789 7656454 109409 70183505 7517465 105863 7000311343 7499968 105801 7946553 X 10 7 7499664 19 The solution is approximately I N 1058 El 24 Prove that the function 13 21 7 cosz 5 has exactly one rooti First f04gt0 and flt7gt777r5lt0i By the Intermediate Value Theorem f has a root between 7 and 0 Suppose f has more than one rooti Then f has at least two roots so let a and b be two roots of By Rolle s theorem f has a critical point between a an i However fz 312 2 sinzi Since sinz Z 71 2 sinz 21 131E312 Z 0 so fz3122sin1201lgt0i Thus f has no critical points Therefore it canlt have more than one rooti It follows that f has exactly one root The best thing for being sad is to learn something MERLYN in Ti H WHITE S The Once and Future King 2008 by Bruce l39kenaga 20

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