MIDTERM 3 CALCULUS 2. Monday, December 3, :15 PM to 6:45 PM. Name PRACTICE EXAM SOLUTIONS

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1 MIDTERM 3 CALCULUS MATH 300 FALL 08 Moday, December 3, 08 5:5 PM to 6:45 PM Name PRACTICE EXAM S Please aswer all of the questios, ad show your work. You must explai your aswers to get credit. You will be graded o the clarity of your expositio! Date: October 7, 08.

2 8 poits. Match the followig fuctios with their correspodig Maclauri series: (a) e x / = (VI) ( x ) (b) cos = (II) (c) ( x) = (III) (d) x arcta(x) = (IV) (I) (II) (III) (IV) (V) (VI) x ( ) x ()! x ( ) x + + ( ) x + ( + )! x! Here are more details o the solutios:.(a). We kow that the Maclauri series for e x is x!. Thus, substitutig x x /, we obtai that the Macluari series for e x / is x!.

3 .(b). We kow that the Maclauri series for cos(x) is ( ) x. Thus, substitutig ()! x x/, we obtai that the Macluari series for cos(x/) is ( ) x ()!..(c). We have the geometric series Usig term-by-term differetiatio, d dx ( x x = ) x = d dx x ( x) = x.(d). The Maclauri series for arcta x is Multiplyig by x, arcta x = ( ) x + + x arcta x = x ( ) x + + = ( ) x + + 3

4 . Cosider the power series (x 5) poits.(a). Fid the radius of covergece of the power series. Show all work i justifyig your aswer. R =.(b). Fid the iterval of covergece. Show all work i justifyig your aswer. [3, 7] Here are more details o the solutios:.(a). Usig the Ratio Test ad applyig limit laws, L = lim (x 5) + + ( + ) (x 5) = lim Settig L <, = x 5 Hece the radius of covergece is. x 5 x 5 lim = = ( + ) x 5 < x 5 < ( + ) x 5.(b). The above iequality gives us a iterval of radius cetered at a = 5. This iterval has x = 3 ad x = 7 as its edpoits so we must check for covergece at these poits. x = 3 (3 5) = ( ) = ( ) The above series coverges by the Alteratig Series Test; eve better, it coverges absolutely by the p-test with p =. 4

5 x = 7 (7 5) = The above series coverges by the p-test with p =. = Sice the power series coverges o both edpoits, the iterval of covergece is [3, 7]. 5

6 3 3. Fid the solutio of the differetial equatio y(x + ) + y = 0 that satisfies the iitial coditio y( ) =. Show all your work. poits The solutio is y = e x / x To see this, observe that the differetial equatio is separable. y(x + ) + y = 0 y = y(x + ) dy = y(x + ) dx dy = (x + )dx y dy y = (x + )dx l y = x / x + C e l y = e x / x+c y = e x / x+c Let K = ±e C. The y = Ke x / x ad pluggig i the iitial coditio, = Ke ( ) / ( ) = Ke + = K Hece the solutio to the differetial equatio with the give iitial coditio is y = e x / x 6

7 4 8 poits 4. Give the followig power series a (x ) we kow that at x = 0 the series coverges ad at x = 8 the series diverges. What do we kow about the followig values? 4.(a). At x = 3 the series a (x ) is: (i) Coverget (ii) Diverget (iii) We caot determie its covergece/divergece with the give iformatio. 4.(b). At x = 4 the series a (x ) is: (i) Coverget (ii) Diverget (iii) We caot determie its covergece/divergece with the give iformatio. 4.(c). At x = 9 the series a (x ) is: (i) Coverget (ii) Diverget (iii) We caot determie its covergece/divergece with the give iformatio. 4.(d). The followig series a is: (i) Coverget (ii) Diverget (iii) We caot determie its covergece/divergece with the give iformatio. I a little more detail: The give power series is cetered at a =, coverges at x = 0, ad diverges at x = 8. Graphically we have 7

8 a Sice the coverget poit x = 0 is distace from the ceter a =, the radius of covergece is at least. Similarly, sice the diverget poit x = 8 is distace 6 from the ceter a =, the radius of covergece is at most 6. Below we have the gree iterval [0, 4) idicatig poits of guarateed covergece, the red itervals (, 4) [8, ) idicatig poits of guarateed divergece, ad the yellow itervals idicatig poits of ucertaity, where we caot determie covergece or divergece with the give iformatio. Note that the poits x = 4, 4 are i the yellow iterval. a (a) x = 3 is i the gree iterval so the series coverges there. (b) x = 4 is i the yellow iterval so we caot determie covergece. (c) x = 9 is i the red iterval so the series is diverget there. (d) Observe that a = a () = a (3 ). This meas we are lookig for covergece of the poit x = 3. Clearly x = 3 is i the gree iterval so the series coverges there. 8

9 5 poits 5.(a). Write the defiitio for the th degree Taylor polyomial of a fuctio f (x) cetered at x = a. 5.(b). Fid the secod degree Taylor polyomial for f (x) = l(sec(x)) cetered at π/4. (a) This is just writig the formula T (x) = f (a) + f (a)(x a) + f (a)! (b) The secod degree Taylor polyomial has the formula (x a) + f (3) (a) (x a) 3 + 3! T (x) = f (a) + f (a)(x a) + f (a) (x a)! For the specific fuctio f (x) = l(sec(x)), we have to fid f (π/4), f (π/4), ad f (π/4). f (π/4) = l(sec(π/4)) = l(/ ) f (x) = sec(x) ta(x) = ta(x) sec(x) f (π/4) = ta(π/4) = f (x) = sec (x) f (π/4) = sec (π/4) = (/ ) = Hece the Taylor polyomial is T (x) = l(/ ) + (x π/4) +! (x π/4) = l(/ ) + (x π/4) + (x π/4) 9

10 6 poits 6.(a). Express the fuctio f (x) = l( + x 3 ) as a power series cetered about x = 0. ( ) x 3 6.(b). Express the defiite itegral ( ) (3 + ) 0 l( + x 3 ) dx as a ifiite series. (a) The Maclauri series for l( + x) is Substitutig x x 3, l( + x) = ( ) x l( + x 3 ) = ( ) (x 3 ) = ( ) x 3 (b) Usig the series from part (a) ad applyig term-by-term itegratio, 0 l( + x 3 ) dx = 0 ( ) x 3 dx = = = = 0 ( ) x 3 dx ( ) x 3 dx 0 ( ) [ x ( ) 3 + ] 0 = ( ) (3 + ) 0

11 7 poits 7.(a). Fill i the blaks to complete the statemet of Taylor s Iequality: If f (+) (x) M o the iterval betwee the ceter, a, ad the poit of approximatio x, the the remaider, R (x), of the th degree Taylor polyomial T (x), satisfies the iequality: R (x) M x a + ( + )! 7.(b). Use Taylor s iequality to determie the umber of terms of the Maclauri series for e x that should be used to estimate the umber e with a error less tha 0.6. Clearly justify your choice of M. 3 or more terms The Maclauri series for f (x) = e x is e x = x! Sice we wat to approximate e = e = f (), x is equal to. Taylor s Iequality for the above Maclauri series gives us R () M ( + )! 0 + = M ( + )! To fid M, ote that f (+) (x) = e x for all positive itegers. The M f (+) () = e = e Iroically, fidig a good choice for M requires us to guess how big e ca be. Nevertheless, we will choose M = 3 to avoid ay circular argumets. Lastly we boud the Taylor s Iequality by our error margi of 0.6: R () 3 ( + )! < 0.6 = 3 5

12 Solvig the iequality, we get 3 ( + )! < < ( + )! Sice 5 < 6 = ( + )!, choice of guaratees that the th degree Taylor polyomial T () approximates e to withi our error margi of 0.6. Sice T (x) cotais + terms, we eed at least three terms of the Maclauri series.

13 8 8 poits 8. Each of the followig slope fields represets oe of the followig differetial equatios. Match each slope field to the correspodig dfferetial equatio. 8. (8 (a) dy poits) dx = xy Each of the followig slope fields represets oe of the followig di eretial (II) equatios. Match each slope field to the correspodig di eretial equatio. (b) dy (a) dx = dy dx y = xy x (III) (b) (c) dy dy dx = dx x = + y x (I) (d) dy (c) dy dx = dx = x + ex (IV) (d) dy dx = ex (I) y (III) y x x (II) y (IV) y x x 3

14 Here are more details o the solutios: 8.(a). Whe x = 0 or y = 0, dy/dx = 0 ad we see that (II) has slope of 0 alog the x ad the y axis. 8.(b). Alog the diagoal lie y = x, dy/dx = ad we see that (III) has the slope fixed at. 8.(c). Alog the vertical lie x =, dy/dx = 0 ad we see that (I) has a fixed slope of 0. 8.(d). Observe that the equatio for dy/dx i (d) has o y. We see that the graphs (I) ad (IV) have the same slope for a fixed value of x. Hece we are dow to two choices (I) ad (IV). But it ca t be (I) sice its slopes are 0 whe x is a egative umber. e x is ever 0 so we elimiate (I) as a possibility ad we have (IV) as our aswer. 4

15 9 9. Fid the sum of the series ( ) + = poits = π/4 Recall that the Maclauri series for arcta x is Pluggig i x =, we get arcta = ad arcta = π/4. arcta x = ( ) + ( ) x+ + + = ( ) + = ( ) + 5

16 0. Assume we approximate the sum of the series 0 0 poits by usig the first 3 terms. Give a upper boud for the error ivolved i the approximatio by usig the Remaider Estimate for the Itegral Test. R 3 3 Let f (x) =. To apply the Remaider Estimate for the Itegral Test, we first check the x coditios ecessary. Firstly, x is differetiable sice f (x) = 4 ad so it is cotiuous. x3 f (x) is also positive for ay positive value of x ad it is decreasig sice it is a reciprocal of x /, a icreasig fuctio. Lastly, we kow that the series coverges via the p-test with p = >. We are usig the first three terms so we wat to estimate the error associated with s 3, the partial sum up to = 3. The the Itegral Test gives us 3+ f (x) dx R 3 3 f (x) dx Sice we are oly iterested i the upper boud, we compute the itegral o the right side. t R 3 dx = lim 3 x t 3 x dx t = lim t 3 x dx = lim [ x ] t t 3 [ = lim t t ] 3 [ = 0 ] = 3 3 Hece R

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