Math 12 Final Exam 2009

DEPARTMENT OF MATHEMATICS Math 12 Final Exam May 13, 2010 Answer Key 1. [10 Points] Evaluate each of the following limits. Please justify your answers. Be clear if the limit equals a value, +∞ or −∞, or Does Not Exist. (a) arctan(x − 3) = x2 − 9 lim x→3   0 L’H = lim x→3 0 1 x “ 1” ln (ex +1) x lim (ex + 1) = lim e (b) x→∞ x→∞ „∞« L’H lim = e ∞ = ex→∞ 2. [15 Points] (a) Z ln 7 0 ex ex +1 1 1 (x−3)2 +1 2x = 1 6   1 lim ln (ex + 1) x = ex→∞ ex lim x L’H x→∞ e = e1 = e =e ∞ = e „∞« Evaluate each of the following integrals. sinh x dx = cosh x Z 25 7 1 1 du = ln |u| u u = cosh x 25 7 = ln 1  25 7  − ln 1 = ln du = sinh xdx (b)  25 7  x = 0 ⇒ u = cosh 0 = 1 Substitute Z ln(ex + 1) x = ex→∞ lim x = ln 7 ⇒ u = cosh(ln 7) = eln 7 + 2 1 eln 7 = 7+ 2 1 7 = 25 7 Z Z x2 1 x2 x2 1 sin2 θ p √ x arcsin x dx = arcsin x − arcsin x − · cos θ dθ dx = 2 2 2 2 1 − x2 1 − sin2 θ Z Z Z 1 sin2 θ x2 1 sin2 θ x2 1 arcsin x− √ ·cos θ dθ = arcsin x− ·cos θ dθ = arcsin x− sin2 θdθ 2 2 2 2 2 cos θ cos2 θ Z Z 1 1 − cos(2θ) x2 1 arcsin x − dθ = arcsin x − 1 − cos(2θ)dθ 2 2 2 4   1 1 x2 1 1 arcsin x − θ − sin(2θ) + C = arcsin x − θ + sin(2θ) + C 4 2 2 4 8 = x2 2 = x2 2 = x2 2 = x2 1 1 x2 1 1 p arcsin x − θ + 2 sin θ cos θ + C = arcsin x − arcsin x + x 1 − x2 + C 2 4 8 2 4 4 1 u = arcsin x du = √ dv = xdx x2 1 dx v = 2 1 − x2 1 x = sin θ x θ √ 1 − x2 Trig. Substitute dx = cos θdθ 3. [20 Points] For each of the following improper integrals, determine whether it converges or diverges. If it converges, find its value. (a) Z ∞ 7 1 dx = lim 2 t→∞ x − 8x + 19 Substitute = lim Z t−4 t→∞ 3 x=7⇒u=3 du = dx x=t⇒u=t−4 1 1 du = lim 2 t→∞ 3 u +3 √u 3 dv = √ √1 du 3 3dv = du 1 = lim √ arctan v t→∞ 3 = 7 1 dx = lim 2 t→∞ x − 8x + 19 u=x−4 v= Substitute t Z t−4 √ 3 √ 3 Z t−4 3  u √ 3 u=3⇒v= 1 2 √3 3 u=t−4⇒v = 1 = lim √ t→∞ 3  arctan  = +1 √ Z t 7 1 dx (x − 4)2 + 3 square √ Z t−4 √ 3 1 3 dv du = lim √ 2 t→∞ 3 3 v +1 3 t−4 √ 3 t−4 √ 3 π √ 6 3  √  1 π π  1 π  − arctan 3 = √ =√ − 3 3 2 3 6 4 Z t Z 4 3x + 2 3x + 2 3x + 2 (b) dx = dx = lim dx 2 − x − 12 − x (x − 4)(x + 3) (x − 4)(x + 3) t→4 0 0 0 Z t t 2 1 = lim + dx = lim 2 ln |x − 4| + ln |x + 3| x+3 t→4− 0 x − 4 t→4− 0 Z complete the = lim (2 ln |t − 4| + ln |t + 3|) − (2 ln 4 + ln 3) = 2 ln 0+ + ln 7 − 2 ln 4 − ln 3 = −∞ Diverges t→4− Partial Fractions Decomposition: A B 3x + 2 = + (x − 4)(x + 3) x−4 x+3 2 Clearing the denominator yields: 3x + 2 = A(x + 3) + B(x − 4) 3x + 2 = (A + B)x + (3A − 4B) so that A + B = 3 and 3A − 4B = 2 Solve for A = 2 and B = 1 4. [10 Points] (a) Find the sum of each of the following series (which do converge): ∞ X (−1)n 4n n=1 9n−1 = −4 + 16 64 − + ... 9 81 Here we have a nice geometric series with a = −4 and r = − As a result, the sum is given by (b) ∞ X 3n n=0 (c) ∞ X n=0 = n! a = 1−r 4 9 9 36 −4 −4  = = −4 · = − 13 4 13 13 1− − 9 9 = e3 (−1)n π 2n 62n (2n + 1)! = ∞ X n=0 π  6 1 6 3 sin = · = π 6 π 2 π (−1)n  π 2n 6 (2n + 1)! = ∞ X n=0 (−1)n  π 2n  π  6  6 π (2n + 1)! 6 = 6 π ∞ X (−1)n n=0  π 2n+1 6 (2n + 1)! 5. [20 Points] In each case determine whether the given series is absolutely convergent, conditionally convergent, or diverges. Justify your answers. (a) ∞ X (−1)n 3n + 1 n=1 First, we show the absolute series is divergent. Note that ∞ X n=1 ∞ X1 1 ≈ which is a divergent 3n + 1 n n=1 p-series with p = 1. Next, 1 n 1 = which is finite and non-zero. Therefore, these two series share Check: lim 3n + 1 = 1 n→∞ 3n + 1 3 n ∞ ∞ X X 1 1 the same behavior. Since is divergent, then is also divergent by Limit Comparison n 3n + 1 n=1 n=1 Test. As a result, we have no chance for Absolute Convergence. 3 Secondly, we are left to examine the original alternating series with the Alternating Series Test. 1 >0 •bn = 3n + 1 1 =0 n→∞ 3n + 1 • lim • 1 bn+1 < 1 1 1 1 1 because < or < bn 3(n + 1) + 1 3n + 1 3n + 4 3n + 1 Therefore, the original series converges by the Alternating Series Test. Finally, we can conclude the original series is Conditionally Convergent (b) ∞ n X e +9 n=1 n+1 Diverges by nth term Divergence Test en + 9 ∞ L’H en = =∞= 6 0. = lim n→∞ n + 1 n→∞ 1 ∞ since lim (c) ∞ X (−1)n n! n=1 nn Try Ratio Test: (−1)n+1 (n + 1)! nn (−1)n+1 (n + 1)! (n + 1)n+1 = lim · · lim n n→∞ n→∞ (−1) n! (−1)n n! (n + 1)n+1 n n n  (n + 1)n! 1 nn n nn = lim = <1 · = lim = lim n n n→∞ n→∞ (n + 1) n→∞ n + 1 n! (n + 1)(n + 1) e The series is Absolutely Convergent by the Ratio Test . (d) ∞ X n=1 (−1)n n3 + 8 n7 − 9 First examine the absolute series ∞ X n3 + 8 n=1 n7 − 9 ≈ ∞ X 1 which is a convergent p-series with p = 4 > 1. n4 n=1 n3 + 8 7 n7 + 8n4 = 1 which is finite and non-zero. Therefore, these two series Check: lim n − 9 = lim 1 n→∞ n→∞ n7 − 9 n4 ∞ X 1 converges, then the absolute series also share the same behavior. Since n4 n=1 Converges by Limit Comparison Test . As a result, we have Absolute Convergence, which implies the original series is Absolutely Convergent . 4 6. [10 Points] Find the Interval and Radius of Convergence for the following power series. Analyze carefully and with full justification. ∞ X (−1)n (3x − 1)n n2 · 2n n=0 Use Ratio Test. (−1)n+1 (3x − 1)n+1 n2 2n |3x − 1| (3x − 1)n+1 (n + 1)2 2n+1 lim = lim · · = n n n→∞ n→∞ (−1) (3x − 1) (3x − 1)n (n + 1)2 2n+1 2 2 n n 2 |3x − 1| < 1 or |3x − 1| < 2. 2 1 That is −2 < 3x − 1 < 2 =⇒ −1 < 3x < 3 =⇒ − < x < 1 3 Endpoints: The Ratio Test gives convergence for x when •x = 1 The original series becomes ∞ X (−1)n 2n n=0 Convergence Test since its absolute series n2 2n ∞ X 1 n=0 n2 = ∞ X (−1)n n=0 n2 which is a convergent by Absolute is a convergent p-series with p = 2 > 1. ∞ ∞ X X (−1)n (−2)n (−1)n (−1)n 2n 1 1 The original series becomes = = which is a 2 n 2 n 3 n 4 n 2 n2 n=0 n=0 n=0 convergent p-series with p = 2 > 1.   1 2 Finally, Interval of Convergence I = − , 1 with Radius of Convergence R = . 3 3 •x = − 7. [5 Points] ∞ X Consider the function f (x) that satisfies the following f (3) = 2 f ′ (3) = −3 f ′′ (3) = 6 7 f ′′′ (3) = −1 Find the Taylor polynomial of degree 3 for f (x) centered at a = 3. 5 T3 (x) = f (3) + f ′ (3)(x − 3) + = 2 − 3(x − 3) + 6 7 2! f ′′ (3) f ′′′ (3) (x − 3)2 + (x − 3)3 2! 3! (x − 3)2 − 1 (x − 3)3 3! 1 3 = 2 − 3(x − 3) + (x − 3)2 − (x − 3)3 7 6 8. [10 Points] MacLaurin Series: Please analyze with detail and justify carefully. (a) Find the MacLaurin series representation for f (x) = x ln(1 + x). ∞ X Your answer should be in sigma notation . n=0 1 = 1−x First, ∞ X xn . n=0 ∞ ∞ X X 1 1 n Next, = = (−x) = (−1)n xn 1+x 1 − (−x) n=0 Then, ln(1 + x) = Z 1 dx = 1+x Finally, x ln(1 + x) = x n=0 Z X ∞ n+1 = ∞ X (−1)n xn+1 n=0 n=0 ∞ X (−1)n xn+1 n=0 (−1)n (x)n dx = n+1 ∞ X (−1)n xn+2 n=0 n+1 (b) Use the MacLaurin Power Series representation for f (x) = x ln(1 + x) from Part(a) to Estimate Z 1 x ln(1 + x) dx with error less than 0 1 . 10 X (−1)n xn+3 1 (−1)n xn+2 Now, x ln(1 + x) dx = dx = n+1 (n + 3)(n + 1) 0 0 0 n=0 n=0   1 x3 1 1 5 x4 x5 1 1 1 = − + + ... = − + + . . . − (0 − 0 + 0 + . . .) ≈ − = 3 4·2 5·3 3 8 15 3 8 24 0 Z 1 Z ∞ 1X ∞ Note this is an alternating series. Use the Alternating Series Estimation Theorem. If we approximate the actual sum with only the first two terms, the error from the actual sum will be at most 1 1 1 the absolute value of the next term, . Here < as desired. 15 15 10 6 9. [15 Points] Volumes of Revolution (a) Consider the region bounded by y = ex, y = ln x, x = 1, and x = 2. Rotate the region about the y–axis . Compute the volume of the resulting solid using the Cylindrical Shells Method. Sketch the solid, along with one of the approximating cylindrical shells. See me for a sketch. Z Z 2 2π radius height dx = V = 1 2 x 2πx(e − ln x) dx = 2π 1 Double Integration By Parts: u = ln x dv = ex dx u=x du = dx v = ex = 2π Z 2 1 xex dx − 2 xex = 2π du = 1 Z 2 1 2 − ex 1 − Z 2 1 xex − x ln x dx dv = xdx 1 x2 dx v = x 2  x ln x dx = 2π xex 2 x2 x2 ln x − 2 4 1 2 1 !! 2 1 − Z 1 2 ex dx − 2 x2 ln x − 2 1 Z 2 1 x dx 2 !!    1 1 2 1 2 1 = 2π 2e − e − (e − e ) − 2 ln 2 − ln 1 − (1 − 2 4     3 1 = 2π e2 − 2 ln 2 + = 2π 2e2 − e1 − e2 + e1 − 2 ln 2 + 0 + 1 − 4 4 (b) Consider the same region bounded by y = ex, y = ln x, x = 1, and x = 2. Rotate the region about the vertical line x = −3 . Set up, BUT DO NOT EVALUATE!!, the integral to compute the volume of the resulting solid using the Cylindrical Shells Method. See me for a sketch. Z Z 2 2π radius height dx = V = 1 10. [15 Points] 2 1 2π(x + 3)(ex − ln x) dx Consider the Parametric Curve represented by x = et − t and y = 4et/2 . (a) Find the arclength of this parametric curve for 0 ≤ t ≤ 1. dx dy = et − 1 and = 2et/2 . dt dt s Z 1  2  2 Z 1p Z 1q
2 dx dy L= (et − 1)2 + 2et/2 dt = e2t − 2et + 1 + 4et dt + dt = dt dt 0 0 0 Z 1p Z 1q = e2t + 2et + 1 dt = (et + 1)2 dt First, 0 = Z 0 1 0 1 et + 1 dt = et + t 0
= (e + 1) − e0 + 0 = e − +1 − 1 = e 7 (b) Set up, BUT DO NOT EVALUATE!!, the definite integral representing the surface area of the solid obtained by rotating this curve about the x-axis, for 0 ≤ t ≤ 1. s   2 Z 1 Z 1 q
2 dy dx 2 2πy(t) (et − 1)2 + 2et/2 dt + dt = 2πy(t) S.A. = dt dt 0 0 Z 1  
2π 4et/2 et + 1 dt from part(a) above. = 0 11. [15 Points] For each part, sketch the Polar curve(s), and answer the related questions: (a) Compute the area enclosed by the cardioid r = 1 + sin θ. It’s important to note that one full cycle of the cardioid closes up on itself as θ ranges from θ = 0 to θ = 2π. Z Z Z 1 2π 2 1 2π 1 2π 2 A= r dθ = (1 + sin(θ)) dθ = 1 + 2 sin θ + sin2 θ dθ 2 0 2 0 2 0   Z Z 1 2π 1 cos(2θ) 1 − cos(2θ) 1 2π dθ = 1 + 2 sin θ + 1 + 2 sin θ + − dθ = 2 0 2 2 0 2 2   Z cos(2θ) 1 3 sin(2θ) 2π 1 2π 3 + 2 sin θ − dθ = θ − 2 cos θ − = 2 0 2 2 2 2 4 0    1 sin(4π) = 3π − 2 cos(2π) − − (0 − 2 cos 0 − 0) 2 4 3π 1 1 ((3π − 2 − 0) − (0 − 2 − 0)) = (3π − 2 + 2) = 2 2 2 8 (b) Set up, BUT DO NOT EVALUATE!!, the definite integral representing the area bounded inside r = 2 + 2 cos θ and outside r = 3. These two polar curves intersect when 2 + 2 cos θ = 3 ⇒ 2 cos θ = 1 ⇒ cos θ = 21 ⇒ θ = − π3 or θ = π3 . Using symmetry, we will integrate from θ = 0 to θ = π3 and double that area. ! Z π Z π

3 1 3 2 2 Area = A = 2 (outer r) − (inner r) dθ = (2 + 2 cos θ)2 − 32 dθ 2 0 0 12. [10 Points] Find the general solution for each of the following differential equations. dy = (1 + y 2 )ex dx Z Antidifferentiate: 1 dy = ex dx 1 + y2 Separable (a) 1 dy = 1 + y2 Z ex dx arctan y = ex + C Finally, y = tan (ex + C) (b) dy 2 + 2xy = e−x dx dy 2 + 2xy = e−x dx Z Z P (x) dx 2x dx 2 Integrating Factor: I(x) = e =e = ex (already in) Linear Form: 2 Take I(x) = ex . Multiply Diff. Eq. in its linear form by I(x): ex 2 dy 2 2 2 + 2xex y = e−x ex = e0 = 1 dx 9 Recognize left side as a product rule derivative:  2 ′ ex y = 1 Antidifferentiate: 2 ex y = x + C Finally, y = x+C ex2 10