Ed Dewey Stephen Neal Dae Han Kang Sharath Prased Alisha Zachariah. Chris Janjigian Animesh Anand Reese Johnston Jeremy Schwend Alex Troesch

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1 MATH 222 (2 and 4) Fall 213 Practice Final Solutions Circle your TA s name from the following list. Ed Dewey Stephen Neal Dae Han Kang Sharath Prased Alisha Zachariah Chris Janjigian Animesh Anand Reese Johnston Jeremy Schwend Alex Troesch Please inform your TA if you find any errors in the solutions. Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Score Problem 7 Problem 8 Problem 9 Problem 1 Problem 11 Problem 12 Score Instructions Write neatly on this exam. If you need extra paper, let us know. You must show all of your work, except on Problem 1. All problems graded out of 1. No calculators, books, or notes (except for those notes on your 3 5 notecard.)

2 Note: Everything on this page will appear on the actual exam as well. Formulas cos(arcsin x) = p 1 x 2 sec(arctan x) = p 1+x 2. tan(arcsec x) = p x 2 1. csc(arcsin x) = 1 x cot(arcsin x) = p 1 x 2 x Bound for remainder term If f is a n + 1 di erentiable function on an interval containing x = and if we have a constant M n such that f (n+1) (t) apple for all t between and x then R n f(x) apple M n x n+1 (n + 1)!

3 1. For each statement below, CIRCLE true or false. You do not need to show your work. (a) (b) (c) (d) (e) True False True False True False True False True False (a) R e 3x dx R e x2 dx. (b) (x 2 + x 3 ) 2 = o(x 3 ). (c) P 1 n=1 1 n 4 +5 is a finite number. (d) Let ~a and b be any space vectors and let t be any number. Then a (b + ta) =a b + ta 2. (e) = 46. Solution: (a) True. (b) True. (c) True. (d) False. (e) False.

4 2. Below are four direction fields and two equations. Match the equation to the appropriate direction field. (5 points each). (a) dy dx =sinx +cosy. Answer: (b) dy dx = xy2. Answer: I. II. III. Solution: (a) IV (b) I IV.

5 3. Below you will find a number of mathematical expressions. Circle 5 1 those which are nonsense. For instance, writing is nonsense since we cannot raise a vector to the fifth power. Let a 1A, b 5A, c 11 A 4, and d = (a) (a b) c (b) a b c (c) 3a +5b d (d) a a (e) (a b)2 a d. Solution: (a) (a b) c Well defined. (b) a b c Nonsense. (c) 3a +5b d Nonsense. (d) a a Well defined. (e) (a b)2 a d. Welldefined.

6 4. Compute R 5x 1 (x+5)(x 2 +1) dx. Solution: We rewrite this as: 5x 1 (x +5)(x 2 +1) = A x +5 + Bx + C x 2 +1 We use the method of equating coe cients to determine A, B and C. This gives 5x 1=A(x 2 +1)+(Bx + C)(x +5) = Ax 2 + A + Bx 2 + Cx +5Bx +5C =(A + B)x 2 +(C +5B)x + A +5C. So we get the system of equations: 8 >< = A + B 5 = C +5B >: 1 = A +5C The first equation yields B = A so it reduces to the system of equations ( 5 = C 5A 1 = A +5C Solving this yields A = 1andC =andhenceb =1. Sowehave Z Z 5x 1 (x +5)(x 2 +1) dx = 1 x +5 + x x 2 +1 dx = ln x ln x C Our final answer is ln x ln x C.

7 5. Compute R cos 3 (5 +1)d. Solution: We have Z Z cos 3 (5 +1)d = (1 sin 2 (5 +1))cos(5 +1)d u =sin(5 +1)sodu =5cos(5 +1)d and Z = (1 u 2 ) du 5 = u u C = 1 5 sin(5 +1) 1 15 sin3 (5 +1)+C The final answer is 1 5 sin(5 +1) 1 15 sin3 (5 +1)+C.

8 6. Compute R 1 x 2 e x dx. (Note: The original copy of the practice exam had xe x instead of x 2 e x. The problems are similar, but I like this one more.) Solution: We have Z 1 x 2 e x dx = lim b!1 Z b x 2 e x dx We use integration by parts with f = x 2 and g = e x so that f =2x and g = e x : = lim b!1 [x 2 ( e x )] b = lim b!1 [x 2 e x ] b +2 Z b Z b 2x( e x )dx xe x dx We then use integration by parts again, with h = x and k = e x so that h = 1 and k = e x yielding: Since b2 e b and b e b = lim [x 2 e x ] b +2 [x( e x )] b b!1 = lim b!1 [x 2 e x ] b +2 [x( e x )] b + Z b Z b 1( e x )dx e x dx = lim b!1 [x 2 e x ] b 2[x(e x )] b + 2[ e x ] b = lim b!1 [b 2 e b ] 2[b(e b ) ] + 2[ e b + 1] both go to as b!1,wethenhave: = 2[ ] + 2[ + 1] = 2 So the final answer is R 1 x 2 e x dx = 2.

9 7. The squirrel population in Madison has a continuous birth rate of 8% and a natural continuous death rate of 3%. In addition, each year 3 squirrels are eaten by foxes and 1 squirrels are run over by cars. There were 1, squirrels in Madison on January 1, 21. We are interested in explicitly finding a function S that models the squirrel population in Madison at a given time. Use the following space to work out your answer, and record the various parts of the problem at the bottom of the page. Variables: Di erential equation and initial condition for S: Solution for P satisfying initial conditions:

10 Solution: Answers below. The di erential equation is obtained by ds dt The solution is as follows: =.8S.3S 3 1 =.5S 4. Z ds =.5S 4 dt Z ds.5s 4 = dt or.5s = 4 2 ln.5s 4 = t + C or.5s = 4 ln.5s 4 = t 2 + C 2 or.5s = 4.5S 4 = e t 2 + C 2 = e C/2 e t/2 or.5s = 4.5S 4 = ±e C/2 e t/2 or.5s = 4 Changing constants, we can rewrite ±e C/2 as a new constant A, where the solution.5s = 4 gets absorbed by the case A = :.5S 4 = Ae t/2 Solving the initial value yields so A = 1..5S = Ae t/2 + 4 S = 2Ae t/2 +8, 1, = S() = 2Ae /2 +8, = 2A +8, Variables: t is time in years since January 1, 21. S(t) is squirrel population in Madison at time t. Di erential equation and initial condition for S: ds dt =.5S 4 and S() = 1,. Solution for P satisfying initial conditions: 2, e t/2 +8,

11 8. Compute a solution to the initial value problem dy dx = x + xy2 2y and y() = p e 2 1 Solution: This is a separable di erential equation. Z dy dx = x1+y2 Z 2y 2ydy 1+y = xdx 2 Since 1 + y 2 cannot equal, we do not need to worry about division by. ln 1+y 2 = x2 2 + C 1+y 2 = e C e x2 2 1+y 2 = ±e C e x2 2 y 2 = ±e C e x2 q y = 2 1 ±e C e x2 2 1 Now to solve for C we use the initial condition q y() = p e 2 1= ±e C e 2 2 1= p ±e C 1 So we choose C =2andthepositivebranch(i.e. the+fromthe±) q yielding. This yields our final answer y = e 2 e x2 2 1.

12 9. Find the Taylor polynomial of degree 14 at x = (i.e. find T 14 ) of the function f(x) = 1x4 (1 x 5 ).Remembertousenotationcorrectly! 2 1x Solution: Since 4 (1 x 5 ) =2 d 2 dx 1 1 x 5 we have: 1x 4 T 1 (1 x 5 ) = T 12 d 1 2 dx 1 x 5 =2 d dx T x 5 So T 14 f(x) =1x 4 +2x 9 +3x 14. =2 d dx 1+x5 + x 1 + x 15 + o(x 15 ) =2 5x 4 +1x 9 +15x 14 + o(x 14 ) =1x 4 +2x 9 +3x 14 + o(x 14 )

13 1. Does P 1 n=1 e n +n+ p n e n +n 3 + 3p n converge? You must justify your answer. Solution: We first use the limit comparison test, comparing with P 1 n=1 n e n.wecheckthatthelimitcomparisontestapplies: lim n!1 e n + n + p n e n + n p n en n = lim n!1 = lim n!1 1 ne n 1 ne n e n n e n e + n 3 n e n e n + n + p n p en e n + n n n + n n + p n n + 3p n e n = =1 e n e n n n Since this limit converges to a positive number, the limit comparison test applies. So the original series converges if and only if the new series P 1 n=1 n e n converges. To check this, we apply the ratio test: L = lim n!1 (n +1)/e n+1 n/e n n +1 = lim n!1 n 1 e =1 1 e < 1. Since L<1, the series P 1 n=1 n e n converges and hence the original series also converges.

14 11. Imagine that you have a function f(x) thatsatisfies f (n+1) (x) apple(n +1) for all n. Show that the Taylor series T 1 f(x) convergestof(x) for any value of x. Youshouldmakeuseofthe boundfortheremainder term on the second page of this exam. Solution: For some fixed x we want to show that T 1 f(x) converges to f(x). It is equivalent to show that for our fixed x we have that lim n!1 R n f(x) =. We use the Bound on the Remainder Term, with M = n +1,toobtain lim R (n +1) x n+1 nf(x) <= lim n!1 n!1 (n +1)! x n = x lim n!1 n! = where the last limit equals because factorial beats exponential.

15 12. Let P be the plane spanned by the points (5,, ), (4, 2, ) and (, 1, 6). Use the space below to compute the following: (a) (4 points). The normal vector n to P. (b) (3 points). An equation (standard or parametric is fine) for P. (c) (3 points). Does the point A =(2, 6, 11) line in P?

16 Solution: To compute the normal vector, we first compute two vectors lying in P: a A 2 A and b A 1 A 6 6 We get a normal vector to our plane by taking the cross product of a and b So n 6 A. n = a b 1 i 1 5 =det@ j 2 1A k 6 = ( 1k +i 6j)+(12i 1k +j) 1 12 =12i +6j +k 6 A Based on the normal vector, we know that a standard equation for P has the form 12x +6y +z = c for some constant c. Plugginginthe point (5,, ) yields 12(5)+6()+() = c so c =6andourequation is 12x +6y =6. To check if A =(2, 6, 11) lies on P we plug into our equation, yielding 12(2) + 6(6)? =6. Sincethisistrue, we see that A lies on P.

Carolyn Abbott Tejas Bhojraj Zachary Carter Mohamed Abou Dbai Ed Dewey. Jale Dinler Di Fang Bingyang Hu Canberk Irimagzi Chris Janjigian

Carolyn Abbott Tejas Bhojraj Zachary Carter Mohamed Abou Dbai Ed Dewey. Jale Dinler Di Fang Bingyang Hu Canberk Irimagzi Chris Janjigian Practice Final a Solutions (/7 Version) MATH (Lectures,, and 4) Fall 05. Name: Student ID#: Circle your TAs name: Carolyn Abbott Tejas Bhojraj achary Carter Mohamed Abou Dbai Ed Dewey Jale Dinler Di Fang