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

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1 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 Bingyang Hu Canberk Irimagzi Chris Janjigian Tao Ju Ahmet Kabakulak Dima Kuzmenko Ethan McCarthy Tung Nguyen Jaeun Park Adrian Tovar Lopez Polly Yu Problem Problem Problem 3 Problem 4 Problem 5 Score Problem 6 Problem 7 Problem 8 Problem 9 Problem 0 Score Instructions Write neatly on this exam. If you need extra paper, let us know. Please read the instructions on every problem carefully. On Problems 4 only the answer will be graded. On Problems 5 0 you must show your work and we will grade the work and your justification, and not just the final answer. Each problem is worth ten points. No calculators, books, or notes (except for those notes on your 3 inch by 5 inch notecard.) Please simplify any formula involving a trigonometric function and an inverse trigonometric function. For example, please write cos(arcsin x) = p x. Note that we have provided some formulas on the next page to help with this.

2 Formulas You may freely quote any algebraic or trigonometric identity, as well as well as any of the following formulas or minor variants of those formulas. Integrals cos(arcsin x) = p x sec(arctan x) = p +x. tan(arcsec x) = p x. R x n dx = ( x n+ n+ R e x dx = e x + C R cos xdx =sinx + C + C when n 6= ln x + C when n = R sin xdx = R tan xdx = cos x + C ln cos x + C R cot xdx =ln sin x + C R sec xdx =ln sec x + tan x + C. R csc xdx = ln csc x + cot x + C. R +x dx = arctan(x)+c. Taylor series T e x = P xk T sin x = P xk+ ( )k (k+)! T cos x = P xk ( )k (k)! T x = P xk T +x = P ( )k x k T ( + x) b = P b k xk where b k = b(b )(b ) (b k+) Other sin( ) =sin cos cos( ) = cos ( ) sin ( ) lim n! ( + n )n = e.

3 . 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) The integral R x 3 + dx apple R (b) R p x 3 dx is finite. x +5 dx. (c) The series P k +e k k + p k converges. (d) The series P is finite. k k (e) sin(x ) x is o(x 4 ). (a) False. This integral has two improprieties: a vertical asymptote at x = 3 and an infinite bound. We can rewrite it as: 3 4 p dx + p dx + p dx. x 3 3 x 3 4 x 3 The first are finite but the last one is not. (b) True. A bigger denominator yields a smaller function and thus a smaller integral. And x 3 + x + 5 for all x. (c) True. Try Limit Comparison Test with P k k followed by the Ratio Test. (d) True, though this one s tricky. Try the Ratio Test. (e) True. sin(x )=x x6 3! + = x + o(x 4 ). Thus sin(x ) x is o(x 4 ).

4 . On this page only the answer will be graded. (a) Compute R x (x+) dx. Solution: This is a partial fractions problem. We rewrite x (x + ) = A x + B x + C x +. Equating coe cients we get = Ax(x+)+B(x+)+Cx =(A+C)x +(A+B)x+B. This yields the system of equations: 8 >< 0=A + C 0=A + B >: =B So A = and C = and we get x (x + ) dx = x + x + x + So our answer is ln x x +ln x + + C. dx = ln x x +ln x + + C. (b) Compute R x 03 ln(x)dx. Solution Use integration by parts with f =ln(x) and g 0 = x 03.Thenf 0 = x and g = x04 04 and we get: x 03 ln(x)dx = = fg fg 0 f 0 g = ln(x)x04 04 = ln(x)x04 04 = ln(x)x04 04 x x04 04 dx x 03 dx 04 x C

5 3. On this page only the answer will be graded. (a) Compute T 4 ( p ex4 ) +x 3 Solution: We start with the fact that T e x4 ( + x 3 ) / = T e x4 T ( + x 3 ) /. We know that T e u =+u + u +... which yields T e x4 =+x 4 + o(x 4 ). We also know that T ( + u) / =+ / u + / u which yields T ( + x 3 ) / = +( /)x 3 + / x 6 + =+( /)x 3 + o(x 4 ). We thus obtain: T e x4 ( + x 3 ) / =(+x 4 + o(x 4 ))( x3 + o(x 4 )) = We conclude that T 4 e x4 ( + x 3 ) / equals x3 + x 4. x3 + x 4 + o(x 4 ) (b) For which values of b does the Taylor series for converge at x = b? 3+x Solution Since T +u = P ( u)k, we see that T 3+x = 3 T + = 3 x 3 X ( 3 x ) k = X 3 ( 3 x ) k Plugging in x = b we get: T 3+x x=b = X 3 ( 3 b ) k This is a geometric series P ark with a = 3 and r = 3 b. The problem only asks about convergence. The Geometric Series Test shows that a geometric series converges if and only if r <. This series thus converges if and only if 3 b < which is equivalent to asking that b < 3 or that b < q 3.

6 4. On this page, only the answer will be graded. (a) Let P be the plane through the points (,, ), (3,, 0) and (0, 4, 0). Compute the normal vector to P Solution: We first compute two vector lying in the plane: ~a A A and ~ b 4A A. We then compute ~a ~ b A 3A. Thus A is our normal vector. Note that any scalar multiple of this vector would 7 also be a correct answer. (b) Let ~a =(, ). Compute ~v? with respect to ~a where ~v =(3, 4). We first compute ~v // ~v ~a = ~a = 3+ 4 ~a = ~a + 5.Thus x = 3x + ~v? = ~v = ~v // 3 = = = x +t (c) At what point (x, y) doestheline` : = intersect the line y =3x +? y 3+4t Solution: We plug in the values. Combining the equation y = 3x + withthe parametric equation for ` we get: ( x =+t +t ) 3+4t 3x +=3+4t We then solve this system of equations. This yields 3( + t)+=3+4t and thus 5+6t =3+4t and so = t and t =. It follows that x = and y = 3( )+ =. Note that this point is also on `. So the lines intersect at the point (, ).

7 5. On this page, you must show your work to receive full credit. (a) Let f(x) =x 3 sin(x ). Compute f (409) (0). Solution: By definition, we have T f(x) = P f (j) (0) j=0 j! x j. Since T sin(u) = P ( uk+ )k (k+)!, we can comput T f(x) another way as T x 3 sin(x )=x 3 T sin(x ) X = x 3 ( ) k (x ) k+ (k + )! X = x 3 ( ) k x 4k+ (k + )! X = ( ) k x 4k+5 (k + )! Relating these two expressions for T f(x) we see that the coe cient of x 409 must be equal. On the hand this coe cient is when j = 409 which is f (409) (0) 409!. On the other hand, if k = 0 then we also get a coe cient of x 409 and in that case it equals ( ) 0 ( 0+)! = 03!. Thus we get: f (409) (0) 409! = 03! and so f (409) (0) = 409! 03!. (b) Let t denote time in years since January, 00 and let S(t) denote the number of squirrels in Madison at time t. This squirrel population has a continuous birth rate of 8% and a natural continuous death rate of %. In addition, each year 50 squirrels are eaten by foxes and 50 squirrels are run over by cars. Write down a di erent equation for S(t). DO NOT SOLVE THE DIFFERENTIAL EQUATION. Solution: ds dt =(.08.0)S +( 50 50) =.06S 400.

8 6. On this page, you must show your work to receive full credit. (a) dy dx x +x y =+x and y(0) = 7. Solution: This is a linear di erential equation with a(x) = x +x.thus R m(x) =e We then get x +x dx = e ln +x =(e ln +x ) = +x y = m(x) =(+x ) =(+x ) m(x)k(x)dx +x ( + x )dx dx =(+x )(x + C). =(+x ). (b) Solving for the initial condition we get 7 = ( + 0 )(0 + C) and thus C = 7 and our final answer is y =(+x )(x + 7). dy dx = ex y + and y(0) = Solution: y +dy = e x dx y + y = e x + C Rearranging we get: y + y (e x + C) =0 Use the quadratic formula to get two solutions: Now solve for C: y = ± p + 4(e x + C) = ± p + 4( + C) ) = ± p 5+4C So we choose the + sign and get C =. The final answer is y = +p +4(e x +).

9 7. On this page, you must show your work to receive full credit. Compute px x dx. For your final answer, you should simplify any expression that combines a trigonometric and inverse trigonometric function (e.g. cos(arcsin x) = p x ). Solution We complete the square to get x x = (x ) and we have the integral: px p x dx = (x ) dx We perform the trig substitution x =sin and dx = cos d to get p = sin cos d = cos d We now use the double angle formula to get: = ( + cos( ))d = ( + sin( )) + C Now we need to replace the s by the original variables. We have = arcsin(x ). For sin( ) we use the double angle formula sin( ) = sin cos = sin(arcsin(x )) cos(arcsin(x )) = (x ) p (x ). Plugging this all in yields: = arcsin(x ) + (x ) p (x ) + C.

10 8. On this page, you must show your work and justify your answer to receive full credit. Let a>0 be a real number, and let I n = R x n e ax dx. Derive a reduction formula for I n. Solution: We have: I n = x n e ax dx Using integration by parts with f = x n and g 0 = e ax we get f 0 = nx n and g = a eax. This yields: = fg 0 = fg f 0 g = a xn e ax (nx n = a xn e ax n a = a xn e ax n a I n Thus our reduction formula is I n = a xn e ax n a I n. x n )( a eax )dx e ax dx

11 9. On this page, you must show your work and justify your answer to receive full credit. Let f(x) =e x + x 5. Find B so that R f(x) appleb for all apple x apple. If using the Error Bound Formula, you must clearly indicate your chosen value for M and explain why this choice of M is valid for the desired range. Solution: We will use the Error Bound Formula with c = and n =. We must compute M and for this we will compute f (3) (x). We have: f 0 (x) =( )e x +5x 4 f 00 (x) =( ) e x + 0x 3 f (3) (x) =( ) 3 e x + 60x. We wish to find M where M f (3) (x) for all apple x apple. By the Triangle Inequality ( ) 3 e x + 60x apple ( ) 3 e x + 60x =8e x + 60x. Since e x is a decreasing function, the maximum of 8e x occurs at the left endpoint, and is given by 8e ( ) =8e 4. Since 60x is an increasing function its maximal value occurs at the right endpoint, and is 60 = 40. Thus we have: So we choose M =8e f (3) (x) apple8e for all apple x apple. Using the Error Bound Formula, we then obtain that R f(x) apple Mcn+ (n + )! = (8e4 + 40) 3 3! So B = (8e4 +40) 3 3! is our final answer. for all apple x apple.

12 0. On this page, you must show your work and justify your answer to receive full credit. You must clearly state any convergence test that you use and explicitly verify each hypothesis for that test. For which values of b does the series P k= ekb + p k converge? Solution: Our solution will have two cases depending on the value of b. Case : b>0 We first assume b>0 so that e kb is an exponential growth function in k and thus e kb is the term which grows the fastest in the numerator. We define a k = ekb + p k and c k = ekb and we will apply the Limit Comparison Theorem to these two series. We check the hypotheses: a k = ekb + p k and c k = ekb are positive for all k since the numerator and denominator are built from positive numbers for all k. a We check that lim k e c k =lim kb + p k e =lim kb + p p k e kb =lim e kb + k = e kb +0 = since exponential growth function grows faster than any polynomial function. Since this limit is a positive real number, it satisfies the second hypothesis. We may thus apply the Limit Comparison Theorem to conclude that either both series converge or both diverge. We next apply the Ratio Test to P k= c k. We compute L =lim c k+ c k =lim e(k+)b (k+)! e =lim b e kb k+ = 0. Since L =0<, the Ratio Test implies that P k= c k converges. Thus the original series converges as well whenever b>0. Case : b apple 0 We now assume b apple 0 so that e kb does not grow with k. Then p p k is the term which grows the fastest in the numerator. We define d k = k and we will apply the Limit Comparison Theorem to these two series. We check the hypotheses: a k = ekb + p p k and d k = k are positive for all k since the numerator and denominator are built from positive numbers for all k. We check that lim a k d k e =lim kb + p k p e k =lim kb + p k p e k =lim kb p k + = 0+ since p k grows faster than e kb when b apple 0. Since this limit is a positive real number, it satisfies the second hypothesis. We may thus apply the Limit Comparison Theorem to conclude that either both series converge or both diverge. Finally, we apply the Ratio Test to P k= d k. We compute L =lim d p k+ d k =lim k+ p p k =lim k+ p k k+ = 0 = 0. Since L =0<, the Ratio Test implies that P k= d k converges. We concdlue that the original series also converges whenver b apple 0. Thus in conclusion, our original series converges for all values of b. (k+)!

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

Final a MATH 222 (Lectures,2, and 4) Fall 205. Name: Student ID#: Circle your TAs name: Carolyn Abbott Tejas Bhojraj Zachary Carter Mohamed Abou Dbai Ed Dewey Jale Dinler Di Fang Bingyang Hu Canberk Irimagzi