Fall 2013 Hour Exam 2 11/08/13 Time Limit: 50 Minutes

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1 Math 8 Fall Hour Exam /8/ Time Limit: 5 Minutes Name (Print): This exam contains 9 pages (including this cover page) and 7 problems. Check to see if any pages are missing. Enter all requested information on the top of this page. You may not use your books, notes, or any electronic device including cell phones. The following rules apply: Mysterious or unsupported answers will not receive full credit. A correct answer, unsupported by calculations, explanation, or algebraic work will receive little to no credit; an incorrect answer supported by substantially correct calculations and explanations will receive partial credit. You must support your answers in all limit problems by a calculation or a brief explanation. For each series, you must clearly indicate which test you are using. You must also provide a proper conclusion. Problem Points Score Total:

2 Math 8 Hour Exam - Page of 9 /8/. ( points) Let C be the curve given in polar coordinates by r = tan θ for θ < π. (a) Express C as an equation of the form y = f(x). (b) Sketch the curve C. (c) Find the area of the region bounded by C and the line θ = π 4. (a) Squaring both sides of the equation yields r = tan θ. We now replace r with x + y and tan θ with y x and then simplify to obtain: r = tan θ x + y = y x x 4 + x y = y y x y = x 4 y ( x ) = x 4 y = y = x4 x x x (b) A sketch of the curve is shown below. y x (c) The area of the region bounded by r = tan θ and θ = π 4 is A = A = β α π/4 π/4 f(θ) dθ tan θ dθ A = (sec θ ) dθ A = ] π/4 tan θ θ A = ( π ) 4

3 Math 8 Hour Exam - Page of 9 /8/. ( points) Decide whether each of the following series converges conditionally, converges absolutely, or diverges. (a) k= ( ) k k + k (b) m= sin(m) (a) We test for absolute convergence by considering the series of absolute values: ( ) k k + k = k + k. (c) j= ( ) j j + j Let a k = k+. We use the Limit Comparison Test with the series b k k = k. The value of L in the LCT is L = lim k a k b k = lim k k+ k k = lim k k k + k = lim k + k =. Since < L < and k diverges (p-series with p = ), the series k+ k diverges by the LCT. Thus, the alternating series is not absolutely convergent. We now use the Alternating Series Test. The sequence a k is decreasing for k and lim a k = lim k k k + k =. Thus, the series converges and is conditionally convergent since it is not absolutely convergent. (b) The sequence a m = sin(m) has infinitely many positive and infinitely many negative terms. Thus, we test for absolute convergence by considering the series of absolute values: sin(m) = sin(m). Since sin(m) for all m and the series (p = > ), the series sin(m) sin(m) is absolutely convergent. (c) Let a j = ( )j j + j. Then, lim a j = lim j j ( ) j j + j = lim is a convergent p-series converges by the Comparison Test. Thus, the series j j ( )j j + j = lim j ( )j ( )j + which does not exist. The expression ( )j + tends to as j but ( )j alternates between and.] Since the limit is not zero, the series diverges by the Divergence Test.

4 Math 8 Hour Exam - Page 4 of 9 /8/. ( points) Let R be the region bounded by the curves y = x and y = x. (a) Determine where these curves cross and sketch the region R. (b) Find the area of R. (c) Find the volume of the solid obtained by revolving R around the vertical line x =. (a) The curves intersect when x = x. The solutions to the equation are x = and x =. The corresponding y-values are y = and y = 4, respectively. (b) The area of R is A = (x x ) = x x ] = = 4. (c) We calculate the volume using shells. The corresponding formula is V = b a π( x)(f(x) g(x)) where a =, b =, f(x) = x, and g(x) = x. Thus, the volume is V = π( x)(x x ) = π (6x 5x +x ) = π x 5 x + ] 4 x4 = 6π 4 y x

5 Math 8 Hour Exam - Page 5 of 9 /8/ 4. ( points) Find the interval of convergence of the power series n= (x + ) n 5 n n. We use the Ratio Test to find the interval of convergence. convergence we have r = lim n a n+ a n r = lim (x + ) n+ n 5 n+ (n + ) 5 n n (x + ) n ( ) n r = lim x + n 5 n + r = ( ) n x + lim 5 n n + }{{} = r = x + 5 Testing for absolute According to the Ratio Test, the series will converge when r = 5 x + <, i.e. x + < 5 x + < 5 5 < x + < 5 6 < x < 4 However, the test is inconclusive when r = 5 x + =, i.e. when x = 6 or x = 4. When x = 6, the power series becomes n= ( 6 + ) n 5 n n = n= ( 5) n 5 n n = ( ) n n. n= This is an absolutely convergent series because the series of absolute values convergent p-series. When x = 4, the power series becomes n is a n= This is a convergent p-series. (4 + ) n 5 n n = Thus, the interval of convergence is 6 x 4. (5) n 5 n n = n. n= n=

6 Math 8 Hour Exam - Page 6 of 9 /8/ 5. (5 points) Find the third order Taylor polynomial for f(x) = x centered at x = 4. f and its first three derivatives evaluated at x = 4 are f(x) = x / f( 4 ) = ( 4 )/ = f (x) = x / f ( 4 ) = ( 4 ) / = f (x) = 4 x / f ( 4 ) = 4 ( 4 ) / = f (x) = 8 x 5/ f ( 4 ) = 8 ( 4 ) 5/ = The third order Taylor polynomial of f centered at 4 is p (x) = f( 4 ) + f ( 4 )(x 4 ) + f ( 4 )! (x 4 ) + f ( 4 )! (x 4 ) p (x) = + (x 4 ) +! (x 4 ) +! (x 4 ) p (x) = + (x 4 ) (x 4 ) + (x 4 ) y 4 4 x Figure : Plots of f(x) = x (blue) and p (x) (red)

7 Math 8 Hour Exam - Page 7 of 9 /8/ 6. (5 points) Let C be the parametrized curve x = sin(t), y = sin(t) for t π. (a) Find an expression for dy as a function of t. (b) Find all the points of C where the tangent line is horizontal and all the points where it is vertical. (a) The derivatives dt and dy dt are dt = cos(t), dy dt = cos(t). Thus, the derivative dy in terms of t is dy dy = dt dt = cos(t) cos(t) = cos(t) cos(t). (b) The tangent line is horizontal when the derivative is zero, i.e. when cos(t) =. The solutions on the interval t π are t = π 4, π 4, 5π 4, 7π 4. The corresponding points on the curve C are (, ), (, ), (, ), (, ). The tangent line is vertical when the derivative is undefined, i.e. when cos(t) =. The solutions on the interval t π are t = π, π. The corresponding points on the curve C are (, ), (, ). y - - x -

8 Math 8 Hour Exam - Page 8 of 9 /8/ 7. ( points) Evaluate the following integrals: (a) (b) (c) 4 8 e x + 8x x 9x + 4 ln(x) x (a) The integration technique is partial fractions. follows: x + 8x = After clearing denominators we obtain A x + + The integrand may be decomposed as B x. = A(x ) + B(x + ). When x = we have B = and when x = we have A =. Thus, the integral becomes 4 4 ( 8 x + 8x = 8 x + + ) x = ln x + + ln x ] = ln(4) + ln() ] 4 8 ln(8) + ln(6) = ln() + ln(8) ln(4) ln(6) ( ) 8 = ln 4 6 ( ) = ln (b) The integration technique is the trigonometric substitution. Let x = tan θ. Then = sec θ. These substitutions yield the result x 9x + 4 = sec θ ( 9( tan θ) tan θ) + 4 = sec θ 4 9 tan θ 4 tan θ + 4 dθ = sec θ 4 9 tan θ sec θ dθ = sec θ 4 tan θ dθ = cos θ 4 sin θ dθ ]

9 Math 8 Hour Exam - Page 9 of 9 /8/ If we let u = sin θ and du = cos θ dθ then we obtain x 9x + 4 = cos θ 4 sin θ dθ = 4 u du = 4 ( ) + C = u 4 sin θ + C Since x = tan θ we have tan θ = x adjacent. If we draw a right triangle then we take x as the side opposite θ and as the side adjacent to θ. = opposite 9x + 4 x θ opposite x 9x +4 Thus, sin θ = hypotenuse = Theorem. Finally, the integral is (c) We begin by rewriting the integral as where the hypotenuse is obtained using Pythagoras x 9x + 4 = 9x 4 sin θ + C = + 4 4x e ln(x) x = e x ln(x). + C. Letting u = ln(x) and dv = x yields du = x and v = x. The integration by parts formula yields: u dv = uv v du x ln(x) = ( x ln(x) ) ( ) x x = x ln(x) + x = x ln(x) + ] x = x ln(x) 4 x The value of the definite integral on the interval, e] is e ln(x) x = x ln(x) ] e 4 x = e 4 e + 4 = 4 e + 4

Math 142, Final Exam. 12/7/10.

Math 142, Final Exam. 12/7/10. Math 4, Final Exam. /7/0. No notes, calculator, or text. There are 00 points total. Partial credit may be given. Write your full name in the upper right corner of page. Number the pages in the upper right