Solutions to MATH 271 Test #3H

Transcription

1 Solutions to MATH 71 Test #3H This is the :4 class s version of the test. See pages 4 7 for the 4:4 class s. (1) (5 points) Let a k = ( 1)k. Is a k increasing? Decreasing? Boune above? Boune k below? Convergant (an if so, to what?) Solution: The sequence a k oes not come from a nice real function, so we write out a few terms, starting with k = 1: 1, 1, 1 3, 1 4, 1, etc. This sequence is 5 not increasing (since a < a 1 ), an it is not ecreasing, either (since a 3 > a ). The sequence a k is boune below (a k 1, for all k) an boune above (a k 1, for all k). Lastly, lim a k =, so the sequence converges to. k + Graing: +5 points for each right answer; + points for each wrong answer; + points if a question wasn t answere. () (1 points) Fin the slope of the tangent line to the curve r = θ + 1 at θ = π. Note that these are polar coorinates. Solution: This is a polar coorinates question, so you must parametrize the curve with respect to θ. y x = y/θ x/θ = (r sin θ) θ (r cos θ) θ = ((θ + 1) sin θ) θ ((θ + 1) cos θ) θ You coul also have use the polar coorinate-specific formulas to get to this point. y ((θ + 1) sin θ) x = θ (θ + 1) cos θ + sin θ = (θ + 1) sin θ + cos θ ((θ + 1) cos θ) θ (π + 1) ( 1) + = (π + 1) + ( 1) = π + 1. Graing for partial creit: +3 points (total) for taking the erivative of θ + 1; +5 points (total) for the correct answer but no work. 1

2 (3) Consier the parametric curve { x = sin t y = sin t }, where t π/. (a) (1 points) Sketch the curve that these parametric equations represent. Solution: This can actually be one without a graphing calculator. This is because y = (sin t) = x, so the curve is part of the graph of y = x. Which part is etermine by the starting an ening values of t. When t =, (x, y) = (, ), an when t = π/, (x, y) = (1, 1). As t increases from to π/, the points trace out go from (, ) to (1, 1) without oubling back. So the graph looks like: y (, ).. (1, 1).. x (b) (1 points) Set up, but o not evaluate, an integral which gives the length of this curve. Solution: The length of the curve is s that has x/t an y/t in it: s, an here we nee to use the version of s = s = π/ (x ) + t ( ) y π/ t = cos t + ( sin t cos t) t t. Graing: +3 points for the basic formula, +4 points for the proper form of s, +3 points for substituting. (c) (1 points) Set up, but o not evaluate, an integral which gives the surface area of the surface obtaine by rotating this curve aroun the x-axis. Solution: Since the curve is being rotate aroun the x-axis, the raius is y, an the surface area is πy s = π/ π sin t cos t + ( sin t cos t) t.

3 (4) Consier the series ( ln(k + 1) ln k). k=1 (a) (15 points) Fin the secon, thir, an fourth partial sums of this series. Solution: If we let a k = ln(k + 1) ln k, then these sums are s = a 1 + a = ( ln ln 1) + ( ln 3 ln ) = ln 3 s 3 = a 1 + a + a 3 = ( ln ln 1) + ( ln 3 ln ) + ( ln 4 ln 3) = ln 4 s 4 = a 1 + a + a 3 + a 4 = ( ln ln 1) + ( ln 3 ln ) + ( ln 4 ln 3) + ( ln 5 ln 4) = ln 5 Graing: +5 points for each. Graing for partial creit: 5 points for the general form, without substituting for a k ; +5 points (total) for just items of the sequence. (b) (1 points) Base on your answer to (a), oes the series converge? If so, to what? Solution: It appears that s n = ln(n + 1) (in fact this coul be proven by inuction or noticing that the series is a telescoping series), an the limit of s n as n approaches infinity is +, so the series oes not converge to any real number. Thus the series iverges. Graing: +5 points for recognizing the pattern, +5 points for answering the question. Graing for partial creit: +9 points (total) for missing part (a), but eucing correctly from those (incorrect) answers. (5) (1 points) Set up, but o not { evaluate, an } integral that gives the area of the region x = t 3 boune above by the curve, where t is a real number, an boune y = (1 t ) 4 below by the x-axis. A sketch of the curve is given below. Solution: The area of a region below a parametrize curve an above the x-axis is y x t, where LEFT is the value of t that gives the left-han bounary, LEFT t an RIGHT is efine similarly. We nee to fin these. The parametrize curve crosses the x-axis when y =, or when (1 t ) 4 =. This happens when 1 t =, or when t = ±1. When t = 1, the point on the curve is (x( 1), y( 1)) = ( ( 1) 3, ) = (1, ); when t = 1, the point on the curve is (x(1), y(1)) = ( 1, ). Thus, LEFT = 1 an RIGHT = 1, so the area is 1 y x LEFT t t = (1 t ) 4 ( 3t ) t. 1 Graing: +4 points for fining the left an right bounaries, +3 points for the basic area formula, +3 points for substitution. Graing for common mistakes: points for 1 1 ; 7 points for no work. 3

4 Solutions to MATH 71 Test #3J (1) (5 points) Let a k = ln k. Is a k increasing? Decreasing? Boune above? Boune below? Convergant (an if so, to what?) Solution: The sequence a k comes from the function f(x) = ln x. Since f(x) is strictly increasing (f (x) = 1/x > ), a k is increasing an not ecreasing. The sequence a k is boune below (a k, for all k) but not boune above. Lastly, it grows without boun, so the sequence iverges. Graing: +5 points for each right answer; + points for each wrong answer; + points if a question wasn t answere. AFTER CONSIDERING WHAT INCREASING AND DECREASING MEAN, SEMANTICALLY, I HAVE DECIDED TO GIVE EVERYONE AN EXTRA +5 POINTS. () (1 points) Fin the slope of the tangent line to the curve r = 3θ at θ = π. Note that these are polar coorinates. Solution: This is a polar coorinates question, so you must parametrize the curve with respect to θ. y x = y/θ x/θ = (r sin θ) θ (r cos θ) θ = ((3θ ) sin θ) θ ((3θ ) cos θ) θ You coul also have use the polar coorinate-specific formulas to get to this point. y ((3θ ) sin θ) x = θ (3θ ) cos θ + 3 sin θ = (3θ ) sin θ + 3 cos θ ((3θ ) cos θ) θ ( ) 3(π) = ( = 6π. 3(π) ) Graing for partial creit: +3 points (total) for taking the erivative of θ + 1; +5 points (total) for the correct answer but no work. 1

5 (3) Consier the parametric curve { x = cos t y = cos 3 t }, where t π/4. (a) (1 points) Sketch the curve that these parametric equations represent. Solution: This can actually be one without a graphing calculator. This is because y = (cos t) 3 = x 3, so the curve is part of the graph of y = x 3. Which part is etermine by the starting an ening values of t. When t =, (x, y) = (1, 1), an when t = π/4, (x, y) (.77,.353). As t increases from to π/4, the points trace out go from (1, 1) to (.7,.353) without oubling back. So the graph looks like:. y. (.77,.353) (1, 1).. x (b) (1 points) Set up, but o not evaluate, an integral which gives the length of this curve. Solution: The length of the curve is s that has x/t an y/t in it: s, an here we nee to use the version of s = s = π/4 (x ) + t ( ) y π/4 t = ( sin t) + (3 cos t t sin t) t. Graing: +3 points for the basic formula, +4 points for the proper form of s, +3 points for substituting. (c) (1 points) Set up, but o not evaluate, an integral which gives the surface area of the surface obtaine by rotating this curve aroun the y-axis. Solution: Since the curve is being rotate aroun the y-axis, the raius is x, an the surface area is πx s = π/4 π sin t ( sin t) + (3 cos t sin t) t.

6 (4) Consier the series (e (k+1) e k ). k=1 (a) (15 points) Fin the secon, thir, an fourth partial sums of this series. Solution: If we let a k = e (k+1) e k, then these sums are s = a 1 + a = (e e 1 ) + (e 3 e ) 9.9 s 3 = a 1 + a + a 3 = (e e 1 ) + (e 3 e ) + (e 4 e 3 ) s 4 = a 1 + a + a 3 + a 4 = (e e 1 ) + (e 3 e ) + (e 4 e 3 ) + (e 5 e 4 ) Graing: +5 points for each. Graing for partial creit: 5 points for the general form, without substituting for a k ; +5 points (total) for just items of the sequence. (b) (1 points) Base on your answer to (a), oes the series converge? If so, to what? Solution: It appears that s n approaches as n approaches infinity, Thus the series iverges. Graing: +5 points for recognizing the pattern, +5 points for answering the question. Graing for partial creit: +9 points (total) for missing part (a), but eucing correctly from those (incorrect) answers. BECAUSE THE SERIES WAS SUPPOSED TO BE (e (k+1) e k ), A TELESCOPING SERIES, BUT k=1 I MADE A TYPO, I GAVE 1 BONUS POINTS TO EVERYONE. (5) (1 points) Set up, but o not evaluate, an integral that gives the area of the region x = tan t boune above by the curve ( ) π y = 16, where t is a real number, an t boune below by the x-axis. A sketch of the curve is given below. Solution: The area of a region below a parametrize curve an above the x-axis is y x t, where LEFT is the value of t that gives the left-han bounary, LEFT t an RIGHT is efine similarly. We nee to fin these. The parametrize curve crosses the x-axis when y =, i.e., when ( π 16 t 3 ) =. =

7 This happens when π 16 t =, or when t = ±π/4. When t = π/4, the point on the curve is (x( π/4), y( π/4)) = ( tan( π/4), ) = (1, ); when t = π/4, the point on the curve is (x(π/4), y(π/4)) = ( 1, ). Thus, LEFT = π/4 an RIGHT = π/4, so the area is LEFT y x π/4 ( ) π t t = π/4 16 t ( sec t) t. Graing: +4 points for fining the left an right bounaries, +3 points for the basic area formula, +3 points for substitution. Graing for common mistakes: points for π/4 π/4 ; 7 points for no work. 4