Calculus III - Exam 2- Answers. Summer A 2013

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1 Calculus III - Exam 2- s. Summer A 2013 Instructions: Please show all your work in an organized manner. If the valid steps (reasoning) are not shown, the final answer does not carry any credits. Write your answers on a separate sheet/s indicating the correct part and question number. Calculators are only allowed for basic arithmetics. Once you choose a question you must complete all the subparts. Complete all 5 parts. [Duration: 2 hours] Part I. one question from part I 1. Water traveling along a starlight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at the banks. Consider a long straight stretch of river flowing north, with parallel banks 60 m apart. If the maximum water speed is 4 m/s, the speed of flow x units from the west bank can be model by f(x) = (x 60)x Suppose we would like to pilot a boat to land at the point B on the east bank directly opposite starting point A. If we maintain a constant speed of 15 m/s and a constant heading, find the angle at which the boat should head. Velocity of boat on calm water u(t) = 15cos(θ)i+15sin(θ)j Velocity of boat on river v(t) = 15cos(θ)i+[15sin(θ) (1/225)[15cos(θ)t 60](15cos(θ)t)]j v(t) = 15cos(θ)i+[15sin(θ) cos 2 (θ)t 2 +4cos(θ)t]j Position vector of boat. Obtained using integration and using initial point r(0) = 0i+0j r(t) = 15cos(θ)ti+[15sin(θ)t cos 2 (θ)t 3 /3+2cos(θ)t 2 ]j Since the horizontal speed is 15cos(θ) and horizontal distance of travel is 60 m, time to reach point B is 60 t = (15cos(θ)) = 4 cos(θ) We want the y component of r(t) to be zero when t = 4 cos(θ) y(t) = [15sin(θ)( cos(θ) ) cos2 (θ)( cos(θ) )3 /3+2cos(θ)( cos(θ) )2 ] = 0 4 y(t) = ( cos(θ) )[15sin(θ) 16/3+8] = 0 15sin(θ) 16/3+8 = 0 θ = deg Boat should keep head of deg with positive x axis to reach point B.

2 2. An amusement park ride consists of large flat, horizontal wheel. Customers board the wheel while it is stationary and try to stay on as long as possible as it begins to rotate. Suppose the wheel rotates at constant angular speed of ω radians per second and that a volunteer weighing M lbs stands 15 ft from the center of the wheel, as shown. (a) Find the position vector of volunteer assuming counterclockwise rotation and initial point (15, 0). (b) Find a T and a N (c) Find centripetal force F N (in terms of ω and M) required to stay on the disk. (d) If the frictional force of the volunteer is 0.12M, find the largest value of ω that will allow the volunteer to stay in place? (a) Position vector of volunteer r(t) =< 15cos(ωt),15sin(ωt) > (b) v(t) = r (t) =< 15ωsin(ωt),15ωcos(ωt) > a(t) = v (t) = r (t) =< 15ω 2 cos(ωt), 15ω 2 sin(ωt) > Since the disk rotate at constant speed a T = 0 and a N = a(t). a N = (15ω 2 ) 2 (cos 2 ωt+sin 2 ωt) = 15ω 2. (c) F N = ma N = M 32 15ω2 (d) 0.12M == M 32 15ω2 max = ωmax ω max = 15

3 Part II. two questions from part II 1. Given sin(x 2 +y 2 ) f(x,y) = x 2 +y 2, for (x,y) (0,0) 1, for (x,y) = (0,0) (a) Find f(x,y), if exists. (b) Determine the set of points which the function f(x, y) is continuous. (a) Let x = rcos(θ), y = rsin(θ) Then f(x,y) = r 0 sin(r 2 ) r 2 = 1 (b) x 2 +y 2 is polynomial and continuous everywhere. sin(x 2 +y 2 ) is a continuous function of a continuous function, which is continuous everywhere. So sin(x2 +y 2 ) x 2 +y 2 is continuous everywhere except possibly at (0, 0). But f(x,y) = f(0,0). So f(x,y) is continuous everywhere. 2. Use it definition to show x 2 +y 2 = 0 Show for any given ǫ > 0 there exists δ > 0 such that x 2 +y 2 0 < ǫ whenever 0 < x 2 +y 2 < δ consider x 2 +y 2 0 < 3x 2 y x 2 +y 2 < y 3x2 x 2 = 3y = 3 y 2 <= 3 Let δ = ǫ/3 then for any given ǫ > 0 there exists δ > 0 such that x 2 +y 2 0 < ǫ whenever 0 < x 2 +y 2 < δ = ǫ/3 3. Find the it, if it exists, or show that it does not exist. Path x = y 2 y 0 y 2 y 2 (y 2 ) 2 +y 4 y 0 y 4 2y 4 = 1 2 xy 2 x 2 +y 4 y 2 +x 2 < ǫ

4 Path y = 0 0 x 0 x 2 = using L Hpital 0 s = x 0 2x = using L Hpital 0 s = x 0 2 = 0 Since different paths gives different its. Limit does not exist. Part III. one questions from part III 1. Show that the production function P(l,k) = l α k β, where α and β are constants, satisfies P = αk β l α 1 l P k = βlα k β 1 l P l +k P k = (α+β)p l P l +k P k = l(αkβ l α 1 )+k(βl α k β 1 ) = (α+β)(l α k β ) 2. Find the partial derivatives x and y sin(x 2 yz)+ln(x+y +z) = 3 Let F(x,y,z) = sin(x 2 yz)+ln(x+y +z) 3 F = x2 ycos(x 2 1 yz)+ x+y +z F x = 1 2xyzcos(x2 yz)+ x+y +z F y = x2 zcos(x 2 1 yz)+ x+y +z Then x = F x = 2xyzcos(x2 yz)+ 1 F z x 2 ycos(x 2 yz)+ 1 y = F y = x2 zcos(x 2 yz)+ 1 F z x 2 ycos(x 2 yz)+ 1 Part IV. one question from part IV 1. The period T of a pendulum of length L is T = 2π L g, where g is the acceleration due to gravity. A pendulum is moved from the Central Zone, where g = 32 ft/s 2, to Greenland, where g = ft/s 2. Because of the change in temperature, the length of the pendulum changes from 2 ft to 1.97 ft. Use total differential to approximate the change in the period of the pendulum.

5 g = 32 and g = 0.13 L = 2 and L = 0.03 T L = 2π1 2 (L/g) 1/21 g = π gl T L g = 2π1 2 (L/g) 1/2 L( 1)g 2 = π g 3/2 dt = T T L+ L g g π L dt = ( ) L+( π gl g 3/2) g π 2 dt = ( )( 0.03)+( π /2)(0.13) = An annular cylinder has an inside radius of r 1 and an outside radius of r 2. Its moment of inertia is I = M 2 (r2 1 +r2 2 ) where M is the mass. Two radii are increasing at of 2 m/s. Assume mass M is constant. (a) Write a formula for rate of change of moment of inertia. (b) Find the rate at which I is changing at the instant the radii are r 1 = 6 m and r 2 = 8 m. (a) I t = I r 1 r 1 t + I r 2 r 2 t I = M r 1 2 2r 1 = Mr 1 I = M r 2 2 2r 2 = Mr 2 r 1 t = 2 r 2 t = 2 I t = Mr 12+Mr 2 2 = 2M(r 1 +r 2 ) (b) I = 2M(6+8) = 28M t Part V. one question from part V 1. Use Chain rule 2 to find r x = 2rs, y = sin(r +s). r = x x r + y y r 4x and, where z = s 2 y

6 x = 4x ln(4) 2 y y = 4 x (2 y) 2 x r = 2s x s = 2r y = cos(r +s) r y = cos(r +s) s r = ln(4) (4x 2 y )2s+( 4 x (2 y) 2)cos(r+s) = ( 4 2rs ln(4) 2 sin(r +s) )2s+( 4 2rs (2 sin(r +s)) 2)cos(r+ s) s = ln(4) (4x 2 y )2r+( 4 x (2 y) 2)cos(r+s) = ( 4 2rs ln(4) 2 sin(r +s) )2r+( 4 2rs (2 sin(r +s)) 2)cos(r+ s) 2. Find 2 z s2, if z = f(x,y)+g(y), where x = t 3s and y = 2t+s. s = x x s + y y s s = [f x +0]( 3)+[f y +g y ](1) Let P = s = [f x +0]( 3)+[f y +g y ](1) 2 z s 2 = P s = P x x s + P y y s 2 z s 2 = P s = [ 3f xx +f yx +0]( 3)+[ 3f xy +f yy +g yy ](1)