buckley (jdb4687) - HW06 - beshaj - (54110) This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page - find all choices before answering. 001 1. y = 1 8/x e 2 2. y = 1 x/8 e 2 3. y = 1 4/x e 4 4. y = 1 x/4 e 2 5. y = 1 x/4 e 4 6. y = 1 x/2 e 4 10.0 points Determine A so that the curve y = 9x + 44 can be written in parametric form as x(t) = t 5 , y(t) = At 1 . 1. A = 8 2. A = 9 1 3. A = 8 004 4. A = 9 Find parametric equations representing the line segment joining P (1, 4) to Q(4, 2) as 5. A = 10 6. A = 10 (x(t), y(t)) , 002 10.0 points Find a Cartesian equation for the curve given in parametric form by x(t) = 4 cos 3t , 1. 2. 25x2 + 16y 2 = 400 3. 16x + 25y 2 = 400 x2 y 2 1 4. = 25 16 400 2 2 x y 1 5. = 16 25 400 6. 25x2 16y 2 = 400 003 0 t 1, where P = (x(0), y(0)) , Q = (x(1), y(1)) y(t) = 5 sin 3t . 1 x2 y 2 + = 25 16 400 2 10.0 points and x(t), y(t) are linear functions of t. \u0010 \u0011 2 t 2 t 1. 1 + 5 sin , 2 + 6 cos 2 2 2. (4 5t2 , 2 + 6t2 ) 3. \u0010 4 5 sin2 t \u0011 t , 4 6 cos2 2 2 4. (4 5t, 2 + 6t) 5. (1 + 5t2 , 4 6t2 ) 10.0 points Determine a Cartesian equation for the curve given in parametric form by x(t) = 2 ln(16t) , y(t) = t . 6. (1 + 5t, 4 6t) 005 10.0 points buckley (jdb4687) - HW06 - beshaj - (54110) Find the path (x(t), y(t)) of a particle that moves once counter-clockwise around the curve x2 + (y 2)2 = 16 , 4 0t 2. (4 cos t, 2 + 4 sin t), 0 t 2 3. (4 cos t, 2 4 sin t), 0 t 2 4. ( 4 cos t, 2 4 sin t), 5. (4 cos t, 2 4 sin t), u2 t 6. ( 4 cos t, 2 4 sin t), 006 0 t 2 x2 y 2 + = 1 25 16 in parametric form (x(t), y(t)). 1. (5 sin t, 4 cos t), 0 t 2 2. (5 sin t, 4 cos t), 0t 2 3 0t 4. (5 cos 2t, 4 sin t), 0 t 2 5. (5 sin 2t, 4 cos t), 0t 10.0 points Locate the points given in polar coordinates by 2 \u0011 P 2, , 3 among 1 \u0011 Q 2, , 4 \u0010 2 \u0010 1 \u0011 R 1, 2 c b 2 4 4 1. P : b t Q: u c R: b 2. t P : u Q: b c R: b 3. P : b c Q: b t R: u 4. c P : b Q: b t R: u 5. t P : u c Q: b R: b 6. c P : b t Q: u R: b 008 3. (5 sin 2t, 4 cos 3t), \u0010 u b 2 10.0 points Find an equation for the ellipse 007 4 0t 0t r rs starting at (4, 2). 1. (4 cos t, 2 + 4 sin t), 2 10.0 points Find the Cartesian coordinates, (a, b), of the point given in polar coordinates by P (2, /3). 1. (a, b) = ( 3, 1) 2. (a, b) = (1, 3) 3. (a, b) = (2, 3) 4. (a, b) = (2 3, 2) 5. (a, b) = (1, 2) 6. (a, b) = (2, 2 3) 7. (a, b) = ( 3, 1) buckley (jdb4687) - HW06 - beshaj - (54110) 8. (a, b) = (1, 009 3 at = /4. 3) 1. slope = e/4 + 1 10.0 points Find a polar equation for the curve given by the Cartesian equation y = 4x2 . 1. r = 4 csc cot 2. 4r = sec cot 2. slope = 3. slope = 1 1 e/4 1 e/4 + 1 4. slope = e/4 5. slope = e/4 3. r = 4 sec tan 4. 4r = csc cot 6. slope = e/4 1 012 5. r = 4 csc tan 6. 4r = sec tan 010 10.0 points Find a Cartesian equation for the curve given by the polar equation r + 2 sin = 0 . 1. x2 + (y + 1)2 + 1 = 0 2. (x + 1)2 + y 2 = 1 3. x2 + (y 1)2 = 1 4. x2 + (y 1)2 + 1 = 0 5. (x 1)2 + y 2 + 1 = 0 6. (x + 1)2 + y 2 + 1 = 0 7. (x 1)2 + y 2 = 1 Find the slope of the tangent line to the graph of r = 1 2 sin at = /6. 1 3 3 32 2. slope = 1+2 3 3. slope = 2 3 1. slope = 4. slope = 3 3 32 5. slope = 2 31 3+2 6. slope = 1+2 3 8. x2 + (y + 1)2 = 1 011 10.0 points Find the slope of the tangent line to the graph of r = e 2 10.0 points 013 10.0 points Find an equation for the tangent line to the graph of r = 3 cos sin at = /4. buckley (jdb4687) - HW06 - beshaj - (54110) at the point P = (x(1), y(1)). 2 1. y = 3x 3 2. y = 2 1 x+ 3 3 4 3 3. y = 3x + 4. y = 1. y = 3 x+7 2e 2. y = 3 x5 e 3 3. y + x = 7 e 4 1 x+ 3 3 1 2 5. y = x 3 3 4. y = 2 6. y = 3x + 3 5. y + 014 Find 3 x3 2e 3 x = 4 2e 3 6. y + x = 5 e 10.0 points dy when dx 016 x(t) = t ln t , 1. 4 y(t) = sin7 t . 10.0 points Find the x-intercept of the tangent line at P (4, 2) to the graph of the curve defined parametrically by 1 + ln t dy = dx 7 sin6 t cos t x(t) = 8 cos 2t , y(t) = 4 sin t . dy 1 + ln t 2. = dx 7 cos6 t sin t 1. x-intercept = 12 dy 7 sin6 t cos t 3. = dx 1 + ln t 2. x-intercept = 6 6 4. 6 sin t cos t dy = dx 1 + ln t 5. dy 1 + ln t = dx 6 sin6 t cos t 3. x-intercept = 12 4. x-intercept = 3 5. x-intercept = 3 dy 6 cos6 t sin t 6. = dx 1 + ln t 7. 6. x-intercept = 6 1 + ln t dy = dx 6 cos6 t sin t 017 d2 y Find 2 for the curve given parametrically dx by 7 cos6 t sin t dy = 8. dx 1 + ln t 015 10.0 points x(t) = 1 + 2t2 , Find an equation for the tangent line to the curve given parametrically by x(t) = 2e t , 10.0 points y(t) = t ln t 4 3 d2 y = 1. 2 dx 2t y(t) = t2 + 2t3 . buckley (jdb4687) - HW06 - beshaj - (54110) d2 y 3t = 2 dx 8 2. t = , 3 6 3 d2 y = 3. 2 dx 8t 3. t = , 4 3 2. 4. t = , 3 3 d2 y 5 4. = dx2 3t 5. t = , 6 6 5t d2 y = 5. 2 dx 3 6. t = , 4 4 d2 y 4t 6. = 2 dx 3 018 5 020 10.0 points Determine all values of t for which the curve given parametrically by x = t3 3t2 + 1 , y = 3t3 + t2 2t has a vertical tangent? 10.0 points The curve traced out by a point P on the circumference of a circle as the circle rolls along a straight line shown in y P 2 1. t = 0 , 9 2 2. t = 0 , 9 x 3. t = 2 is called a Cycloid and the shaded region is the region, A, below an arch. If the circle has radius R, the cycloid is given parametrically by 4. t = 2 5. t = 2 9 6. t = 0 , 2 x(t) = R(t sin t) , 2 7. t = 9 y(t) = R(1 cos t) . Find the area of A when R = 4. 8. t = 0 , 2 1. area(A) = 46 019 10.0 points Find all values of t, /2 < t < /2, for which the tangent line to the graph of x(t) = 4t + 3 tan t , is vertical. 1. no values of t y(t) = 4t 3 tan t , 2. area(A) = 44 3. area(A) = 48 4. area(A) = 47 5. area(A) = 45 021 10.0 points buckley (jdb4687) - HW06 - beshaj - (54110) Which one of the following integrals gives the length of the parametric curve x(t) = 2t2 , Z 1. I = 2 0 8p 4 2. I = Z 3. I = Z 8p 8 4. I = 2 Z 4 5. I = 2 Z 0 6. I = 0 4. length = 5 sinh(1) feet 5. length = 10 cosh(1) feet 6. length = 5 cosh(1) feet 16t2 + 1 dt 023 10.0 points The shaded region shown in 16t2 + 1 dt y 3 sinh t |16t2 + 1| dt 0 Z 0 t 8. |16t2 + 1| dt 0 0 y(t) = t , 6 |16t2 + 1| dt 4p 16t2 + 1 dt 022 x 10.0 points 3 A cable suspended between two towers as shown in 3 cosh t is bounded by the x-axis, the upper half of the curve given parametrically by x(t) = 3 cosh t , y(t) = 3 sinh t , and the line x = 3 cosh t. Find the area of this shaded region. 1. area = 9 t 2 forms a catenary modeled by 2. area = 9 (sinh 2t + t) 4 x y = 5 cosh , 5 3. area = 9 9 sinh 2t + t 4 2 4. area = 9 (sinh 2t) t 2 5. area = 9 9 sinh 2t t 4 2 5 5 5 x 5 , where x and y are measured in feet. Find the length of the suspended cable. 1. length = 5e feet 2. length = 10e feet 3. length = 10 sinh(1) feet