HW 8 Gagan Solutions

HW 8 Gagan Solutions - Nanda, Gagan – Homework 8 – Due:...

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Unformatted text preview: Nanda, Gagan – Homework 8 – Due: Oct 14 2004, 3:00 am – Inst: R Gompf 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the rate at which the surface area of a cube is changing with respect to its side length x when x = 4 cm. 1. rate = 16 π cm 2 / cm 2. rate = 32 π cm 2 / cm 3. rate = 12 cm 2 / cm 4. rate = 48 cm 2 / cm correct 5. rate = 8 cm 2 / cm Explanation: For a cube, surface area = 6 x 2 . Now the rate at which the surface area is changing with respect to x is the derivative with respect to x . Thus rate = 12 x. Consequently, rate = 48 cm 2 / cm . 002 (part 1 of 1) 10 points A ladder 10 feet long rests against a vertical wall. Let θ be the angle between the top of the ladder and the wall and let x be the distance from the bottom of the ladder to the wall. If the bottom of the ladder slides away from the wall, how fast does x change with respect to θ when θ = π 3 ? 1. 4 ft/rad 2. 8 ft/rad 3. 3 ft/rad 4. 1 ft/rad 5. 5 ft/rad correct Explanation: x θ 10 From the diagram we can see that sin θ = x 10 x = 10 sin θ We want to find the rate of change of x with respect to θ , that is dx dθ : dx dθ = 10 cos θ So when θ = π 3 , dx dθ = 10 cos π 3 = 10 µ 1 2 ¶ = 5 ft / rad 003 (part 1 of 1) 10 points Determine the value of dy/dt at x = 2 when y = 3 x 3- 4 x and dx/dt = 2. 1. dy dt fl fl fl x =2 = 63 2. dy dt fl fl fl x =2 = 65 3. dy dt fl fl fl x =2 = 66 4. dy dt fl fl fl x =2 = 62 Nanda, Gagan – Homework 8 – Due: Oct 14 2004, 3:00 am – Inst: R Gompf 2 5. dy dt fl fl fl x =2 = 64 correct Explanation: Differentiating implicitly with respect to t we see that dy dt = ( 9 x 2- 4 ) dx dt = 2 ( 9 x 2- 4 ) . At x = 2, therefore, dy dt = 2(32) = 64 . 004 (part 1 of 2) 10 points A point is moving on the graph of 5 x 3 + 4 y 3 = xy. When the point is at P = ‡ 1 9 , 1 9 · , its x-coordinate is decreasing at a speed of 5 units per second. (i) What is the speed of the y-coordinate at that time? 1. speed y-coord =- 11 units/sec 2. speed y-coord = 9 units/sec 3. speed y-coord = 11 units/sec 4. speed y-coord =- 10 units/sec 5. speed y-coord = 10 units/sec correct Explanation: Differentiating 5 x 3 + 4 y 3 = xy implicitly with respect to t we see that 15 x 2 dx dt + 12 y 2 dy dt = y dx dt + x dy dt . Thus dy dt = ‡ 15 x 2- y x- 12 y 2 · dx dt . Now at P , 15 x 2- y = ‡ 15 (9) 2- 1 9 · = 1 (9) 2 (6) , while x- 12 y 2 = ‡ 1 9- 12 (9) 2 · =- 1 (9) 2 (3) . Hence, at P , dy dt =- 2 dx dt . When the x-coordinate at P is decreasing at a rate of 5 units per second, therefore, dy dt = 5 ‡ 2 · , so the speed y-coord = 10 units/sec....
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This note was uploaded on 02/28/2010 for the course M 56200 taught by Professor Radin during the Fall '09 term at University of Texas.

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HW 8 Gagan Solutions - Nanda, Gagan – Homework 8 – Due:...

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