A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position, as shown in the figure.

Question

The winch pulls in rope at a rate of −0.1meter per second (ds/dt = −0.1 m/sec).

Find the rate of vertical change and the rate of horizontal change at the end of the pipe when y = 4. (Round your answers to three decimal places.)

horizontal ?m/sec

vertical ?m/sec

Answer 1

For the bottom right triangle, applying the Pythagorean Theorem yields x² + y² = 12² or x² = 144 - y²

For the upper right triangle, applying the Pythagorean Theorem yields x² + (12 - y)² = s² and substituting "144 - y²" for x² (from the bottom right triangle), we get:

(144 - y²) + (12 - y)² = s²

144 - y² + 144 - 24y + y² = s²

288 - 24y = s²

4(72 - 6y) = s²

s = √[4(72 - 6y)]

s = 2√(72 - 6y)

s = 2(72 - 6y)^1/2

Taking the derivative with respect to time "t" we get:

ds/dt = 1/2(2)(72 - 6y)^-1/2(-6)(dy/dt)

ds/dt = -6(dy/dt)/(72 - 6y)^1/2

(ds/dt)(72 - 6y)^1/2/(-6) = dy/dt

Plugging in s = -0.1 and y = 4 we get:

dy/dt = (-0.1)(72 - 6(4))^1/2/(-6)

dy/dt = (-0.1)(72 - 24)^1/2/(-6)

dy/dt = (0.1)(48)^1/2/6

dy/dt ≈ 0.11547 or, rounding: dy/dt = 0.115 m/sec

To find dx/dt. let's go back to x² + y² = 144 and take the derivative implicitly:

2x•dx/dt + 2y•dy/dt = 0

x•dx/dt + y•dy/dt = 0

x•dx/dt = -y•dy/dt

dx/dt = -y/x•dy/dt

If y = 4, we can solve for x to get x = √(144 - 4²) = 11.3137 so:

dx/dt = -4/11.3137•( 0.11547) = -0.04082 or, rounding: dx/dt = -0.041 m/sec