[Related Rates] A ladder 20 feet long leans up against a house. The bottom of the ladder starts to slip away from the house at 0.17 feet per second.

Question

A ladder 20 feet long leans up against a house. The bottom of the ladder starts to slip away from the house at 0.17 feet per second. How fast is the tip of the ladder along the side of the house slipping when the ladder is 7.2 feet away from the house? (Round to 3 decimal places.)   Consider the angle the bottom of the ladder makes with the ground.

How fast is the angle changing (in radians) when the ladder is 7.2 feet away from the house?

Answer 1

The wall, the floor and the ladder have made a right triangle. Therefore we can use the Pythagorean Theorem to solve the problem:

r² = x² + y²

Since the length of the ladder (r) is fixed, we have to put it in the equation as constant.

20² = x² + y²

400 = x² + y²

At x = 7.2 ft., the value of y is:

400 = (7.2)² + y²

y² = 400 - (7.2)²

y ≈ 18.659 ft.

Derive the equation with respect to time (t):

0 = 2x (dx/dt) + 2y (dy/dt)

Given that dx/dt = 0.17 ft/sec, we can now solve dy/dt which is the rate of the tip of the ladder along the side of the house slipping down.

0 = 2(7.2) (0.17) + 2(18.659...) (dy/dt)

-2(18.659...) (dy/dt) = 2(7.2) (0.17)

dy/dt. ≈ -0.0656 ft/sec

(It's negative because the direction of the tip of the ladder is going down)