A ladder 36 feet long is leaning against a vertical wall. If the bottom of the ladder is pulled away from the base of the building at a rate of 3 ft/sec, how fast is the top of the ladder sliding down the wall when it is 28 feet above the ground? Round answer to nearest tenth.
A) 2.4 ft/sec
B) 2.2 ft/sec
C) 2.0 ft/sec
D) 1.8 ft/sec

Question

Quizplus · Accepted Answer

Using the Pythagorean theorem, the distance of the bottom of the ladder from the wall is sqrt(36^2 - 28^2) = sqrt(1296 - 784) = sqrt(512) = 22.6 feet. Differentiating both sides of the Pythagorean theorem with respect to time gives 2x(dx/dt) + 2y(dy/dt) = 0, where x is the distance from the wall, y is the height of the ladder on the wall, dx/dt = 3 ft/sec (rate at which the bottom is moving away from the wall), and dy/dt is what we're solving for. Plugging in the values gives 2(22.6)(3) + 2(28)(dy/dt) = 0. Solving for dy/dt gives dy/dt = -2.4 ft/sec, indicating the top of the ladder is sliding down at 2.4 ft/sec.

A Ladder 36 Feet Long Is Leaning Against a Vertical