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 2 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) 1.6 ft/sec
B) 1.4 ft/sec
C) 1.2 ft/sec
D) 1.0 ft/sec

Question

Quizplus · Accepted Answer

Using the Pythagorean theorem, we find the distance of the bottom of the ladder from the wall when the top is 28 feet above the ground. Let $x$ be the distance from the wall to the bottom of the ladder, and $y$ be the height of the ladder on the wall. Given $y = 28$ feet and the ladder's length $l = 36$ feet, we have $x^2 + y^2 = l^2$. Substituting the given values, $x^2 + 28^2 = 36^2$, we find $x = 20$ feet. Differentiating both sides of $x^2 + y^2 = l^2$ with respect to time $t$, we get $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$. Given $\frac{dx}{dt} = 2$ ft/sec (the rate at which the bottom of the ladder moves away from the wall), we solve for $\frac{dy}{dt}$ (the rate at which the top of the ladder slides down the wall) when $x = 20$ feet and $y = 28$ feet. Substituting the values, we get $2(20)(2) + 2(28)\frac{dy}{dt} = 0$, which simplifies to $\frac{dy}{dt} = -\frac{80}{56} = -\frac{10}{7}$ ft/sec. The negative sign indicates the direction of motion (downward). The magnitude is approximately $1.4$ ft/sec, which, when rounded to the nearest tenth, is $1.4$ ft/sec.

A Ladder 36 Feet Long Is Leaning Against a Vertical