Question

A 25-ft ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.18 ft/sec, how fast, in ft/sec, is the top of the ladder sliding down the wall, at the instant when the bottom of the ladder is 20 ft from the wall? Answer with 2 decimal places.

Answer 1

Let $x$ be the distance between the base of the ladder and the bottom of the wall, and $y$ the distance between the top of the ladder and the bottom of the wall, so that

$x^2+y^2=(25\,\mathrm{ft})^2$

Differentiate both sides with respect to time $t$:

$2x\dfrac{\mathrm dx}{\mathrm dt}+2y\dfrac{\mathrm dy}{\mathrm dt}=0$

When $x=20\,\rm ft$, the top of the ladder is

$y=\sqrt{(25\,\mathrm{ft})^2-(20\,\mathrm{ft})^2}=15\,\mathrm{ft}$

above the ground. Then, given that the bottom of the ladder slides away from the wall at a rate of $\dfrac{\mathrm dx}{\mathrm dt}=0.18\dfrac{\rm ft}{\rm s}$, we have

$2(20\,\mathrm{ft})\left(0.18\dfrac{\rm ft}{\rm s}\right)+2(15\,\mathrm{ft})\dfrac{\mathrm dy}{\mathrm dt}=0\implies\dfrac{\mathrm dy}{\mathrm dt}=-0.24\dfrac{\rm ft}{\rm s}$

That is, the top of the ladder is sliding downward at a rate of 0.24 ft/s.