Question

A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding away from the wall at the rate of 3 feet per second, find the rate at which the top of the ladder is sliding down when the top of the ladder is 8 feet from the ground.

Answer 1

Answer:

6.87 ft/s is the rate at which the top of ladder slides down.

Explanation:

Given:

Length of the ladder is, $L=20\ ft$

Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.

Now, from triangle ABC,

AB² + BC² = AC²

$h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1$

Differentiating the above equation with respect to time, 't'. This gives,

$\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2$

In the above equation the term $\frac{dh}{dt}$ is the rate at which top of ladder slides down and $\frac{db}{dt}$ is the rate at which bottom of ladder slides away.

Now, as per question, $h=8\ ft, \frac{db}{dt}=3\ ft/s$

Plug in $h=8$ in equation (1) and solve for $b$. This gives,

$8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft$

Now, plug in all the given values in equation (2) and solve for $\frac{dh}{dt}$

$8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s$

Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.