# A ladder 10ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a speed of 2ft/s, how fast is the angle between the top of the ladder and the wall changing when the angle is pi/4 rad?

Apr 6, 2018

$\frac{\sqrt{2}}{5}$ $r a d$/$s$

#### Explanation:

Let $x$ be the distance between the base of the wall and the bottom of the ladder. And let $\theta$ be the angle between the top of the ladder and the wall,

Then $\frac{x}{10} = \sin \theta$, so $x = 10 \sin \theta$.

Differentiating with respect to time $t$ gets us

$\frac{\mathrm{dx}}{\mathrm{dt}} = 10 \cos \theta \frac{d \theta}{\mathrm{dt}}$

We were told that $\frac{\mathrm{dx}}{\mathrm{dt}} = 2$ $f t$/$s$ and we seek $\frac{d \theta}{\mathrm{dt}}$ when $\theta = \frac{\pi}{4}$

$2 = 10 \left(\cos \left(\frac{\pi}{4}\right)\right) \frac{d \theta}{\mathrm{dt}}$

$2 = 10 \left(\frac{\sqrt{2}}{2}\right) \frac{d \theta}{\mathrm{dt}} = 5 \sqrt{2} \frac{d \theta}{\mathrm{dt}}$

$\frac{d \theta}{\mathrm{dt}} = \frac{2}{5 \sqrt{2}} = \frac{\sqrt{2}}{5}$ $r a d$/$s$