Question

A ladder $5m$ long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2cm/s$ How fast is its height on the wall decreasing when the foot of the ladder is $4cm$ away from the wall.

Answer 1

Let $ym$ be the height of the wall at which the ladder touches. Also, let the foot of the ladder be $xm$ away from the wall.
Then, by Pythagoras theorem, we have:
$x^2+y^2=25$ [Length of the ladder $=5m$]
$\Rightarrow y=\sqrt{25-x^2}$
Then, the rate of change of height $\left(y\right)$ with respect to time $\left(t\right)$ is given by,
$\frac{dy}{dt}=\frac{-x}{\sqrt{25-x^2}}\cdot \frac{dx}{dt}$
It is given that $\frac{dx}{dt}=2cm/s$
$\therefore \frac{dy}{dt}=\frac{-2x}{\sqrt{25-x^2}}$
Now, when $x=4m$, we have:
$\frac{dy}{dt}=\frac{-2\times 4}{\sqrt{25-4^2}}=-\frac{8}{3}$
Hence, the height of the ladder on the wall is decreasing at the rate of $\frac{8}{3}$ cm/s.