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 $2 cm/s$. How fast is its height on the wall decreasing when the foot of the ladder is $4 m$ away from the wall?

Answer 1

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