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 \(\frac{8}{3} cm/s\)