 Question 10

# A ladder 5 m 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?

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:

x2 + y2 = 25 [Length of the ladder = 5 m]

\begin{align}\Rightarrow y = \sqrt{25 - x^2}\end{align}

Then, the rate of change of height (y) with respect to time (t) is given by,

\begin{align}\frac{dy}{dx} = \frac{-x}{\sqrt{25 - x^2}}.\frac{dx}{dt}\end{align}

It is given that

\begin{align}\frac{dx}{dt}= 2 \; cm/s.\end{align}

\begin{align}\therefore\frac{dy}{dt} = \frac{-2x}{\sqrt{25 - x^2}}\end{align}

Now, when x = 4 m, we have:

\begin{align}\therefore\frac{dy}{dt} = \frac{-2 \times 4}{\sqrt{25 - 4^2}}=-\frac{8}{3}\end{align}

Hence, the height of the ladder on the wall is decreasing at the rate of

\begin{align}\frac{8}{3}\end{align}