# A ladder 5 m long is leaning against a wall.

Question:

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?

Solution:

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be maway from the wall.

Then, by Pythagoras theorem, we have:

$x^{2}+y^{2}=25$ [Length of the ladder $\left.=5 \mathrm{~m}\right]$

$\Rightarrow y=\sqrt{25-x^{2}}$

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

$\frac{d y}{d t}=\frac{-x}{\sqrt{25-x^{2}}} \cdot \frac{d x}{d t}$

It is given that $\frac{d x}{d t}=2 \mathrm{~cm} / \mathrm{s}$.

$\therefore \frac{d y}{d t}=\frac{-2 x}{\sqrt{25-x^{2}}}$

Now, when x = 4 m, we have:

$\frac{d y}{d t}=\frac{-2 \times 4}{\sqrt{25-4^{2}}}=-\frac{8}{3}$

Hence, the height of the ladder on the wall is decreasing at the rate of $\frac{8}{3} \mathrm{~cm} / \mathrm{s}$.