Question

A laddre 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 m/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

Let AB= 5m be the ladder and y be the height of the wall at which the ladder touches. Also, let the foot of the ladder be at B whose distance from the wall is x.

Given that the bottom of ladder is pulled along the ground at 2 cm/s, so $\frac{dx}{dt}=2m/s$
As we know that $\Delta ABC$ is right angled, so by pythagoras theorem, we have $x^2+y^2=5^2$ ....(i)
when x=4, then
$y^2=5^2-4^2\Rightarrow y=\sqrt{25-16}\Rightarrow y=3m$
Differentiating Eq. (i) w.r.t time t on both sides,
we get
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\Rightarrow x\frac{dx}{dt}+y\frac{dy}{dt}=0\Rightarrow 4\times 2+3\times \frac{dy}{dt}=0[\because x=4and\frac{dx}{dt}=2]$
$\Rightarrow$ The rate of fall of height on the wall $\frac{dy}{dt}=-\frac{8}{3}cm/s$
[Negative sign shows that height of ladder on the wall is decreasing at the rate of -$\frac{8}{3}cm/s$.]