Question

A ladder $5\text{m}$ long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2\text{cm/s}$. How fast is its height on the wall decreasing when the foot of the ladder is $4\text{m}$ away from the wall?

Answer 1

Consider, the below figure for the given conditions,
Here, $AB=5\text{m}$


Assuming $BC=x,AC=y$

Also,

$\frac{d}{dt}\left(BC\right)=\frac{dx}{dt}=2\text{cm/s}$

Using Pythagoras theorem, we get

$x^2+y^2=5^2$

Differentiating w.r.t to time, we get

$\Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

$\Rightarrow 4x+2y\frac{dy}{dt}=0[\because \frac{dx}{dt}=2]$

$\Rightarrow \frac{dy}{dt}=-\frac{2x}{y}$

$\Rightarrow \frac{dy}{dt}=-\frac{2x}{\sqrt{25-x^2}}$

$[\because x^2+y^2=25]$

When $x=4\text{m}$

$\Rightarrow \frac{dy}{dt}=-\frac{8}{\sqrt{25-4^2}}$

$\Rightarrow \frac{dy}{dt}=-\frac{8}{\sqrt{9}}=-\frac{8}{3}\text{cm/s}$

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

$\frac{8}{3}\text{cm/s}$