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?

Answer 1

The height is decreasing at the rate of $2.67$cms${}^{-1}$

For this we have to use Pythagoras theorem:

$x^2+y^2=h^2$ ...............(1)

differentiate equition 1 with respect to 't'.

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

Since we have

$\frac{dx}{dt}=2$cms${}^{-1}$

$x=4m$

so,

The length of wall is when the foot of the ladder is $400$cm away is:

$y=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3m=300cm$

Thus,

The decreasing of the wall is:

Erom equation (1) we get

$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

$=-\frac{400cm}{300cm}\times 2$cms${}^{-1}$

$=2.67$cms${}^{-1}$

This is the required answer.