Question

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

Answer 1

Since a right angle triangle is formed with the bottom as base $(x=4m)$, ladder as hypotenuse $(L=5m)$ and the wall as altitude $\left(y\right)$ then

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

$4^2+y^2=5^2$

$y=3m$

Now, differentiating equation (1) in respect to time $t$, we get

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ {$L$ is constant here}

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

$\frac{dy}{dt}=\frac{-4}{3}\times 0.02$ [$\frac{dx}{dt}=2cm/s=0.02m/s$]

$\frac{dy}{dt}=-0.0267m/sec$

The height is decreasing at the rate of $2.67cm/s$.