# A ladder is leaning against a wall, and the floor and slipping. If the bottom of the ladder is slipping at 30 cms^(-1) then how fast is the top of the ladder sliding down the wall when the ladder is at 45^o?

Sep 29, 2017

It is sliding down at a rate (speed) of $30 \setminus m {s}^{-} 1$.

#### Explanation: Let us define the following variables:

 { (x, "Distance of bottom of ladder from the wall " (cms^-1)), (y, "Distance of top of ladder from the floor "(cm)), (t, "Time "(s)), (theta, "Angle between Ladder and floor "angle ABO " (radians)") :}

We are told that $\frac{\mathrm{dx}}{\mathrm{dt}} = 30 \setminus c m {s}^{- 1}$ (constant)
We aim to find $\frac{\mathrm{dy}}{\mathrm{dt}}$ when $\theta = {45}^{o} = \frac{\pi}{4}$

The ladder is a fixed length, $l \setminus \left(m\right)$ say, so by Pythagoras;

${x}^{2} + {y}^{2} = {l}^{2}$

Differentiating Implicitly wrt $t$ we get:

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

$\frac{\mathrm{dx}}{\mathrm{dt}} = 30 \implies 30 x + y \frac{\mathrm{dy}}{\mathrm{dt}} = 0$

$\therefore \frac{\mathrm{dy}}{\mathrm{dt}} = - 30 \frac{x}{y}$ ..... [A]

Using trigonometry, we have:

$\tan \setminus \angle A B O = \frac{y}{x} \implies \tan \theta = \frac{y}{x}$
$\theta = \frac{\pi}{4} \implies \frac{y}{x} = 1 \implies \frac{x}{y} = 1$

With $\frac{x}{y} = 1$ in Eq[A] then:

$\implies \frac{\mathrm{dy}}{\mathrm{dt}} = - 30$

The minus sign denotes that the ladder is sliding down, i.e., the height $y$ is decreasing. It is sliding down at a rate (speed) of $30 \setminus m {s}^{-} 1$.