# A ladder 5m long is leaning against a wall. The bottom of the 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??

Jul 30, 2017

The height is decreasing at the rate of $= 2.67 c m {s}^{-} 1$

#### Explanation:

Here , we apply Pythagoras

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

Differentiating with respect to $t$

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

$\frac{\mathrm{dx}}{\mathrm{dt}} = 0.02 m {s}^{-} 1$

$x = 4 m$

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

Therefore,

$\frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{x}{y} \frac{\mathrm{dx}}{\mathrm{dt}} = - \frac{4}{3} \cdot 0.02 = - 0.0267 m {s}^{-} 1$