Q. A ladder, $5$ meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of $10 \,cm/sec$, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is $2$ metres from the wall is

Application of Derivatives Report Error

Solution:

Given $ \frac{dy}{dt} =10\, cm/s$,
$x = 2m = 200\, cm$ and Length of ladder $= 5\, m = 500 \,cm$
$\Rightarrow $ Let $∠ACB = \theta$
Now $sin \,\theta = \frac{y}{500}$
Diff w.r.t. $t$, we get
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$cos\,\theta \frac{d\theta}{dt} = \frac{1}{500} \frac{dy}{dt}$
$\Rightarrow \frac{200}{500} \times \frac{d\theta }{dt}$
$= \frac{1}{500} \times 10$
$\Rightarrow \frac{d\theta }{dt} = \frac{10}{200}$
$\Rightarrow \frac{d\theta }{dt} = \frac{1}{20}$ rad/s.

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