# A ladder, 5 metre long, standing on a horizontal floor,

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Question:

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

Solution:

Length of the ladder $=500 \mathrm{~cm}$

Let the horizontal length covered between the wall and the ladder be $x$ and vertical length covered between the wall and the ladder be $y$

And let the angle between the floor and ladder be $\theta$.

Then, $\sin \theta=\frac{y}{500}$

On differentiating with respect to $t$, we get

$\cos \theta \frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{1}{500} \frac{\mathrm{d} y}{\mathrm{~d} t}$         ......(1)

It is given that $\frac{\mathrm{d} y}{\mathrm{~d} t}=-10 \mathrm{~cm} / \mathrm{sec} .$         ......(2)

Also,

$\cos \theta=\frac{x}{500}$

When $x=200 \mathrm{~cm}, \quad \cos \theta=\frac{200}{500}=\frac{2}{5}$         .....(3)

Substituting $(2)$ and $(3)$ in $(1)$, we get

$\frac{2}{5} \frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{1}{500}(-10)$

$\frac{\mathrm{d} \theta}{\mathrm{d} t}=-\frac{1}{20} \operatorname{radian} /$ second

Hence, the angle between the floor and the ladder is decreasing at the rate of $\frac{1}{20}$ radian/second.