Question

The top of a ladder 6 metres long is resting against a vertical wall on a level pavement, when the ladder begins to slide outwards. At the moment when the foot of the ladder is 4 metres from the wall, it is sliding away from the wall at the rate of 0.5 m/sec. How fast is the top-sliding downwards at this instance?
How far is the foot from the wall when it and the top are moving at the same rate?

Answer 1

Let the bottom of the ladder be at a distance of x m from the wall and its top be at a height of y m from the ground.

Here,

$$\begin{gathered} x^2+y^2=36 \\ \Rightarrow 2x\frac{dx}{dt}=-2y\frac{dy}{dt}...\left(1\right) \\ \mathrm{When}x=4,y=\sqrt{36-16}=2\sqrt{5} \\ \Rightarrow 2\times 4\times 0.5=-2\times 2\sqrt{5}\frac{dy}{dt}\left[\because \frac{dx}{dt}=0.5\mathrm{m}/\sec \right] \\ \Rightarrow \frac{dy}{dt}=\frac{-1}{\sqrt{5}}\mathrm{m}/\sec \\ \text{From eq. (1), we get} \\ 2x\frac{dx}{dt}=-2y\frac{dy}{dt}\left[\because \frac{dx}{dt}=\frac{dy}{dt}\right] \\ \Rightarrow x=-y \\ \text{Substituting x=-}\mathit{\text{y}}\text{in}{\mathit{\text{x}}}^2\text{+}{\mathit{\text{y}}}^2\text{=36, we get} \\ {x}^2+x^2=36 \\ \Rightarrow x^2=18 \\ \Rightarrow x=3\sqrt{2}\mathrm{m} \end{gathered}$$