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?

Explanation

`x^{2}+y^{2}=36`


`Rightarrow 2 x frac{d x}{d t}=-2 y frac{d y}{d t} ldots(1)`


When `x=4, y=sqrt{36-16}=2 sqrt{5}`


`Rightarrow 2 times 4 times 0.5=-2 times 2 sqrt{5} frac{d y}{d t}[because frac{d x}{d t}=0.5 m / {sec}]`


`Rightarrow frac{d y}{d t}=frac{-1}{sqrt{5}} m / {sec}`


From eq `cdot(1)`, we get


`2 x frac{d x}{d t}=-2 y frac{d y}{d t}[because frac{d x}{d t}=frac{d y}{d t}]`


`Rightarrow x=-y`


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