A ladder 13m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5m/sec. How fast is the angle `theta` between the ladder and the ground is changing when the foot of the ladder is 12m away from the wall.

Explanation

`tan theta=frac{y}{x}` and


`x^{2}+y^{2}=(13)^{2}`


`Rightarrow x^{2}(1+tan ^{2} theta)=169`


`Rightarrow sec ^{2} theta=frac{169}{x^{2}}`


`Rightarrow 2 sec ^{2} theta tan theta frac{d theta}{d t}=169(frac{-2}{x^{3}}) frac{d x}{d t}`


`Rightarrow frac{d theta}{d t}=frac{-338 times 1.5}{(12)^{3} 2 sec ^{2} theta tan theta} ldots(1)`


When `x=12, y=sqrt{169-144}=5 m`


So,


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