# A ladder 13 m long is leaning against a vertical wall. The bottom of the ladder is dragged away from the wall along the ground at the rate of 2 cm/sec. How fast is the height on the wall decreasing when the foot of the ladder is 5 m away from the wall ?

consider the ladder of 13m length of AB.
Also set at any line t, the lower end of the laddet be at a distance of xm from the wall BC and at that moment, its height on the wall be ym.
consider the figer show.

we have $\frac{d}{dt}= 2cm/sec$
As  $x^{2}+y^{2}= 13^{2}$
$\Rightarrow 2x\frac{dx}{dt}+2y\times \frac{dy}{dt}= 0$
$\Rightarrow \frac{dy}{dt}= \frac{-x}{y}\times \frac{dx}{dt}$
$\Rightarrow \frac{dy}{dt}= \frac{-x}{\sqrt{169-x^{2}}}\times 2$
$\therefore \frac{dy}{dt}]_{when x= 5m}\; = \frac{-5}{\sqrt{169-5^{2}}}\times 2= \frac{-5}{6}\, cm/sec$
Here the height on the wall is decreasing at the rate of $\frac{5}{6}\, cm/sec$

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