$\begin{array}{1 1}(A) \;\frac{8}{3}cm/s \\ (B)\frac{-8}{3}cm/s \\ (C)\;\frac{8}{3}cm^2/s \\ (D)\;\frac{8}{3}cm^2/s \end{array} $

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- If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
- $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$

Step 1:

Given : The length of the ladder is $l=5m$

$\large\frac{dx}{dt}=$$2cm/s$

The distance of the foot of the ladder from the foot $x=4m$

The height of the wall is $y=\sqrt{5^2-4^2}=3$

(By pythagoras theorem)

When $x=5,y=3$

Also $x^2+y^2=5^2$

Step 2:

Differentiating w.r.t $t$ on both sides we get,

$2x\times \large\frac{dx}{dt}$$+2y.\large\frac{dy}{dt}$$=0$

$\Rightarrow x.\large\frac{dx}{dt}$$+y.\large\frac{dy}{dt}$$=0$

Step 3:

Now substituting the values for $x,y$ and $\large\frac{dx}{dt}$ we get,

$4\times 2+3\times \large\frac{dy}{dt}$$=0$

$\Rightarrow \large\frac{dy}{dt}$$=-\large\frac{8}{3}$$cm/s$

Hence the height of the ladder on the wall decrease at the rate of $\large\frac{8}{3}$$cm/s$

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