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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?

$\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 $l=5m$
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$
answered Jul 8, 2013 by sreemathi.v

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