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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The length \(x\) of a rectangle is decreasing at the rate of $5\; cm/minute$ and the width y is increasing at the rate of $4\; cm/minute$. When \(x = 8\)cm and \(y = 6\)cm, find the rates of change of (a) the perimeter

This is (a) part of the multi-part question q7

$\begin{array}{1 1} (A)\;-2cm/min. \\(B)\;2cm/min. \\ (C)\;-3cm/min. \\ (D)\;3cm/min.\end{array} $

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1 Answer

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Toolbox:
  • 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:
Rate of decrease in length is $\large\frac{dx}{dt}$$=-5cm/min$.
Negative sign shows it is decreasing.
Rate of increase in width is $\large\frac{dy}{dt}$$=4cm/min$.
Given : $x=8cm.$ and $y=6cm$
Perimeter of the rectangle is $P=2x+2y$
Differentiating w.r.t $t$ on both sides we get,
$\large\frac{dp}{dt}=$$2.\large\frac{dx}{dt}+$$2.\large\frac{dy}{dt}$
Step 2:
Now substituting the values for $\large\frac{dx}{dt}$ and $\large\frac{dy}{dt}$ we get,
$\large\frac{dp}{dt}=$$2(-5)+2(4)$
$\quad\;\;=-2$cm/min.
Hence the perimeter decreases at the rate of $-2$cm/min.
answered Jul 5, 2013 by sreemathi.v
 

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