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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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If the area of a circle increases at a uniform rate,then prove that perimeter varies inversely as the radius.

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  • If $ y = f(x)$, then $ \large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $ x $.
  • $ \bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
Area of the circle is
$A= \pi r^2$
Differentiating w.r.t $t$ on both sides we get,
$ \large\frac{dA}{dt}= \pi.2r.\large\frac{dr}{dt}$
It is given that the area is increasing at a uniform rate.
$ \therefore \large\frac{dA}{dt}=k\: \: \Rightarrow 2\pi r.\large\frac{dr}{dt}=k$
Step 2
where $k$ is a constant $ \therefore \large\frac{dr}{dt}=\large\frac{k}{2\pi r}$
Perimeter of the circle is
$ p=2\pi r$
Differentiatily w.r.t $t$ we get
$ \large\frac{dp}{dt}=2\pi.\large\frac{dr}{dt}$
Substituting for $ \large\frac{dr}{dt}$ we get
$ \large\frac{dp}{dt}=2\pi.\large\frac{k}{2\pi r}$
$ \large\frac{dp}{dt}=\large\frac{k}{r}$
$ \Rightarrow \large\frac{dp}{dt}\: \alpha\: \large\frac{1}{r}$
Hence this proves that the perimeter varies inversely as the radius.
answered Jul 30, 2013 by thanvigandhi_1

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