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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the surface.Prove that the radius is decreasing at a constant rate.

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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 $.
  • $ \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
Volume of the sphere = $ \large\frac{4}{3} \pi r^3$
Differentiating w.r.t $t$ we get
$ \large\frac{dv}{dt}=-\large\frac{4}{3} \pi\: \: 3r^2.\large\frac{dr}{dt}$
( -ve sign ) indicate that the volume is decreasing.
$ \Rightarrow \large\frac{dv}{dt}=-4\pi r^2\: \large\frac{dr}{dt}$
But the surface area of the sphere is $ 4\pi r^2$
i.e., $A=4 \pi r^2$
$ \therefore \large\frac{dv}{dt}=-A \large\frac{dr}{dt}$
Step 2
It is given that the rate of decrease of volume is proportional to its surface area.
i.e., $ \large\frac{dv}{dt} \: \alpha\: A$
$ \Rightarrow \large\frac{dv}{dt}=-kA$ where $ k $ is the constant of proportionality.
$ \therefore kA=-A \: \large\frac{dr}{dt}$
$ \Rightarrow \large\frac{dr}{dt}=-k$
Hence the radius is decreasing at a constant rate.

 

answered Jul 30, 2013 by thanvigandhi_1
edited Aug 14, 2013 by thanvigandhi_1
 

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