Class 12 Mathematics Application of Derivatives: NCERT Solutions for Question 8
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
The volume of a sphere (V) with radius (r) is given by,
\begin{align} V=\frac{4}{3}\pi r^2\end{align}
∴Rate of change of volume (V) with respect to time (t) is given by,
\begin{align} \frac{dV}{dt} =\frac{dV}{dr}.\frac{dr}{dt}\;\;\;[By\; Chain\; Rule]\end{align}
\begin{align} =\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right).\frac{dr}{dt}\end{align}
\begin{align} =4\pi r^2.\frac{dr}{dt}\end{align}
It is given that
\begin{align} \frac{dV}{dt}=900\; cm^3/s\end{align}
\begin{align} \therefore 900=4\pi r^2.\frac{dr}{dt}\end{align}
\begin{align} \Rightarrow \frac{dr}{dt}=\frac{900}{4\pi r^2}=\frac{225}{\pi r^2}\end{align}
Therefore, when radius = 15 cm,
\begin{align} \frac{dr}{dt}=\frac{225}{\pi (15)^2}=\frac{1}{\pi }\end{align}
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is
\begin{align} \frac{1}{\pi }\end{align}
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