Question

# A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is 10cm.

Hint: Take radius and volume of spherical balloons r and v. Find the volume of balloon and differentiate it w.r.t r. Then find $\dfrac{dv}{dr}$when r = 10cm. Thus find the rate at which the volume is increasing.

We know that a balloon is in spherical shape. Let us consider ‘r’ as the radius of the balloon which is spherical.
Let ‘v’ be the volume of the balloon.
Here we need to find the rate at which the balloon’s volume is increasing when the radius r is 10cm.
i.e. here we need to find the change of volume with respect to the radius, when r = 10.
$\therefore$We need to find $\dfrac{dv}{dr}$when r = 10cm.
We know the volume of the sphere is given by the formula $\dfrac{4}{3}\pi {{r}^{3}}$.
$\therefore$Volume of sphere$=\dfrac{4}{3}\pi {{r}^{3}}$
i.e. $V=\dfrac{4}{3}\pi {{r}^{3}}$
Let us differentiate the above equation w.r.t radius r.
\begin{align} & \dfrac{dv}{dr}=\dfrac{4}{3}\pi \dfrac{d}{dr}{{\left( r \right)}^{3}} \\ & \Rightarrow \dfrac{dv}{dr}=\dfrac{4}{3}\pi \times 3{{r}^{2}} \\ \end{align}
Cancel out the like terms and we get,
$\dfrac{dv}{dr}=4\pi {{r}^{2}}$
We need to find the value of $\dfrac{dv}{dr}$when r = 10.
\begin{align} & \therefore \dfrac{dv}{dr}=4\times \pi \times {{10}^{2}} \\ & \dfrac{dv}{dr}=4\times \pi \times 10\times 10 \\ & \dfrac{dv}{dr}=400\pi \\ \end{align}
The volume is in $c{{m}^{3}}$and the radius is in cm.
So, $\dfrac{dv}{dr}=400\pi \dfrac{c{{m}^{3}}}{cm}$
Hence the volume is increasing at the rate of $400\pi \dfrac{c{{m}^{3}}}{cm}$when r = 10cm.

Note: Here the rate of increase in volume of the balloon signified the change in the shape as well. So the rate of change in the increase in the volume w.r.t the original shape of the balloon. Thus, as it changes the volume of the balloon changes, which is spherical in shape. That’s why we consider the formula of a sphere to differentiate it w.r.t radius r.