Question

How fast is the radius of the basketball increasing when the radius is 16cm if air is pumped into a basketball at a rate of $100c{{m}^{3}}/\sec$?

Answer 1

In this problem, we have to find the rate of change of the radius of a basketball if air is pumped into a basketball at a rate of $100c{{m}^{3}}/\sec$ .Here the basketball is a perfect sphere. We are given the radius of the basketball is 16cm. We can now assume the radius as r, then its rate of change will be $\dfrac{dr}{dt}$. We can see that we are given $c{{m}^{3}}$ which indicates the rate of change in volume per second. We have to find $\dfrac{dr}{dt}$ by differentiating the given values to find the answer.

Complete step-by-step solution:
We know that the given radius of the basketball is 16cm.
We have to find the rate of change of the radius $\dfrac{dr}{dt}$.
We know that the given rate of change of volume is,
$\dfrac{dV}{dt}=100c{{m}^{3}}/\sec$ …….. (1)
Here the basketball is a perfect sphere, whose volume is
$\Rightarrow V=\dfrac{4}{3}\pi {{r}^{3}}$
We can now differentiate the volume, V with respect to time, t, we get
$\Rightarrow \dfrac{dV}{dt}=\dfrac{4}{3}\pi 3{{r}^{2}}\dfrac{dr}{dt}$
We can now simplify the above step, we get
$\Rightarrow \dfrac{dV}{dt}=4\pi {{r}^{2}}\dfrac{dr}{dt}$
We can now substitute the given radius value and the (1) in the above step, we get
$\Rightarrow 100=4\pi {{\left( 16 \right)}^{2}}\dfrac{dr}{dt}$
We can now write the above step as,
$\Rightarrow \dfrac{dr}{dt}=\dfrac{100}{4\pi {{16}^{2}}}=\dfrac{25}{\pi {{\left( 16 \right)}^{2}}}cm/\sec$
Therefore, the radius of the basketball will be increasing at a speed of $\dfrac{25}{\pi {{\left( 16 \right)}^{2}}}cm/\sec$.

Note: We should know that $\dfrac{dV}{dt}$ is the rate of change of volume with respect to time and $\dfrac{dr}{dt}$ is the rate of change of radius with respect to time. Here the given basketball is nothing but a sphere whose volume is $\Rightarrow V=\dfrac{4}{3}\pi {{r}^{3}}$. We should also mention the unit in the answer part.