Question

In a competition, a brave child tries to inflate a huge spherical balloon bearing slogans against child labor at the rate of 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon is increasing when its radius is 15 cm.

Answer 1

First, find the derivative of volume with respect to time i.e., $\dfrac{{dV}}{{dt}}$. After that put it equal to the rate of inflation of the spherical balloon. Then substitute the value of radius and find the derivative of the radius with respect to the time i.e., $\dfrac{{dr}}{{dt}}$.

Complete step-by-step answer:
Given:- The rate of change of volume is $\dfrac{{dV}}{{dt}} = 900\,c{m^3}/s$
The radius of the balloon is, $r = 15\,cm$
As the balloon is spherical. Then, the volume of the balloon will be,
$V = \dfrac{4}{3}\pi {r^3}$
Now, differentiate the function with respect to time t. Then,
$\dfrac{{dV}}{{dt}} = \dfrac{d}{{dt}}\left( {4\pi {r^3}} \right)$
Since the value $4\pi$ is constant. So, it will remain outside, the only term which will differentiate is the radius with respect to time t,
$\dfrac{{dV}}{{dt}} = 4\pi \left( {3{r^2}} \right)\dfrac{{dr}}{{dt}}$
Now, put $\dfrac{{dV}}{{dt}} = 900$ and r= 15 cm in the equation. So,
$900 = 12\pi {\left( {15} \right)^2}\dfrac{{dr}}{{dt}}$
Square the number 15 and then multiply the result with 12 on the right side of the equation,
$900 = 900\pi \times \dfrac{{dr}}{{dt}}$
Now, divide both sides by $900\pi$, So that the coefficient on $\dfrac{{dr}}{{dt}}$ side must be equal to 1. Then, we get,
$\dfrac{{900\pi }}{{900\pi }} \times \dfrac{{dr}}{{dt}} = \dfrac{{900}}{{900\pi }}$
Cancel out the common factors on both sides of the equation to get the final answer,
$\dfrac{{dr}}{{dt}} = \dfrac{1}{\pi }$cm/s
Hence, the rate at which the radius of the balloon is increasing when its radius is 15 cm is $\dfrac{1}{\pi }$cm/s.

Note: Derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations.
The formula to remember is,
$\dfrac{d}{{dx}}\left( {a{y^n}} \right) = a{y^{n - 1}}\dfrac{{dy}}{{dx}}$
where y is a variable and a is a constant.