Question

The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of increasing of its surface area, when the radius is 2 cm.

Answer 1

Hint: Use the formula of volume of sphere, i.e., $\dfrac{4}{3}\pi {{r}^{3}}$ and differentiate both sides of the equation with respect to time. The rate of change of volume is given, so put the value in the differential equation and find the rate of change of radius in terms of the radius at that point of time. Now apply the same method to find the differential equation of change of total surface area and substitute rate of change of r and solve to get the answer.

Complete step-by-step answer:
We know that the volume of the sphere is given by:
$\text{V}=\dfrac{4}{3}\pi {{r}^{3}}$
Now if we differentiate both sides of the equation with respect to time t. Also, we know that$\dfrac{d{{r}^{3}}}{dt}=3{{r}^{2}}\dfrac{dr}{dt}$ .
$\Rightarrow \dfrac{dV}{dt}=\dfrac{d\left( \dfrac{4}{3}\pi {{r}^{3}} \right)}{dt}$
$\Rightarrow \dfrac{dV}{dt}=\dfrac{4}{3}\pi \dfrac{d{{r}^{3}}}{dt}$
$\Rightarrow \dfrac{dV}{dt}=\dfrac{4}{3}\pi \times 3{{r}^{2}}\dfrac{dr}{dt}$
Now, it is given in the question that the rate of change of volume, i.e., $\dfrac{dV}{dt}=3c{{m}^{3}}{{s}^{-1}}$ .
 $3=\dfrac{4}{3}\pi \times 3{{r}^{2}}\dfrac{dr}{dt}$
$\Rightarrow \dfrac{3}{4\pi {{r}^{2}}}=\dfrac{dr}{dt}............(i)$
Now let us move to the total surface area of the sphere.
$T=4\pi {{r}^{2}}$
Now if we differentiate both sides of the equation with respect to time t.
$\dfrac{dT}{dt}=\dfrac{d\left( 4\pi {{r}^{2}} \right)}{dt}$
$\Rightarrow \dfrac{dT}{dt}=4\pi \dfrac{d{{r}^{2}}}{dt}$
$\Rightarrow \dfrac{dT}{dt}=4\pi \times 2r\dfrac{dr}{dt}$
Now we will substitute the value of $\dfrac{dr}{dt}$ from equation (i). On doing so, we get
$\Rightarrow \dfrac{dT}{dt}=4\pi \times 2r\times \dfrac{3}{4\pi {{r}^{2}}}=\dfrac{6}{r}$
Now the value of r at that point of time as given in the question, i.e., r=2, we get
 $\Rightarrow \dfrac{dT}{dt}=\dfrac{6}{2}=3c{{m}^{2}}{{s}^{-1}}$
Therefore, we can conclude that the rate of change of total surface area of the sphere is equal to 3 sq cm per second.

Note: Always remember that when you differentiate the equation with respect to time, the variables are taken at any time t, i.e., the variables are also changing continuously with time as we saw for r in the above question. Also, remember that the rate of change of volume, surface area and curved surface area of a geometrical figure purely depends on the rate of change of its dimensions.