Question

The volume of a sphere is increasing at the rate of $8\,c{m^3}/s$ . Find the rate at which its surface area is increasing when the radius of the sphere is 12cm.

Answer 1

in this question, a sphere is there, where volume is increasing at a certain rate and the radius of the sphere is given. You will have to find the rate at which the surface area of the sphere is increasing volume and surface area for a sphere. Let's get started.

Step wise solution:
Given data: The rate at which, the volume of the sphere is increasing is $\dfrac{{dv}}{{dt}} = 8\,c{m^3}/s$
The radius of the sphere is r= 12 cm.
We need to find out the rate at which the surface area of the sphere is increasing, i.e.: $\dfrac{{ds}}{{dt}} = ?$
To find out the rate at which the radius of the sphere is increasing.
We know, the volume of a sphere is given by $V = \dfrac{4}{3}\pi {r^3}$
Differentiating both sides with respect to time (t), we get,

\dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}(\dfrac{4}{3}\pi {r^3})\\\ \Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{{4\pi }}{3}\dfrac{d}{{dt}}({r^3})\\\ \Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{{4\pi }}{{3}}({3}{r^2})\dfrac{{dr}}{{dt}}\\\ \Rightarrow \dfrac{{dv}}{{dt}} = 4\pi {r^2}\dfrac{{dr}}{{dt}} .....(i) \\\ $$ As we are given $$\dfrac{{dv}}{{dt}} = 8\,c{m^3}/s$$ and r=12 cm, Equation (i) can be written as

\Rightarrow 8 = 4\pi {(12)^2}\dfrac{{dr}}{{dt}}\\
\Rightarrow \dfrac{{dr}}{{dt}} = \dfrac{8}{{4\pi {{(12)}^2}}}\\
\Rightarrow \dfrac{{dr}}{{dt}} = \dfrac{1}{{72}},,cm/s \\ $$

Is the rate at which the radius of the sphere is increasing.
To find out the rate at which the surface are of the sphere is increasing.
Now, the sphere area of a sphere can be given by $S = 4\pi {r^2}$
Differentiating both sides with respect to t, we get, the rate of the surface area of the sphere at any time t, $\dfrac{{ds}}{{dt}} = \dfrac{d}{{dt}}(4\pi {r^2})$ .

i.e.:$\dfrac{{ds}}{{dt}} = 4\pi \dfrac{d}{{dt}}({r^2})\\\ \Rightarrow \dfrac{{ds}}{{dt}} = 4\pi (2r)\dfrac{{dr}}{{dt}}\\\ \Rightarrow \dfrac{{ds}}{{dt}} = 8\pi r\dfrac{{dr}}{{dt}} .....(ii) \\\$

We are given that r= 12cm and $\dfrac{{dr}}{{dt}} = \dfrac{1}{{72}}\,\,cm/s$
I,e. equation (ii) can be written as,

\Rightarrow \dfrac{{ds}}{{dt}} = \dfrac{4}{3}\,\,\,c{m^2}/s \\\ $$ Hence, the surface area of the sphere is increased at rate of $$\dfrac{4}{3}\,\,\,c{m^2}/s$$ . **Note:** Students often make mistakes in the formula of surface area and volume of sphere. Try to avoid putting the value of $$\pi \,\,as\,\,\dfrac{{22}}{7}\,\,or\,\,3.14$$ , keep it as Pi only to have ease in computing the numbers.