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Question: The volume of a sphere increases at the rate of \[20\,{\text{c}}{{\text{m}}^{\text{3}}}{{\text{s}}^{...

The volume of a sphere increases at the rate of 20cm3s - 120\,{\text{c}}{{\text{m}}^{\text{3}}}{{\text{s}}^{{\text{ - 1}}}} . Find the rate of change of its surface area when its radius is 5cm5\,{\text{cm}} .

Explanation

Solution

Hint : We are asked to find the rate of change of surface area of the sphere. First, recall the formula for sphere area of a sphere. Recall how rate of change of a quantity is found with respect to time, use this to find the rate of surface area of the sphere using the value of radius and rate of change of volume

** Complete step-by-step answer** :
Given, rate of change volume is dVdt=20cm3s - 1\dfrac{{dV}}{{dt}} = 20\,{\text{c}}{{\text{m}}^{\text{3}}}{{\text{s}}^{{\text{ - 1}}}}
Radius of the sphere is r=5cmr = 5\,{\text{cm}}
We are asked to find the rate of change of surface area of the sphere.
Surface area of a sphere is written as,
S=4πr2S = 4\pi {r^2}
To find the rate of change of surface area, differentiate SS with respect to time tt .
dSdt=ddt(4πr2)\dfrac{{dS}}{{dt}} = \dfrac{d}{{dt}}\left( {4\pi {r^2}} \right)
dSdt=4π×2r×drdt\Rightarrow \dfrac{{dS}}{{dt}} = 4\pi \times 2r \times \dfrac{{dr}}{{dt}}
dSdt=8πrdrdt\Rightarrow \dfrac{{dS}}{{dt}} = 8\pi r\dfrac{{dr}}{{dt}} (i)
Volume of a sphere is written as,
V=43πr3V = \dfrac{4}{3}\pi {r^3}
For rate of change of volume, we differentiate VV with respect to time tt .

\Rightarrow \dfrac{{dV}}{{dt}} = \dfrac{4}{3}\pi \times 3{r^2} \times \dfrac{{dr}}{{dt}} \\\ \Rightarrow \dfrac{{dV}}{{dt}} = 4\pi {r^2}\dfrac{{dr}}{{dt}} $$ $$ \Rightarrow \dfrac{{dV}}{{dt}} = \dfrac{r}{2} \times 8\pi r\dfrac{{dr}}{{dt}}$$ Using equation (i) the term $$8\pi r\dfrac{{dr}}{{dt}}$$ can be written as $$\dfrac{{dS}}{{dt}}$$ $$\dfrac{{dV}}{{dt}} = \dfrac{r}{2} \times \dfrac{{dS}}{{dt}}$$ $$ \Rightarrow \dfrac{{dS}}{{dt}} = \dfrac{2}{r} \times \dfrac{{dV}}{{dt}}$$ Putting the value of $$\dfrac{{dV}}{{dt}}$$ and $$r$$ , we get $$\dfrac{{dS}}{{dt}} = \dfrac{2}{5} \times 20 \\\ \Rightarrow \dfrac{{dS}}{{dt}} = 8\,{\text{c}}{{\text{m}}^{\text{2}}}{{\text{s}}^{{\text{ - 1}}}} $$ Therefore, rate of change of surface is $$8\,{\text{c}}{{\text{m}}^{\text{2}}}{{\text{s}}^{{\text{ - 1}}}}$$ . **So, the correct answer is “$$8\,{\text{c}}{{\text{m}}^{\text{2}}}{{\text{s}}^{{\text{ - 1}}}}$$”.** **Note** : Regarding shapes like sphere, cylinder, cone, cube there are some important formulas that one should always remember that is their formula for volume and surface area. In the above question we have discussed the sphere. For cylinder the formulas for volume and surface area are $$V = \pi {r^2}h$$ and $$S = 2\pi rh + 2\pi {r^2}$$ respectively, where $$r$$ is the radius and $$h$$ is the height of the cylinder. For cone we have $$V = \dfrac{1}{3}\pi {r^2}h$$ and $$S = \pi rl + \pi {r^2}$$ , where $$l$$ is the slant height and $$r$$ is the radius of the cone. For the cube we have $$V = {a^3}$$ and $$S = 6{a^2}$$ , where $$a$$ is the side of the cube.