Question

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

Answer 1

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 \[\dfrac{{dV}}{{dt}} = 20\,{\text{c}}{{\text{m}}^{\text{3}}}{{\text{s}}^{{\text{ - 1}}}}\]
Radius of the sphere is \[r = 5\,{\text{cm}}\]
We are asked to find the rate of change of surface area of the sphere.
Surface area of a sphere is written as,
 \[S = 4\pi {r^2}\]
To find the rate of change of surface area, differentiate \[S\] with respect to time \[t\] .
 \[\dfrac{{dS}}{{dt}} = \dfrac{d}{{dt}}\left( {4\pi {r^2}} \right)\]
 \[ \Rightarrow \dfrac{{dS}}{{dt}} = 4\pi \times 2r \times \dfrac{{dr}}{{dt}}\]
 \[ \Rightarrow \dfrac{{dS}}{{dt}} = 8\pi r\dfrac{{dr}}{{dt}}\] (i)
Volume of a sphere is written as,
 \[V = \dfrac{4}{3}\pi {r^3}\]
For rate of change of volume, we differentiate \[V\] with respect to time \[t\] .
 \[\dfrac{{dV}}{{dt}} = \dfrac{d}{{dt}}\left( {\dfrac{4}{3}\pi {r^3}} \right) \\
   \Rightarrow \dfrac{{dV}}{{dt}} = \dfrac{4}{3}\pi \times 3{r^2} \times \dfrac{{dr}}{{dt}} \\
   \Rightarrow \dfrac{{dV}}{{dt}} = 4\pi {r^2}\dfrac{{dr}}{{dt}} \]
 \[ \Rightarrow \dfrac{{dV}}{{dt}} = \dfrac{r}{2} \times 8\pi r\dfrac{{dr}}{{dt}}\]
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.