Question

The surface area of a spherical balloon is increasing at the rate of \[2~c{{m}^{2}}/sec\]. At what rate is the volume of the balloon increasing when the radius of the balloon is 6 cm?

Answer 1

Hint: Here we have to find the rate of change of volume with respect to time. For that, we will first equate the rate of change of surface area with \[2~c{{m}^{2}}/sec\] and then we will apply the formula of surface area. From there, we get the value of the rate of change of radius with respect to time. Then we will find the rate of change of volume by using the formula of volume of sphere. We will substitute the value of the rate of change of radius with time to get the required value.

Formula used:
We will use the following formulas:
1. The formula of volume of the sphere, \[V = \dfrac{4}{3}\pi {r^3}\], where \[r\] is the radius and \[V\] is the volume.
2. The formula of surface area of sphere, \[S = 4\pi {r^2}\], where \[r\] is the radius and \[S\] be the surface area of the sphere

Complete step by step solution:
Let \[r\] be radius of the spherical volume at time \[t\], \[V\] be the volume of the spherical volume at time \[t\] and let \[S\] be the surface area of the spherical balloon at time \[t\].
It is given that the rate of change of surface area with respect to time is \[2~c{{m}^{2}}/sec\].
Therefore, we can it as;
\[ \Rightarrow \dfrac{{dS}}{{dt}} = 2\]
Substituting \[S = 4\pi {r^2}\] in the above equation, we get
\[ \Rightarrow \dfrac{{d\left( {4\pi {r^2}} \right)}}{{dt}} = 2\]
Differentiating the terms, we get
\[ \Rightarrow 8\pi r\dfrac{{dr}}{{dt}} = 2\]
On further simplification, we get
\[ \Rightarrow \dfrac{{dr}}{{dt}} = \dfrac{1}{{4\pi r}}\] ……… \[\left( 3 \right)\]
We have to find the rate of change of volume with respect to time i.e. \[\dfrac{{dV}}{{dt}}\] .
Substituting \[V = \dfrac{4}{3}\pi {r^3}\] in the above expression, we get
\[\Rightarrow \dfrac{{dV}}{{dt}} = \dfrac{{d\left( {\dfrac{4}{3}\pi {r^3}} \right)}}{{dt}}\]
Differentiating the terms, we get
\[\Rightarrow \dfrac{{dV}}{{dt}} = 4\pi {r^2}\dfrac{{dr}}{{dt}}\]
Now, we will substitute the value of \[\dfrac{{dr}}{{dt}}\] from equation \[\left( 3 \right)\] in the above equation, we get
\[\Rightarrow \dfrac{{dV}}{{dt}} = 4\pi {r^2} \times \dfrac{1}{{4\pi r}}\]
On further simplification, we get
\[\Rightarrow \dfrac{{dV}}{{dt}} = r\]
The value of radius given in the question is 6 cm.
Therefore,
\[\Rightarrow \dfrac{{dV}}{{dt}} = 6\]
Therefore, the rate of change of volume of the spherical balloon with time is \[6c{m^3}/\sec \] .

Note: We have got the positive value of rate of change of volume which means that volume is increasing with time. However, if we get the negative value of the rate of change of volume, then that means that the volume is decreasing with time.