Find the rate of increase of radius of sphere when the radius is 4cm and when the radius is 8cm respectively.

Question

A spherical balloon is being inflated. The rate of increase of the volume is $30\frac{\text{cm}^3}{\text{s}}$. Find the rate of increase of the radius of the sphere when the radius is $4$cm (a) and $8$cm (b).

$\frac{dr}{dt} = \frac{dv}{dt} \frac{dr}{dv}$

$\frac{dr}{dv}= \frac{1}{4 \pi } \left(\frac{3V}{4\pi}\right)^{-2/3}$

By substituting the value of $r=4$ into the formula for the volume of the sphere, $V= (4/3)\pi r^3$, I obtained $V= 256\pi /3$. Substituting $V= 256\pi /3$ into $\frac{dr}{dv} =\frac{1}{4\pi} (\frac{3V}{4\pi})^{-2/3}$,

$\frac{dr}{dv} = \frac{1}{64\pi}$.

Hence, $\frac{dr}{dt}= 30\frac{1}{64\pi}= 0.149$ ($3$s.f.).

Here is my answer for this question, not sure if it's correct. Does anyone else have the same answer? For (b) I used the same method to obtain $\frac{dr}{dt}= 0.0373$ ($3$s.f.) for when $r=8$cm.

Answer 1

For a sphere $V= \frac{4}{3}\pi r^3$. $\frac{dV}{dr}= 4\pi r^2$ which we can write as $dV= (4\pi r^2)dr$. You are told that dV= 30.

When r= 4 that becomes $30= (4\pi (16))dr= 64\pi dr$ so $dr= \frac{30}{64\pi}= \frac{30}{201}= 0.149$ cm/sec.

When r= 8 that is $30= (4\pi (64)dr= 256\pi dr$ so $dr= \frac{30}{256\pi}=0.0373 cm/sec$.

Yes, the numbers you have are correct. My only comment is that your answers really should have the units, "cm/sec", not just numbers.

Answer 2

Your results are correct (except for the missing units). But you can make your calculations easier by differentiating $V = \frac 43 \pi r^3$ directly wrt. $t$:

$$\frac {dV}{dt} = 4\pi r^2\frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac 1{4\pi r^2}\frac {dV}{dt}$$

Now, plug in the given quantities for $\frac {dV}{dt}$ and $r$.