# Related rate questions and the intuition behind them

I am trying to do the following question from the Schaum Calculus book.

Gas is escaping from a spherical balloon at the rate of $2$ ft$^3$/min. How fast is the surface area shrinking when the radius is $12$ ft?

A sphere of radius $r$ has volume $V = \frac{4}{3}\pi r^3$ and surface area $S = 4\pi r^2$. So I understand I need to know these equations as they show me the relationship between volume and surface area ??

So first off we are told the rate of volume is changing at $2$ cubic ft/min. This I am okay with But why do we pay interest to the rate of change of the radius over time ? Is it because it is the only thing in the equation that can actually change ?

Now I need to take the derivative of both related equations ($V = \frac{4}{3}\pi r^3$ and surface area $S = 4\pi r^2$). Derivative of Volume is $\frac{dV}{dt} = 2~\frac{\text{ft}^3}{\text{min}}$ based on the info given at the start. So this is a loss so actually $= -2$

The derivative of the surface area confuses me . The book has it as $4\pi r^2\frac{dr}{dt}$. I don't get why it isn't $4\pi (2r) = 8\pi r$. Why do we put $\frac{dr}{dt}$ in it ?

The book then says based on this $\frac{dV}{dt}=-2 = 4\pi r^2 \frac{dr}{dt}$ and, therefore, $\frac{dr}{dt} = -\frac{1}{2\pi r^2}$.

But then now it takes the derivative of the surface area as $\frac{dS}{dt} = 8\pi r \frac{dr}{dt}$.

And then $\frac{dS}{dt} = -\frac{8\pi r}{2\pi r^2} = -\frac{4}{r}$. So, when $r = 12$, $\frac{dS}{dt} = −\frac{4}{12}=-3~\frac{\text{ft}^2}{\text{min}}$.

I can't quite understand how it was known what to differentiate when.

What kind of chain of steps do people follow to solve these ?

thanks

• It is the derivative of the volume that is $$4\pi r^2\frac{dr}{dt}$$ not the derivative of the surface area. Mar 21, 2015 at 10:06
• On the third line from the bottom the question reads $-\frac{4}{12}=-3$. Mar 21, 2015 at 11:24

It is chain rule. The rate of change of volume is

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

The rate of change of surface area is

$$\frac{dS}{dt}=\frac{dS}{dr}\cdot\frac{dr}{dt}=8\pi r \frac{dr}{dt}$$

Since rate of change of volume is $-2$,

$$4\pi r^2\frac{dr}{dt}=-2$$

You find $\frac{dr}{dt}$ from this, then plugging it into the surface area derivative gives you the rate of change of surface area.

• Thanks. I don't quite understand why the equation pays so much attention to the radius when we are actually interested in the volume and the surface area?? Mar 21, 2015 at 9:32
• I also didn't get why dV/dt wasn't the actually derivative of V i.e they provide us 2ft^3/min instead of taking the derivative of the volume formula Mar 21, 2015 at 9:33
• The radius is good because both the volume and surface area are functions of the radius. So it is the most convenient parameter. The equations are much messier using the volume and surface area - they have lots of cube and square roots. Mar 21, 2015 at 9:36
• It gives you an idea of how the rate of change of volume is related to the rate of change of radius, hence how it affects the rate of change of surface area. Since the radius is given, you will have to use that information to derive the others. And since the formula of $V$ is in terms of $r$, which is itself a variable, the derivative of $V$ has to involve the derivative of $r$ using chain rule. Mar 21, 2015 at 9:38

Express rates with respect to time as primes. Let $4 \pi = k$ for calculation ease.Essentially it is :

$$V = k r^3/3 , V^{'} = k r^2 r^{'} , A = k r^2, A' = 2 k r r^{'} ,\rightarrow \frac{A'}{V'}=\frac{2}{ r }$$

$${A'}= \frac{2\cdot 2 }{12}= \frac{1}{3}.$$