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
Best Answer
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.