Calculus Made Easy -Exercises IX, Question 10 (rates of change)

Question

I am working with this exercise, which should be pretty easy, as a similar one is exemplified in the book.

10. A spherical balloon is increasing in volume. If, when its radius is r feet, its volume is increasing at the rate of 4 cubic feet per second, at what rate is its surface then increasing?

This is what i did: the formula for the volume of a sphere is $\ \frac43^3\;$.

Rate of change of volume with respect to time is 4, so $$\frac{dv}{dr}\cdot\frac{dr}{dt} = 4\implies 4r^2\cdot\frac{ dr}{dt} = 4$$, and $$\frac{dr}{dt} = \frac{1}{r^2}.$$

As what is being asked is rate of change of surface area, I need to calculate now $\dfrac{dS}{dr} \cdot\dfrac{ dr}{dt}$. Surface area of a sphere is $4r^2$, its derivative with respect to $r$ is $8r$, and multiplied with the previously found $\dfrac{dr}{dt} = \dfrac{1}{r^2}$, it gives a result of $\dfrac{8}{r}$.

When I check the solutions of the book -rather bare, with no steps- it says the answer should be 8 times the square root of .

What have I done wrong?

Answer 1

Your answer is correct. The answer in the book cannot be correct since it has the wrong units for the rate at which the surface area is changing.

Setting up the problem in a different way: \begin{align*} V & = \frac{4}{3}\pi r^3\\ \frac{dV}{dt} & = 4\pi r^2 \frac{dr}{dt}\\ \frac{1}{4\pi r^2} \frac{dV}{dt} & = \frac{dr}{dt}\\ \frac{1}{4\pi r^2} 4~\frac{\text{ft}^3}{\text{s}} & = \frac{dr}{dt}\\ \frac{1}{\pi r^2}~\frac{\text{ft}^3}{\text{s}} & = \frac{dr}{dt} \end{align*} We will substitute this result into the rate at which the surface area is changing. \begin{align*} S & = 4\pi r^2\\ \dfrac{dS}{dt} & = 8\pi r \frac{dr}{dt}\\ & = 8\pi r \cdot \frac{1}{\pi r^2}~\frac{\text{ft}^3}{\text{s}}\\ & = \frac{8}{r}~\frac{\text{ft}^3}{\text{s}} \end{align*} The units make sense since $r$ is measured in feet, leaving us with units of $\text{ft}^2/\text{s}$ for the rate at which the surface area is changing.

The units in the book's answer do not make sense since the lack of an $r$ in the denominator would mean that the units for $dS/dt$ would be $\text{ft}^3/\text{s}$.