# How can rate of change be with respect to time if you're not differentiating with respect to time?

A high school Calculus textbook asks: Determine the instantaneous rate of change in the surface area of a spherical balloon (as it is inflated) at the point in time when the radius reaches 10 cm.

My solution: The function relating surface area of a sphere to its radius is $A=4\pi r^2$. So, instantaneous rate of change of surface area with respect to radius is $A'=8\pi r$. When $r=10$, $A'=80\pi$. This answer is in agreement with the text, but what's confusing me are the units. The textbook gives the units as $\mathrm{cm}^2$/unit of time. However, I would think that the units would be $\mathrm{cm}^2/\mathrm{cm}$ because I'm finding rate of change of surface area with respect to the radius, not time. On the other hand, it does seem plausible that the rate of change of surface area would depend on time (as in, how quickly the balloon is being blown up). Also, the question uses the phrase "at the point in time" This is really confusing me.

Is the textbook mistaken, or am I? Any help is appreciated.

• If it's expanding and you want to know how fast, you need a rate of change with respect to time. – MasterOfBinary Nov 4 '13 at 17:59

• The question is confusing. As you (@yroc) point out, $dA \over dr$ is quite different from $dA \over dt$. The former is $\frac {dA}{dr} = 8 \pi r$ whereas the latter is $\frac {dA}{dt} = 8 \pi r \frac {dr}{dt}$. – Randall Blake Aug 10 '18 at 2:08