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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.

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    $\begingroup$ If it's expanding and you want to know how fast, you need a rate of change with respect to time. $\endgroup$ – MasterOfBinary Nov 4 '13 at 17:59
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You are correct. The units are cm. The phrasing of the question is quite alright though "at the point in time" where the balloon is 10 cm, the rate of change is 80 pi, regardless of how fast you blew it full.

However, they are not correct in stating that the units are cm2/unit of time, unless the balloon is increasing in size by precisely 1 cm per unit of time.

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    $\begingroup$ Thanks Jolle. The hardest thing to understand is that rate of change of SA is the same regardless of how fast you blow up the balloon, but I guess the key is seeing that rate of change of surface area WRT radius and rate of change of surface area WRT time are two different things. $\endgroup$ – yroc Nov 4 '13 at 17:55
  • $\begingroup$ 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}$. $\endgroup$ – Randall Blake Aug 10 '18 at 2:08

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