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I'm confused about what sign to use for an inflating or deflating sphere. The question describes a deflating spherical balloon and asks for the instantaneous rate of change of its volume when its radius is 5 cm.

Using the definition of instantaneous rate of change, $\lim_{h\to0}\frac{\frac{4}{3}\pi(5+h)^3-\frac{4}{3}\pi(5)^3}{h}$, which evaluates to $100\pi$

The fact that I'm getting a positive rate of change seems a bit counterintuitive since the balloon is deflating, but on the other hand it makes sense because even though volume is decreasing during deflation, radius is also decreasing, so the two negatives "cancel out".

There are other signs that the rate of change should be positive. Like the fact that the derivative function of $V(r)=\frac{4}{3}\pi r^3$ is $V'(r)=4\pi r^2$, which is always positive, and that the tangent to $V(r)=\frac{4}{3}\pi r^3$ is always positive.

However the textbook gives the answer $-100\pi$, so I'm not sure.

The only other thing I tried was to use $-h$ in the definition, so $\lim_{h\to0}\frac{\frac{4}{3}\pi(5-h)^3-\frac{4}{3}\pi(5)^3}{h}$. When I did this, I got $-100\pi$, but I always thought the variable $h$ could be positive or negative by itself (i.e., $h\in R$). What is the implication of using $+h$ vs. $-h$ in the definition?

Edit: Here's the exact wording of the question:

Determine the instantaneous rate of change in the volume of a spherical balloon, as it's deflated, at the point in time when the radius reaches 5 cm

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The key thing is that you want the rate of change of the volume with respect to the time. What you have found in the limit is the rate of change with respect to the radius. So you want $$ V'(t) = \frac{dV}{dt} $$ and not $$ V'(r) = \frac{dV}{dr}. $$

The volume of a sphere is $V(r) = \frac{4}{3}\pi r^3$. Here the radius is a function of time, so we can write $r(t)$. Then the volume as a function of time is: $$ V(t) = \frac{4}{3}\pi r(t)^3. $$ Now taking the derivative (with respect to $t$ since you want to know how fast the volume is changing with respect to the time) you use the chain rule and get $$ V'(t) = \frac{4}{3}\pi 3r(t)^2r'(t) $$ Now, when the radius is $5$, then the instantaneous rate of change is $$ V'(t_0) = \frac{4}{3}\pi 3\cdot 5^2 r'(t_0) = 100\pi r'(t_0). $$ Here $t_0$ is the time corresponding to the radius being $5$. Now since the sphere is deflating, you have $r'(t_0)$ negative (I am guessing that $r'(t_0) = -1?$), and so $V'(t_0)$ is negative.

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  • $\begingroup$ Thanks for your answer. I edited my post to include the exact wording of the question in the text. Does it affect your impression that it's change in volume WRT time that the question is after? $\endgroup$ Commented Dec 30, 2013 at 16:41
  • $\begingroup$ @yroc: I would say the problem is worded badly. You always find the rate of change with respect to something. If it is with respect to $r$, then you get $V'(5) = 100\pi$ which is positive. If you find it with respect to time, then you get $v'(t_0) = 100\pi r'(t_0)$. You would need to know that $r'(t_0)$ is in order to find a value. $\endgroup$ Commented Dec 30, 2013 at 16:44
  • $\begingroup$ Unfortunately a problem in a lot of high school texts, I think. I'm pretty sure they intended rate of change WRT radius, because finding rate WRT time is too advanced at this point in our learning. Thanks so much! $\endgroup$ Commented Dec 30, 2013 at 16:49

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