I have a problem. Volume of the sphere is given as: $$V(r) = \frac{4}{3}\pi r^3$$ I am then told (part A of the question) to find the average rate of change of V with respect to r, when r changes from 5 ft to 8 ft. (exact quote)

I know we're going to form up the equation: $$\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}$$

After deriving $V(r)$, we get: $$\frac{dV}{dr} = 4\pi r^2$$ Is this correct so far?

Now, I don't have a RATE of change with respect to the radius. I have a change from 5 feet to 8 feet, but it does not indicate a duration of time over which this occurs. Is this missing or am I misunderstanding what I'm needing to require.

The answer to this question states that part A is $387 ft^3$

The problem is that I have no idea how they got to the answer.

I THINK that I'm supposed to use a time range of 1 minute for part A, meaning $\frac{dr}{dt} = 3 ft/min$. However, $$\frac{dV}{dr}\frac{dr}{dt} = 4\pi (5^2)(3 ft/min) = 942.4778$$ which is DEFINITELY not the answer.

Where am I going wrong?

  • 1
    $\begingroup$ If you are looking for the "average" rate of change, couldn't you just calculate the volume at r=5 and r=8 and not use calculus? I've never seen a question ask for the average rate of change in this context. $\endgroup$ – turkeyhundt Oct 29 '16 at 2:07

You're thinking far too hard. A derivative finds instantaneous rate-of-change - which is explicitly not what the question was asking for. You have correctly computed the instantaneous rate of change with respect to time at the moment that $r = 5$; this is more or less irrelevant to the average rate of change.

Average rate of change is much simpler to compute, and takes no calculus whatsoever. For example, if I travel $500$ miles in $8$ hours, the average rate of change of my position is just $500/8 = 62.5 mph$, even though at some points in time my instantaneous rate of change may be much higher.

Secondly, you computed $\frac{dV}{dt}$; the question is asking with respect to r, which means that even if it were asking for instantaneous rate of change you'd want $\frac{dV}{dr}$ instead.

So what you want is just $\frac{\Delta V}{\Delta r}$ - which is just $\frac{V(8) - V(5)}{8 - 5}$.

  • $\begingroup$ In which case, I should just use $\frac{4}{3}\pi r^3$ substituting 5 and 8 respectively, then dividing by 3. This gives me 540.3539. $\endgroup$ – UtahJarhead Oct 29 '16 at 2:37
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    $\begingroup$ It looks to me like there's an error in the solutions. $387$ happens to be $8^3 - 5^3$; oddly, they seem to have stopped there and neglected both the division by $3$ and the coefficient of $\frac{4}{3}\pi$. It's worth noting that this was suspicious to begin with - the rate of change of something involving $\pi$ is almost never a whole number. I'd recommend asking your teacher or professor to confirm this. $\endgroup$ – Reese Oct 29 '16 at 2:42
  • $\begingroup$ Is this math correct and what you get as the correct answer? $\frac{4}{3}\pi 5^3 = 523.599$, $\frac{4}{3}\pi 8^3 = 2144.66$. $(2144.66 - 523.599)/3 = 540.3536$ $\endgroup$ – UtahJarhead Oct 29 '16 at 2:44
  • $\begingroup$ Also, you are correct. There are MANY instances of us correcting him during in-class demonstrations, so this would not surprise me. $\endgroup$ – UtahJarhead Oct 29 '16 at 2:45
  • 1
    $\begingroup$ Yes, that solution's correct. $\endgroup$ – Reese Oct 29 '16 at 2:45

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