What is the association between two separate rates of change in finding the derivative of the volume of a sphere?

Question

Imagine a balloon being inflated. The balloon has a volume $V$ and a radius $r$, which are both functions of time $t$. Finally, suppose $\frac{dV}{dt}$ is constant.

Is the following fact true, and if so why? (Or why not?)

$$\frac{dV}{dr} = \bigg(\frac{dV}{dt}\bigg)\bigg(\frac{dV}{dr}\bigg)$$

I understand how $\frac{dV}{dr}$ represents the rate of change of $V$ with respect to $r$, but not how both quantities, as functions of time, can be multiplied in that manner to produce $\frac{dV}{dr}$.

Thank you.

Answer 1

I think what you're confused about is the idea that a seemingly uninvolved intermediary variable can be used to determine a relationship between two other variables. The following example may help clear things up:

Suppose that Bob sells apples for $5$ dollars each. Bob also loves to eat cookies, and suppose a cookie costs $1$ dollar. Then, for each apple Bob sells, he makes $5$ dollars. For each dollar Bob earns, he can buy one cookie. If the number of apples sold is $A$, the number of dollars Bob earns is $D$, and the number of cookies Bob can buy is $C$, then:$$\text{5 Cookies per Apple}=\frac{dC}{dA}=\frac{dC}{dD}\cdot\frac{dD}{dA}=\text{($1$ Cookie per Dollar)$\cdot$($5$ Dollars per Apple)}$$

Now, let's see what this means in the context of your problem. $V$ is the volume of the balloon, and $r$ is the radius, so $t$ is our "intermediary variable" (much like how "Dollars=$D$" was in the example). Then, $\frac{dV}{dr}$ is "volume per radius"—in other words, "how much space the balloon takes up based on its radius". $\frac{dV}{dt}$ is the "volume per time"— in other words, how fast the balloon is growing per time unit (e.g. cubic meters per second). Lastly, $\frac{dt}{dr}$ is "seconds per radius"— in other words, if the radius grows by $R$ length units, then we have been inflating the balloon for $T$ time units. Bringing it all together, we have $$\frac{dV}{dr}=\frac{dV}{dt}\cdot\frac{dt}{dr}$$

Answer 2

By chain rule we have

$$\frac{dV}{dr} = \bigg(\frac{dV}{dt}\bigg)\bigg(\frac{dt}{dr}\bigg)$$

indeed recall that

$$V=\frac43 \pi r^3 \implies \frac{dV}{dr} =4\pi r^2$$

and

$$\frac{dV}{dt} =4\pi r^2\frac{dr}{dt}$$