0
$\begingroup$

I know mechanically how to solve related rate problems. What I'm trying to do is understand the motivation or rationale behind taking the derivative with respect to time (or some other variable) when solving related rate problems.

For example, the equation for the volume of a sphere relates volume to its radius:

$$V = \frac{4}{3} \pi r^3$$

Because there's a relation between $V$ and $r$, we can infer that the rate at which $V$ changes is related to the rate at which $r$ changes and vice versa.

It makes sense to me that if we were given either the rate of change for volume or radius that we could solve for the other's rate of change.

What I don't fully understand is why take derivative of the equation with respect to time when it seems like $V$ is a function of $r$ (or at least related to it), so wouldn't it make sense to take the derivative with respect to the independent variable $r$ instead?

Also, taking the derivative of both sides of the equation with respect to $t$ assumes that the variables $V$ and $r$ are functions of $t$ (parametric equations?). But I don't understand the motivation for assuming this since the equation only shows the relation between $V$ and $r$.

As a recap my two questions are:

  1. Why take the derivative of an equation or function with respect to time instead of the independent variable?

  2. Why assume the variables are functions of time or some other variable not explicitly present in the equation?

$\endgroup$
2
  • $\begingroup$ Just my humble comment. Suppose you have a sphere full of water. If you empty it as time passes, the volume will change and the radius will also change as a function of the time. $\endgroup$ – Sebastiano Aug 21 '20 at 11:33
  • $\begingroup$ Because the "independent" variable $r$ depends on $t$. Also, parameterization is an important technique, even under some circumstances where the notion of the "third" variable is implicit. $\endgroup$ – justadzr Aug 21 '20 at 11:34
1
$\begingroup$

When concerned with rate related problems, it is often assumed that the functions that we want to study are dependent on time, although these time dependencies are sometimes omitted, or are obtained via a generalisation.

Now, it is true that for a given sphere, its volume $V$ depends on its radius $r$ via the relation $$V = \frac{4}{3}\pi r^3,$$ although this equation holds true for some fixed value of $r$, which in turn yields a fixed value of $V$.

Now, we can assume the existence of time in our equation, and then assume that our radius $r$ varies with time $t \in [0,T]$ for some $T > 0$. Then, indeed $r$ is a function of time, i.e. $r : [0,T] \to \mathbb{R}$. Since we know what the volume of a sphere is with respect to its radius for a fixed value of radius, we can generalise the above volume equation to: $$V(t) = \frac{4}{3}\pi (r(t))^3,$$ where the time dependence of the volume came about due to the time dependence of the radius. That is, we can apply the time independent equation for each time $t \in [0,T]$.

Thus, if we want to study the rate of change of the variable volume $V(t)$, we would differentiate with respect to its time variable, to obtain $$\frac{d V(t)}{dt} = 4\pi \frac{dr(t)}{dt} (r(t))^2.$$

If say we have a function $r : [0,T] \to \mathbb{R}$, but the radius is constant with respect to $t \in [0,1]$, i.e. $r(t) = r_0$ for some $r_0 \in \mathbb{R}$, then $\frac{dr(t)}{dt} = 0$ for all $t$, thus giving us that $\frac{dV(t)}{dt} = 0 $ for all $t$. Therefore $V$ is constant, i.e. $V(t) = V_0$ for some $V_0 \in \mathbb{R}$ for all $t \in [0,T]$.

Looking back at the generalised equation for the volume (with respect to time), we therefore obtain $$V_0 = \frac{4}{3}\pi r_0^3$$ which is precisely the time independent equation we have described above.

To answer the questions more precisely:

  1. It does not make sense to take the derivative of $V$ with respect to $r$ to determine the rate of change of $V$ with respect to time, $t$. Unless say you wanted to determine the rate of change of $V$ with respect to its radius, in which case you can indeed differentiate with respect to $r$.

  2. This is done as to bring the time variable that we want to study into our problem, I've described this above.

$\endgroup$
3
  • $\begingroup$ This makes more sense now. Especially the part about $r$ varies with time means that $r$ is a function of time, $r(t)$, and since $V$ is a function of $r$ then it's implied that $V$ must be a function of time as well. Thinking back it makes sense now that the rates of change were often given as per units of time. For example, the radius changing at a rate of 5cm/sec, which mathematically would be $\frac{dr}{dt}=5$. And this notation informs us that $r$ must be a function of $t$ since derivatives are taken with respect to the independent variable. Thank you for your help! $\endgroup$ – Slecker Aug 21 '20 at 12:19
  • $\begingroup$ This explanation also ties in nicely with the chain rule. That a change in $t$ will cause a change $r$ which will then cause a change $V$. $\endgroup$ – Slecker Aug 21 '20 at 12:28
  • $\begingroup$ Glad it helped :) $\endgroup$ – Spaceman Aug 21 '20 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.