I am trying to better understand related rates problems. Prior to being exposed to related rates problems, the derivatives I solved were always with respect to the independent variable of the equation, e.g. $y=x^2, \frac{dy}{dx}=2x$. With related rates, the problems are solved by taking the derivatives with respect to time, i.e.$\frac{dy}{dt}=2x\frac{dx}{dt}$. Although I am able to solve the problems, I am not comfortable with the switch from taking a derivative with respect to the independent variable x to taking the derivative with respect to a third variable time. I’ve thought long and hard about my issue and I think I have a solution. Would it be correct to say that the instantaneous rate of change of a function with respect to its independent variable, i.e. its derivative, gives the ratio of related rates? In other words, I can determine the ratio of related rates at a specific point by finding the instantaneous rate of change of the function at that point.

For example, if I wanted to know the rate of change with respect to time of the area of a circle given the rate of change in radius with respect to time, I would start by taking the derivative of the area equation $A=πr^2$ with respect to r to find the instantaneous rate of change of A with respect to r; then to find the rate of change of area with respect to time at a specified radius, I would find the instantaneous rate of change at that radius and multiply it by the rate of change of radius with time, i.e. $\frac{dA}{dr}=\frac{d}{dr}[\pi r^2]=2\pi r$, $\frac{dA}{dt}=\frac{dA}{dr} \frac{dr}{dt}$, $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$.

If the problem involved more than one independent variable, I could find the contribution from each independent variable separately and sum them together. For example, if I wanted to know the rate of change in the volume of a cylinder when both the height and radius are changing with time. I could find the instantaneous rate of change of the volume with respect to the cylinder height and use it to find the rate of change of volume with time due to the change of height with time, i.e. $\frac{dV}{dh}=\frac{d}{dh}[\pi r^2h]=\pi r^2$, $\frac{dV}{dt}=\frac{dV}{dh} \frac{dh}{dt}$, $\frac{dV}{dt}=\pi r^2\frac{dh}{dt}$; and then find the instantaneous rate of change of the volume with respect to the cylinder radius and use it to find the rate of change of volume with time due to the change of radius with time, i.e. $\frac{dV}{dr}=\frac{d}{dr}[\pi r^2h]=2\pi rh$, $\frac{dV}{dt}=\frac{dV}{dr} \frac{dr}{dt}$, $\frac{dV}{dt}=2\pi rh\frac{dr}{dt}$. The total rate of change of volume with time would be the sum of the change due to the height changing with time plus the change due to the radius changing with time, i.e. $\frac{dV}{dt}=\pi r^2\frac{dh}{dt}+2\pi rh\frac{dr}{dt}$.

Am I thinking about the math correctly? Thank you for your insight.

  • $\begingroup$ have a look at the Chain Rule of differentiation en.wikipedia.org/wiki/Chain_rule and the Product Rule en.wikipedia.org/wiki/Product_rule $\endgroup$ Commented Apr 2, 2018 at 16:31
  • $\begingroup$ @David Quinn, Hi David, I looked at the information in wikipedia and found it very useful. I believe it helped me get on the right track. As a result, I tried to answer my own question. Please let me know if I made any sense. It is difficult to write about math. $\endgroup$
    – SLax
    Commented Apr 6, 2018 at 2:19

2 Answers 2


I am answering my own question for two reasons: (1) I believe I now understand my problem and (2) because I believe it is not easy for readers of my question to determine the source of my misunderstanding. What I failed to fully understand and apply was the chain rule. As long as the problems I worked were in terms of a single variable, i.e. $y=(sin(x))^2$, I could successfully apply the chain rule. Although, when faced with related rates problems, I failed to realize that despite the base equation not being explicitly in terms of time $t$, it could be if the time based equations were given. But many related rates problems provide the rates of change rather than an equation describing the parameters as a function of time. Take for example, the following problem:

The radius of a cylinder is decreasing at the rate of 9 mm/hr while its height is increasing at the rate of 2 mm/hr. What is the rate of change in the surface area of the cylinder when the radius is 8 mm and the height is 3 mm?

The base equation is:$$A=2\pi r^2+2\pi rh$$ What must be understood is that the area function is a composite of two sub functions r(t) and h(t). If, rather than giving the rate of change of radius, rate of change of height, radius, and height, the time based equations for radius and height were given, the governing function could be written solely in terms of time t. For example, if the following equations were given, $$r(t)=-9t+44$$ $$h(t)=2t-5$$ (note: at $t=4$, $r=8$, $h=2$, $\frac{dr}{dt}=-9$, $\frac{dh}{dt}=2)$

then the base equation could be written solely in terms of time $t$. $$A(t)=2\pi (-9t+44)^2+2\pi (-9t+44)(2t-5)$$ $$A(t)=126\pi t^2-1318\pi t+3432$$ This equation can be differentiated without the chain rule, i.e. $$\frac{dA}{dt}=252\pi t - 1318\pi$$ At time t=4, $$\frac{dA}{dt}=252\pi (4)-1318\pi = -310\pi$$ But the chain rule gives us an easier way to solve the problem. If we think of the base function as the following: $$A(t)=2\pi (r(t))^2+2\pi (r(t))(h(t))$$ then $A(t)$ can be seen as a composite of two functions $r(t)$ and $h(t)$. The chain rule allows the derivative to be taken in several steps. For example, $$\frac{dA}{dt}=2\pi \frac{d}{dt}[(r(t))^2]+2\pi\frac{d}{dt}[(r(t))(h(t))]$$ $$\frac{dA}{dt}=2\pi (2r(t))\frac{d}{dt}[r(t)]+2\pi\bigl(\frac{d}{dt}[r(t)]h(t)+r(t)\frac{d}{dt}[h(t)]\bigr)$$ $$\frac{dA}{dt}=4\pi r(t)\frac{d}{dt}[r(t)]+2\pi h(t)\frac{d}{dt}[r(t)]+2\pi r(t)\frac{d}{dt}[h(t)]$$ $$\frac{d}{dt}[r(t)]=-9$$ $$\frac{d}{dt}[h(t)]=2$$ $$\frac{dA}{dt}=4\pi r(t)(-9)+2\pi h(t)(-9)+2\pi r(t)(2)$$ $$\frac{dA}{dt}=-36\pi r(t)-18\pi h(t) + 4\pi r(t)$$ $$\frac{dA}{dt}=-36\pi(-9t+44)-18\pi(2t-5)+4\pi(-9t+44)$$ $$\frac{dA}{dt}=252\pi t - 1318\pi$$ At time t = 4 hrs, $$\frac{dA}{dt}=252\pi(4)-1318\pi=-310\pi$$ Now back to the original problem. Because the problem didn't give the equations for r(t) and h(t), but rather gave the values at a snap shot in time, i.e. $r=8 mm$, $\frac{dr}{dt}=-9 mm/hr$, $h=3 mm$, $\frac{dh}{dt}=2 mm/hr$, it is necessary to use the chain rule. $$\frac{dA}{dt}=2\pi\frac{d}{dt}[r^2]+2\pi\frac{d}{dt}[rh]$$ $$\frac{dA}{dt}=2\pi(2r)\frac{dr}{dt}+2\pi(\frac{dr}{dt}h+r\frac{dh}{dt})$$ $$\frac{dA}{dt}=4\pi r\frac{dr}{dt}+2\pi h\frac{dr}{dt}+2\pi r\frac{dh}{dt}$$ $$\frac{dA}{dt}=4\pi(8)(-9)+2\pi(3)(-9)+2\pi(8)(2)$$ $$\frac{dA}{dt}=-310\pi$$ So the moral of the story is that related rates problems are not anything new, they are applying the chain rule to a composite function of time.

  • $\begingroup$ That is correct. $\endgroup$ Commented Apr 6, 2018 at 16:33

While you get the correct results, you don't offer any reasoning that explains your motivation for doing this, which means we cannot tell if you are thinking about the math correctly.

Quite frankly, it makes me suspicious that you are looking at the chain rule (single and multi-variable) - which can be found in many places - and just tossing operations together that will get you to them. If so, then you are most certainly not thinking about the math correctly. These are not pulled out of thin air. They are developed from principles.

First of all, it is not "dependent $y$, independent $x$, third variable $t$". Instead, $t$ is independent, $x$ is dependent on $t$, and $y$ is dependent on $x$ (which makes it dependent on $t$ as well).

(What follows is "heuristic", which means it gives only the broad outline, skipping over fiddly details that are important to serious mathematics, but get in the way of following the overall logic.)

So what happens if we make a change $\Delta t$ in the value of $t$? Then $x$ changes by some amount $\Delta x$, and in turn, this causes the value of $y$ to change by some amount $\Delta y$. The ratio of the change in $y$ with respect to the change in $t$ is $$\frac{\Delta y}{\Delta t}$$ Assuming that $\Delta x \ne 0$, we can multiply and divide by it without changing the overall value, and do some rearranging: $$\begin{align}\frac{\Delta y}{\Delta t} &= \frac{\Delta y\Delta x}{\Delta t\Delta x}\\&= \frac{\Delta y\Delta x}{\Delta x\Delta t} \\&= \frac{\Delta y}{\Delta x}\frac{\Delta x}{\Delta t}\end{align}$$ If we let $\Delta t \to 0$, then $\frac{\Delta y}{\Delta t} \to \frac {dy}{dt}$ and $\frac{\Delta x}{\Delta t} \to \frac {dx}{dt}$. This only converges, though, if $\Delta x \to 0$. But when that holds then, $\frac{\Delta y}{\Delta x} \to \frac {dy}{dx}$. Thus the previous calculation becomes the chain-rule: $$\frac {dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}$$ Note that even though math teachers will warn you not to think of derivatives as a simple division, the chain rule really does come from a cancellation of the "dx" parts (but before the limit is taken).

In the multivariate case, functional notation is useful: $z = f(x, y)$ (two variables is easier to follow, but it scales up to any number of variables). Now we suppose that $x$ and $y$ are themselves dependent on some variable $t$ (and so $z$ will be dependent on $t$ as well). Then a change in $t$ induces a change in each of the other variables:

$$\Delta z = f(x + \Delta x, y + \Delta y) - f(x,y)$$ By adding and subtracting $f(x, y + \Delta y)$ we can get changes in one variable at a time: $$\Delta z = [f(x + \Delta x, y + \Delta y) - f(x, y + \Delta y)] + [f(x, y + \Delta y) - f(x,y)]$$ Similarly treating each of the two differences as before: $$\frac{\Delta z}{\Delta t} = \frac{f(x + \Delta x, y + \Delta y) - f(x, y + \Delta y)}{\Delta t} + \frac{f(x, y + \Delta y) - f(x,y)}{\Delta t}\\=\frac{f(x + \Delta x, y + \Delta y) - f(x, y + \Delta y)}{\Delta x}\frac{\Delta x}{\Delta t} + \frac{f(x, y + \Delta y) - f(x,y)}{\Delta y}\frac{\Delta y}{\Delta t}$$

Now as $\Delta y \to 0$, $$\frac{f(x, y + \Delta y) - f(x,y)}{\Delta y} \to \frac{\partial z}{\partial y}$$ where we use the "partial derivative" $\partial$ to remind you that the value of this derivative depends not only on how $z$ changes with $y$, but also on the particular variable $x$ that was the "partner" of $y$. I.e., if we had another variable $u$ dependent on $t$ such that $z(t) = f(x(t), y(t)) = g(u(t), y(t))$ for the same $y$, we would find in general that $$\frac{\partial f}{\partial y} \ne \frac{\partial g}{\partial y}$$ even though both can be expressed as $\frac{\partial z}{\partial y}$. Even though it isn't explicitly mentioned, $\frac{\partial z}{\partial y}$ depends on whether $y$ is being paired with $x$ or with $u$.

Similarly as $\Delta x \to 0$, $$\frac{f(x + \Delta x, y + \Delta y) - f(x,y + \Delta y)}{\Delta y} \to \frac{\partial z}{\partial x}(x,y + \Delta y)$$ which, when $\Delta y \to 0$ as well, goes to $\frac{\partial z}{\partial x}(x,y)$.

So letting $\Delta t \to 0$, which also sends $\Delta x \to 0$ and $\Delta z \to 0$, we get:

$$\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$$

If you think of a rectangle in the $xy$ plane with sides of $dx$ and $dy$, then $dz$ jumps from the value at the corner $(x, y)$ to the diagonally opposite corner $(x+dx, y+dy)$. But $\frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy$ goes around the side, breaking the path into two parts, when first only $x$ changes, then only $y$ changes. In the end, you get to the same place, but by a different path.

  • $\begingroup$ I appreciate the time and effort you put into your answer. If I appear to be tossing operations together and pulling them from thin air, then it is obvious that I am not understanding. I was attempting to get an intuitive understanding of related rates problems by thinking about the different rates of change in the problem. For example, in my circle area example, trying to understand the relationship between the rate of change of area with radius and the rate of change of radius with time. $\endgroup$
    – SLax
    Commented Apr 3, 2018 at 21:08
  • $\begingroup$ I am trying to gain a more intuitive feel for what is happening. Perhaps I just need to do more real-world problems involving the chain rule. $\endgroup$
    – SLax
    Commented Apr 3, 2018 at 21:09
  • $\begingroup$ Sorry i wasn't more help. It is often a guessing game as to what a poster needs. $\endgroup$ Commented Apr 3, 2018 at 23:46
  • $\begingroup$ Hi Paul, I think I figured out my problem. Knowing that my question was hard to decipher, I tried to answer my own question. Please let me know if it makes sense. Writing about math is very difficult. I admire you for answering questions like mine. $\endgroup$
    – SLax
    Commented Apr 6, 2018 at 2:16

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