# What can go wrong with applying chain rule to angular velocity of circular motion?

Lets say I have a circular motion, like this:

I know that:

$$\omega = \frac{\text{d} \phi}{\text{d}t}$$

Mathematically, what I am doing wrong, when I attempt to apply the chain rule in the following way?

$$\omega = \frac{\text{d}\phi}{\text{d}t}=\frac{\text{d}r}{\text{d}t}\frac{\text{d}\phi}{\text{d}r}(=0?)$$

Which result seems wrong because $$\frac{\text{d} r}{\text{d} t}$$ is zero, since my $$r$$ coordinate is not changing. So $$\omega$$ would be $$0$$ for all circular motion. What am I doing wrong?

This is not a homework problem, but inspired by one.

• this doesn’t work because $\phi$ is not a function of r – Eli Feb 27 at 22:32
• The chain rule is used when you have functions which are constructed as the composition of functions and or variables. Additionally you can see the product rule and division are simple cases of it when you have either $f(x)*g(x)$ or $f(x)/g(x)$. The most general way I like to do it is, draw a graph or lattice of all the functions. There is a way to do it with that but I forget it exactly – marshal craft Feb 28 at 11:05

The reason is that $$r(t)$$ and $$\phi(t)$$ are your independent variables, so you are not allowed to use the chain rule for them, because $$\phi$$ does not depend on $$r$$. An analogy would be the following: $$$$\frac{d}{dx}x=1$$$$ However, if I introduce another independent variable $$y$$, it would be of course wrong to write $$$$\frac{d}{dx}x=\frac{dy}{dx}\frac{d}{dy}x=0$$$$

Though $$\frac{\mathrm{d}r}{\mathrm{d}t}$$ is $$0$$, $$\frac{\mathrm{d}\phi}{\mathrm{d}r}$$ also tends to infinity. So this is a zero into infinity form and its limit will result in a finite quantity which will be equal to the angular velocity calculated without chain rule .

• If "finite" is changed to "indefined", this answer will be "the" answer. The answer by Ruben Campos Delgado and user256872 are not wrong, but they are not directly-to-the-point and may cause confusion to newbies. – verdelite Feb 27 at 16:55
• I think this is the correct answer. The reason for which $d\phi/dr$ is infinity is that $d\phi/dr=(dr/d\phi)^{-1}$ where $dr/d\phi=0$ since the radius does not depend on the angle – pp.ch.te Mar 1 at 20:04

The meaning of pushing an intermediate variable is to relate the dependencies of change. So, the expression

$$\frac{d\phi}{dt} = \frac{d \phi}{dr} \frac{dr}{dt}$$

Assumes that the angle $$(\phi)$$ can be written as some function of the radius from the origin i.e: $$\phi(r(t))$$ but is that really possible in this case? We can say that there is no 'differentiable' map from $$r \to \phi$$.

I mean think about it, how would you construct a function which associates the radius , a fixed variable, with the angle which changes with time? It would not even be a function because you would have to span all the angles with a single radius value.

• Seem the angle could depend on radius or anything else. – marshal craft Feb 28 at 11:17
• I didn't get you, could you rephrase your point @marshalcraft – Buraian Feb 28 at 11:18
• Hmm I dlwas commenting cause I down voted, perhaps I did not read your answer closely enough and was in error, however the system has locked the vote. – marshal craft Feb 28 at 13:22

The problem is that $$r$$ is a constant, so its rate of change relative to any other variable must be zero. Thus $$\frac{\text{d}r}{\text{d}t} = 0$$ and $$\frac{\text{d}r}{\text{d}\phi} = 0$$

So $$\frac{\text{d}\phi}{\text{d}r}$$ is undefined. When you try to apply the chain rule in

$$\omega = \frac{\text{d}\phi}{\text{d}t}=\frac{\text{d}r}{\text{d}t}\frac{\text{d}\phi}{\text{d}r} = \frac{\text{d}r}{\text{d}t} / \frac{\text{d}r}{\text{d}\phi}$$

you're attempting to divide zero by zero, which is an indeterminate form.

$$d\phi/dr$$ is undefined. $$\phi$$ is only defined for 1 value of $$r$$ and is thus not differentiable. So you end up with $$\omega$$ being zero x undefined quantity.

Since you've got vectors on your diagram, I think it's worth noting that you could write $$\omega = \frac{\mathrm{d}\phi}{\mathrm{d}t} = \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \cdot \vec{\nabla}\phi$$ because $$\vec{r}$$ is a function of $$t$$.

$$dr/ dt ≠0$$ as direction of $$r$$ is continuously changing. $$r$$ means both direction as well as magnitude, you are only considering the magnitude, which is wrong.

• here $r$ is not a vector: it’s the (scalar) distance from the origin. – ZeroTheHero Feb 28 at 2:13
• Typically in physics its a vector i do think, but yeah shows just an scalar, and then the basis vectors show little hat for vector typically. And them you see on basis vector has r as subscript. So not sure exactly what is really going on with the askers drawing myself. – marshal craft Feb 28 at 11:16