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In solving some differential equations in physics, one has to do some peculiar manipulation of differentials to wind up with a desired function. My professor has warned up that doing so, "Will make math people very mad, but in this case, it's all that works."

The example I am referring to is if one has two first order differential equations:

$$\frac{d\vec{r}}{dt}=something$$ $$\frac{d\theta}{dt}=something_2$$

If one wants to express $\theta$ as a function of $\vec{r}$, the technique we learned is the following:

$$\frac{d\theta}{d\vec{r}}=\frac{something_2}{something}$$

I am wondering what, if anything is wrong with treating the differentials in this manner. In the past (Calculus 1}, we learned that one can treat the differential as an operator, and it is used to perform logarithmic differentiation.

So, I do not see any problems that this use may cause, but my professor is adamant that it is not mathematically correct.

If anyone could provide their stance on this, it would be appreciated.

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You have something upside down. If $\theta$ is on top, then $something_2$ should be on top as well. –  John Habert Jan 29 at 2:00
    
Indeed I did. Nice catch. –  Richard P Jan 29 at 2:01
    
the notation $\frac{d}{d\vec{r}}$ refers to a directional derivative, so $\frac{d\theta}{d\vec{r}}$ has to be computed via $\frac{d\theta}{d\vec{r}}=\vec{r}\cdot{\rm grad}\theta$ –  janmarqz Jan 29 at 2:40
    
@janmarqz No, it just means $grad\theta$. –  Keshav Srinivasan Jan 29 at 3:08
    
@KeshavSrinivasan.. why? –  janmarqz Jan 29 at 3:10

3 Answers 3

up vote 3 down vote accepted

There are two big clues that something rather suspect is going on:

The first is that you're ostensibly taking the derivative with respect to something that isn't a scalar variable. What might you mean by that? Sense can be made of such things... but just doing it blindly without at least a notion of what you think such a thing might mean is a bad idea.

The second, bigger red flag is that on the right hand side, you are trying to divide a scalar by a vector. It seems unlikely that can be made meaningful at all -- any reasonable results you might get from doing that are almost certainly because you were doing something else in your head and you completely ignored what you actually did.

If you would care to post (ideally in a new question) the context where you had the idea to do a manipulation like that, I'm sure someone can suggest a more reasonable way to do what you're trying to do. (or, if appropriate, explain why whatever you're trying to do is inherently flawed)


Although after some thought, I bet I can guess what you're trying to do. I bet you are considering a trajectory through $3$-space, and you are trying to do single-variable on that trajectory rather than doing multi-variable calculus in $3$-space.

If that is what you are really doing, then everything could be made to make sense -- but I'm out of practice and don't see immediately how to sort through the details.

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Your skills in deduction are excellent. It is indeed a trajectory through 3-space. And it can be treated as a single variable on the trajectory. Thank you for the informative post, as I said, it gives the correct solution, just in a convoluted way. –  Richard P Jan 29 at 22:02

Taking the derivative of a scalar with respect to a vector may look strange at first glance, but all it means is taking the vector obtained by taking the derivative of the scalar with respect to each component of the vector. (It's used by Landau and Lifshitz, for instance.) When the vector in question is the position vector $\vec{r}$, this is of course just the gradient. So $$\frac{df}{dt} = \frac{df}{d\vec{r}} \cdot \frac{d\vec{r}}{dt}$$ is just another way of saying

$$\frac{df}{dt} = \nabla f \cdot \frac{d\vec{r}}{dt},$$

which is a version of the multivariable chain rule.

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if you are considering that $\vec{r}$ depends on $t$ then $\vec{r}$ is a curve, and $\frac{d\vec{r}}{dt}$ is a tangent vector (field) along that curve so the formula for $\frac{df}{dt}$ means how $f$ varies in the direction $\frac{d\vec{r}}{dt}$. –  janmarqz Jan 29 at 14:38

It looks like you are just doing a variation of the Chain Rule and, as such, is perfectly fine mathematically. And based on downvotes, it seems that handwaving like your professor is doing does annoy people.

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Downvotes without comments are exactly useful. –  John Habert Jan 29 at 3:01
    
I am not one of the downvoters, but Hurkyl's answer might help to understand why they did so. Best wishes. –  Mark Fantini Jan 29 at 3:07
    
Hurkyl's answer wasn't there before. Thanks. I also honestly wasn't watching the vector notation carefully so it is my own fault. –  John Habert Jan 29 at 3:08

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