Proving that the line integral of a vector field $\vec F$ along a curve of length $c$ to which it is always tangent is $||\vec F|| * c$

For a vector field $$\vec F$$ and a curve $$C$$ with length $$c$$, if $$\vec F$$ is tangent to $$C$$ at every point, then the line integral of $$\vec F$$ along $$C$$ is:

$$\int_C \vec F \cdot d\vec r = \int_C ( ||\vec F|| *||d\vec r||)= ||\vec F|| *\int_C||d\vec r|| =||\vec F|| *c$$

I use $$\cdot$$ to mean the dot product and $$*$$ to mean multiplication.

I get to $$||\vec F|| * ||d\vec r||$$ because $$\vec F$$ is always tangent to $$C$$, which means $$\cos\theta$$ in the dot product formula is $$1$$. I then take $$||\vec F||$$ out of the integral, since it's a scalar.

The next step is the dubious one: I have an integral $$\int_C||d\vec r||$$. Is this technically a valid integral? I don't know if I'm allowed to play around with the differential in the integral like that.

Furthermore, would it even be correct to say that $$\int_C||d\vec r|| =$$ (length of the curve $$C$$) ?

Any help is greatly appreciated!

• The answer given below says everything that's needed, and just to emphasise another mistake: $\vec{F}$ being tangent to the curve doesn't mean $\cos$ of the angle is $1$. It could be $-1$ (the vectors could be parallel, but pointing in opposite directions). Dec 8 '19 at 14:20
• @peek-a-boo Indeed, forgot about that, thank you! Dec 9 '19 at 14:04

$$||\vec{F}||$$ need not be constant, so it is invalid to move it out of the integral. Consider $$\vec{F}(r,\theta, \phi) = (r,\theta, \phi)$$ with a path that starts at a point and proceeds radially outward, so that $$\vec{F}$$ is tangent to every point on the path. Here, $$||\vec{F}(r,\theta, \phi)|| = r$$, which is not a constant and so does not migrate out of the integral with respect to $$r$$. (Also, since $$||\vec{F}||$$ is not constant, which of its many values are you writing in front of the integral?)
Also, you should be careful with $$|\mathrm{d}\vec{r}|$$. If the path $$C$$ starts at a point, proceeds radially outwards, stops, then proceeds radially inward back to its starting point, $$\int F \cdot \mathrm{d}\vec{r} = 0$$, but $$\int \vec{F} \cdot |\mathrm{d}\vec{r}| = 0$$ is twice the integral along the part of the path that is just radially outward.
• That all makes sense, thank you for the help! So I guess the only way what I wrote makes sense is if $\vec F$ is a constant vector field (or would it sitll not make sense?) Also, so $\int \vec F \cdot |d\vec r|$ is a valid integral (that's equal to the distance traveled)? Isn't it a breach of notation to wrap the differential of an integral in magnitude bars like that? Dec 9 '19 at 14:03
• Recall that $\mathrm{d}\vec{r}$ is the placeholder for the $\Delta r$ in the Riemann sum underlying the integral. One can also take $|\Delta r|$, where the magnitude of the change in $r$ is what one wants (for whatever quantity one is calculating, for instance while rectifying a curve). It is probably not what you want in a dot product since $|\mathrm{d}r|$ is a scalar and, depending on your definitions, either the dot product of a vector and a scalar is undefined or is zero. Dec 9 '19 at 23:42
• Hm okay, I'll keep that in mind. Last question (sorry to keep bothering): I'm now not sure as to when I can move $||\vec F||$ out of the integral. If $||\vec F||$ is not constant, is it always invalid to move it out of $\int_C ||\vec F|| \cdot ||d\vec r||$? Or can you still move $||\vec F|| out in some cases, even if it's not constant? If so, which cases would those be? Dec 10 '19 at 16:32 • Generally, never. Use explicit notation: $$\int_C ||\vec{F}(r)|| \cdot ||\mathrm{d}\vec{r}|| = ||\vec{F}(\text{what goes here?})|| \int_C ||\mathrm{d}\vec{r}|| \text{.}$$ When you suppress the arguments, it is easier to wrongly believe that a varying function is a constant, but this is a very confused thing to do. Dec 10 '19 at 16:42 • Got it! Seeing it explicitly like that helps a lot. I was always thinking of$\vec F(x,y)$, but I realize now that$d\vec r$is just$dxdy$, and of course you can't move$\vec F(x,y)$out of$\int \vec F(x,y) dxdy\$. Thanks again for all the help! Dec 10 '19 at 16:44