So I'm trying to understand the derivation for the line element in spherical coordinates as well as the derivation for the basis vectors. Both hinge on the question posed in the title. So here's what I have so far in figuring this out:
In general, $\mathrm d\mathbf r = \frac{\partial\mathbf r}{\partial r}\mathrm dr+\frac{\partial\mathbf r}{\partial\theta}\mathrm d\theta+\frac{\partial\mathbf r}{\partial\varphi}\mathrm d\varphi$, but if we make $\mathrm d\theta=\mathrm d\varphi=0$ then the equation becomes $\mathrm d\mathbf r = \frac{\partial\mathbf r}{\partial r}\mathrm dr$, and also $\mathrm d\mathbf r$ points in the $\hat{\boldsymbol r}$ direction, since there is no change in the $\theta$ or $\varphi$ directions of the infinitessimal displacement. The next step would be to make the infinitesimal displacement a unit vector, and we accomplish this by diving by the magnitude. And here's where I'm getting stuck. I feel like the magnitude of the infintessimal displacement should be $\left\lvert\frac{\partial\mathbf r}{\partial r}\mathrm dr\right\rvert$ since $\mathrm d\mathbf r = \frac{\partial\mathbf r}{\partial r}\mathrm dr$, but of course the book is writing the magnitude simply as $\left\lvert\frac{\partial\mathbf r}{\partial r}\right\rvert$. What am I missing here?
Okay, that was a stupid question, and the answer was staring me in the face the whole time. If $\frac{\partial\mathbf r}{\partial\mathrm r}\mathrm dr$ points in the same direction as $\hat{\boldsymbol r}$ then so does $\frac{\partial\mathbf r}{\partial\mathrm r}$ and therefore $\hat{\boldsymbol r}=\frac{\frac{\partial\mathbf r}{\partial\mathrm r}\mathrm dr}{\left\lvert\frac{\partial\mathbf r}{\partial\mathrm r}\mathrm dr\right\rvert}=\frac{\frac{\partial\mathbf r}{\partial\mathrm r}}{\left\lvert\frac{\partial\mathbf r}{\partial\mathrm r}\right\rvert}$.