Proof of change in position vector in spherical coordinates

Question

I have found it hard to proof that ${d\vec r=dr\hat r+rd\theta\hat \theta}$ in spherical coordinates. Also it would be great if somebody can explain what ${d\vec r}$ is because I read different things in papers and I got stuck with all these notations and fundamental definitions.

Answer 1

${\vec r=\big(r\cos\theta\sin\phi,r\sin\theta\sin\phi,r\cos\phi\big)}$ is position vector in spherical coordinates. $${d\vec r=\frac{\partial \vec r}{\partial r}dr+\frac{\partial \vec r}{\partial \theta}d\theta+\frac{\partial \vec r}{\partial \phi}d\phi=\frac{\frac{\partial \vec r}{\partial r}}{\bigg|\frac{\partial \vec r}{\partial r}\bigg|}\bigg|\frac{\partial \vec r}{\partial r}\bigg|dr+\frac{\frac{\partial \vec r}{\partial \theta}}{\bigg|\frac{\partial \vec r}{\partial \theta}\bigg|}\bigg|\frac{\partial \vec r}{\partial \theta}\bigg|d\theta+\frac{\frac{\partial \vec r}{\partial \phi}}{\bigg|\frac{\partial \vec r}{\partial \phi}\bigg|}\bigg|\frac{\partial \vec r}{\partial \phi}\bigg|d\phi=\hat r\bigg|\frac{\partial \vec r}{\partial r}\bigg|dr+\hat \theta\bigg|\frac{\partial \vec r}{\partial \theta}\bigg|d\theta+\hat \phi\bigg|\frac{\partial \vec r}{\partial \phi}\bigg|d\phi}$$ and since we can find the magnitudes of vectors in parentasis, we get: $${d\vec r=dr\hat r+rd\theta\hat\theta+r\sin\theta d\phi\hat\phi}$$ This is the derivation in three dimensions.