Dr/dt = (dr/du)(du/dt) + (dr/dv)(dv/dt)

[jamesb1]

Sometimes I see the equation in the style as seen in the topic title (the d is the partial derivative) during vector analysis lectures and I get confused.

Its like saying dr/dt = dr/dt + dr/dt which doesn't make a lot of sense is it not?

 

[Solkar]

if $$r \equiv r(u,v)$$ then
$$dr = \frac{\partial r}{\partial u} du + \frac{\partial r}{\partial v} dv$$
is the total differential.

Divide that symbolically by dt.

 

[WannabeNewton]

Its like saying dr/dt = dr/dt + dr/dt which doesn't make a lot of sense is it not?

No it isn't and the reason is because you can't just divide out $\partial u$ and $du$ like that. The former is a nonsense expression. This is where the horrible practice of "dividing out" differentials from single variable calculus comes back to bite you. The expression in the title comes out of the chain rule.

For a proof of this chain rule, see here (Theorem 6): http://math.bard.edu/belk/math461/MultivariableCalculus.pdf

 

[jamesb1]

Wow I just realized I made a very silly misconception with regards to the partial and total derivatives. I fully understand now! Thank you for your help :)