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

1. Apr 26, 2013

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?

2. Apr 26, 2013

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.

3. Apr 26, 2013

WannabeNewton

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

Last edited: Apr 26, 2013
4. Apr 27, 2013

jamesb1

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