# Multivariable Calculus - chain rule on vectors

1. Sep 10, 2013

### Nikitin

Hi! I do not understand the math used in the beginning of this video:

In example 1 (4 minutes in the video), why is it wrong to simply solve the problem like this:

$$\vec{V} = [x,-y] \Rightarrow \frac{d\vec{V}}{dt} = [\frac{dx}{dt},-\frac{dy}{dt}] = \vec{a} = [V_x,-V_y]$$, where V_x and V_y are the velocity-components in the x and y directions, respectively.

I thought you'd only use the chain-rule on non-vector multivariable functions??

EDIT: I'm farily sure the guy did some mistakes.. did he not? Look at his work 5:00 minutes in.

Last edited by a moderator: Sep 25, 2014
2. Sep 10, 2013

### HallsofIvy

Staff Emeritus
Perhaps you have not learned Calculus on vector valued functions? (It is typically introduced in "Calculus III" but there are whole books on the subject.) If (f(t), g(t), h(t)) is a (three dimensional) vector valued function of the variable t, then differentiation is defined "component-wise". That is, the derivative with respect to t is <df/dt, dg/dt, dh/dt>. That is precisely what is being done in your example (though I see NO use of the "chain rule" in what you give).

(I looked at the you-tube from 4:45 to 5:15 and saw no error or anything unusual.)

3. Sep 10, 2013

### Nikitin

I've been through multivariable calculus, but they never talked much about vector-differentiation. It was kinda obvious that you just differentiate term by term. So I've never heard about using the chain-rule directly on a vector. Though I suppose there is nothing wrong in doing that.

Anyway, concerning the video, if you look closely at his calculations from 1:00 to 5:00, you will see that the guy uses the chain rule (wrongly) and ends up with: $$\vec{a} = \frac{d}{dt}(x*\vec{i}-y*\vec{j}) = [x,y]$$ Which is obviously wrong. He does everything, including his use of the chain-rule, wrong...

Duno why I wasted those 10 minutes on that guy's videos. Need to find some decent lecture-videos on fluid mechanics.

Last edited: Sep 10, 2013
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