Multivariable Calculus - chain rule on vectors

  • Thread starter Nikitin
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  • #1
Nikitin
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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:

[tex]\vec{V} = [x,-y] \Rightarrow \frac{d\vec{V}}{dt} = [\frac{dx}{dt},-\frac{dy}{dt}] = \vec{a} = [V_x,-V_y][/tex], 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.
 
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  • #2
HallsofIvy
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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
Nikitin
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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: [tex] \vec{a} = \frac{d}{dt}(x*\vec{i}-y*\vec{j}) = [x,y][/tex] 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.
 
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