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Verify derivative of a dot product.

by yungman
Tags: derivative, product, verify
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yungman
#1
May11-11, 12:44 PM
P: 3,883
Let [tex] \vec w(t) \;,\; \vec v (t) [/tex] be 3 space vectors that is a function of time t. I want to verify that:

[tex] \frac {d(\vec w \cdot \vec v)}{dt} = \vec v \cdot \frac { d\vec w}{dt} \;+\; \vec w \cdot \frac { d\vec v}{dt} [/tex]

I work through the verification by splitting w and v into x, y, z components, do the dot product and take the derivative to verify already. Just want to run this by the expert to confirm.

Thanks

Alan
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arildno
#2
May11-11, 01:58 PM
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That will work.

Note, however, that by definition, the dot product has the distributive property of multiplication:
[tex](u+du)\cdot(v+dv)=u\cdot{v}+u\cdot{dv}+v\cdot{du}+du\cdot{dv}[/tex]
For all vectors u,du,v and dv.

Thus, the result for the derivative ought to be apparent..
yungman
#3
May11-11, 02:39 PM
P: 3,883
Quote Quote by arildno View Post
That will work.

Note, however, that by definition, the dot product has the distributive property of multiplication:
[tex](u+du)\cdot(v+dv)=u\cdot{v}+u\cdot{dv}+v\cdot{du}+du\cdot{dv}[/tex]
For all vectors u,du,v and dv.

Thus, the result for the derivative ought to be apparent..
Thanks

But I don't see how

[tex](u+du)\cdot(v+dv)=u\cdot{v}+u\cdot{dv}+v\cdot{du}+du\cdot{dv}[/tex]


relate to my original question. Please explain.

Thanks

Alan

arildno
#4
May11-11, 03:05 PM
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P: 12,016
Verify derivative of a dot product.

It shows that a dot product works "just like" a normal product, and thus, the differentiation rule "ought" to be the same (i.e, your result).

However, for rigorous verification, you should do as you've done.
yungman
#5
May11-11, 03:15 PM
P: 3,883
Thanks


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