verify derivative of a dot product.

by yungman
Tags: derivative, product, verify
 P: 3,842 Let $$\vec w(t) \;,\; \vec v (t)$$ be 3 space vectors that is a function of time t. I want to verify that: $$\frac {d(\vec w \cdot \vec v)}{dt} = \vec v \cdot \frac { d\vec w}{dt} \;+\; \vec w \cdot \frac { d\vec v}{dt}$$ 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
 Sci Advisor HW Helper PF Gold P: 12,016 That will work. Note, however, that by definition, the dot product has the distributive property of multiplication: $$(u+du)\cdot(v+dv)=u\cdot{v}+u\cdot{dv}+v\cdot{du}+du\cdot{dv}$$ For all vectors u,du,v and dv. Thus, the result for the derivative ought to be apparent..
P: 3,842
 Quote by arildno That will work. Note, however, that by definition, the dot product has the distributive property of multiplication: $$(u+du)\cdot(v+dv)=u\cdot{v}+u\cdot{dv}+v\cdot{du}+du\cdot{dv}$$ 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

$$(u+du)\cdot(v+dv)=u\cdot{v}+u\cdot{dv}+v\cdot{du}+du\cdot{dv}$$

relate to my original question. Please explain.

Thanks

Alan

HW Helper
PF Gold
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.
 P: 3,842 Thanks

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