Verify derivative of a dot product.

[yungman]

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

 

[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..

 

[yungman]

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

 

[arildno]

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]

Thanks