Definition of the derivative of a vector-valued function?1. The problem statement, all variables and given/known data
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \vec v (t))= \vec u (t)' \vec v(t)+\vec u(t) \vec v(t)'[/itex]
I know this is a product rule on the RHS but how does one prove it?
Hmm, what do you mean by [itex]x[/itex]?OK, I guess I shouldnt have left out the x ie
[itex]\displaystyle \frac{d}{dt} (\vec u (t) x \vec v (t))= \vec u (t)' x\vec v(t)+\vec u(t) x\vec v(t)'[/itex]
Yes, I meant the cross product! I didnt know the correct LaTex :-)Hmm, what do you mean by [itex]x[/itex]?
Did you mean the cross product (also called outer product):
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'[/itex]
Or did you mean the dot product (also called inner product):
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \cdot \vec v (t))= \vec u (t)' \cdot \vec v(t)+\vec u(t) \cdot \vec v(t)'[/itex]
(Or did you mean yet another unusual product? )
Either way, I think the problem asks you to write out the products in the components of the vectors and apply the regular product rule.
Let [itex]\vec u =(x(t),x_2(t),x_3(t)...x_n(t))[/itex] and similarly for v henceDid you mean the cross product (also called outer product):
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'[/itex]
Not sure I understand what you are saying I.L.S. Is it not correct now that we know u and v are in R^3The cross product is defined only for vectors in R^{3}.
Sorry, but no.Not sure I understand what you are saying I.L.S. Is it not correct now that we know u and v are in R^3
Suppose that vector u(t) and v(t) take values in R^3. Prove the following identityWe are now 10 posts into this problem about vector multiplication (11 with this post), but still don't know what kind of multiplication is intended. Please give us the complete problem statement, which should include the space the vectors are in, and the kind of multiplication that is meant.
What do you mean with [itex]\times[/itex]?? How is it defined??Suppose that vector u(t) and v(t) take values in R^3. Prove the following identity
[itex]\frac{d}{dt}(\vec u(t) \times \vec v (t))= \vec u'(t) \times \vec v(t)+ \vec u(t) \times \vec v'(t)[/itex]
I think I see my mstake. I should not have x_n and y_n because the vectors are defined in 3-D space.....
According to my notes, its the cross product....What do you mean with [itex]\times[/itex]?? How is it defined??
The dot product isI see. So the real problem seems to be that you're not yet aware of the various vector products.
Can you tell us what the dot product is?
And can you perhaps find the definition (that I already gave) of the cross product in your notes?
Good, you have the dot product down (that also works in n dimensions)!The dot product is
A= a_x i + a_y j +a_ k, B=b_x i+b_y j+b_z k
A dot B = a_xb_x+ay_by+a_zb_z
The cross product is as your post 10. It looks like I did the dot product...?
No... if you're supposed to use the cross product, you have to substitute the definition of the cross product, apply the regular product rule, and write it back into vector form.Are you saying my answer in #6 is using the component wise product....but this is what is in my notes for a similar porblem. I dont know any other easier way.
So is #6 correct?
Ok, will try that. but what is the difference between your above and my dot product in #16. They look the same to me except with the addition of the basis i,j and k...?Good, you have the dot product down (that also works in n dimensions)!
And no, you did the component-wise product, which isn't usually used:
[itex]\vec a \vec b = a_x b_x \vec i+a_y b_y \vec j+a_z b_z \vec k[/itex]
Yes... those i, j, and k.Ok, will try that. but what is the difference between your above and my dot product in #16. They look the same to me except with the addition of the basis i,j and k...?
Let's not call this "A dot B"... it will confuse everyone (including me)!A dot B including the basis is the 'component wise product' method and gives a vector. (new to me!)
[itex]=u' \times v+u \times v'=\begin{pmatrix}Hmm, what do you mean by [itex]x[/itex]?
Did you mean the cross product (also called outer product):
[itex]\displaystyle \frac{d}{dt} (\vec u (t) \times \vec v (t))= \vec u (t)' \times \vec v(t)+\vec u(t) \times \vec v(t)'[/itex]