Prove the Following Vector Identities Part 1

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  • #1
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Homework Statement


[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?

Thanks
 

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  • #2
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Homework Statement


[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?
Definition of the derivative of a vector-valued function?

Also, for the sake of clarity, what kind of multiplication is implied here?
 
  • #3
bugatti79
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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]
 
  • #4
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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]

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? :wink:)

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.
 
  • #5
bugatti79
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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? :wink:)

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.

Yes, I meant the cross product! I didnt know the correct LaTex :-)
Will look at it tomorrow. Thanks!
 
  • #6
bugatti79
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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]

Let [itex]\vec u =(x(t),x_2(t),x_3(t)...x_n(t))[/itex] and similarly for v hence

[itex]\displaystyle \frac{d}{dt} \vec u (t) \times \vec v(t)=\frac{d}{dt} (x(t) \times y(t)), \frac{d}{dt}(x_2(t) \times y_2(t))......\frac{d}{dt} (x_n(t) \times y_n (t))[/itex]

[itex]=x(t) y'(t)+y(t) x'(t), x_2(t)y_2'(t)+y_2(t)x_2'(t).....[/itex]

[itex]=(x(t)y'(t), x_2(t)y_2'(t),x_n(t)y_n'(t))+(y(t)x'(t), y_2(t)x_2'(t)....)[/itex]

[itex]=(x_n(t)y_n'(t))+ (y_n(t)x_n'(t))[/itex]

[itex]\displaystyle =\frac{d}{dt} (x_n(t) \times y_n(t))[/itex]

[itex]\displaystyle =\frac{d}{dt} (\vec u(t) \times \vec v (t))[/itex]..........?
 
  • #7
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The cross product is defined only for vectors in R3.
 
  • #8
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What you did would be for yet another (unusual) vector product defined by component-wise multiplication.
As such, your proof would be correct (although your notation could be better).

But are you sure we're talking about a component-wise product? :confused:
(It would fit better with leaving out the multiplication sign completely.)
 
  • #9
bugatti79
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The cross product is defined only for vectors in R3.

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
 
  • #10
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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

Sorry, but no.

Let me refresh the definition of the cross product of u and v.

[tex]\vec u \times \vec v
= \begin{pmatrix}
u_y v_z - u_z v_y\\
u_z v_x - u_x v_z \\
u_x v_y - u_y v_x\end{pmatrix}[/tex]

Did you mean this vector product?
 
  • #11
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We 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.
 
  • #12
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We 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.

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.....
 
  • #13
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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.....

What do you mean with [itex]\times[/itex]?? How is it defined??
 
  • #14
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I 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?
 
  • #15
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What do you mean with [itex]\times[/itex]?? How is it defined??

According to my notes, its the cross product....
 
  • #16
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I 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?

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...?
 
  • #17
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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...?

Good, you have the dot product down (that also works in n dimensions)! :smile:

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]
 
  • #18
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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?
 
  • #19
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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?

No... :rolleyes: 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.
 
  • #20
bugatti79
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Good, you have the dot product down (that also works in n dimensions)! :smile:

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]

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...?
 
  • #21
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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...?

Yes... those i, j, and k.

The difference is that the one is a scalar (just a number) and the other is a vector, which are pretty much very different things!

As it happens, the product rule for derivatives holds for all three types of vector products. :tongue2:
 
  • #22
bugatti79
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To summarise

A dot B gives a number.

A dot B including the basis is the 'component wise product' method and gives a vector. (new to me!)

A Cross B gives a vector

I will be back with the right answer ;-) hopefully!
 
  • #23
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A dot B including the basis is the 'component wise product' method and gives a vector. (new to me!)

Let's not call this "A dot B"... it will confuse everyone (including me)!
As I said, it is an unusual vector product, but there are applications for it.
If you have to, simply call it "A B" and always explain which product you mean!
 
  • #24
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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]

[itex]=u' \times v+u \times v'=\begin{pmatrix}
i & j & k\\
u'_1 &u'_2 &u'_3 \\
v_1&v_2 &v_3
\end{pmatrix}+\begin{pmatrix}
i & j & k\\
u_1 &u_2 &u_3 \\
v'_1&v'_2 &v'_3
\end{pmatrix}[/itex]

Is this attempt correct so far?
 
  • #25
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What you wrote it correct.
 
  • #26
bugatti79
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What you wrote it correct.

So I just evaulate this, tidy it up and show that it equals the RHS of question?

Cheers
 
  • #27
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Uhh... what you have is the RHS of the question.

What you need to do is take the LHS of the question, evaluate it, and show that it is equal to what you just wrote (aka the RHS of the question).
 
  • #28
bugatti79
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Uhh... what you have is the RHS of the question.

What you need to do is take the LHS of the question, evaluate it, and show that it is equal to what you just wrote (aka the RHS of the question).

Hmm...How would you evaluate the LHS? I thought the RHS matrix could be simplified further to arrive at something that resembles the LHS :-)
 
  • #29
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The left side is the derivative of the cross product u(t) X v(t). Calculate the cross product, and then take the derivative. If that turns out to be equal to what you have in post #24, you have shown that d/dt(u(t) X v(t)) = u'(t) X v(t) + u(t) X v'(t).
 
  • #30
bugatti79
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The left side is the derivative of the cross product u(t) X v(t). Calculate the cross product, and then take the derivative. If that turns out to be equal to what you have in post #24, you have shown that d/dt(u(t) X v(t)) = u'(t) X v(t) + u(t) X v'(t).

Thank you Mark and I.L.S
 

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