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Prove the Following Vector Identities Part 1

  1. Nov 4, 2011 #1
    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?

    Thanks
     
  2. jcsd
  3. Nov 4, 2011 #2

    Mark44

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    Definition of the derivative of a vector-valued function?

    Also, for the sake of clarity, what kind of multiplication is implied here?
     
  4. Nov 4, 2011 #3
    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]
     
  5. Nov 4, 2011 #4

    I like Serena

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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.
     
  6. Nov 4, 2011 #5
    Yes, I meant the cross product! I didnt know the correct LaTex :-)
    Will look at it tomorrow. Thanks!
     
  7. Nov 5, 2011 #6
    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]..........?
     
  8. Nov 5, 2011 #7

    Mark44

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    The cross product is defined only for vectors in R3.
     
  9. Nov 5, 2011 #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.)
     
  10. Nov 5, 2011 #9
    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
     
  11. Nov 5, 2011 #10

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    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?
     
  12. Nov 5, 2011 #11

    Mark44

    Staff: Mentor

    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.
     
  13. Nov 5, 2011 #12
    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.....
     
  14. Nov 5, 2011 #13

    micromass

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    What do you mean with [itex]\times[/itex]?? How is it defined??
     
  15. Nov 5, 2011 #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?
     
  16. Nov 5, 2011 #15
    According to my notes, its the cross product....
     
  17. Nov 5, 2011 #16
    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...?
     
  18. Nov 5, 2011 #17

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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]
     
  19. Nov 5, 2011 #18
    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?
     
  20. Nov 5, 2011 #19

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    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.
     
  21. Nov 5, 2011 #20
    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...?
     
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