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Product rule

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Look up, figure out, or make an intelligent guess at the product rule for the scalar
    product. That is, a rule of the form

    d/dt [a(t).b(t)] =?+?

    Verify your proposed rule on the functions
    a(t) = ti + sin(t)j + e^(t)k and b(t) = cos(t)i - t^(2)j - e^(t)k:

    2. Relevant equations



    3. The attempt at a solution

    I'm thinking that i can differentiate each function separately and then take the scalar product of these two vectors?

    which gives me d/dt (a) =i - cos(t)j - e^(t)k
    and d/dt (b) = sin(t)j - 2tj - e^(-t)k

    and then i would take the scalar product of these two vectors? but how do i verify that this result is correct?

    Thanks Danny
     
    Last edited: Feb 6, 2012
  2. jcsd
  3. Feb 6, 2012 #2

    Mark44

    Staff: Mentor

    The above is almost right. What's the derivative of et?
    Two of the terms above are incorrect.
    First off, you need to "look up, figure out, or make an intelligent guess" about what the derivative of the dot product of two vectors is.
    Second, you are given two example vector funtions, a(t) and b(t). Take their dot product, and then differentiate that product.

    Compare your result with what you found, figured out, guessed in the first part. If they are the same, then good.
     
  4. Feb 6, 2012 #3
    whoops,
    a'(t) = i + cost(t)j -e(t)k
    b'(t) = -sin(t)i -2tj-e^(-t)k
     
    Last edited: Feb 6, 2012
  5. Feb 6, 2012 #4

    Mark44

    Staff: Mentor

    Now you're just guessing. You'll never be able to get this problem until you have the basic differentiate rules down pat.
     
  6. Feb 6, 2012 #5
    Can you point out the terms that are incorrect after i differentiated.
     
  7. Feb 6, 2012 #6

    Mark44

    Staff: Mentor

    In each of a'(t) and b'(t), the exponential term is wrong. What is d/dt(et)?
     
  8. Feb 6, 2012 #7
  9. Feb 6, 2012 #8

    Mark44

    Staff: Mentor

    Yes, so why are you showing a'(t) = <omitted terms> -et?
    And why are you showing b'(t) = <omitted terms> -e-t?
     
  10. Feb 6, 2012 #9
    Ah no! I typed the question in wrong my apologies.

    a(t) = ti+sin(t)j-e^(t)k
    b(t) = cos(t)i-t^(2)j+e^(-t)
     
  11. Feb 6, 2012 #10
    So taking the dot product of the two derivatives i get
    a'(t) . b'(t) = -sin(t)-2tcos(t)-1
     
  12. Feb 6, 2012 #11

    Mark44

    Staff: Mentor

    Well, that makes a difference. You and I were working different problems. With that change, your derivatives look fine.

    That's NOT what I said back in post #2. Your formula is for the derivative of the dot product, which may or not be the same as the dot product of the derivatives.
     
  13. Feb 6, 2012 #12
    Ok well doing that I get,

    cos(t)-tsin(t)-2tsin(t)-t^(2)cos(t)-t

    Which does not match the other way.
     
  14. Feb 6, 2012 #13

    Mark44

    Staff: Mentor

    You have a mistake above . What did you get for a(t)[itex]\cdot[/itex]b(t)?
     
  15. Feb 6, 2012 #14
    tcos(t)-t^(2)sin(t)-1
     
  16. Feb 6, 2012 #15

    Mark44

    Staff: Mentor

    Yes, now what do you get for the derivative of that dot product?
     
  17. Feb 6, 2012 #16
    cos(t)-tsin(t)-2tsin(t)-t^(2)cost(t)-t
     
  18. Feb 6, 2012 #17

    Mark44

    Staff: Mentor

    You are getting the relatively complicated stuff right, but are completely falling down on some really simple stuff. What is d/dt(-1)?
     
  19. Feb 6, 2012 #18
    ah mustn't be my day :(
     
  20. Feb 6, 2012 #19
    So the answer is,
    cos(t)-tsin(t)-2tsin(t)-t^(2)cos(t)
     
  21. Feb 6, 2012 #20
    nvm I looked it up.
     
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