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Homework Help: Differentiation Problem

  1. Aug 10, 2010 #1
    1. The problem statement, all variables and given/known data

    a=(t^2 + 1)i - j + tk

    Find da/dt

    I multiplied out the brackets so a= it^2 + i - j + tk

    then got as far as da/dt= 2it + 0 + 0 + ?

    a) is this right so far?
    b) how do i integrate tk with respect to t?

    Any help would be greatly appreciated.
  2. jcsd
  3. Aug 10, 2010 #2

    Char. Limit

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    Gold Member

    I would assume a constant k, in which it's simple.
  4. Aug 10, 2010 #3
    Thinking about it, the question says "Given a=(t^2)i -j + tk find:"


    Do you think that the bold is relevant? There are another two parts one for b and db/dt then the third part is d/dt(a.b). The whole question is worth 7. Any ideas?
  5. Aug 10, 2010 #4


    Staff: Mentor

    Yes, the bolded things are very relevant. i, j, and k on the right side are unit vectors. da/dt will also be a vector quantity.
  6. Aug 10, 2010 #5


    Staff: Mentor

    For both a and b, differentiation and integration of vector functions is done component-wise.

    As for the integration, here's an example:
    [tex]\int 2t\bold{i} dt = t^2\bold{i} + \bold{C}[/tex]
  7. Aug 10, 2010 #6
    so would da/dt= 2ti + 0j + 1k ?
  8. Aug 10, 2010 #7


    Staff: Mentor

    Yes. This could also be written as da/dt = 2ti + k
  9. Aug 10, 2010 #8
    Thank you for the help! But just to check I understand, I'm going to do the second and third parts.

    b=2tj - k

    so db/dt=2j ?

    and the third part, d/dt(a.b) how would I times the two together?

    Would a.b= t^2i - 2tj -tk?

    So d/dt(a.b)=2ti - 2j -k ?
  10. Aug 10, 2010 #9
    Another question whilst I'm at it, if f=5x^5.y

    is df/dx= 25x^4.y + 5x^5.ydx?
  11. Aug 10, 2010 #10


    Staff: Mentor

    What does 5x^5.y mean?
    Is it 5x5* y or 5x5y?
    Is f a function of two independent variables, x and y?
  12. Aug 10, 2010 #11
    Sorry for being unlclear, it's (5x^5)* y. And yes, it is f(x,y).
  13. Aug 10, 2010 #12


    Staff: Mentor

    A better way to write it would be f = 5yx5.

    So you're trying to find fx, or [tex]\frac{\partial f}{\partial x}[/tex].

    You don't need to use the product rule. Since x and y are independent variables, y is considered a constant when differentiating with respect to x.

    Why do you have dx in one part of your answer? It should not be there.
  14. Aug 10, 2010 #13
    I just remembered something from school, I thought you had to put something like that in when differentiating y with respect with x. So df/dx would just be 25yx^4?
  15. Aug 10, 2010 #14
    Implicit differentiation? Or do you not do that when the function is f(x,y)?
  16. Aug 10, 2010 #15


    Staff: Mentor

    That's the answer, but you should indicate that you're doing a partial differentiation. There are two (at least) kinds of notation: one with subscripts, which in this case would be fx, and another is similar to df/dx but with different letters -- [tex]\frac{\partial f}{\partial x}[/tex]

    No, not then either.

    If you have z = f(x, y) and you want the total differential, it will involve differentials of x and y.

    dz = fx dx + fy dy

    [tex]dz = \frac{\partial f}{\partial x}~dx + \frac{\partial f}{\partial y}~dy [/tex]
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