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Math Help - Vector Calculus.

  1. #1
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    Vector Calculus.

    If A = t2i - tj + (2t + 1)k and B = (2t 3)i + j - tk, evaluate at t = 1,

    (i)
    d/dt(A x B)

    (ii)
    d/dt(A xdB/dt)


    i, j, k are vectors


    It's the working I need help with. I get confused when it comes to differentiating the vectors when there's no scalar.

    Thanks.
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  2. #2
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    Am I right in assuming this is a cross product? if so, you need to use the formula:


    and then differentiate the remaining function with respect to t. so for example, when multiplied the i vector becomes 2ti and hence, when differentiated becomes 2i. then repeat using the formula for j and k above and differentiate each term (remember that e1 is i, e2 is j ad e3 is k). this should give you the answer to the first part of the question.

    For the second part, differentiate B : so B= 6t^2i - k

    Then multiply this by A using the formula above. And then differentiate each term as in the first part.

    Does this make sense? I have deliberately left the answer out so that you can have a go but if it gets difficult I can give you more detailed working
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  3. #3
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    We use matrices for cross product , but that way seems a lot easier. I've done them both out, but I've got answers that don't look right.

    For the first one I've got -i+(9t^2-2t-6)j+(3t^2+6t-3)k

    What really gets me is, say for example I have 2ti+j-k (which is what i got for dB/dt) and i have to differentiate it, I don't know do I get 2i as the answer or 2i+j-k?
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  4. #4
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    Quote Originally Posted by Daithi View Post
    We use matrices for cross product , but that way seems a lot easier. I've done them both out, but I've got answers that don't look right.

    For the first one I've got -i+(9t^2-2t-6)j+(3t^2+6t-3)k

    What really gets me is, say for example I have 2ti+j-k (which is what i got for dB/dt) and i have to differentiate it, I don't know do I get 2i as the answer or 2i+j-k?
    You're expected to know that the derivative with respect to t of a_1(t) i + a_2(t) j + a_3(t) k is \frac{d a_1}{dt} i + \frac{d a_2}{dt} j + \frac{d a_3}{dt} k ...... And you know the derivative of a constant is zero, right?
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  5. #5
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    So for example

    d/dx(2xi+3x^2j+2k)
    =
    2i+6xj+0k
    =
    2i+6xj


    Is that correct?
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  6. #6
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    Quote Originally Posted by Daithi View Post
    So for example

    d/dx(2xi+3x^2j+2k)
    =
    2i+6xj+0k
    =
    2i+6xj


    Is that correct?
    Yes.
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  7. #7
    Member Greengoblin's Avatar
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    Quote Originally Posted by Daithi View Post
    We use matrices for cross product , but that way seems a lot easier.
    Hi, that formula is what you get from calculating the determinant of the matrix containing unit vectors, i,j,k, and components of your vectors, that is:

    \left| \begin{array}{ccc} \bf{i} & \bf{j} & \bf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{array} \right| = \bf{i} \left| \begin{array}{cc} a_y & a_z \\ b_y & b_z \end{array} \right| - \bf{j} \left| \begin{array}{cc} a_x & a_z \\ b_x & b_z \end{array} \right| + \bf{k} \left| \begin{array} {cl} a_x & a_y \\ b_x & b_y \end{array} \right|

    =\bf{i}\ (a_yb_z-a_zb_y)-\bf{j}\ (a_xb_z-a_zb_x)+\bf{k}\ (a_xb_y-a_yb_x)

    (except only i,j,k should display in bold on that last part, instead of all of it, but I can't get the latex to work properly )
    Last edited by Greengoblin; December 11th 2008 at 04:05 AM.
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