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Thread: Cross product

  1. #1
    Junior Member
    Joined
    Jul 2011
    Posts
    28

    Cross product derivative

    Find 1st and 2nd derivative.

    $\displaystyle f(t) = (\bold a \times t \bold b) \times (\bold a + t^2 \bold b)$

    $\displaystyle (\bold a \times t\bold b)' = (\bold a \times \bold b)$
    $\displaystyle (\bold a + t^2 \bold b)' = 2t \bold b $

    so...

    $\displaystyle f'(t) = (\bold a \times t\bold b) \times (\bold a + t^2\bold b)' + (\bold a \times t\bold b)' \times (\bold a + t^2\bold b)$
    $\displaystyle f'(t) = (\bold a \times t\bold b) \times (2t\bold b) + (\bold a \times \bold b) \times (\bold a + t^2\bold b)$

    Here's where I'm having trouble.

    $\displaystyle f''(t) = [(\bold a \times t\bold b) \times (2t\bold b)' + (\bold a \times t\bold b)' \times (\bold a + t^2\bold b) ] + [(\bold a \times \bold b) \times (\bold a + t^2\bold b)' + (\bold a \times \bold b)' \times (\bold a + t^2\bold b)] \;?$

    Or should I use distributive property on that 2nd part?

    $\displaystyle f''(t) = [(\bold a \times t\bold b) \times (2t\bold b)' + (\bold a \times \bold b)' \times (\bold a + t^2\bold b) ] + [(\bold a \times \bold b) \times (\bold a) + (\bold a \times \bold b) \times (t^2\bold b)]$
    Last edited by deezy; Sep 11th 2011 at 07:34 AM.
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  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280

    Re: Cross product

    We may write

    $\displaystyle f(t) = (\bold a \times t \bold b) \times (\bold a + t^2 \bold b)=$

    $\displaystyle =t \; (\bold a \times \bold b) \times \bold a+t^3\; (\bold a \times \bold b) \times \bold b.$
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