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Math Help - Proof that a parametrized curve orthogonal to a certain vector v

  1. #1
    Senior Member x3bnm's Avatar
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    Proof that a parametrized curve orthogonal to a certain vector v

    Let \alpha\colon I\rightarrow R^3 be a parametrized curve and let \mathbf{v} \in R^3 be a fixed vector. Assume that \alpha'(t) is orthogonal to \mathbf{v}
    for all t \in I and that \alpha(0) is also orthogonal to \mathbf{v}

    Prove that \alpha(t) is orthogonal to \mathbf{v} for all t \in I




    We are given that \alpha'(t) and \alpha(0) is orthogonal to \mathbf{v}

    Let \mathbf{v} = (v_1, v_2, v_3) and \alpha(t) = (x(t), y(t), z(t))

    \therefore \alpha'(t) = (x'(t), y'(t), z'(t))

    So we have:
    \alpha'(t) \cdot\,  \mathbf{v} = (x'(t) * \, v_1 +\,\, y'(t) * v_2 + \,\, z'(t) * v_3) = 0 \text{    [Given]}

    \alpha(0) \cdot\,  \mathbf{v} = (x(0)  *\, v_1 +\,\, y(0)  * v_2 +\,\, z(0)  * v_3) = 0  \text{    [Given]}


    Now how do I use this info and prove that \alpha(t) is orthogonal to \mathbf{v} for all t \in I?

    Is it possible for anyone to kindly give some hints on how to prove this?
    Last edited by x3bnm; April 19th 2012 at 09:21 AM.
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Proof that a parametrized curve orthogonal to a certain vector v

    I think I solved it. For those who are interested:

    Since

    \alpha'(t) \cdot\, \mathbf{v} = (x'(t) *\, v_1)+\,\,( y'(t) * v_2) +\,\, (z'(t) * v_3) = 0 \text{    [Given]}


    Particularly:

    (x'(t) *\, v_1)+\,\, (y'(t) * v_2) +\,\, (z'(t) * v_3) = 0


    Taking integration of both sides:

    (v_1 * x(t)) + C_1 +\,\,( v_2 * (y(t)) + C_2 +\,\, (v_3 * z(t)) + C_3 = C


    By taking all constants to the right side we get:

    (v_1 * x(t)) + (v_2 * y(t)) + (v_3 * z(t)) = C -  C_1 -  C_2 -  C_3  \text{  ................(1)}



    We also know that:

    \alpha(0) \cdot\,  \mathbf{v} = (x(0)  *\, v_1) +\,\, (y(0)  * v_2) +\,\, (z(0)  * v_3) = 0  \text{    [Given]}


    So pluging in t = 0 into eq(1) we get:

    C -  C_1 -  C_2 -  C_3 = 0

    therefore:

    (v_1 * x(t)) + (v_2 * y(t)) + (v_3 * z(t)) = 0

    But this means: \alpha(t) \cdot\,\, \mathbf{v} = 0

    So \alpha(t) is orthogonal to \mathbf{v}.[As was to be shown]
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  3. #3
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    Re: Proof that a parametrized curve orthogonal to a certain vector v

    Looks good but I think I would go the "other way"- that is, differentiate \alpha rather than integrate \alpha'.

    Let u(t)= \alpha(t)\cdot v. Then, since v is a constant vector, u'= \alpha\cdot v= 0. Since the derivative of u is 0, u is a constant. And since u(0)= \alpha(0)\cdot v= 0 it follows that u(t)= \alpha(t)\cdot v= 0 for all t.
    Thanks from x3bnm and skrzatus
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  4. #4
    Senior Member x3bnm's Avatar
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    Re: Proof that a parametrized curve orthogonal to a certain vector v

    Quote Originally Posted by HallsofIvy View Post
    Looks good but I think I would go the "other way"- that is, differentiate \alpha rather than integrate \alpha'.

    Let u(t)= \alpha(t)\cdot v. Then, since v is a constant vector, u'= \alpha\cdot v= 0. Since the derivative of u is 0, u is a constant. And since u(0)= \alpha(0)\cdot v= 0 it follows that u(t)= \alpha(t)\cdot v= 0 for all t.

    Thank you for showing another way of solving this problem. I believe your way is the correct and better method. Again thanks.
    Last edited by x3bnm; April 19th 2012 at 04:40 PM.
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