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Math Help - Parallel Vectors functions

  1. #1
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    Parallel Vectors functions

    Hello,

    I have to find every value t such that the vector r'(t) is parallel to the yz plane, given that r(t) = <cos(t^2), e^(sqrt(t)), t>

    Well, I know how to find the derivative of that, but then don't know what to do from there.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    Hello,

    I have to find every value t such that the vector r'(t) is parallel to the yz plane, given that r(t) = <cos(t^2), e^(sqrt(t)), t>

    Well, I know how to find the derivative of that, but then don't know what to do from there.
    Any unit vector parallel to the yz plane will take the form:
    <0, cos(phi), sin(phi)>
    So r(t) must be proportional to one of these in general.

    Comparing the x components we see that
    cos(t^2) = 0
    Thus
    t^2 = (2n + 1)pi/2 for n = 0, 1, 2, ...

    t = (+/-)sqrt{(2n + 1)pi/2}

    Additionally, dividing the z component by the y component gives us:
    e^(sqrt(t))/t = tan(phi)

    Since tan(phi) can be any angle t can be any value, so this doesn't give any restrictions on t. So the solution is:

    t = (+/-)sqrt{(2n + 1)pi/2}

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Any unit vector parallel to the yz plane will take the form:
    <0, cos(phi), sin(phi)>
    So r(t) must be proportional to one of these in general.

    Comparing the x components we see that
    cos(t^2) = 0
    Thus
    t^2 = (2n + 1)pi/2 for n = 0, 1, 2, ...

    t = (+/-)sqrt{(2n + 1)pi/2}

    Additionally, dividing the z component by the y component gives us:
    e^(sqrt(t))/t = tan(phi)

    Since tan(phi) can be any angle t can be any value, so this doesn't give any restrictions on t. So the solution is:

    t = (+/-)sqrt{(2n + 1)pi/2}

    -Dan
    Where in the world is the 2n+1 from?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ideasman View Post
    Where in the world is the 2n+1 from?
    cos(t^2) = 0

    So
    t^2 = pi/2, 3pi/2, 5pi/2, etc.

    Now, 2n + 1 = 1, 3, 5, ... for n = 0, 1, 2, ...

    So
    t^2 = (2n+1)pi/2 for n = 0, 1, 2, ...

    -Dan
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