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Math Help - a particle is moving.. w/ equation.

  1. #1
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    a particle is moving.. w/ equation.

    A particle is moving so that the position at time t is given by:
    r(t) = e^t(cost i - sint j)

    Find the velocity vector at time t:

    find the acceleration vector at time t:

    find the speed at time t:

    i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?

    i wasn't sure on the derivatives of a natural log.

    thanks for any help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rcmango View Post
    A particle is moving so that the position at time t is given by:
    r(t) = e^t(cost i - sint j)

    Find the velocity vector at time t:

    find the acceleration vector at time t:

    find the speed at time t:

    i think i need to find the first and second derivatives for velocity and acceleration, not sure about speed?

    i wasn't sure on the derivatives of a natural log.

    thanks for any help.
    you are right about first and second derivatives for velocity and acceleration. speed is just velocity without direction, so i believe you just remove the i and j. derivative of lnx is 1/x, of course if x is a function you have to multiply by it's derivative
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    speed is just velocity without direction, so i believe you just remove the i and j.
    Yes, and no.

    Speed in this case is the magnitude of the velocity vector. So if you have a velocity:
    v = ai + bj
    then the speed would be:
    |v| = sqrt(a^2 + b^2)

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Yes, and no.

    Speed in this case is the magnitude of the velocity vector. So if you have a velocity:
    v = ai + bj
    then the speed would be:
    |v| = sqrt(a^2 + b^2)

    -Dan
    A yes, now i remember. speed = |v|. There i go running my mouth off again without knowing the facts
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  5. #5
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    might need some help fixing my differentiation skills..

    for the first derivative my guess would be to use the chain rule to differentiate?

    so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)

    and for the 2nd derivative i guessed to use the chain and product rule in combination: so i got something like:- 1/x^2(cost i - sin j)(sint i - cost j)[(sint i - cost j)(sint i - cost j) + (cost i - sint j)(cost i + sint i)]

    how am i doing.

    thanks.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by rcmango View Post
    A particle is moving so that the position at time t is given by:
    r(t) = e^t(cost i - sint j)

    Find the velocity vector at time t:

    find the acceleration vector at time t:

    find the speed at time t:
    Quote Originally Posted by rcmango View Post
    might need some help fixing my differentiation skills..

    for the first derivative my guess would be to use the chain rule to differentiate?

    so the first derivative is: -1/x(cost i - sin j)(sint i - cost j)
    What is x??

    I think you are making this too hard. Given h(x) = f(x)*g(x)
    h'(x) = f'(x)*g(x) + f(x)*g'(x)

    So:
    r(t) = e^t*(cos(t)i - sin(t)j)
    r'(t) = e^t*(cos(t)i - sin(t)j) + e^t*(-sin(t)i - cos(t)j)
    r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)

    r''(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j) + e^t*([-sin(t) - cos(t)]i + [-cos(t) + sin(t)]j)
    r''(t) = e^t*(-2sin(t)i + -2cos(t)j)
    r''(t) = -2e^t*(sin(t)i + cos(t)j)

    For the speed:
    r'(t) = e^t*([cos(t) - sin(t)]i + [-sin(t) - cos(t)]j)

    So
    |r'(t)| = sqrt{e^(2t)[cos(t) - sin(t)]^2 + e^(2t)[-sin(t) - cos(t)]^2}

    |r'(t)| = e^t*sqrt{cos^2(t) + sin^2(t) - 2sin(t)cos(t) + sin^2(t) + cos^2(t) + 2sin(t)cos(t)}

    |r'(t)| = e^t*sqrt{1 + 1}

    |r'(t)| = e^t*sqrt{2}

    -Dan
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