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Math Help - Parametric equations

  1. #1
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    Parametric equations

    So I don't seem to understand how to find a parametric equation from a vector function or to find an equation in x and y whose graph is the path of the particle.

    Ok first. Let's just say we have v (t) = (sin t)i + t^2j + e^tk .

    I thought all you do is x = sin t etc... but that's not the case apparently.

    I thought you did something similar for my latter inquiry... but that's not the case apparently.
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  2. #2
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    Quote Originally Posted by Oneiromancy View Post
    So I don't seem to understand how to find a parametric equation from a vector function or to find an equation in x and y whose graph is the path of the particle.

    Ok first. Let's just say we have v (t) = (sin t)i + t^2j + e^tk .

    I thought all you do is x = sin t etc... but that's not the case apparently.

    I thought you did something similar for my latter inquiry... but that's not the case apparently.
    That is how I have always done it

    Let \bold{r(t)}=\left\langle{f(t),g(t),h(t)}\right\ran  gle

    Then we parametrize it by

    x=f(t)
    y=g(t)
    z=h(t)
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  3. #3
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    Quote Originally Posted by Oneiromancy View Post
    So I don't seem to understand how to find a parametric equation from a vector function or to find an equation in x and y whose graph is the path of the particle.

    Ok first. Let's just say we have v (t) = (sin t)i + t^2j + e^tk .

    I thought all you do is x = sin t etc... but that's not the case apparently.

    I thought you did something similar for my latter inquiry... but that's not the case apparently.
    It is the case.

    Post the exact question you're working on.
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  4. #4
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    Ok I have 2 problems that I can't do.

    1) Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of t.

    r(t) = (t + 1)i + (t^2 - 1)j , t = 1

    The answer is y = x^2 - 2x

    I know how to get v and a.


    2) Find parametric equations for the line that is tangent to the given curve at the given parameter value t_0 = 0

    r(t) = (sin t)i + (t^2 - cos t)j + (e^t)k , t_0 = 0

    The answer should be x = t , y = -1 , z = 1 + t

    I thought all you do is find dr/dt then just plug in 0 but that didn't work.

    These problems seem way too easy. I know I'm simply doing it completely wrong.
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  5. #5
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    Quote Originally Posted by Oneiromancy View Post
    ]
    1) Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of t.

    r(t) = (t + 1)i + (t^2 - 1)j , t = 1

    The answer is y = x^2 - 2x

    I know how to get v and a.
    So
    x = t + 1
    and
    y = t^2 - 1

    So from the x equation we know that
    t = x - 1

    Inserting this into the y equation:
    y = (x - 1)^2 - 1

    Now simplify.

    -Dan
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  6. #6
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    Quote Originally Posted by Oneiromancy View Post
    Ok I have 2 problems that I can't do.

    1) Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of t.

    r(t) = (t + 1)i + (t^2 - 1)j , t = 1

    The answer is y = x^2 - 2x

    I know how to get v and a.


    2) Find parametric equations for the line that is tangent to the given curve at the given parameter value t_0 = 0

    r(t) = (sin t)i + (t^2 - cos t)j + (e^t)k , t_0 = 0

    The answer should be x = t , y = -1 , z = 1 + t

    I thought all you do is find dr/dt then just plug in 0 but that didn't work.

    These problems seem way too easy. I know I'm simply doing it completely wrong.
    1) x = t + 1 => t = x - 1 .... (1)
    y = t^2 - 1 .... (2)

    Substitute (1) into (2): y = (x - 1)^2 - 1 = x^2 - 2x + 1 - 1 = x^2 - 2x.


    2) dr/dt at t = 0 gives a vector v in the direction of the line. And a point on the line is (0, -1, 1). So a vector equation of the line is:

    r = 0 i - 1 j + 1 k + t v = -j + k + t v. (You can put in what v is).

    Therefore a parametric equation is x = ...., y = ....., z = ......
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  7. #7
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    Got it now, thank you.
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