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Math Help - Parametric Derivative problem

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    Parametric Derivative problem

    Find the equation of the tangent line to the graph C at the point (-3,8) for the parametric equations: x(t)=t^2-4t+1 and y(t)=t^3

    For this, The derivative is a vertical line, where x=-3. Can you have a vertical derivative? Is the answer no solution or x=-3?

    Thank You.
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    Quote Originally Posted by nic42991 View Post
    Find the equation of the tangent line to the graph C at the point (-3,8) for the parametric equations: x(t)=t^2-4t+1 and y(t)=t^3

    For this, The derivative is a vertical line, where x=-3. Can you have a vertical derivative? Is the answer no solution or x=-3?

    Thank You.
    ok, if x = -3, then what is t? you will need this value to find the derivative

    now, \frac {dy}{dx} = \frac {dy}{dt} \cdot \frac {dt}{dx} = \frac {\frac {dy}{dt}}{\frac {dx}{dt}}

    and you plug in the value of t to find your slope.

    then all you have to do is find the equation of the line with the above slope passing through (-3, 8), which i suppose you know how to do
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    I plugged in the point (-3,8) into the x(t) and y(t) equations, and I got:

    x(3)=2
    y(8)=2

    for dx/dy I got (3t^2)/(2t-4) and if you plug in 2, you get 12/0, which is undef.

    Did I do it correctly?
    Can the derivative be a vertical line?
    Since the derivative is a vertical line, and it goes through the point (-3,8), Is the equation for the derivative x=-3?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nic42991 View Post
    I plugged in the point (-3,8) into the x(t) and y(t) equations, and I got:

    x(3)=2
    y(8)=2

    for dx/dy I got (3t^2)/(2t-4) and if you plug in 2, you get 12/0, which is undef.

    Did I do it correctly?
    Can the derivative be a vertical line?
    Since the derivative is a vertical line, and it goes through the point (-3,8), Is the equation for the derivative x=-3?
    no, i told you to solve for t, how is it that you can assume you already have it? and how is it you have two different t values for the same point (you assumed t = 3 for x and t = 8 for y)?

    we have x = -3 and y = 8

    that means t^2 - 4t + 1 = -3 and t^3 = 8

    solve for t that satisfies both equations.

    then continue to do the problem.
    Last edited by Jhevon; October 12th 2008 at 07:43 PM.
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    t=2 for both equations.

    Correct me if I'm wrong, but I plugged in 2 for the derivative for \frac {\frac {dy}{dt}}{\frac {dx}{dt}} which is (3t^2)/(2t-4) so you get 12/0, which is undefined.

    Is this correct so far?
    Last edited by nic42991; October 12th 2008 at 07:30 PM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nic42991 View Post
    t=2 for both equations.

    Correct me if I'm wrong, but I plugged in 2 for the derivative for \frac {\frac {dy}{dt}}{\frac {dx}{dt}} which is (3t^2)/(2t-4) so you get 12/0, which is undefined.

    Is this correct so far?
    yes, you are correct so far.

    of course, you realize an undefined slope means a vertical line, right?
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    Quote Originally Posted by Jhevon View Post
    yes, you are correct so far.

    of course, you realize an undefined slope means a vertical line, right?
    Yes. Therefore, the equation of the derivative is x=-3 since it passes through the point (-3,8) right? I just wasn't sure if the derivative could be a vertical line
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nic42991 View Post
    Yes. Therefore, the equation of the derivative is x=-3 since it passes through the point (-3,8) right?
    yes
    I just wasn't sure if the derivative could be a vertical line
    the derivative gives the slope of the tangent line. it is possible for the tangent line to be vertical.
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    Quote Originally Posted by Jhevon View Post
    yes the derivative gives the slope of the tangent line. it is possible for the tangent line to be vertical.
    Thank you very much for all your help! You're the one that always answers my questions!
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    [quote=nic42991;202183]Thank you very much for all your help![/quore]you're welcome
    You're the one that always answers my questions!
    really? i hadn't noticed

    take care
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