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Math Help - Parametric Equations - Horizontal and Vertical Tangent

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    Parametric Equations - Horizontal and Vertical Tangent

    Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. (Enter your answers as a comma-separated list of ordered pairs.)

    x = t^{3} - 3t

    y = t^{2} - 3

    I'm assuming setting the parametric derivative to 0 would get the horizontal tangent. But how about finding the vertical tangent (undefined)?
    Last edited by Jason76; July 26th 2014 at 02:26 AM.
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    Re: Parametric Equations - Horizontal and Vertical Tangent

    \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}

    A fraction is 0 when the numerator, here \frac{dy}{dt}, is 0.

    It does not exist (is undefined) when the denominator, here \frac{dx}{dt}, is 0.
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    Re: Parametric Equations - Horizontal and Vertical Tangent

    I came up with (-2,2) for the horizontal tangent and (0,-3) for the vertical one (but they are wrong on homework) ???

    Horizontal

    y = t^{2} - 3

    \dfrac{dy}{dt} = 2t

    2t = 0

    t = 0

    y = (0)^{2} - 3 = -3

    x = (0)^{3} - 3(0) = 0

    (0,-3)

    Vertical

    x = t^{3} -3t

    \dfrac{dx}{dt}  = 3t^{3} - 3t

    3t^{2} - 3 = 0

    t = 1

    y = (1)^{2} - 3 = -2

    x = (1)^{3} - 3(1) = -2

    (-2,-2)
    Last edited by Jason76; July 27th 2014 at 09:55 AM.
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    Re: Parametric Equations - Horizontal and Vertical Tangent

    Quote Originally Posted by Jason76 View Post
    I came up with (-2,2) for the horizontal tangent and (0,-3) for the vertical one (but they are wrong on homework) ???

    Horizontal

    y = t^{2} - 3

    \dfrac{dy}{dt} = 2t

    2t = 0

    t = 0

    y = (0)^{2} - 3 = -3

    x = (0)^{3} - 3(0) = 0

    (0,-3)

    Vertical

    x = t^{3} -3t

    \dfrac{dx}{dt}  = 3t^{3} - 3t

    3t^{2} - 3 = 0

    t = 1

    y = (1)^{2} - 3 = -2

    x = (1)^{3} - 3(1) = -2

    (-2,-2)
    I get horizontal tangent at (0,-3)

    Vertical tangents at (-2,-2) and (2,-2)

    $(-1)^2=1$ also.

    Parametric Equations - Horizontal and Vertical Tangent-clipboard01.jpg
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    Re: Parametric Equations - Horizontal and Vertical Tangent

    Quote Originally Posted by Jason76 View Post
    I came up with (-2,2) for the horizontal tangent and (0,-3) for the vertical one (but they are wrong on homework) ???

    Horizontal

    y = t^{2} - 3

    \dfrac{dy}{dt} = 2t

    2t = 0

    t = 0

    y = (0)^{2} - 3 = -3

    x = (0)^{3} - 3(0) = 0

    (0,-3)

    Vertical

    x = t^{3} -3t

    \dfrac{dx}{dt}  = 3t^{3} - 3t
    This is wrong. Possibly a typo but then you continue with it. The derivative of t^3 is 3t^2 not 3t^3.

    3t^{2} - 3 = 0

    t = 1

    y = (1)^{2} - 3 = -2

    x = (1)^{3} - 3(1) = -2

    (-2,-2)
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    Re: Parametric Equations - Horizontal and Vertical Tangent

    Quote Originally Posted by romsek View Post
    I get horizontal tangent at (0,-3)

    Vertical tangents at (-2,-2) and (2,-2)

    $(-1)^2=1$ also.

    Click image for larger version. 

Name:	Clipboard01.jpg 
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ID:	31353
    Are you sure you don't have those backwards? The horiz tangent is vertical one etc..
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    Re: Parametric Equations - Horizontal and Vertical Tangent

    No idea on this one. Nothing seems to work.
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