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Math Help - Parametrics

  1. #1
    Senior Member topher0805's Avatar
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    Parametrics

    These are normally really easy for me and this one appears to be incredibly simple but I can't seem to get the correct answer.

    Given x = e^t - 6 and y = e^{2t}

    Eliminate the parameter to find a Cartesian equation of the curve.

    Isolate t:

    x+6=e^t

    t=\ln {(x+6)}

    Plug it into the y equation:

    y=e^{2\ln {(x+6)}} , x> {-6}
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  2. #2
    o_O
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    Error in one of your steps:

    e^{t} = x + 6
    \ln e^{t} = \ln (x + 6)
    t = \ln (x + 6)

    6 is inside the logarithm as well.
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  3. #3
    Senior Member topher0805's Avatar
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    That was just a typo by me. It still isn't the right answer and I have no idea why.
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  4. #4
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    Oooh. You just need to simplify a bit more. Remember your logarithm rules? You can bring the 2 up:

    y = e^{\ln(x+6)^{2}}

    And hopefully you'll see how the e and the ln disappears

    Or you could've realized:
    y = e^{2t} = \left(e^{t}\right)^{2} = (x + 6)^{2}
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  5. #5
    Senior Member topher0805's Avatar
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    I don't know how I missed that.

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  6. #6
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    Hello, topher0805!

    Given: . \begin{array}{cccc}x& = & e^t - 6 & {\color{blue}[1]}\\ y &= &e^{2t} & {\color{blue}[2]}\end{array}

    Eliminate the parameter to find a Cartesian equation of the curve.

    From [1], we have: . x+6)^2" alt="e^t \:=\:x+6\quad\Rightarrow\quad e^{2t} \:=\x+6)^2" />

    Substitute into [2]: . x+6)^2" alt="y \:=\x+6)^2" />

    This is an up-opening parabola with vertex (-6, 0).


    However, note the domains of x\text{ and }y\!:\;\;\begin{Bmatrix} x &\in &(\text{-}6,\,\infty) \\ y &\in &(0,\,\infty) \end{Bmatrix}
    Code:
                         |*
                         |
                         |
                         *
                         |
                        *|
                       * |
                     *   |
          --------o------+-----
                 -6      |

    The graph is only the "right half" of the parabola.

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