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Math Help - elimination

  1. #1
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    elimination

    x = e^(2t), y=e^(3t)+8
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  2. #2
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    Hello, pourcarlos!

    Eliminate the parameter: . \begin{array}{ccc}x \:=\:e^{2t} & [1] \\ y\:=\:e^{3t}+8 & [2]\end{array}

    From [1]: . x \:=\:e^{2t} \quad\Rightarrow\quad 2t \:=\:\ln x \quad\Rightarrow\quad t \:=\:\tfrac{1}{2}\ln x

    Substitute into [2]: . y \;=\;e^{3(\frac{1}{2}\ln x)}+8 \;=\;e^{(\frac{3}{2}\ln x)} + 8 \;=\;e^{(\ln x^{\frac{3}{2}})} + 8

    Therefore: . y \;=\;x^{\frac{3}{2}} + 8


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  3. #3
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    oh great, thank you
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  4. #4
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    Quote Originally Posted by pourcarlos View Post
    x = e^(2t), y=e^(3t)+8
    An alternative to the perfectly correct solution provide by Soroban:

    x^3 = e^{6t} and (y - 8)^2 = e^{6t}. Therefore ....

    How much of the curve is required will obviously depend on the values of t that are allowed. If t \geq 0 then the curve starts at (1, 9) ....
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