Results 1 to 4 of 4

Thread: elimination

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    20

    elimination

    x = e^(2t), y=e^(3t)+8
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, pourcarlos!

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

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

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

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


    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    20
    oh great, thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    Quote Originally Posted by pourcarlos View Post
    x = e^(2t), y=e^(3t)+8
    An alternative to the perfectly correct solution provide by Soroban:

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

    How much of the curve is required will obviously depend on the values of t that are allowed. If $\displaystyle t \geq 0$ then the curve starts at (1, 9) ....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Sep 12th 2011, 09:03 PM
  2. Replies: 1
    Last Post: Feb 16th 2011, 02:06 PM
  3. Elimination
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jan 27th 2010, 11:50 PM
  4. Elimination
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 19th 2009, 05:28 PM
  5. Elimination
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 18th 2008, 08:57 PM

Search Tags


/mathhelpforum @mathhelpforum