Parametrics

• Apr 8th 2008, 07:42 PM
topher0805
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 $\displaystyle x = e^t - 6$ and $\displaystyle y = e^{2t}$

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

Isolate t:

$\displaystyle x+6=e^t$

$\displaystyle t=\ln {(x+6)}$

Plug it into the y equation:

$\displaystyle y=e^{2\ln {(x+6)}}$ , $\displaystyle x> {-6}$
• Apr 8th 2008, 07:47 PM
o_O
Error in one of your steps:

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

6 is inside the logarithm as well.
• Apr 8th 2008, 07:49 PM
topher0805
That was just a typo by me. It still isn't the right answer and I have no idea why.
• Apr 8th 2008, 07:51 PM
o_O
Oooh. You just need to simplify a bit more. Remember your logarithm rules? You can bring the 2 up:

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

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

Or you could've realized:
$\displaystyle y = e^{2t} = \left(e^{t}\right)^{2} = (x + 6)^{2}$
• Apr 8th 2008, 07:52 PM
topher0805
I don't know how I missed that.

• Apr 8th 2008, 08:12 PM
Soroban
Hello, topher0805!

Quote:

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

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

From [1], we have: .$\displaystyle e^t \:=\:x+6\quad\Rightarrow\quad e^{2t} \:=\:(x+6)^2$

Substitute into [2]: .$\displaystyle y \:=\:(x+6)^2$

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

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

                    |*                     |                     |                     *                     |                     *|                   * |                 *  |       --------o------+-----             -6      |

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