1. ## parametric equation

Hey all, sorry if this is in the wrong section. I'm in a precalc class but we're jumping between trig and this so not really sure anymore. Any help would be appreciated!

a. Eliminate parameter t to find the equation in rectangular form.
Sketch the curve defined by:
$
x = \sec t,
y = \tan^2 t,
0\leq t \leq \pi
$

So far I made a chart with 3 columns, labeled: t, x, and y. For the t column I chose the values:
$
\frac {\pi}{3},
\frac {\pi}{4},
\frac {\pi}{6},
\frac {3\pi}{4},
\frac {2\pi}{3},
\frac {5\pi}{6}
$

Am I supposed to convert? If so, how do I convert $tan^2 t$?
$
\sec t = \frac {1}{\cos t}?
$

2. Originally Posted by kevin11
Hey all, sorry if this is in the wrong section. I'm in a precalc class but we're jumping between trig and this so not really sure anymore. Any help would be appreciated!

a. Eliminate parameter t to find the equation in rectangular form.
Sketch the curve defined by:
$
x = \sec t,
y = \tan^2 t,
0\leq t \leq \pi
$

So far I made a chart with 3 columns, labeled: t, x, and y. For the t column I chose the values:
$
\frac {\pi}{3},
\frac {\pi}{4},
\frac {\pi}{6},
\frac {3\pi}{4},
\frac {2\pi}{3},
\frac {5\pi}{6}
$

Am I supposed to convert? If so, how do I convert $tan^2 t$?
$
\sec t = \frac {1}{\cos t}?
$
$\tan^2{t} = \sec^2{t} - 1$

$y = x^2 - 1$

3. Hello, kevin11!

(a) Eliminate parameter $t$ to find the equation in rectangular form.

(b) Sketch the curve defined by: . $\begin{Bmatrix}x &=& \sec t \\ y &=& \tan^2\!t\end{Bmatrix}\quad 0\leq t \leq \pi$
skeeter is absolutely correct!

The equation is: . $y \:=\:x^2-1\quad\hdots$ a parabola.
. . It opens upward, vertex (0,-1), x-intercepts (±1, 0)
But there's more . . .

I too made a chart and got these values (some approximate).

. . $\begin{array}{c|cc}
t & x & y \\ \hline \\[-4mm]
0 & 1 & 0 \\ \\[-4mm]
\frac{\pi}{6} & 1.15 & 0.33 \\ \\[-4mm]
\frac{\pi}{4} & 1.41 & 1 \\ \\[-4mm]
\frac{\pi}{3} & 2 & 3 \end{array}$

L . $\begin{array}{c|cc}
\frac{\pi}{2} & \infty & \infty \\ \\[-4mm]
\frac{2\pi}{3} & \text{-}2 & 3 \\ \\[-4mm]
\frac{3\pi}{4} & \text{-}1.41 & 1 \\ \\[-4mm]
\frac{5\pi}{6} & \text{-}1.15 & 0.33 \\ \\[-4mm]
\pi & \text{-}1 & 0 \end{array}$

So the graph looks like this:

Code:
                  |
*          3+           *
|
*          |          *
*        |        *
- - - - * - - + - - * - - - -
-2    -1     |     1     2
|

4. Hey guys, thanks for your help, really helped me out!