parametric equation

Apr 2010
15
0
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:
\(\displaystyle
x = \sec t,
y = \tan^2 t,
0\leq t \leq \pi
\)


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

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

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
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:
\(\displaystyle
x = \sec t,
y = \tan^2 t,
0\leq t \leq \pi
\)


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

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

\(\displaystyle y = x^2 - 1\)
 
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Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, kevin11!

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

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

The equation is: .\(\displaystyle y \:=\:x^2-1\quad\hdots\) 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).

. . \(\displaystyle \begin{array}{c|cc}
t & x & y \\ \hline \\[-4mm]
0 & 1 & 0 \\ \\[-4mm]
\frac{\pi}{6} & 1.15 & 0.33 \\ \\[-4mm]
\frac{\pi}{4} & 1.41 & 1 \\ \\[-4mm]
\frac{\pi}{3} & 2 & 3 \end{array}\)
L . \(\displaystyle \begin{array}{c|cc}
\frac{\pi}{2} & \infty & \infty \\ \\[-4mm]
\frac{2\pi}{3} & \text{-}2 & 3 \\ \\[-4mm]
\frac{3\pi}{4} & \text{-}1.41 & 1 \\ \\[-4mm]
\frac{5\pi}{6} & \text{-}1.15 & 0.33 \\ \\[-4mm]
\pi & \text{-}1 & 0 \end{array}\)


So the graph looks like this:


Code:
                  |
      *          3+           *
                  |
       *          |          *
         *        |        *
    - - - - * - - + - - * - - - -
     -2    -1     |     1     2
                  |
 
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Apr 2010
15
0
Hey guys, thanks for your help, really helped me out!