Question

Graphing a parametric

Answer 1

This is part 2 to the last question I asked...
x = tan^2 theta
y = sec theta
-pi/2 < theta < pi/2

I need to graph it. I made a t-chart, but I'm a little confused by the endpoints. pi/2 is undefined for tan, right?

Answer 2

So would the endpoints be asymptotes?

Answer 3

Desmos is showing pi/2 as defined...

Answer 4

You have just eliminted the parameter, so you can just use that to graph it.

Answer 5

So I wouldn't need to make a t-chart?
Wouldn't pi/2 still be undefined for x though? According to the original x equation

Answer 6

When we eliminated \(\theta\), we got: \(\displaystyle y^2-x=1\).
Alternatively, \(\displaystyle y^2=x+1\).

Note that we had: \(\displaystyle \theta \in \left(-\frac{\pi}{2},~\frac{\pi}{2}\right)\).
so take that into account. (Sorry I hate to disconnect)

Answer 7

(basically you are graphing \(y^2=x+1\) over \(x\in(0,\infty)\).

Answer 8

I still don't get how pi/2 is defined... Shouldn't making a t-chart give me the same graph as eliminating the parameter?
Also, shouldn't I be graphing it from -pi/2 to pi/2? Not from 0 to infinity?

Answer 9

The graph is NOT showing t=pi/2 defined.
It's showing x=pi/2 defined.

Do you see the difference?

Answer 10

The fact that you labeled your x-axis with trig values is maybe what's making things confusing.

Answer 11

Oh, I think I see what you're saying...
But wouldn't x be undefined at t = pi/2 ?

Answer 12

\[\large\rm \lim_{t\to \pi/2}\tan^2t=\infty\]This value t=pi/2 is what's extending our function infinitely in the x-direction, ya?

When t is near pi/2, x is really really big.

Answer 13

So say I was trying to do this problem without eliminating the parameter, just by doing a t-chart. When I plug in pi/2 for x, it's undefined. So wouldn't there be an asymptote or something there? Sorry, this is really confusing me :/