Question

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

Answer 1

\[x=\tan^2 \theta,~~~y=\sec \theta, ~~~~ -\pi/2 < \theta < \pi/2\]

Answer 2

I'm thinking trig identities...?

Answer 3

mhmm

Answer 4

tan^2+1=sec^2

Answer 5

Yeah, so let x = tanx and y = secx?

Answer 6

\[1+x^2=y^2 \implies y = \sqrt{1+x^2}\]

Answer 7

`sec` takes both positive and negative values

Answer 8

But the interval only allows us to take positive square roots, yeah?

Answer 9

yeah i think you're right
but im not sure if we need to preserve the range of parameter after switching to cartesian form

Answer 10

For example :

x = t, y = t^2, 0<t<1

Answer 11

eliminating parameter gives you y = x^2

do you worry about the range of "t" after switching to cartesian ?

Answer 12

we need to loose that info in cartesian form


y = sqrt(1+x^2) looks wrong to me because somehow you're trying to forcibly use the range of values of t hmm

Answer 13

\[1+x=y^2 \implies y = \sqrt{1+x},~~ x \ge 0\]

Answer 14

Another example

\(x = r\cos \theta \), y = \(r\sin \theta\) ; \(\large 0\le \theta \le \pi\)


Is the cartesian form \(\large x^2 + y^2 = r^2\)
or \(\large y = \sqrt{r^2-x^2}\)

?

Answer 15

Ah that makes sense so you want to define the same curve by restricting the domain

Answer 16

i need to review the general convention, but you make more sense :)

Answer 17

Yeah, haha, I'm still kind of confused by this, maybe it's just this question is like the hardest one from the book, and I haven't done enough practice.

Answer 18

you're right, start and endpoints matter in cartesian form too even through we loose the orientation of curve information

http://tutorial.math.lamar.edu/Classes/CalcII/ParametricEqn.aspx

Answer 19

Thanks, for the clarification :)

Answer 20

:D