MHB Curves defined by parametric curves

[ineedhelpnow]

eliminate the parameter to find a cartesian equation of the curve.

$x=sin\frac{1}{2} \theta$
$y=cos\frac{1}{2} \theta$
$-\pi \le \theta \le \pi$

$x=e^t-1$
$y=e^{2t}$

 

[MarkFL]

1.) Hint: Pythagorean identity.

2.) Hint: Square the first equation...

 

[ineedhelpnow]

is the second one $y=x^2+1$

 

[MarkFL]

is the second one $y=x^2+1$

No, the first equation may be written as:

$$e^t=x+1$$

Square:

$$e^{2t}=(x+1)^2$$

$$y=(x+1)^2$$

 

[ineedhelpnow]

oh i see. i just disregarded the sign for some reason. and that second one would be $x^2+y^2=1$, right? square both sides and then use identity which simplifies one side of the equation to 1

 

[MarkFL]

oh i see. i just disregarded the sign for some reason. and that second one would be $x^2+y^2=1$, right? square both sides and then use identity which simplifies one side of the equation to 1

Yes...but do you need to account for the restriction on $\theta$?

 

[ineedhelpnow]

i don't know. only when graphing it?

 

[MarkFL]

only when graphing it?

Well, since the graph represents the set of points that satisfy the given equation, it's not only the graph you need worry about. Does every point on the circle satisfy all of the original conditions? If not, can we write the equation is such a way that all initial conditions are satisfied?

 

[ineedhelpnow]

isnt it just $x^2+y^2=1$ $y \ge 0$

 

[MarkFL]

isnt it just $x^2+y^2=1$ $y \ge 0$

Yes, or you could just write:

$$y=\sqrt{1-x^2}$$

This describes the curve and ensures $0\le y$ all in one statement.

 

[ineedhelpnow]

oh ok. i see what you mean now.