# Homework Help: How to interpret parametric equations

1. Oct 24, 2014

### TheKracken

1. The problem statement, all variables and given/known data
Eliminate the parameter to find a description of the following circles or circular arc's in terms of x and y
and find the center and radius and indicate the positive orientation

x=cos(t) , y = 3sin(t) ; 0< t < pi/2 (should be less than or equal to)

2. Relevant equations
Not sure?

3. The attempt at a solution
When I find t = arccos(x) I then plug it in y=3 sin(t) and I result in
y= sqrt(9-x^2)

The book doesn't even give an answer for the equation part but it does tell me the origin, and radius and the
orientation but I figured you could rewrite my equation to y^2 + x^2 = 9 and that tell me the radius of 3 (sqrt (9) right?) then the orgin is 0,0 because nothing it being done to the x and y

but here is what trips me up, how do they know it is the lower half of a circle going counter clockwise? I mean it has to do something with the t's but I just dont see it.

Last edited by a moderator: Oct 24, 2014
2. Oct 24, 2014

### Staff: Mentor

No. y2 + x2 = (3 sin(t))2 + (cos(t))2 = 9sin2(t) + cos2(t). Are you sure you wrote the problem correctly?
Based on what you wrote in the problem description, your curve is not a circle.

3. Oct 24, 2014

### TheKracken

I solved for t and pluged it in for x. Yes I am positive I wrote it down correctly.

4. Oct 24, 2014

### TheKracken

I should probably mention this is a parametric equation? I am sure you are correct, I may have gotten the right answer by doing something incorrectly by coincidence. Not the first time ive done that :P

5. Oct 24, 2014

### Staff: Mentor

Please show what you did to get $y = \sqrt{9 - x^2}$. Your mistake is in that work.

The parametric equations x = cos(t), y = 3sin(t) do NOT represent a circle.

6. Oct 24, 2014

### GFauxPas

To find the orientation, put t = 0 and t = pi/2 (or put more points if you'd like), and see which way you're going.

7. Oct 24, 2014

### TheKracken

I want to apologize, I did type the equation incorrectly, it is 3cos(t). I was looking at the next problem it looks like when I typed it out, but my work was for this problem.

In that case is my work correct?

8. Oct 24, 2014

### Staff: Mentor

Instead of solving for t in one of the parametric equations, I would eliminate the parameter directly. Since (now) x = 3cos(t) and y = 3sin(t), x2 + y2 = ?

Keep in mind that $0 \leq t \leq \pi/2$. As t increases from 0, what do the points on the curve do? What is the starting point (i.e., when t = 0)? What is the ending point (when t = $\pi/2$)?