Question

Help?
Eliminate the parameter.
x = 4 cos t, y = 4 sin t

Answer 1

@whpalmer4

Answer 2

@sourwing

Answer 3

Alternatively, you can take the fact that the parametric equations describe a circle with radius 4, so you have
(4cost)2+(4sint)2=42x2+y2=16

Answer 4

Okay, wait... Can you do that in the equation thing? I can't tell what exactly you did.

Answer 5

I know we have to square both sides first..

Answer 6

@sexyperson

Answer 7

Where did you get the 42?

Answer 8

look at this example from openstudy

Answer 9

http://openstudy.com/study#/updates/51756758e4b0be9997e012d3

Answer 10

I still don't get it..

Answer 11

Is that supposed to be (4cost)^2+(4sint)^2=4^2+x^2+y^2=16?

Answer 12

How do you know it equals 16?

Answer 13

@Light&Happiness

Answer 14

If I square both 4's inside the parenthesis, I don't get 16, I get 32..???

Answer 15

Okay, I don't do parametric equations very often, haven't for many years, but yes, this describes a circle with radius 4. Here's a plot of the two parametric equations: remember that one is the x coordinate, and the other is the y coordinate...

Answer 16

So you can see that as \(t\) goes from \(0\) to \(2\pi\), the y coordinate goes from 4 to 0 to -4 and back to 0 and then to 4, and the x coordinate simultaneously goes from 0 to 4 to 0 to -4 to 0

Answer 17

If you look at the graph of a circle with radius 4, that's what happens as you start at the top and go clockwise around the circle, right?

Answer 18

Yes..?

Answer 19

Here it is showing the angle from vertical (in degrees) as the x coordinate.

Answer 20

I'm confused on this part:(4cost)2+(4sint)2=42x2+y2=16

Answer 21

:/

Answer 22

then ignore it, it looks wrong to me.

Answer 23

A circle with radius 4 with center at the origin will have what equation?

\[(x-h)^2 + (y-k)^2 = r^2\]

Answer 24

I thought so.. So we need to use that? Do we need to divide by the 4's?

Answer 25

what would your equation for a circle with radius 4 centered at the origin be?

Answer 26

formula gives you a circle with radius \(r\) centered at \((h,k)\). We want center at \((0,0)\) so \(h = 0, k = 0\) and radius \(4\) so \(r = 4\):

\[(x-0)^2 + (y-0)^2 = 4^2\] or\[x^2+y^2=16\]

Yes, I realize that I have not eliminated the parameter in the way they want you to do. I make no pretense about doing so — I'm trying to get you to learn how to reject bogus information.

Answer 27

(x-0)^2+(y-0)^2=4^2?

Answer 28

Okay, whatever works. :)

Answer 29

Yes. Now, if you know what you're supposed to be getting, you have a better shot at evaluating whether the method someone is telling you how to do this is going to work, right?

Answer 30

I guess so... But that's my answer?... x^2+y^2=16. What happened to my t? I'm lost..

Answer 31

you're eliminating the parametric variable, right? t is the parametric variable. So yes, x^2+y^2=16 is the answer here, but to the extent you're supposed to show your work and know how to do this, you're not exactly done yet :-)

Answer 32

Here, if you read this webpage, partway down the page they do an equivalent problem: http://colalg.math.csusb.edu/~devel/precalcdemo/param/src/param.html

Answer 33

Thanks, I'll check it out! God bless you! :D