Question

Eliminate the parameter.

x = 3 cos t, y = 3 sin t

Would someone mind helping me with this problem? I don't know how to do this at all

Answer 1

hint : \(\large \sin^2t + \cos^2t = 1\)

Answer 2

eliminating parameter simply means cooking up a new equation having only x's and y's

Answer 3

.... from the given parametric equaitons

Answer 4

Okay, what do I do with the hint you gave? I'm sorry if I sound like a fool. sin, cos, and tan confuse me terribly

Answer 5

you only need to remember that identity for doing these problems...

Answer 6

x = 3 cos t, y = 3 sin t

x^2 = ?
y^2 = ?

Answer 7

x = 3 cos t,
so
x^2=(3cos t)^2=??

Answer 8

9 cos^2 t??

Answer 9

Yes, and what about squaring y=3 sin t?

Answer 10

9 sin^2 t

Answer 11

Mia: We started with x=3cos t; squaring that produces x^2 = 9 (cos t)^2.
Squaring y = 3 sin t produces y^2 = 9 (sin t)^2.

Please look at the identity that ganeshie8 shared with us earlier. (sin t)^2 + (cos t)^2 = ?

Use this to simplify your expression.

Answer 12

Why are they being added?

Answer 13

x^2 + y^2 = 9 (sin t)^2 + 9 (cos t)^2 = 9 ( ?? ) = 9

Good question. The answer revolves around ganeshie8's identity:

(sin t)^2 + (cos t)^2 = 1. our job was to eliminate that t, right?

Answer 14

that x^2 + y^2 (on the far left) equals 9 (on the far right. And x^2 + y^2 = r^2, where r is the radius of the circle in question. We end up with r^2=9. What is r?

Answer 15

Would that be the 1 from the identity?

Answer 16

Yes, that's right.

r=radius. r^2 = 9 = 3^2. What is r?

Answer 17

What I meant was, is r=1?

Answer 18

Oh, I'm sorry. what we get is r-squared on the left and 9, or 3 squared, on the right.

Taking the sqrt of both sides, r=3, OK? Think: What does " r " represent?

Answer 19

You said that r is the radius of the circle