Find two different parametric descriptions for the circle of radius 4 centered at (-3,2).

I can only think of one

(x,y) = ( -3+ 4sint, 2+ 4cost )

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- Jul 10th 2011, 05:30 AMVeronica1999finding parametric equations
Find two different parametric descriptions for the circle of radius 4 centered at (-3,2).

I can only think of one

(x,y) = ( -3+ 4sint, 2+ 4cost ) - Jul 10th 2011, 05:39 AMPlatoRe: finding parametric equations
- Jul 10th 2011, 05:51 AMVeronica1999Re: finding parametric equations
Thanks for your corrections.

Could you give me some approaches on getting the other equation? - Jul 10th 2011, 06:03 AMArchie MeadeRe: finding parametric equations
- Jul 10th 2011, 06:06 AMPlatoRe: finding parametric equations
- Jul 10th 2011, 06:06 AMVeronica1999Re: finding parametric equations
In my text book I find equations written as (x,y) = ( 5sin12t , 6-5cos12t)

Is this an incorrect notation too? - Jul 10th 2011, 06:09 AMVeronica1999Re: finding parametric equations
Thank You!!!

- Jul 10th 2011, 06:23 AMPlatoRe: finding parametric equations
Well for that author it is not incorrect. But as far as I am concerned it is.

The cosine is a function. It's symbol is $\displaystyle \cos$.

Like all functions we write $\displaystyle \cos(t)$ as its value at t.

**That is definitely incorrect period.**

The $\displaystyle \cos(t)$ is associated with the-coordinate and $\displaystyle \sin(t)$ is associated with the*x*-coordinate.*y*

So in ordered pair notation the first circle would be:

$\displaystyle \left( {4\cos (t) - 3,4\sin (t) + 2} \right)$. - Jul 10th 2011, 07:02 AMVeronica1999Re: finding parametric equations
Please excuse my ignorance, but I don't understand why the equation (x,y) = (5sin (12t) , 6- 5cos(12t)) is incorrect.

Can't I look at x as a function of values of 5sin(12t) and y as a function of the values of 6-5cos(12t)

If I plug in t=1

x= 1.039558454 y= 1.11

t=2

x = 2.033683215 y= 1.432273

If I keep on going for different values of t and connect all the dots

I will end up with the equation x squared + ( y-6 ) squared = 25 - Jul 10th 2011, 07:38 AMSironRe: finding parametric equations
Like plato already said, $\displaystyle \cos(t)$ is associated with the x-coordinate and $\displaystyle \sin(t)$ with the y-coordinate.

Remark:

If you don't see that you can draw a circle with a chosen radius r centered at (0,0), take a point P on the circle and project this on the x-axis and y-axis. With the proposition of Phytagoras you'll find the coordinates $\displaystyle r\cos(t)$ and $\displaystyle r \sin(t)$ so you get $\displaystyle P(r \cos(t), r \sin (t))$ and not $\displaystyle P(r \sin(t), r \cos (t))$ - Jul 10th 2011, 07:58 AMArchie MeadeRe: finding parametric equations
- Jul 10th 2011, 08:03 AMSironRe: finding parametric equations
@ Archie Meade:

Indeed! I've forgotten to define P, thanks for the remark. - Jul 10th 2011, 08:14 AMArchie MeadeRe: finding parametric equations
For example, for simplicity take a point on the circumference in the 1st quadrant.

Draw a right-angled triangle from that point to the point on the x-axis directly below it and the circle centre.

Now if we label the angle in the triangle where it touches the circumference as "t",

then we obtain an alternative parametric representation. - Jul 10th 2011, 08:26 AMOpalgRe: finding parametric equations
I disagree. There is nothing wrong with the parametrisation $\displaystyle (x,y) = ( 5\sin(12t) , 6-5\cos(12t))$ (for a circle centred at (0,6) with radius 5). It is

*unusual*to have sin associated with the x-coordinate and cos with the y-coordinate. But it is certainly not*incorrect*. In fact, the question asks for alternative parametrisations, and that would be a perfectly reasonable way of obtaining one. - Jul 10th 2011, 08:42 AMVeronica1999Re: finding parametric equations