1. finding 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 )

2. Re: finding parametric equations

Originally Posted by Veronica1999
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 )
Please get into the habit of using correct notation.
$\displaystyle \left\{ {\begin{array}{*{20}c} {x(t) = 4\cos (t) - 3} \\ {y(t) = 4\sin (t) + 2} \\ \end{array} } \right.$

Note that it is $\displaystyle \cos(t)$ and not $\displaystyle \cos t$.

3. Re: finding parametric equations

Could you give me some approaches on getting the other equation?

4. Re: finding parametric equations

Originally Posted by Veronica1999
Could you give me some approaches on getting the other equation?
$\displaystyle (x+3)^2+(y-2)^2=4^2=4^2(1)=4^2[sin^2(t)+cos^2(t)]$

$\displaystyle \Rightarrow\ (x+3)^2+(y-2)^2=[4sin(t)]^2+[4cos(t)]^2$

Hence

$\displaystyle x+3=4sin(t);\;\;\;y-2=4cos(t)$

or

$\displaystyle x+3=4cos(t);\;\;\;y-2=4sin(t)$

5. Re: finding parametric equations

Originally Posted by Veronica1999
Could you give me some approaches on getting the other equation?
I do not know what that question could mean.
I suspect that author has a definite idea what he/she does mean, but I do not. Perhaps the meaning is in your lecture notes/textbook.

6. Re: finding parametric equations

In my text book I find equations written as (x,y) = ( 5sin12t , 6-5cos12t)
Is this an incorrect notation too?

Thank You!!!

8. Re: finding parametric equations

Originally Posted by Veronica1999
Is this an incorrect notation too?
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.
Originally Posted by Veronica1999
In my text book I find equations written as (x,y) = ( 5sin12t , 6-5cos12t)
That is definitely incorrect period.
The $\displaystyle \cos(t)$ is associated with the x-coordinate and $\displaystyle \sin(t)$ is associated with the y-coordinate.

So in ordered pair notation the first circle would be:
$\displaystyle \left( {4\cos (t) - 3,4\sin (t) + 2} \right)$.

9. Re: finding parametric equations

Originally Posted by Plato
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 x-coordinate and $\displaystyle \sin(t)$ is associated with the y-coordinate.

So in ordered pair notation the first circle would be:
$\displaystyle \left( {4\cos (t) - 3,4\sin (t) + 2} \right)$.
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

10. Re: 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 $\displaystyle P(r \sin(t), r \cos (t))$

11. Re: finding parametric equations

Originally Posted by Siron
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 $\displaystyle P(r \sin(t), r \cos (t))$
The parameter "t" has not been defined yet.
You would be correct if "t" is the angle that the point on the circle
makes with the positive part of the x-axis.
However, "t" may be defined differently.

12. Re: finding parametric equations

Indeed! I've forgotten to define P, thanks for the remark.

13. Re: 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.

14. Re: finding parametric equations

Originally Posted by Plato
Originally Posted by Veronica1999
In my text book I find equations written as (x,y) = ( 5sin12t , 6-5cos12t)
Is this an incorrect notation too?
That is definitely incorrect period.
The $\displaystyle \cos(t)$ is associated with the x-coordinate and $\displaystyle \sin(t)$ is associated with the y-coordinate.

So in ordered pair notation the first circle would be:
$\displaystyle \left( {4\cos (t) - 3,4\sin (t) + 2} \right)$.
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.

15. Re: finding parametric equations

Originally Posted by Opalg
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.
Thank you!!!!
I am so happy now.
I was extremely frustrated for the past hour trying to understand what was wrong with
x=5sin(12t)