I have some questions about parametric equations in Calculus. [see details, not enough space here]?

Question

1.) If you are given two equations and bounds, how do you eliminate the parameter to find a Cartesian equation of the curve?
2.) If you again are given two equations and bounds, how do you describe the motion of a particle with position (x, y) as t varies in the given interval?

I'm not asking you to do them for me, but how do you set them up to solve them?

Examples:
1.) x = 3 cosx , y = 4 sinx , -pi/2 < x < pi/2

2.) x = 2 + cos t, y = 3 + sin t, 0 < t < 2
The motion takes place on a unit circle centered at (___,___). As t goes from 0 to 2, the particle makes (2, 1, 4, or 3) complete (counterclockwise or clockwise) rotation(s) around the circle, starting and ending at (_____,_____).

For the second one, do you have to convert them into radians first?

Answer 1

you probably wanted
x=3 cos @ and y=4 sin @ -pi/2<@ x/3= cos @
y/4= sin @
square and sum
x^2/9 +y^2/16=1 x>=0 as cos@ is positive in the given intervall
(it is half of an ellipse with the mayor axis on the y axis)
2) x-2= cos t
y-3= sin t
sqare and sum
(x-2)^2+(y-3)^2 = 1 circle center (2,3) and radius=1
At t=0 the point is at (3,3)Extreme of an horizontal diameter
as t increases x decreases and y increases. So the point is moving counterclockwise
To see if it makes turns we have to solve
3=2+cos t and 3 = 3+sin t and see how many are in the given range
cos t= 1 t=0 t=2pi
sin t=0 t=0 t=2pi
so in the given time interval the point does not return to its initial position as 2pi >2
The point ends at (2+cos 2,3+sin 2) without having turned
t is a real number so it is in radians
Maybe the given interval is 0

Answer 2

There are various ways to eliminate a parameter depending on the given equations. Suppose that you are given x = f(t) and y = g(t). One way is to try to rearrange the first to give
t = f(-1)(x) and then put this in the g(t).
Sometimes you can't do this, but there are other dodges too difficult to explain here in much detail.
Example x = 4t^3 + t, y = 4t^3 - t.
You notice that x + y = 8t^3 and x - y = 2t
Therefore x + y = (x - y)^3
Experience and insight help a lot!

2. Most advanced work involving trig functions is automatically in radians to start with.

Answer 3

x=rcostheta, y=rsintheta, and r=x^2+y^2.
and usually for area it is r1(^2)-r2(^2)