plzzzzzzzzzzz help?

Question

I have a summer hw, i don't know how to do this problem and i have to do 7 like it .. plz explain this to me so i can do the others.
A parametrization is given for a curve.
(a) what are the initial and terminal points, if any? Indicate the direction in which the curve is traced.
(b) Find a Cartesian equation for a curve that contains the parametrized curve. What portion of the graph of the Cartesian equation is traced by the parametrized curve ?
Is it right that the initial point is (1,0) and the terminal point is (-1,0)??
and i also have to graph it, i can do that by using a grapher but is there a way that i can do it without a grapher ??
Thanks soooooooo much

Im sooooooo sorry i forgot to put the problem..
x = cos t
y = sine t
t: [ 0, py]

Answer 1

Paramerization describes the motion of each coordinate individually, usually with respect to another variable...such as time, t.

So If I said, x=t and y=t is the parameterization of a curve, you'd go through and plug in a bunch of different values for t in and find a bunch of different (x,y) points.
so...
t=0 x=0 y=0 (0,0)
t=1 x=1 y=1 (1,1)
t=2 x=2 y=2 (2,2)

it is obvious that this is the parameterization of the line y=x.

Now answering part A:
Usually the question will state that there are limits on the input, e.g. "The parameterized curve x=t, y=t where 0<=t<=8."
The initial point would be where t=0 and the terminal point is where t=8.

Part B.
To find a cartesian equation for the curve, it is necessary to determine a relation between x and y using t as a translational variable. In the example above one would write.
x=t, y=t, therefore x=y
if x=2t and y=3t
x/2=t and y/3=t
x/2=y/3
3x/2=y
Easy!

Part C:
The best way to graph the curve would be to deparameterize it by the method in part B then plot the curve as you would in cartesian coordinates.

Answer 2

A parametrized curve is normally based on a parameter that varies (often, t for time). It may be represented as an equation or a Cartesian parametrized coordinate, for example: the point A(t) = (1.2 t, 17/t), with t belonging to an interval (a,b).

You can read the above as:

The location of the point A, at time t, is, on the x-axis, 1.2 times t, and on the y-axis, 17 divided by t; with t being a value between a and b.

If (a,b) is a continuous interval, then the start point is found by replacing t with the "start value" (in my example, that is a); the terminal point is found by replacing b with the final value in the interval (in my example, b). Unless there are indications to proceed otherwise, the curved is traced by starting with the lower number (a).

To find the cartesian equation, there are indeed two ways, the first being to graph the curve (Oh, look! I get a circle with centre at (m,n) and a radius of r).

Otherwise, you have to find a relationship between the two coordinates (x and y). Sometimes it is easy. Sometimes not.

x = 1.2*t
then t = x/1.2
y = 17/t
then t = 17/y

we can try t - t = 0 = x/1.2 - 17/y

17/y = x/1.2
y/17 = 1.2/x
y = 20.4/x (for x not equal 0)

(a hyperbolic branch)

we could also try t/t = 1

(which, of course, whould give the same result).

The domain of x depend on the original domain of t. In this case, the domain of x = (1.2a, 1.2b)

Answer 3

What is the parameterization? It might look like this:

x(t) = 2t - 1
y(t) = t + 3