(With a TI-89 calculator)
Graph x(t) = 3t^2 and y(t)=2-2t on [0,1]. Write a Cartesian equation that models this graph. Identify the initial and terminal points.
I graphed both equations, but I didn't really get what [0,1] meant. I think it's an interval. Also, with the Cartesian equation, I tried doing it algebraically:
y(t) = 2-2t
y-2 = -2t
t=(y-2)/-2
x=3[(y-2)/-2] ?
I'm also not sure about the terminal and initial point. Help please?
Thank you!
[0,1] means that you only need to graph the equation from t=0 to t=1
The initial point can be found by (x(0), y(0)); the terminal point is (x(1), y(1))
Thanks! That really helps.
However, what about the Cartesian equation? Am I on the right track with that?
Yes, [0,1] represents an interval. In this case, it means you need to consider the values of t between 0 and 1 to graph the equations.
To write a Cartesian equation that models these equations, you need to eliminate the parameter t. Let's start with the given equations:
x(t) = 3t^2
y(t) = 2 - 2t
To eliminate t, we can solve the second equation for t and substitute it into the first equation. Here's how:
1. Start with y(t) = 2 - 2t.
2. Move the 2 to the other side of the equation: y(t) - 2 = -2t.
3. Divide both sides by -2 to isolate t: t = (2 - y(t)) / 2.
4. Substitute t = (2 - y(t)) / 2 into x(t): x(t) = 3[(2 - y(t)) / 2]^2.
Simplifying x(t), we get:
x(t) = 3[(2 - y(t))^2 / 4]
x(t) = (3/4) * (2 - y(t))^2
So, the Cartesian equation that models this graph is x = (3/4) * (2 - y)^2. This equation represents a parabola in the x-y plane.
Now, let's find the initial and terminal points. In this case, the initial point corresponds to t = 0, and the terminal point corresponds to t = 1.
For t = 0:
x(0) = (3/4) * (2 - y(0))^2
x(0) = (3/4) * (2 - 2)^2
x(0) = (3/4) * 0^2
x(0) = 0
So, when t = 0, the initial point is (0, const) on the graph.
For t = 1:
x(1) = (3/4) * (2 - y(1))^2
x(1) = (3/4) * (2 - (2 - 2))^2
x(1) = (3/4) * (2 - 0)^2
x(1) = (3/4) * 2^2
x(1) = (3/4) * 4
x(1) = 3
So, when t = 1, the terminal point is (3, const) on the graph.
Therefore, the initial point is (0, const) and the terminal point is (3, const).