sketch + describe graph of the curve parametrized by:
x=1-3cos^(2)t
y=-2+4sin^(2)t
-infinity less than t less than infinity
how would i go about solving this.
i know that usually i can solve for t and than plug that in. But for this, i tried using cos^2 + sin^2 = 1... but i'm stuck
cos^2(t)=(1-x)/3
sin^2(t)=(y+2)/4
Add two equations
1=(1-x)/3+(y+2)/4
12=4-4x+3y+6
y=(4/3)x+2/3
i don't really understand parametrized curves. but isn't y=(4/3)x +(2/3) a straight line. So, it doesn't have to be a curve?
Straight line is a special case of the curve just like the natural numbers are a special case of integers
To sketch and describe the graph of the curve parametrized by x = 1 - 3 cos^2(t) and y = -2 + 4 sin^2(t), you can start by manipulating the equations and using trigonometric identities.
1. Start with the equation cos^2(t) + sin^2(t) = 1, which is the fundamental trigonometric identity.
2. Let's begin with the x equation: x = 1 - 3 cos^2(t).
To make this equation easier to work with, you can use the identity cos^2(t) = 1 - sin^2(t).
Substitute cos^2(t) with its equivalent expression in the x equation:
x = 1 - 3(1 - sin^2(t))
x = 1 - 3 + 3 sin^2(t)
x = -2 + 3 sin^2(t)
3. Now, let's focus on the y equation: y = -2 + 4 sin^2(t).
This equation already has sin^2(t) directly, so no further substitution is needed.
4. The parametric equations are now simplified as:
x = -2 + 3 sin^2(t)
y = -2 + 4 sin^2(t)
5. To sketch the graph of the parametric equation, you need to examine the range and behavior of the parameter t.
Since t can take any value from negative infinity to infinity, you can focus on a smaller interval, such as 0 ≤ t ≤ 2π, which captures one complete cycle of the trigonometric functions.
6. Plot points by substituting different values of t into the parametric equations.
- Let t = 0:
x = -2 + 3(0)^2 = -2 + 0 = -2
y = -2 + 4(0)^2 = -2 + 0 = -2
The point (x, y) = (-2, -2) is the starting point.
- Let t = π/2:
x = -2 + 3(sin^2(π/2)) = -2 + 3(1) = 1
y = -2 + 4(sin^2(π/2)) = -2 + 4(0) = -2
The point (x, y) = (1, -2) represents the curve at t = π/2.
- Continue calculating for other values of t, such as t = π, 3π/2, 2π, etc.
7. As you calculate and plot more points, you'll start to notice a pattern.
The graph of the curve is a closed loop resembling an elongated figure-eight or the shape of an infinity symbol (∞).
The curve intersects the x-axis at (-2, 0) and the y-axis at (1, -2).
The curve is symmetrical about the x-axis and y-axis.
The entire curve is within the range where -2 ≤ x ≤ 1 and -2 ≤ y ≤ -1.
By following these steps, you can graphically represent the curve parametrized by x = 1 - 3 cos^2(t) and y = -2 + 4 sin^2(t).