x=2sint
y=-3cost
0<[smaller and equal to] t <[larger and equal to] pie
find a cartesian equation for the parametrized curve what portion of the graph of the cartesian equation is traced by the parametrized curve?
To find the cartesian equation for the parametrized curve, we need to eliminate the parameter 't' and express 'y' in terms of 'x'.
Given:
x = 2sin(t) ...(1)
y = -3cos(t) ...(2)
0 ≤ t ≤ π
Let's square equation (1) and (2), and add them together:
x^2 = (2sin(t))^2
= 4sin^2(t) ...(3)
y^2 = (-3cos(t))^2
= 9cos^2(t) ...(4)
Adding equations (3) and (4), we get:
x^2 + y^2 = 4sin^2(t) + 9cos^2(t)
Using the identity sin^2(t) + cos^2(t) = 1, we have:
x^2 + y^2 = 4(1 - cos^2(t)) + 9cos^2(t)
= 4 - 4cos^2(t) + 9cos^2(t)
= 4 + 5cos^2(t)
Simplifying, we have:
x^2 + y^2 = 4 + 5cos^2(t)
This is the cartesian equation for the parametrized curve.
To determine what portion of the graph is traced by the parametrized curve, we can analyze the range of the parameter 't'. In this case, 0 ≤ t ≤ π covers the interval from 0 to π (inclusive).
With this range, the parametrized curve will trace the portion of the graph of the cartesian equation x^2 + y^2 = 4 + 5cos^2(t) for t values between 0 and π, including the endpoints.
Hope this helps! Let me know if you have any further questions.
To find the cartesian equation for the given parametric equations, x = 2sin(t) and y = -3cos(t), we need to eliminate the parameter t.
We can use the trigonometric identity sin^2(t) + cos^2(t) = 1 to eliminate t.
Starting with the equation x = 2sin(t):
Rearrange the equation to get sin(t) = x/2.
Square both sides to get sin^2(t) = (x/2)^2.
Similarly, using the equation y = -3cos(t):
Rearrange the equation to get cos(t) = y/-3.
Square both sides to get cos^2(t) = (y/-3)^2.
Now, using the trigonometric identity sin^2(t) + cos^2(t) = 1, substitute the expressions we obtained for sin^2(t) and cos^2(t):
(x/2)^2 + (y/-3)^2 = 1.
Simplifying this equation gives us the cartesian equation of the parametrized curve:
9x^2 + 4y^2 = 36.
To determine what portion of the graph of the cartesian equation is traced by the parametrized curve, we can analyze the range of t given in the question.
The given range is 0 ≤ t ≤ pi, which corresponds to a half rotation of the circle.
From the cartesian equation 9x^2 + 4y^2 = 36, we can recognize that it represents an ellipse centered at the origin.
Since the range of t is from 0 to pi, the parametrized curve will trace a half-ellipse in the first and second quadrants.
Therefore, the portion of the graph traced by the parametrized curve is the upper half of the ellipse.
(x/2)^2 + (y/3)^2 = 0
Since x and y have opposite signs, you are in the second or fourth quadrant of an ellipse.
The 0 < t < pi requirement restricts you to the second quadrant.