(a) Graph the parameterized curve described by x = 2sint, y = -3cost, 0<= t <= pi. Indicate the direction in which the curve is traced.

(b) Write a Cartesian equation for the parametrized curve. What portion of the graph of the Cartesian equation is traced by the parametrized curve?

(a) To graph the parameterized curve, we can substitute different values of t into the equations x = 2sin(t) and y = -3cos(t) to obtain corresponding points on the curve.

First, let's consider the range of t given in the problem, which is 0 ≤ t ≤ π. This means we need to evaluate the equations for values of t from 0 to π.

Let's start with t = 0:
x = 2sin(0) = 0
y = -3cos(0) = -3

So, the point (0, -3) is on the curve.

Next, let's consider t = π/4:
x = 2sin(π/4) = 2(√2/2) = √2
y = -3cos(π/4) = -3(√2/2) = -3√2/2

So, the point (√2, -3√2/2) is on the curve.

Similarly, you can substitute other values of t within the given range to obtain more points on the curve.

By plotting these points, you can trace out the graph of the parameterized curve.

To indicate the direction in which the curve is traced, you can start with the initial point and see if the subsequent points move in the clockwise or counterclockwise direction.

(b) To write a Cartesian equation for the parametrized curve, we can eliminate the parameter t from the equations x = 2sin(t) and y = -3cos(t).

For x = 2sin(t), we can solve for t:
t = arcsin(x/2)

Substituting t back into the equation y = -3cos(t), we get:
y = -3cos(arcsin(x/2))

Using a trigonometric identity, cos(arcsin(z)) = √(1 - z^2), where z = x/2:
y = -3√(1 - (x/2)^2)

This is the Cartesian equation for the parametrized curve.

To determine the portion of the graph of the Cartesian equation traced by the parametrized curve, we consider the range of t mentioned earlier, which is 0 ≤ t ≤ π.

By substituting t = 0 and t = π into the Cartesian equation, we can determine the corresponding values of x and y. The resulting points will represent the endpoints of the portion of the graph traced by the parametrized curve.

In this case, when t = 0, we have x = 0 and y = -3, representing the starting point, and when t = π, we have x = 2 and y = 0, representing the ending point.

So, the portion of the graph of the Cartesian equation traced by the parametrized curve is the line segment from (0, -3) to (2, 0).