the path of a particle in the xy-plane is vector r = (cos2t, sint) for t for all [-pi/2, pi] where t represents time. Sketch the path. Is it a smooth curve?

Question

How do i sketch the path? do i just plug in random points between [-pi/2, pi]? how would i connect it together? and how do i know if it's a smooth curve? Thank you.

Answer 1

When t = -pi/2, x = cos(-pi) = -1 and y = sin -pi/2 = -1

When t = -pi/4, x = cos(-pi/2) = 0 and y = sin -pi/4 = -0.707

When t = 0, x = 1 and y = 0

When t = pi/4, x = 0 and y = .707

When t = pi/2, x = -1 and y = 1

When t = 3pi/4, x = cos 3pi/2 = 0 and y = sin3pi/4 = 0.707
When t = pi, x = 1 and y = 0

The particle follows a smooth curve. The curve looks like a parabola that is open to the left.

You can use a trigonometric identity to show that the curve is
x = 1 - 2y^2, which is indeed a parabola.

Answer 2

wow thx a lot man