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?
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.
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.
wow thx a lot man
To sketch the path of the particle described by the vector r = (cos2t, sint), you can start by plugging in specific values of t within the given interval [-π/2, π]. It's recommended to choose a few key values of t to get a sense of the shape of the curve.
For example, you can calculate the position of the particle at t = -π/2, t = 0, and t = π. The values you obtain by evaluating r = (cos2t, sint) at these specific t values will give you corresponding (x, y) coordinates for the points on the path.
Once you have enough points, you can connect them together to form the curve. In this case, since t is continuous between -π/2 and π, the resulting curve will be smooth and continuous.
To determine if a curve is smooth, you can observe if there are any abrupt changes or discontinuities. If the path has no sharp corners or sudden changes in direction, it can be considered a smooth curve. In this case, since r = (cos2t, sint) provides the position of the particle at any given t within the interval [-π/2, π], the resulting curve will be smooth.
Note that you can also plot the curve using graphing software or online graphing tools to visualize it more accurately.