Answer
Circle of radius $2$, centered at the origin
Work Step by Step
Step 1
From a vector-valued function: \[ \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} \] the parametric equations can be obtained: \[ \begin{align*} x &= x(t) \\ y &= y(t) \end{align*} \] The graph can be identified by remembering certain forms, using trigonometric relationships, converting to Cartesian form, or by plotting some points and drawing a smooth curve through them.
Step 2
From: \[ \mathbf{r} = 2\sin(3t)\mathbf{i} - 2\cos(3t)\mathbf{j} \] we have parametric equations: \[ \begin{align*} x &= 2\sin(3t) \\ y &= -2\cos(3t) \end{align*} \] This looks very similar to the parametric equations of a circle centered at the origin: \[ \begin{align*} x &= a\sin(t) \\ y &= a\cos(t) \end{align*} \]
Step 3
If we find: \[ x^2 + y^2 \] then we can use the Pythagorean Identity to get: \[ x^2 + y^2 = (2\sin(3t))^2 + (-2\cos(3t))^2 = 4\sin^2(3t) + 4\cos^2(3t) = 4(\sin^2(3t) + \cos^2(3t)) = 4(1) = 4 \] This is the equation of a circle centered at the origin: \[ x^2 + y^2 = r^2 \] So, the radius is: \[ r = 2 \] Result : Circle of radius $2$, centered at the origin.
