Please have a look at the curve below:
This is part of a line integral question. The solution to the parametrization of the curve from $(-3,0)$ to $(3,0)$ is $r(t) = (3sint, 3cost)$, $t \in [\frac{\pi}{2}, \frac{-\pi}{2}]$.
How is this true? Could someone please help me understand this parametrization? Thanks
First, since $(3\sin(t))^2+(3\cos(t))^2=9$, the points that you're getting are on the unit circle. If we try a few points, say $t=\frac{\pi}{2}$, then $r(t)=(3,0)$; $t=0$ gives $(0,3)$; and $t=-\frac{\pi}{2}=(-3,0)$. These are, indeed, three points on the curve that you drew.
Now, the way that $t$ is written is a little unusual; using this notation, I would assume that $t$ starts at $\frac{\pi}{2}$ and decreases to $-\frac{\pi}{2}$. However, this seems to plot the three points that we computed above in the wrong order (it starts on the right and ends on the left).
We can, additionally, check the signs of $\sin$ and $\cos$ to check that we're on the right parts of the curve. For example, for $t\in(0,\frac{\pi}{2})$, both $3\sin(t)$ and $3\cos(t)$, so we are in the first quadrant. For $t\in(-\frac{\pi}{2},0)$, $3\sin(t)$ is negative, but $3\cos(t)$ is positive, so we are in the second quadrant (and on the circle of radius $3$, as illustrated above).