Answer
See graph and explanations.
Work Step by Step
Step 1. Given the first derivative $y'=sin(t), 0\leq t\leq2\pi$, we have $y''=cos(t)$
Step 2. Possible critical points are $t=0, \pi, 2\pi$. Determine the increasing $y'\gt0$ and decreasing $y'\lt0$ regions by signs of $y'$: $(0)..(+)..(\pi)..(-)..(2\pi)$; thus the function increases on $(0,\pi)$, decreases on $(\pi,2\pi)$. Local maximum is at $t=\pi$ and local minima at $t=0,2\pi$
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $t=\frac{\pi}{2}, \frac{3\pi}{2}$: $(0)..(+)..(\frac{\pi}{2})..(-)..(\frac{3\pi}{2})..(+)..(2\pi)$. Thus the function is concave down on $(\frac{\pi}{2},\frac{3\pi}{2})$ and concave up on $(0,\frac{\pi}{2}), (\frac{3\pi}{2}, 2\pi)$ with $t=\frac{\pi}{2}, \frac{3\pi}{2}$ as the inflection points.
Step 4. Sketch the function based on the above results as shown.
