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