Answer
\(a)\) Graphed below
\(b)\) \(\vec{r}'(t)=4\cos t \vec{i} +2\sin t \vec{j}\)
\(c)\) Graphed below
Work Step by Step
\(a)\) Graph.
\(b)\) \(\vec{r}(t)=\langle 4 \sin t, - 2 \cos t \rangle\) finding the derivative on both sides we have:
\[
\vec{r}'(t)=\langle 4 \cos t, 2 \sin t \rangle
\]
\(c)\) At \(t=3\pi/4\) we have:
\[
\vec{r}(3\pi/4)=\langle 2\sqrt{2},\sqrt{2}\rangle
\] \[
\vec{r}'(3\pi/4)= \langle-2\sqrt{2}, \sqrt{2} \rangle
\]
