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