Answer
\(a)\) Graphed below
\(b)\) \(\vec{r}'(t)=\langle1,2t\rangle\)
\(c)\) Graphed below
Work Step by Step
\(a)\) We can plug in values to sketch a rough graph of the function:
At \(t=-1\):
\[ \vec{r}(t)=\langle -3,2 \rangle \]
At \(t=0\):
\[ \vec{r}(t)=\langle -2,1 \rangle \]
At \(t=1\):
\[ \vec{r}(t)=\langle -1,2 \rangle \]
\(b)\) \(\vec{r}(t)=\langle t-2,t^2+1 \rangle\) To find the derivative of \(\vec{r}\), we have:
\[\vec{r}'(t)=\langle1,2t\rangle\]
\(c)\) At \(t=-1\) we have:
\[
\vec{r}(-1)=\langle (-1)-2,(-1)^2+1 \rangle \\
\vec{r}(-1)=\langle -3,2 \rangle \\
\] \[
\vec{r}'(-1)=\langle1,2(-1)\rangle \\
\vec{r}'(-1)=\langle1,-2\rangle
\]
