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Given the plane curve $\vec{r}(t) = \langle t-2, t^2 - 1 \rangle$\\
Sketch the position vector $\vec r (t)$ and the tangent vector $\vec r \ '(t)$ at $t = 1$
| 23 views

Given the plane curve $\vec{r}(t) = \langle t-2, t^2 - 1 \rangle$\\
Sketch the position vector $\vec r (t)$ and the tangent vector $\vec r \ '(t)$ at $t = 1$

\begin{align*}
&\begin{matrix}
t & x & y \\
0 & -2 & -1 \\
1 & -1 & 0 \\
2 & 0 & 3 \\
3 & 1 & 8 \\
4 & 2 & 15 \\
5 & 3 & 24 \\
\end{matrix} && \text{Table of values}\\
&\includegraphics[scale=0.3]{{graph_no_vector.png}} && \text{Graph of $\vec r (t)$}\\
&\vec r \ '(t) = \langle 1, 2t \rangle && \text{Take the derivative of $\vec r (t)$}\\
&\vec r \ '(1) = \langle 1, 2 \rangle && \text{Substitute $t = 1$}\\
&\includegraphics[scale=0.3]{{graph_with_vector.png}} && \text{Graph $\vec r \ '(1)$ at $t = 1$}
\end{align*}
by Diamond (54,858 points)