## Related questions with answers

Find r′(t), r(t_0), and r′(t_0) for the given value of t_0. Then sketch the plane curve represented by the vector-valued function, and sketch the vectors r(t_0) and r′(t_0). Position the vectors such that the initial point of r(t_0) is at the origin and the initial point of r′(t_0) is at the terminal point of r(t_0). What is the relationship between r′(t_0) and the curve? r(t) = ⟨e^-t, e^t⟩, t_0 = 0

Solutions

Verified$r'(0)$ is tangent to r(0) at $t=0$

$\color{#4257b2}{\mathbf{r}(t)=\left\langle e^{-t}, e^{t}\right\rangle, t_{0}=0}$

$\begin{align*} \mathbf{r}^{\prime}(t) &=\frac{d}{d t}\left\langle e^{-t}, e^{t}\right\rangle \\ &=\left\langle\frac{d}{d t}\left(e^{-t}\right), \frac{d}{d t}\left(e^{t}\right)\right\rangle \\ &=\left\langle- e^{-t}, e^{t}\right\rangle \end{align*}$

$\color{#c34632}{\mathbf{r}^{\prime}(t)=\left\langle- e^{-t}, e^{t}\right\rangle }$

$\hspace*{5mm}$Now is

$\begin{align*} \mathbf{r}(0) &=\left\langle e^{-(0)}, e^{(0)}\right\rangle \\ &=\langle 1,1\rangle \end{align*}$

$\color{#c34632}{\mathbf{r}(0)=\langle 1,1\rangle }$

$\begin{align*} \mathbf{r}^{\prime}(0) &=\left\langle- e^{-(0)}, e^{(0)}\right\rangle \\ &=\langle- 1,1\rangle \end{align*}$

$\color{#c34632}{ \mathbf{r}^{\prime}(0)=\langle- 1,1\rangle }$

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