Question

find r(t), r(t_0), and r(t_0) for the given value of t_0. r(t)=(4 + t)i + t^2j, t_0 = 1

Explanation:

Step1: Differentiate r(t)

Given $\mathbf{r}(t)=(4 + t)\mathbf{i}+t^{2}\mathbf{j}$. Using the power - rule for differentiation, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$, and the derivative of a constant is 0. The derivative of $4 + t$ with respect to $t$ is 1, and the derivative of $t^{2}$ with respect to $t$ is $2t$. So, $\mathbf{r}'(t)=\mathbf{i}+2t\mathbf{j}$.

Step2: Evaluate r(t) at $t_0 = 1$

Substitute $t = 1$ into $\mathbf{r}(t)$. We have $\mathbf{r}(1)=(4 + 1)\mathbf{i}+1^{2}\mathbf{j}=5\mathbf{i}+\mathbf{j}$.

Step3: Evaluate r'(t) at $t_0 = 1$

Substitute $t = 1$ into $\mathbf{r}'(t)$. We get $\mathbf{r}'(1)=\mathbf{i}+2\times1\mathbf{j}=\mathbf{i}+2\mathbf{j}$.

Answer:

$\mathbf{r}'(t)=\mathbf{i}+2t\mathbf{j}$, $\mathbf{r}(t_0)=\mathbf{r}(1)=5\mathbf{i}+\mathbf{j}$, $\mathbf{r}'(t_0)=\mathbf{r}'(1)=\mathbf{i}+2\mathbf{j}$