Question

find r(t), r(t_0), and r(t_0) for the given value of t_0. then sketch the space curve represented by the vector - valued function, and sketch the vectors r(t_0) and r(t_0). r(t)=tmathbf{i}+t^{2}mathbf{j}+\frac{5}{4}tmathbf{k}, t_0 = 2

Explanation:

Step1: Differentiate r(t)

Given $\mathbf{r}(t)=t\mathbf{i}+t^{2}\mathbf{j}+\frac{5}{4}t\mathbf{k}$. Using the power - rule for differentiation, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. So, $\mathbf{r}^\prime(t)=\frac{d}{dt}(t)\mathbf{i}+\frac{d}{dt}(t^{2})\mathbf{j}+\frac{d}{dt}(\frac{5}{4}t)\mathbf{k}=\mathbf{i}+2t\mathbf{j}+\frac{5}{4}\mathbf{k}$.

Step2: Find r(t₀)

Substitute $t_0 = 2$ into $\mathbf{r}(t)$. $\mathbf{r}(t_0)=\mathbf{r}(2)=2\mathbf{i}+2^{2}\mathbf{j}+\frac{5}{4}\times2\mathbf{k}=2\mathbf{i}+4\mathbf{j}+\frac{5}{2}\mathbf{k}$.

Step3: Find r'(t₀)

Substitute $t_0 = 2$ into $\mathbf{r}^\prime(t)$. $\mathbf{r}^\prime(t_0)=\mathbf{r}^\prime(2)=\mathbf{i}+2\times2\mathbf{j}+\frac{5}{4}\mathbf{k}=\mathbf{i}+4\mathbf{j}+\frac{5}{4}\mathbf{k}$.

Answer:

$\mathbf{r}^\prime(t)=\mathbf{i}+2t\mathbf{j}+\frac{5}{4}\mathbf{k}$, $\mathbf{r}(t_0)=2\mathbf{i}+4\mathbf{j}+\frac{5}{2}\mathbf{k}$, $\mathbf{r}^\prime(t_0)=\mathbf{i}+4\mathbf{j}+\frac{5}{4}\mathbf{k}$