starting from the point (2, 5, 4), reparametrize the curve r(t)=⟨2 + t, 5 + 2t, 4 + 3t⟩ in terms of…

Explanation:

Step1: Find the derivative of $\mathbf{r}(t)$

$\mathbf{r}'(t)=\langle 1,2,3\rangle$

Step2: Calculate the magnitude of $\mathbf{r}'(t)$

$\left|\mathbf{r}'(t)\right|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1 + 4+9}=\sqrt{14}$

Step3: Express $t$ in terms of arc - length $s$

We know that $s=\int_{0}^{t}\left|\mathbf{r}'(u)\right|du$. Since $\left|\mathbf{r}'(u)\right|=\sqrt{14}$ for all $u$, then $s=\int_{0}^{t}\sqrt{14}du=\sqrt{14}t$. So, $t = \frac{s}{\sqrt{14}}$

Step4: Substitute $t$ into $\mathbf{r}(t)$ to get $\mathbf{r}(s)$

$\mathbf{r}(s)=\left\langle2+\frac{s}{\sqrt{14}},5 + 2\frac{s}{\sqrt{14}},4+3\frac{s}{\sqrt{14}}\right\rangle$

Answer:

$\mathbf{r}(s)=\left\langle2+\frac{s}{\sqrt{14}},5+\frac{2s}{\sqrt{14}},4+\frac{3s}{\sqrt{14}}\right\rangle$