consider the following.\n\nr(t)=(4 - t)i+(3t - 2)j+6tk, p(3, 1, 6)\n\n(a) find the arc length function s(t)…

Explanation:

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

$\mathbf{r}'(t)=\frac{d}{dt}((4 - t)\mathbf{i}+(3t - 2)\mathbf{j}+6t\mathbf{k})=-\mathbf{i}+3\mathbf{j}+6\mathbf{k}$

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

$|\mathbf{r}'(t)|=\sqrt{(- 1)^2+3^2 + 6^2}=\sqrt{1 + 9+36}=\sqrt{46}$

Step3: Find the arc - length function $s(t)$

We know that $s(t)=\int_{t_0}^{t}|\mathbf{r}'(u)|du$. First, find $t_0$ such that $\mathbf{r}(t_0)=(3,1,6)$. Set $\begin{cases}4 - t_0=3\3t_0-2 = 1\6t_0=6\end{cases}$, solving gives $t_0 = 1$. So $s(t)=\int_{1}^{t}\sqrt{46}du=\sqrt{46}(t - 1)$

Step4: Reparametrize the curve

Solve $s=\sqrt{46}(t - 1)$ for $t$, we get $t=\frac{s}{\sqrt{46}}+1$. Substitute $t$ into $\mathbf{r}(t)$: $\mathbf{r}(t(s))=(4-(\frac{s}{\sqrt{46}}+1))\mathbf{i}+(3(\frac{s}{\sqrt{46}}+1)-2)\mathbf{j}+6(\frac{s}{\sqrt{46}}+1)\mathbf{k}=(3-\frac{s}{\sqrt{46}})\mathbf{i}+(1+\frac{3s}{\sqrt{46}})\mathbf{j}+(6+\frac{6s}{\sqrt{46}})\mathbf{k}$

Step5: Find the point 5 units along the curve

When $s = 5$, $t=\frac{5}{\sqrt{46}}+1$ $x=4-(\frac{5}{\sqrt{46}}+1)=3-\frac{5}{\sqrt{46}}$ $y=3(\frac{5}{\sqrt{46}}+1)-2=1+\frac{15}{\sqrt{46}}$ $z=6(\frac{5}{\sqrt{46}}+1)=6+\frac{30}{\sqrt{46}}$

Answer:

(a) $s(t)=\sqrt{46}(t - 1)$; $\mathbf{r}(t(s))=(3-\frac{s}{\sqrt{46}})\mathbf{i}+(1+\frac{3s}{\sqrt{46}})\mathbf{j}+(6+\frac{6s}{\sqrt{46}})\mathbf{k}$ (b) $(x,y,z)=(3-\frac{5}{\sqrt{46}},1+\frac{15}{\sqrt{46}},6+\frac{30}{\sqrt{46}})$