9. express the arclength of the curve over the given interval as a definite integral. you do not need to…

Explanation:

Step1: Recall arc - length formula for a vector - valued function

The arc - length $L$ of a vector - valued function $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$ over the interval $[a,b]$ is given by $L = \int_{a}^{b}\vert\vec{r}'(t)\vert dt$, where $\vec{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle$ and $\vert\vec{r}'(t)\vert=\sqrt{(x'(t))^{2}+(y'(t))^{2}+(z'(t))^{2}}$.

Step2: Find $\vec{r}'(t)$ for part (a)

If $\vec{r}(t)=\langle 1,t,t^{2}\rangle$, then $\vec{r}'(t)=\langle 0,1,2t\rangle$. And $\vert\vec{r}'(t)\vert=\sqrt{0^{2}+1^{2}+(2t)^{2}}=\sqrt{1 + 4t^{2}}$. The arc - length over the interval $[1,4]$ is $L=\int_{1}^{4}\sqrt{1 + 4t^{2}}dt$.

Step3: Find $\vec{r}'(t)$ for part (b)

If $\vec{r}(t)=\langle e^{t},1,e^{-t}\rangle$, then $\vec{r}'(t)=\langle e^{t},0,-e^{-t}\rangle$. And $\vert\vec{r}'(t)\vert=\sqrt{(e^{t})^{2}+0^{2}+(-e^{-t})^{2}}=\sqrt{e^{2t}+e^{-2t}}$. The arc - length over the interval $[0,5]$ is $L=\int_{0}^{5}\sqrt{e^{2t}+e^{-2t}}dt$.

Step4: Find $\vec{r}'(t)$ for part (c)

If $\vec{r}(t)=\langle\frac{1}{t},t,2\rangle$, then $\vec{r}'(t)=\langle-\frac{1}{t^{2}},1,0\rangle$. And $\vert\vec{r}'(t)\vert=\sqrt{(-\frac{1}{t^{2}})^{2}+1^{2}+0^{2}}=\sqrt{\frac{1}{t^{4}} + 1}$. The arc - length over the interval $[2,8]$ is $L=\int_{2}^{8}\sqrt{\frac{1}{t^{4}}+1}dt$.

Answer:

(a) $\int_{1}^{4}\sqrt{1 + 4t^{2}}dt$ (b) $\int_{0}^{5}\sqrt{e^{2t}+e^{-2t}}dt$ (c) $\int_{2}^{8}\sqrt{\frac{1}{t^{4}}+1}dt$