Find the length of the curve.
\vec{r}(t) =< 8t, t^2,

Danelle Albright

Answered question

2021-12-21

Find the length of the curve.
$\overrightarrow{r}\left(t\right)=<8t,{t}^{2},\frac{1}{12}{t}^{3}>,0\le t\le 1$

Answer & Explanation

esfloravaou

Beginner2021-12-22Added 43 answers

${\overrightarrow{r}}^{\prime}\left(t\right)=<8,2t,\frac{3}{12}{t}^{2}>$ ${\overrightarrow{r}}^{\prime}\left(t\right)=<8,2t,\frac{1}{4}{t}^{2}>$ $\left|\left|{\overrightarrow{r}}^{\prime}\left(t\right)\right|\right|=\sqrt{{\left(8\right)}^{2}+{\left(2t\right)}^{2}+{\left(\frac{1}{4}{t}^{2}\right)}^{2}}$ $\Rightarrow \left|\left|{\overrightarrow{r}}^{\prime}\left(t\right)\right|\right|=\sqrt{64+4{t}^{2}+\frac{1}{16}{t}^{4}}$ $\Rightarrow \left|\left|{\overrightarrow{r}}^{\prime}\left(t\right)\right|\right|=\sqrt{\frac{1}{16}((64\cdot 16)+64{t}^{2}+{t}^{4})}$ $\Rightarrow \left|\left|{\overrightarrow{r}}^{\prime}\left(t\right)\right|\right|=\frac{1}{4}\sqrt{1024+64{t}^{2}+{t}^{4}}$ $\Rightarrow \left|\left|{\overrightarrow{r}}^{\prime}\left(t\right)\right|\right|=\frac{1}{4}\sqrt{{32}^{2}+(2\cdot 32){t}^{2}+{t}^{4}}$ $\Rightarrow \left|\left|{\overrightarrow{r}}^{\prime}\left(t\right)\right|\right|=\frac{1}{4}\sqrt{{(32+{t}^{2})}^{2}}$ $\Rightarrow \left|\left|{\overrightarrow{r}}^{\prime}\left(t\right)\right|\right|=\frac{1}{4}|32+{t}^{2}|$ Now, $t}^{2$ is always positive. So, $(32+{t}^{2})$ is always positive.

Cheryl King

Beginner2021-12-23Added 36 answers

Where is the second part of solution?

RizerMix

Expert2021-12-29Added 656 answers

we know that, the length of the curve between t = 0 and t = 1 can be computed as $\begin{array}{}L={\int}_{t=0}^{t=1}||\overrightarrow{r}(t)||dt\\ \Rightarrow L={\int}_{t=0}^{t=1}\frac{1}{4}(32+{t}^{2})dt\\ \Rightarrow L=\frac{1}{4}{\int}_{t=0}^{t=1}(32+{t}^{2})dt\\ \Rightarrow L=\frac{1}{4}[32+\frac{{t}^{3}}{3}{]}_{t=0}^{t=1}\\ \Rightarrow L=\frac{1}{4}[(32+\frac{1}{3})-(0+0)]\\ \Rightarrow L=\frac{1}{4}(32+\frac{1}{3})\\ \Rightarrow L=\frac{1}{4}\ast \frac{97}{3}\\ \Rightarrow L=\frac{97}{12}\end{array}$