Danelle Albright

2021-12-21

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

esfloravaou

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

Cheryl King

Where is the second part of solution?

RizerMix

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

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