# Find the arclength of the curve r(t) = (-9 sin t, 2t, -9 cos t), -8<= t<=8

Find the arclength of the curve $r\left(t\right)=<-9\mathrm{sin}t,2t,-9\mathrm{cos}t>,-8\le t\le 8$
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Milton Gilmore
We have
$r\left(t\right)=<-9\mathrm{sin}t,2t,-9\mathrm{cos}t>,-8\le t\le 8\phantom{\rule{0ex}{0ex}}{r}^{\prime }\left(t\right)=<-9\mathrm{cos}t,2,9\mathrm{sin}t>\phantom{\rule{0ex}{0ex}}|{r}^{\prime }\left(t\right)|=\sqrt{\left(-9\mathrm{cos}t{\right)}^{2}+{2}^{2}+\left(9\mathrm{sin}t{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{81{\mathrm{cos}}^{2}t+4+81{\mathrm{sin}}^{2}t}\phantom{\rule{0ex}{0ex}}=\sqrt{81+4}\phantom{\rule{0ex}{0ex}}|{r}^{\prime }\left(t\right)|=\sqrt{85}$
Arc length = $16\sqrt{85}$