# Find the length of the curve. r(t)=<t, 3\cos t, 3\sin t>, -5<=t<=5

Find the length of the curve. $$\displaystyle{r}{\left({t}\right)}={<}{t},{3}{\cos{{t}}},{3}{\sin{{t}}}{>},-{5}\le{t}\le{5}$$

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Step 1
We know that the arc length L of a curve r(t) from -a to a is,
$$\displaystyle{L}={\int_{{-{a}}}^{{a}}}{\left|{r}'{\left({t}\right)}\right|}\cdot{\left.{d}{t}\right.}$$
Therefore, let us find the derivative of given vector function component-wise, we get,
$$\displaystyle{r}{\left({t}\right)}={<}{t},{3}{\cos{{t}}},{3}{\sin{{t}}}{>},-{5}\le{t}\le{5}$$
$$\displaystyle{r}'{\left({t}\right)}={<}{1},-{3}{\sin{{t}}},{3}{\cos{{t}}}{>}$$
Step 2
Plug into the arc length formula
$$\displaystyle{L}=\int^{{5}}_{\left\lbrace-{5}\right\rbrace}{\left|{r}'{\left({t}\right)}\right|}{\left.{d}{t}\right.}$$
$$\displaystyle=\int^{{5}}_{\left\lbrace-{5}\right\rbrace}\sqrt{{{1}^{{2}}+{\left(-{3}{\sin{{t}}}\right)}^{{2}}+{\left({3}{\cos{{t}}}\right)}^{{2}}}}{\left.{d}{t}\right.}$$
$$\displaystyle=\int^{{5}}_{\left\lbrace-{5}\right\rbrace}\sqrt{{{1}+{9}{{\sin}^{{2}}{t}}+{9}{{\cos}^{{2}}{t}}}}{\left.{d}{t}\right.}$$
$$\displaystyle=\int^{{5}}_{\left\lbrace-{5}\right\rbrace}\sqrt{{{1}+{9}{\left({{\sin}^{{2}}{t}}+{{\cos}^{{2}}{t}}\right)}}}{\left.{d}{t}\right.}$$
$$\displaystyle=\int^{{5}}_{\left\lbrace-{5}\right\rbrace}\sqrt{{{1}+{9}{\left({1}\right)}}}{\left.{d}{t}\right.}$$
$$\displaystyle=\int^{{5}}_{\left\lbrace-{5}\right\rbrace}\sqrt{{{10}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\left[\sqrt{{{10}{t}}}\right]}^{{5}}_{\left\lbrace-{5}\right\rbrace}$$
$$\displaystyle={5}\sqrt{{{10}}}-{\left(-{5}\sqrt{{{10}}}\right)}$$
$$\displaystyle={10}\sqrt{{{10}}}$$
The length of curve $$\displaystyle{r}{\left({t}\right)}={\left\langle{t},{3}{\cos{{t}}},{\sin{{t}}}\right\rangle}$$ from -5 to 5 is $$\displaystyle{10}\sqrt{{{10}}}$$