Find the length of the curve. r(t)=\langle6t,\ t^{2},\ \frac{1}{9}t^{3}\rangle,\ 0\leq t\leq1

amanf 2021-08-20 Answered
Find the length of the curve.
\(\displaystyle{r}{\left({t}\right)}={\left\langle{6}{t},\ {t}^{{{2}}},\ {\frac{{{1}}}{{{9}}}}{t}^{{{3}}}\right\rangle},\ {0}\leq{t}\leq{1}\)

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Expert Answer

au4gsf
Answered 2021-08-21 Author has 19296 answers
Step 1
\(\displaystyle{r}{\left({t}\right)}={\left\langle{6}{t},\ {t}^{{{2}}},\ {\frac{{{t}^{{{3}}}}}{{{9}}}}\right\rangle}\Rightarrow{r}'{\left({t}\right)}={\left\langle{6},\ {2}{t},\ {\frac{{{t}^{{{2}}}}}{{{3}}}}\right\rangle}\)
\(\displaystyle{\left|{r}'{\left({t}\right)}\right|}=\sqrt{{{36}+{4}{t}^{{{2}}}+{\frac{{{t}^{{{4}}}}}{{{9}}}}}}\)
\(\displaystyle={\frac{{{1}}}{{{3}}}}\sqrt{{{t}^{{{4}}}+{36}{t}^{{{2}}}+{324}}}\)
\(\displaystyle={\frac{{{1}}}{{{3}}}}\sqrt{{{\left({t}^{{{2}}}+{18}\right)}^{{{2}}}}}+{\frac{{{t}^{{{2}}}+{18}}}{{{3}}}}\)
Length of the curve: \(\displaystyle{L}={\int_{{{0}}}^{{{1}}}}{\left|\overline{{{r}}}'{\left({t}\right)}\right|}{\left.{d}{t}\right.}={\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{3}}}}{\left({t}^{{{2}}}+{18}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{{0}}}^{{{1}}}}{\left({\frac{{{t}^{{{2}}}}}{{{3}}}}+{b}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle={{\left[{\frac{{{t}^{{{3}}}}}{{{9}}}}+{b}^{{{t}}}\right]}_{{{0}}}^{{{1}}}}\)
\(\displaystyle={\frac{{{1}}}{{{9}}}}+{b}-{0}={\frac{{{1}}}{{{9}}}}+{6}={\frac{{{55}}}{{{9}}}}\)
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