Find the curvature of r(t) = <9t, t^2, t^3 >

sputavanomr 2021-11-13 Answered
Find the curvature of \(\displaystyle{r}{\left({t}\right)}={<}{9}{t},{t}^{{2}},{t}^{{3}}{>}\) at the point (9, 1, 1).

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Expert Answer

Otigh1979
Answered 2021-11-14 Author has 345 answers
Given:
\(\displaystyle{r}{\left({t}\right)}={<}{9}{t},{t}^{{2}},{t}^{{3}}{>}\) at the point (9,1,1)
The objective is to find the curvature
The curvature us \(\displaystyle{\frac{{{\left|{r}'\times{t}{'''}\right|}}}{{{\left|{r}'\right|}^{{3}}}}}\)
It is given that,
\(\displaystyle{r}{\left({t}\right)}={<}{9}{t},{t}^{{2}},{t}^{{3}}{>}\)
On differentiating,
\(\displaystyle{r}'{\left({t}\right)}={<}{9},{2}{t},{3}{t}^{{2}}{>}\)
On further differentiating,
\(\displaystyle{r}{''}{\left({t}\right)}={<}{0},{2},{6}{t}{>}\)
\(\displaystyle{r}'\times{r}{''}={<}{6}{t}^{{2}},-{54}{t},{18}{>}\)
At t=1
\(\displaystyle{\left|{r}'\times{r}{''}\right|}=\sqrt{{{6}^{{2}}+{54}^{{2}}+{18}^{{2}}}}\)
\(\displaystyle={57.24}\)
\(\displaystyle{\left|{r}'\right|}={9.7}\)
\(\displaystyle{\frac{{{\left|{r}'\times{r}{''}\right|}}}{{{\left|{r}'\right|}^{{3}}}}}={\frac{{{57.24}}}{{{9.7}^{{3}}}}}\)
\(\displaystyle={0.063}\)
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