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

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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Otigh1979
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
$$\displaystyle{\left|{r}'\times{r}{''}\right|}=\sqrt{{{6}^{{2}}+{54}^{{2}}+{18}^{{2}}}}$$
$$\displaystyle={57.24}$$
$$\displaystyle{\left|{r}'\right|}={9.7}$$
$$\displaystyle{\frac{{{\left|{r}'\times{r}{''}\right|}}}{{{\left|{r}'\right|}^{{3}}}}}={\frac{{{57.24}}}{{{9.7}^{{3}}}}}$$
$$\displaystyle={0.063}$$