The curvature given by the vector function r is k(t)=\frac{|r'(t)\timesr

chillywilly12a 2021-09-16 Answered
The curvature given by the vector function r is
\(\displaystyle{k}{\left({t}\right)}={\frac{{{\left|{r}'{\left({t}\right)}\times{r}{''}{\left({t}\right)}\right|}}}{{{\left|{r}'{\left({t}\right)}\right|}^{{{3}}}}}}\)
Use the formula to find the curvature of \(\displaystyle{r}{\left({t}\right)}={\left\langle\sqrt{{{15}{t}}},\ {e}^{{{t}}},\ {e}^{{-{t}}}\right\rangle}\) at the point \(\displaystyle{\left({0},\ {1},\ {1}\right)}\)

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Expert Answer

dessinemoie
Answered 2021-09-17 Author has 22040 answers

Step 1
Formula for curvature is \(\displaystyle{k}{\left({t}\right)}={\frac{{{\left|{\left|\vec{{{r}}}'{\left({t}\right)}\times\vec{{{r}}}{''}{\left({t}\right)}\right|}\right|}}}{{{\left({\left|{\left|\vec{{{r}}}'{\left({t}\right)}\right|}\right|}\right)}^{{{3}}}}}}\)
Find first and second derivatives.
First derivative is \(\displaystyle\vec{{{r}}}'{\left({t}\right)}={\left(\sqrt{{{15}}},\ {e}^{{{t}}},\ -{\frac{{{1}}}{{{e}^{{{t}}}}}}\right)}\)
Second derivative is \(\displaystyle\vec{{{r}}}{''}{\left({t}\right)}={\left({0},\ {e}^{{{t}}},\ {e}^{{-{t}}}\right)}\)
Now, find norm (length) of \(\vec{r}'(t) : ||\vec{r}'(t)||=\sqrt{(\sqrt{15})^{2}+(e^{t})^{2}+\left(-\frac{1}{e^{t}}\right)^{2}}=\sqrt{2\cos h(2t)+15}\)
Thus, \(\displaystyle{\left({\left|{\left|\vec{{{r}}}'{\left({t}\right)}\right|}\right|}\right)}^{{{3}}}={\left({2}{\cos{{h}}}{\left({2}{t}\right)}+{15}\right)}^{{{\frac{{{3}}}{{{2}}}}}}\)
Next, find cross product of first and second derivatives: \(\displaystyle/\vec{{{r}}}'{\left({t}\right)}\times\vec{{{r}}}{''}{\left({t}\right)}={\left({2},\ -{\frac{{\sqrt{{{15}}}}}{{{e}^{{{t}}}}}},\ \sqrt{{{15}}}{e}^{{{t}}}\right.}\)
Now, find norm (lenght) of \(\displaystyle\vec{{{r}}}'{\left({t}\right)}\times\vec{{{r}}}{''}{\left({t}\right)}:\)
\(\displaystyle{\left|{\left|\vec{{{r}}}'{\left({t}\right)}\times\vec{{{r}}}{''}{\left({t}\right)}\right|}\right|}=\sqrt{{{\left({2}\right)}^{{{2}}}+{\left(-{\frac{{\sqrt{{{15}}}}}{{{e}^{{{t}}}}}}\right)}^{{{2}}}+{\left(\sqrt{{{15}}}{e}^{{{t}}}\right)}^{{{2}}}}}=\sqrt{{{30}{\cos{{h}}}{\left({2}{t}\right)}+{4}}}\)
Finally, curvature is \(\displaystyle{k}{\left({t}\right)}={\frac{{\sqrt{{{30}{\cos{{h}}}{\left({2}{t}\right)}+{4}}}}}{{{\left(\sqrt{{{2}{\cos{{h}}}{\left({2}{t}\right)}+{15}}}\right)}^{{{3}}}}}}=\sqrt{{{2}}}\sqrt{{{\frac{{{15}{\cos{{h}}}{\left({2}{t}\right)}+{2}}}{{{\left({2}{\cos{{h}}}{\left({2}{t}\right)}+{15}\right)}^{{{3}}}}}}}}\)
Now, find curvature at specific point
\(\displaystyle{k}{\left({0}\right)}={\frac{{\sqrt{{{2}}}}}{{{17}}}}\)
\(\displaystyle{k}{\left({t}\right)}=\sqrt{{{2}}}\sqrt{{{\frac{{{15}{\cos{{h}}}{\left({2}{t}\right)}+{2}}}{{{\left({2}{\cos{{h}}}{\left({2}{t}\right)}+{15}\right)}^{{{3}}}}}}}},\ {k}{\left({0}\right)}={\frac{{\sqrt{{{2}}}}}{{{17}}}}\approx{0.0831890330807703}\)

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