Erick Wyatt

2022-11-26

At what point does the curve have maximum curvature?
$y=5\mathrm{ln}\left(x\right)$
$\left(x,y\right)=\left(,\right)$
κ(x) approaches ________ as $x\to \mathrm{\infty }$?

Desirae Wu

Expert

Please see the solution and If you have any doubt Please comment
Thanks
$н=5\mathrm{ln}x$
curvature = $\frac{|{f}^{″}\left(x\right)|}{\left[1+{f}^{\prime }\left(x{\right)}^{2}{\right]}^{\frac{3}{2}}}$
$y=f\left(x\right)\phantom{\rule{0ex}{0ex}}{f}^{\prime }\left(x\right)=\frac{5}{x}\phantom{\rule{0ex}{0ex}}{f}^{″}\left(x\right)=-\frac{5}{{x}^{2}}\phantom{\rule{0ex}{0ex}}k\left(x\right)=\frac{|-\frac{5}{{x}^{2}}|}{\left[1+\frac{25}{{x}^{2}}{\right]}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=\frac{5x}{\left({x}^{2}+25{\right)}^{\frac{3}{2}}}$
for max curvature
${k}^{\prime }\left(x\right)=0$
${k}^{\prime }\left(x\right)=\frac{d}{dx}\frac{5x}{\left({x}^{2}+25{\right)}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=\frac{5\left(25-2{x}^{2}\right)}{\left(25+{x}^{2}{\right)}^{\frac{5}{2}}}=0$
${x}^{2}=\frac{25}{2}\phantom{\rule{0ex}{0ex}}x=±\sqrt{\frac{25}{2}}=±\frac{5}{\sqrt{2}}$
$x\ne$ negative so
point where maximum curveture
$\left(\frac{5}{\sqrt{2}},5\mathrm{ln}\left(\frac{5}{\sqrt{2}}\right)\right)=\left(3.5355,6,3143\right)$
$\underset{x\to \mathrm{\infty }}{lim}k\left(x\right)=\underset{x\to \mathrm{\infty }}{lim}\frac{5x}{\left({x}^{2}+25{\right)}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=0$

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