Erick Wyatt

Answered

2022-11-26

At what point does the curve have maximum curvature?

$y=5\mathrm{ln}(x)$

$(x,y)=(,)$

κ(x) approaches ________ as $x\to \mathrm{\infty}$?

$y=5\mathrm{ln}(x)$

$(x,y)=(,)$

κ(x) approaches ________ as $x\to \mathrm{\infty}$?

Answer & Explanation

Desirae Wu

Expert

2022-11-27Added 10 answers

Please see the solution and If you have any doubt Please comment

Thanks

$\u043d=5\mathrm{ln}x$

curvature = $\frac{|{f}^{\u2033}(x)|}{[1+{f}^{\prime}(x{)}^{2}{]}^{\frac{3}{2}}}$

$y=f(x)\phantom{\rule{0ex}{0ex}}{f}^{\prime}(x)=\frac{5}{x}\phantom{\rule{0ex}{0ex}}{f}^{\u2033}(x)=-\frac{5}{{x}^{2}}\phantom{\rule{0ex}{0ex}}k(x)=\frac{|-\frac{5}{{x}^{2}}|}{[1+\frac{25}{{x}^{2}}{]}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=\frac{5x}{({x}^{2}+25{)}^{\frac{3}{2}}}$

for max curvature

${k}^{\prime}(x)=0$

${k}^{\prime}(x)=\frac{d}{dx}\frac{5x}{({x}^{2}+25{)}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=\frac{5(25-2{x}^{2})}{(25+{x}^{2}{)}^{\frac{5}{2}}}=0$

${x}^{2}=\frac{25}{2}\phantom{\rule{0ex}{0ex}}x=\pm \sqrt{\frac{25}{2}}=\pm \frac{5}{\sqrt{2}}$

$x\ne $ negative so

point where maximum curveture

$(\frac{5}{\sqrt{2}},5\mathrm{ln}(\frac{5}{\sqrt{2}}))=(3.5355,6,3143)$

$\underset{x\to \mathrm{\infty}}{lim}k(x)=\underset{x\to \mathrm{\infty}}{lim}\frac{5x}{({x}^{2}+25{)}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=0$

Thanks

$\u043d=5\mathrm{ln}x$

curvature = $\frac{|{f}^{\u2033}(x)|}{[1+{f}^{\prime}(x{)}^{2}{]}^{\frac{3}{2}}}$

$y=f(x)\phantom{\rule{0ex}{0ex}}{f}^{\prime}(x)=\frac{5}{x}\phantom{\rule{0ex}{0ex}}{f}^{\u2033}(x)=-\frac{5}{{x}^{2}}\phantom{\rule{0ex}{0ex}}k(x)=\frac{|-\frac{5}{{x}^{2}}|}{[1+\frac{25}{{x}^{2}}{]}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=\frac{5x}{({x}^{2}+25{)}^{\frac{3}{2}}}$

for max curvature

${k}^{\prime}(x)=0$

${k}^{\prime}(x)=\frac{d}{dx}\frac{5x}{({x}^{2}+25{)}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=\frac{5(25-2{x}^{2})}{(25+{x}^{2}{)}^{\frac{5}{2}}}=0$

${x}^{2}=\frac{25}{2}\phantom{\rule{0ex}{0ex}}x=\pm \sqrt{\frac{25}{2}}=\pm \frac{5}{\sqrt{2}}$

$x\ne $ negative so

point where maximum curveture

$(\frac{5}{\sqrt{2}},5\mathrm{ln}(\frac{5}{\sqrt{2}}))=(3.5355,6,3143)$

$\underset{x\to \mathrm{\infty}}{lim}k(x)=\underset{x\to \mathrm{\infty}}{lim}\frac{5x}{({x}^{2}+25{)}^{\frac{3}{2}}}\phantom{\rule{0ex}{0ex}}=0$

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