8. - / 0.5 points 0/5 submissions used\nat what point does the curve have maximum curvature?\n y =…

Explanation:

Step1: Find first - derivative

The derivative of $y = 3\ln(x)$ is $y'=\frac{3}{x}$ using the formula $\frac{d}{dx}\ln(x)=\frac{1}{x}$.

Step2: Find second - derivative

Differentiate $y'=\frac{3}{x}=3x^{- 1}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we get $y''=- \frac{3}{x^{2}}$.

Step3: Recall the curvature formula

The curvature formula for a function $y = f(x)$ is $\kappa(x)=\frac{|y''|}{(1+(y')^{2})^{\frac{3}{2}}}$. Substitute $y'=\frac{3}{x}$ and $y''=-\frac{3}{x^{2}}$ into the formula: [ \begin{align*} \kappa(x)&=\frac{\left|-\frac{3}{x^{2}}\right|}{\left(1 + (\frac{3}{x})^{2}\right)^{\frac{3}{2}}}\ &=\frac{\frac{3}{x^{2}}}{\left(1+\frac{9}{x^{2}}\right)^{\frac{3}{2}}}\ &=\frac{3}{x^{2}(1 + \frac{9}{x^{2}})^{\frac{3}{2}}}\ &=\frac{3}{(x^{2}+9)^{\frac{3}{2}}} \end{align*} ]

Step4: Find the maximum of the curvature

To find the maximum of $\kappa(x)$, we can take the derivative of $\kappa(x)$ with respect to $x$ using the quotient - rule or by analyzing the function. Let $u = 3$ and $v=(x^{2}+9)^{\frac{3}{2}}$. The quotient - rule $\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}$, where $u' = 0$ and $v'=3x(x^{2}+9)^{\frac{1}{2}}$. [ \begin{align*} \kappa'(x)&=\frac{0\times(x^{2}+9)^{\frac{3}{2}}-3\times3x(x^{2}+9)^{\frac{1}{2}}}{(x^{2}+9)^{3}}\ &=\frac{-9x}{(x^{2}+9)^{\frac{5}{2}}} \end{align*} ] Set $\kappa'(x)=0$, then $-9x = 0$, so $x = \sqrt{6}$. When $x=\sqrt{6}$, $y = 3\ln(\sqrt{6})=\frac{3}{2}\ln(6)$.

Step5: Analyze the limit as $x\to\infty$

[ \begin{align*} \lim_{x\to\infty}\kappa(x)&=\lim_{x\to\infty}\frac{3}{(x^{2}+9)^{\frac{3}{2}}}\ &=0 \end{align*} ]

Answer:

$(x,y)=(\sqrt{6},\frac{3}{2}\ln(6))$ $\kappa(x)$ approaches $0$ as $x\to\infty$