To find the point where the curve $y = 9 \ln(x)$ has maximum curvature, we will follow these steps:

1. Find the first and second derivatives of the function.
2. Compute the curvature formula in terms of $x$.
3. Maximize the curvature by finding the critical points and testing for maximum.

Step 1: First and second derivatives

The function is:

$y = 9 \ln(x)$

First derivative:

$y' = \frac{9}{x}$

Second derivative:

$y'' = -\frac{9}{x^2}$

Step 2: Curvature formula

The curvature $\kappa$ of a curve given by $y = f(x)$ is calculated using the following formula:

$\kappa(x) = \frac{|y''|}{(1 + (y')^2)^{3/2}}$

Substituting $y' = \frac{9}{x}$ and $y'' = -\frac{9}{x^2}$ into the curvature formula:

$\kappa(x) = \frac{|\frac{9}{x^2}|}{\left(1 + \left(\frac{9}{x}\right)^2\right)^{3/2}} = \frac{\frac{9}{x^2}}{\left(1 + \frac{81}{x^2}\right)^{3/2}}$

Step 3: Maximize the curvature

To find the maximum curvature, we need to differentiate the curvature $\kappa(x)$ with respect to $x$ and set it to zero:

$\frac{d\kappa}{dx} = 0$

I'll compute this derivative next.The derivative of the curvature function is:

$\frac{d\kappa}{dx} = 2187 \cdot (x^2 + 81)^{-2.5} - 18 \cdot (x^2 + 81)^{-1.5}$

To find the critical points where maximum curvature occurs, we need to set this derivative equal to zero:

$2187 \cdot (x^2 + 81)^{-2.5} - 18 \cdot (x^2 + 81)^{-1.5} = 0$

I'll solve this equation for $x$.The critical point is approximately $x = 6.36$.

This is the value of $x$ where the curvature is maximized. Therefore, the point on the curve $y = 9 \ln(x)$ with the maximum curvature occurs at $x \approx 6.36$.

Would you like more details on this, or do you have any further questions?

Here are 5 related questions you might find interesting:

1. How can you find the curvature of any function in general?
2. What is the significance of the point of maximum curvature on a curve?
3. How would you interpret the behavior of the curve near its maximum curvature?
4. What does the second derivative test tell us about points of maximum curvature?
5. How do curvature calculations change for parametric equations?

Tip: Curvature gives an indication of how sharply a curve bends. Higher curvature means the curve is bending more at that point.