The problem is asking to find the exact height of the curve given by:

$y = \ln\left(1 - x^2\right)$

over the interval $0 \leq x \leq \frac{\sqrt{5}}{5}$.

Steps to Find the Maximum Height:

1. Understand the Function:

   • The function $y = \ln(1 - x^2)$ is defined as long as $1 - x^2 > 0$, which means $|x| < 1$. This condition is satisfied within the given interval.

2. Differentiate to Find Critical Points:

   • To find the maximum or minimum values of the function, take the derivative with respect to $x$ and set it to zero. $\frac{dy}{dx} = \frac{d}{dx} \ln(1 - x^2) = \frac{1}{1 - x^2} \cdot (-2x) = \frac{-2x}{1 - x^2}$
   • Set $\frac{dy}{dx} = 0$ to find critical points: $\frac{-2x}{1 - x^2} = 0$ This equation is satisfied when $x = 0$.

3. Evaluate the Function at the Critical Points and Endpoints:

   • At $x = 0$: $y = \ln(1 - 0^2) = \ln(1) = 0$
   • At $x = \frac{\sqrt{5}}{5}$: $y = \ln\left(1 - \left(\frac{\sqrt{5}}{5}\right)^2\right) = \ln\left(1 - \frac{5}{25}\right) = \ln\left(\frac{20}{25}\right) = \ln\left(\frac{4}{5}\right)$ Since $\ln\left(\frac{4}{5}\right)$ is negative, the maximum value occurs at $x = 0$.

Conclusion:

The exact height of the curve is 0.

Would you like more details or have any questions? Here are 5 related questions for further exploration:

1. What is the value of $y$ at $x = \frac{\sqrt{5}}{5}$, and what does this tell you about the curve's behavior?
2. How would the function behave if the interval were extended to include negative values of $x$?
3. What would happen to the maximum height if the domain of $x$ were different, say $0 \leq x \leq 1$?
4. Can you derive a general method to find the maximum value of $y = \ln(1 - kx^2)$ for any constant $k$?
5. How does the concavity of the function $y = \ln(1 - x^2)$ help in identifying maximum points?

Tip: Always evaluate the function at the endpoints of the given interval as well as any critical points to find the maximum or minimum values.