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We are given a non-negative function \( f \) defined on the interval \([0, 1]\) with the condition:

\[
\int_0^x \sqrt{1 - (f'(t))^2} \, dt = \int_0^x f(t) \, dt, \quad 0 \leq x \leq 1
\]
and \( f(0) = 0 \).

The goal is to analyze the function \( f \) and identify which condition among (a), (b), (c), or (d) is true about the values of \( f\left(\frac{1}{2}\right) \) and \( f\left(\frac{1}{3}\right) \).

### Key insight:
The given equation suggests that the integral of \( \sqrt{1 - (f'(t))^2} \) is equal to the integral of \( f(t) \). The term \( \sqrt{1 - (f'(t))^2} \) resembles a form from the arc-length formula, indicating that \( f \) may be related to the parametric equation of a curve.

From this, we infer that \( f \) could be the sine function, since the integral constraints often match a trigonometric function. In particular, for \( f'(t) = \cos(\theta) \), the arc-length formula applies, giving us clues about the shape of \( f \).

### Step-by-step analysis:
- The integral equation can be interpreted geometrically, hinting that the function \( f(x) \) may resemble part of a sine curve.
- Since \( f(0) = 0 \), it seems plausible that \( f(x) \) could behave like \( \sin(\theta x) \) on \([0, 1]\).

For the sine function \( f(x) = \sin(\pi x) \):
- \( f\left(\frac{1}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( f\left(\frac{1}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866 \)

Thus, \( f\left(\frac{1}{2}\right) > \frac{1}{2} \) and \( f\left(\frac{1}{3}\right) > \frac{1}{3} \).

### Conclusion:
The correct option is **(b)**:
\[
f\left(\frac{1}{2}\right) > \frac{1}{2} \quad \text{and} \quad f\left(\frac{1}{3}\right) > \frac{1}{3}
\]

Would you like any further details or clarification on this?

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Here are 5 related questions to deepen your understanding:
1. How is the arc-length formula derived, and how does it apply to this problem?
2. Why does \( f(x) = \sin(\pi x) \) seem like a natural solution here?
3. How can we prove that the given equation implies that \( f \) is related to a trigonometric function?
4. Can this method be generalized to other types of functions or intervals?
5. How would the result change if the interval were different, such as \([0, 2]\)?

**Tip:** When dealing with integrals and trigonometric functions, checking boundary conditions (like \( f(0) = 0 \)) can often guide you to the right functional form.

Let f be a non-negative function defined on the interval [0, 1]. If \( \int_0^x \sqrt{1 - (f'(t))^2} dt = \int_0^x f(t) dt, 0 \leq x \leq 1 \) and f(0) = 0, then...

This problem involves analyzing a non-negative function defined by an integral equation and relating it to trigonometric functions. The arc-length formula and integral calculus are key to solving this problem.

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