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This problem involves a relationship between the speed of a car and the curvature of the road, where the maximum speed is inversely proportional to the square root of the curvature.

Let's break it down:

1. The car travels on a curve defined by the equation \( y = \frac{1}{3}x^3 \).
2. You are given that at the point \( (1, \frac{1}{3}) \), the car can safely travel at 30 miles per hour.
3. You are asked to find the safe speed at the point \( \left( \frac{9}{5}, \frac{243}{125} \right) \).

### Steps:

1. **Find the curvature**: The curvature at any point on the curve \( y = \frac{1}{3}x^3 \) is related to its second derivative.
   - First, differentiate \( y \) with respect to \( x \):
     \[
     y' = \frac{d}{dx} \left( \frac{1}{3}x^3 \right) = x^2
     \]
   - Differentiate again to find the second derivative:
     \[
     y'' = \frac{d}{dx} \left( x^2 \right) = 2x
     \]
   - The curvature \( \kappa \) at a given point is related to the second derivative, and is often expressed in terms of \( x \).

2. **Relationship between speed and curvature**: Since speed is inversely proportional to the square root of the curvature, we can set up a ratio using the given data.
   - At \( x = 1 \), the second derivative (curvature) is \( 2(1) = 2 \), and the car can safely travel at 30 mph.
   - At \( x = \frac{9}{5} \), the second derivative (curvature) is \( 2\left( \frac{9}{5} \right) = \frac{18}{5} \).

3. **Set up the proportionality**:
   Since speed is inversely proportional to the square root of the curvature:
   \[
   \frac{v_1}{v_2} = \sqrt{\frac{\kappa_2}{\kappa_1}}
   \]
   Here, \( v_1 = 30 \) mph, \( \kappa_1 = 2 \), and \( \kappa_2 = \frac{18}{5} \). Substituting these values:
   \[
   \frac{30}{v_2} = \sqrt{\frac{\frac{18}{5}}{2}} = \sqrt{\frac{18}{5 \times 2}} = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}}
   \]
   Now solve for \( v_2 \):
   \[
   v_2 = \frac{30}{\frac{3}{\sqrt{5}}} = 30 \times \frac{\sqrt{5}}{3} = 10\sqrt{5} \approx 10 \times 2.236 = 22.36 \, \text{mph}
   \]

### Final Answer:
The car can safely go approximately **22.36 miles per hour** at the point \( \left( \frac{9}{5}, \frac{243}{125} \right) \).

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Would you like more details on any step? Here are five follow-up questions to deepen your understanding:

1. What is the formula for curvature, and how is it derived for parametric curves?
2. How does the proportional relationship between speed and curvature affect real-world driving conditions?
3. What would happen to the speed if the curvature were doubled at the new point?
4. Can you explain why the square root of curvature is used rather than the curvature itself?
5. How can the second derivative provide insight into the curvature of a function?

**Tip:** When solving problems involving proportionality, it's helpful to remember that constants can often simplify equations if set up correctly at the beginning!

The smaller the curvature in the bend of a road, the faster a car can travel. Assume that the maximum speed around a turn is inversely proportional to the square root of the curvature. A car moving on the path y = (1/3)x^3, where x and y are measured in miles, can safely go 30 miles per hour at (1, 1/3). How fast (in mi/hr) can it go at (9/5, 243/125)?

Learn how to solve a problem involving road curvature and maximum speed using inverse proportionality and differentiation. The car travels along the curve y = (1/3)x^3, and the solution covers finding speed at different points using the second derivative.

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