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The problem states that the maximum speed of a car on a curved road is inversely proportional to the square root of the curvature. The curvature is given by the equation \( y = \frac{1}{3}x^3 \), and the speed for a certain point on the curve \((1, \frac{1}{3})\) is given as 30 miles per hour. The task is to find the speed at another point \(\left(\frac{9}{5}, \frac{243}{125}\right)\).

### Step-by-step solution:

1. **Curvature and Proportionality:**
   The speed \( v \) is inversely proportional to the square root of the curvature \( \kappa \):
   \[
   v \propto \frac{1}{\sqrt{\kappa}}
   \]
   This can be written as:
   \[
   v = \frac{k}{\sqrt{\kappa}}
   \]
   where \( k \) is some constant of proportionality.

2. **Curvature of the curve \( y = \frac{1}{3}x^3 \):**
   The curvature \( \kappa \) of a function \( y = f(x) \) is given by the formula:
   \[
   \kappa = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}
   \]
   
   First, calculate the first and second derivatives of \( y = \frac{1}{3}x^3 \):
   \[
   f'(x) = x^2
   \]
   \[
   f''(x) = 2x
   \]

3. **Curvature at \( x = 1 \):**
   \[
   \kappa_1 = \frac{|2(1)|}{(1 + (1)^4)^{3/2}} = \frac{2}{(1+1)^{3/2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}
   \]
   
   Since the car's speed is given as 30 mi/h at \( x = 1 \), we can find \( k \):
   \[
   30 = \frac{k}{\sqrt{\frac{1}{\sqrt{2}}}} \Rightarrow k = 30\sqrt{\frac{1}{\sqrt{2}}}
   \]

4. **Find curvature at \( x = \frac{9}{5} \):**
   
   Substituting \( x = \frac{9}{5} \) into the curvature formula:
   \[
   f'(x) = \left(\frac{9}{5}\right)^2 = \frac{81}{25}
   \]
   and 
   \[
   f''(x) = 2 \times \frac{9}{5} = \frac{18}{5}
   \]
   Therefore:
   \[
   \kappa_2 = \frac{\frac{18}{5}}{\left(1 + \left(\frac{81}{25}\right)\right)^{3/2}} = \frac{\frac{18}{5}}{\left(1 + \frac{81}{25}\right)^{3/2}}
   \]   
   Simplify:
   \[
   \kappa_2 = \frac{\frac{18}{5}}{\left(\frac{106}{25}\right)^{3/2}} = \frac{\frac{18}{5}}{\left(\frac{106^{3/2}}{25^{3/2}}\right)}

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 find the safe speed of a car on a curved road using inverse proportionality to the curvature. The problem involves calculus concepts like curvature calculation and derivatives, with a focus on how speed changes with road curvature. Given that a car moves at 30 mph at point (1, 1/3), find the speed at point (9/5, 243/125). Suitable for students in Grades 11-12.

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