To calculate the length of a parabola arch, we first need to know its equation. The equation of a parabola with its vertex at the origin and opening upwards/downwards is typically in the form:

$y = ax^2$

Given that the parabola is 53 yards wide and 20 yards high, we know:

• The width of the arch is 53 yards, so the parabola spans from $x = -26.5$ to $x = 26.5$ (since it is symmetric).
• The height of the arch is 20 yards, which means the maximum value of $y$ at $x = 0$ is 20.

Step 1: Find the equation of the parabola

Using the point $(26.5, 20)$, we can substitute into $y = ax^2$ to find $a$.

$20 = a(26.5)^2$ $a = \frac{20}{(26.5)^2}$

Let's compute the value of $a$.

$a \approx \frac{20}{702.25} \approx 0.0285$

Thus, the equation of the parabola is approximately:

$y = 0.0285x^2$

Step 2: Calculate the length of the parabola

The length of the parabola from $x = -26.5$ to $x = 26.5$ is given by the arc length formula:

$L = \int_{-26.5}^{26.5} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx$

The derivative $\frac{dy}{dx}$ is:

$\frac{dy}{dx} = 2ax = 2(0.0285)x = 0.057x$

Thus, the arc length formula becomes:

$L = \int_{-26.5}^{26.5} \sqrt{1 + (0.057x)^2} \, dx$

Now, let me compute this integral to find the length of the parabola.

The length of the parabola arch is approximately 69.07 yards.

Would you like more details or have any other questions?

Here are 5 related questions that could further expand this topic:

1. How would the length change if the height of the arch increased to 30 yards?
2. What if the parabola were only half of this arch, how would you calculate the length?
3. How do different types of curves (e.g., catenary curves) compare in terms of length for the same width and height?
4. What happens to the arc length if the parabola opens downward instead of upward?
5. Can the arc length formula be used for other shapes besides parabolas?

Tip: When solving for arc length, breaking the problem into smaller parts (like finding the derivative first) helps simplify the process and avoid mistakes.