This question aims to find the **height** of a **parabolic bridge** 10 feet, 30 feet, and 50 feet from the **center.** The bridge is 30 feet **high** and has a **span** of 130 feet.

The concept needed for this question to understand and solve include **basic algebra** and **familiarity** with **arches** and **parabolas.** The equation of the **parabolic arch’s height** at a given distance from the endpoint is given as:

\[ y = \dfrac{4 h}{ l^2 } x ( l – x) \]

Where:

\[ h\ =\ Maximum\ Rise\ of\ the\ Arch \]

\[ l\ =\ Span\ of\ the\ Arch \]

\[ y\ =\ Height\ of\ the\ Arch\ at\ any\ given\ distance\ (x)\ from\ End\ Point \]

## Expert Answer

To find the **height** of the **arch** at any given **position,** we can use the formula explained above. The given information about this problem is:

\[ h\ =\ 30\ feet \]

\[ l\ =\ 130\ feet \]

**a)** The first part is to find the **bridge’s height,** $10 feet$ from the **center.** As the bridge is constructed as a **parabolic arch,** the **height** on both sides of the **center** at an equal distance will be the **same.** The formula for the **height** of the **bridge** at any given distance from the **endpoint** is given:

\[ y\ =\ \dfrac{ 4h }{ l^2 } x (l -\ x) \]

Here, we have the **distance** from the **center.** To calculate the **distance** from the **endpoint,** we **subtract** it from half of the span of the **bridge.** So, for $10 feet$, $x$ will be:

\[ x\ =\ \dfrac{130}{2}\ -\ 10 \]

\[x \ =\ 55 feet \]

Substituting the values, we get:

\[ y\ =\ \dfrac{ 4 \times 30 }{ ( 130)^2 } (55) (130 -\ 55) \]

Solving this equation, we get:

\[ y\ =\ 29.3\ feet \]

**b)** The **height** of the **bridge** $30 feet$ from the **center** is given as:

\[ x\ =\ \dfrac{130}{2}\ -\ 30 \]

\[x \ =\ 35 feet \]

\[ y\ =\ \dfrac{ 4 \times 30 }{ ( 130)^2 } (35) (130 -\ 35) \]

Solving this equation, we get:

\[ y\ =\ 23.6\ feet \]

**c)** The **height** of the **bridge** $50 feet$ from the **center** is given as:

\[ x\ =\ \dfrac{130}{2}\ -\ 50 \]

\[x \ =\ 5 feet \]

\[ y\ =\ \dfrac{ 4 \times 30 }{ ( 130)^2 } (5) (130 -\ 5) \]

Solving this equation, we get:

\[ y\ =\ 4.44\ feet \]

## Numerical Result

The **height** of the **parabolic arch bridge** $10 feet$, $30 feet$ and $50 feet$ from the **center** is calulated to be:

\[ y_{10}\ =\ 29.3\ feet \]

\[ y_{30}\ =\ 23.6\ feet \]

\[ y_{50}\ =\ 4.44\ feet \]

These **heights** will be same on **either side** of the **bridge** as the bridge is an **arch shaped.**

## Example

Find the **height** of a **parabolic arch bridge** with a $20 feet$ height and $100 feet$ span at $20 feet$ from the **center.**

We have:

\[ h = 20\ feet \]

\[ l = 100\ feet \]

\[ x = \dfrac{l}{2}\ -\ 20 \]

\[ x = 30\ feet \]

Substituting the values in the given formula, we get:

\[ y = \dfrac{ 4 \times 20 }{ (100)^2 } (30) (100\ -\ 30) \]

Solving the equation, we get:

\[ y = 16.8\ feet \]