Chapter 3.4, Problem 14AYU

Architecture A parabolic arch has a span of 120 feet and a maximum height of 25 feet. Choose suitable rectangular coordinate axes and find the equation of the parabola. Then calculate the height of the arch at points 10 feet, 20 feet, and 40 feet from the center.

To determine

To calculate: The height of the arch at points 10 feet, 20 feet and 40 feet from the centre.

Answer to Problem 14AYU

Solution:

The height of the arch 20 meters from the centre is $2.77$ feet.

The height of the arch 10 meters from the centre is $0.69$ feet.

The height of the arch 40 meters from the centre is $11.11$ feet.

Explanation of Solution

Given:

A parabolic arch has a span of 120 feet and a maximum height of 25 feet.

Calculation:

Consider the below diagram. The given data can be represented in this diagram.

The lines joining points A, B, C, D are the rectangle coordinates.

The line AB is the span which is 120 meters wide.

Here, we have $D=(-60,25),C=(60,25)$.

The cables will form the shape of a parabola, which is of the form $y=a{x}^2,a>0$.

Thus, now, in order to find the value of $a$, let us use the equation of the parabola.

Thus, considering the point $C=(60,25)$, we get

$25=a(60{)}^2$

$a=\frac{25}{{60}^2}$

Therefore, the equation of the parabola is $y=\frac{25}{{60}^2}{x}^2$.

Now, we have to find the height of the arch at a point at different distances from the centre.

Therefore, when $x=10$.

Thus, we get

$y=\frac{25}{{60}^2}(10{)}^2$

$=\frac{25}{36}=0.69$

Thus, the height of the arch 10 meters from the centre is $0.69$ feet.

When $x=20$.

Thus, we get

$y=\frac{25}{{60}^2}(20{)}^2$

$=\frac{25}{9}=2.77$

Thus, the height of the arch 20 meters from the centre is $\text{2}\text{.77}$ feet.

When $x=40$.

Thus, we get

$y=\frac{25}{{60}^2}(40{)}^2$

$=\frac{25}{36}=11.11$

Thus, the height of the arch 40 meters from the centre is $11.11$ feet.