Question

An arch is 630 ft high and has 580=ft base. It can be modeled by the parabola =630\left [ 1-\left ( x/290 \right )^2 \right ]. Find the average height of the arch above the ground.
The average height of the arch is __??? ft above the ground.

Answer 1

Answer:

420 ft

Step-by-step explanation:

The given equation of a parabola is

$y=630[1-\left(\frac{x}{290}\right)^{2}]$

An arch is 630 ft high and has 580=ft base.

Find zeroes of the given function.

$y=0$

$630[1-\left(\frac{x}{290}\right)^{2}]=0$

$1-\left(\frac{x}{290}\right)^{2}=0$

$\left(\frac{x}{290}\right)^{2}=1$

$\frac{x}{290}=\pm 1$

$x=\pm 290$

It means function is above the ground from -290 to 290.

Formula for the average height:

$\text{Average height}=\dfrac{1}{b-a}\int\limits^b_a f(x) dx$

where, a is lower limit and b is upper limit.

For the given problem a=-290 and b=290.

The average height of the arch is

$\text{Average height}=\dfrac{1}{290-(-290)}\int\limits^{290}_{-290} 630[1-\left(\frac{x}{290}\right)^{2}]dx$

$\text{Average height}=\dfrac{630}{580}[\int\limits^{290}_{-290} 1dx -\int\limits^{290}_{-290} \left(\frac{x}{290}\right)^{2}dx]$

$\text{Average height}=\dfrac{63}{58}[[x]^{290}_{-290}-\frac{1}{84100}\left[\frac{x^3}{3}\right]^{290}_{-290}]$

Substitute the limits.

$\text{Average height}=\dfrac{63}{58}\left(580-\frac{580}{3}\right)$

$\text{Average height}=\dfrac{63}{58}(\dfrac{1160}{3})$

$\text{Average height}=420$

Therefore, the average height of the arch is 420 ft above the ground.