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

420 ft

Explanation:

The given equation of a parabola is

`y=630[1-\left((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((x)/(290)\right)^(2)]=0`

`1-\left((x)/(290)\right)^(2)=0`

`\left((x)/(290)\right)^(2)=1`

`(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}=(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}=(1)/(290-(-290))\int\limits^(290)_(-290) 630[1-\left((x)/(290)\right)^(2)]dx`

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

`\text{Average height}=(63)/(58)[[x]^(290)_(-290)-(1)/(84100)\left[(x^3)/(3)\right]^(290)_(-290)]`

Substitute the limits.

`\text{Average height}=(63)/(58)\left(580-(580)/(3)\right)`

`\text{Average height}=(63)/(58)((1160)/(3))`

`\text{Average height}=420`

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