Question

An arched drainage culvert can be modelled by the function:

y=(x(191-x))/60

If water fills the culvert to a height of 136.5 mm, use the quadratic formula to calculate how wide the air space is directly above the water level.

Answer 1

Substitute 136.5 for y.

`136.5=(x(191-x))/(60)`

`8190=x(191-x)`

`8190=191x-x^(2)`

`x^(2) -191x+8190=0`

Compare this equation with
`ax^(2) +bx+c=0`

a = 1, b = - 191, c = 8190

`x=\frac{-b+\sqrt{b^(2)-4ac } }{2}` or

`x=\frac{-b-\sqrt{b^(2)-4ac } }{2}`

`x=\frac{191+\sqrt{191^(2)-4(1)(8190) } }{2}` or

`x=\frac{191-\sqrt{191^(2)-4(1)(8190) } }{2}`

`x=(191+√(36481-32760 ) )/(2)` or

`x=(191-√(36481-32760 ) )/(2)`

`x=(191+√(3721 ) )/(2)` or

`x=(191-√(3721 ) )/(2)`

`x=(191+61)/(2)` or

`x=(191-61)/(2)`

`x=(252)/(2)` or

`x=(130)/(2)`

x = 126 or x = 65

Hence, the points on the curve are (65, 136.5) and (126, 136.5).

The width of the air space is the distance between these points.

Width =
`\sqrt{(x_(2)- x_(1))^(2) +( y_(2)- y_(1)) ^(2)`

=
`\sqrt{(126- 65)^(2) +(136.5-136.5) ^(2)`

= 126 - 65

= 61

Hence, width of the air space is 61m.