Question 1035950
You are given:
{{{ h[v] = 21 }}} ft
Make the arch a parabola the vertex of which is
above the center of the roadway. So the vertex is at:
( 0, 21 )
The points on the arch at the edges of the roadway are:
( 42, 15.4 ) 
( -42, 15.4 )
----------------
Let the form of the parabola be:
{{{ h(x) = a*x^2 + b*x + c }}}
When {{{ x = 0 }}} at the center, {{{ h(0) = 21 }}}, so
{{{ 21 = a*0^2 + b*0 + c }}}
{{{ c = 21 }}}
---------------
The formula for the x-value of the vertex is:
{{{ x[v] = -b/(2a) }}}
{{{ 0 = -b/(2a) }}}
{{{ b = 0 }}}
So far I have:
{{{ h(x) = a*x^2 + 21 }}}
-----------------------
Plug in point ( 42, 15.4 )
{{{ h(42) = a*( 42 )^2 + 21 }}}
{{{ h(42) = 1764a + 21 }}}
{{{ h(42) = 15.4 }}}
{{{ 15.4 = 1764a + 21 }}}
{{{ a = -5.6/1764 }}}
{{{ a = -.0031746 }}}
---------------------
Plugging in ( -42, 15.4 ) gives the same result
The equation is:
{{{ h(x) = -.0031746x^2 + 21 }}}
---------------------------
To find the end points of the arch, set
{{{ h(x) = 0 }}}
{{{ 0 = -.0031746x^2 + 21 }}}
{{{ .0031746x^2 = 21 }}}
{{{ x^2 = 21/.0031746 }}}
{{{ x^2 = 6615.007 }}}
{{{ x = 81.3327 }}}
and, also:
{{{ x = -81.3327 }}}
-------------------
{{{ 2*81.3327 = 162.6654 }}}
------------------------
The width of the arch is 162.7 ft
----------------------------
Here's the plot:
{{{ graph( 600, 500, -100, 100, -5, 30, - .0031746x^2 + 21 ) }}}