assume an arch is shaped like a parabola. It is 20ft wide at base and 80 ft high

a. find equation that models arch

b. how wide is arch 20ft above ground?

Guest Nov 28, 2018

edited by
Guest
Nov 28, 2018

#1**0 **

Let's put one base leg at 0,0 and the other leg at 20,0

and write the equation of a parabola in vertex form

the vertex will be at (h,k) = 10,80

Vertex from = a(x-h)^2 + k so:

f(x) = a (x-10)^2 + k

= a (x^2-20x +100) + 80

We know that point 0,0 is on the graph, so let's sub that in to help find 'a'

0 = a(0^2-20(0) +100) + 80

0 = 100a +80

a = -80/100 = -4/5

So the equation is f(x) = -4/5 (x-10)^2 + 80

at f(x) = 20 20 = -4/5(x-10)^2 + 80

60 = 4/5 (x-10)^2

75 = (x-10)^2

+- sqrt 75 = x-10

x = sqrt 75 +10 and - sqrt 75 + 10

x = 18.66 and 1.34 ft The width is the difference between these x coordinates = 17.32 ft

ElectricPavlov Nov 28, 2018

#3**+1 **

Thanks EP....here's another way using symmetry about the y axis

a. Let the ends of the arch be at (-10, 0) and (10,0)

Let the apex of the arch be at (0,80)

So....we just need to find "a" in this function

y = ax^2 + 80

We know the point (10, 0) = (x, y) is on the graph...so....filling in what we know, we have

0 = a(10)^2 + 80

0 = 100a + 80 subtract 80 from both sides

-80 = 100a divide both sides by 100

-80/ 100 = a = - 4/5

So....the function is

y = (-4/5)x^2 + 80

b. At 20 ft above the ground, we have

20 = (-4/5)x^2 + 80 subtract 80 from both sides

-60 = (-4/5)x^2 multiply both sides by -5/4

75 = x^2 take the positve root

sqrt (75) = x = 5sqrt(3) ft

But this is only half the width...so multiplying by 2 we have

Width = 10sqrt(3) ≈ 17.3 ft

CPhill Nov 28, 2018