A parabolic arch has a height of 20 m and a width of 36 m at the base. If the vertex of the parabola
is at the top of the arch, at which height above the base is it 18 m wide?
Thu
Well, if the base of the arch is 36 m wide and the vertex is at the top, then the arch is perfectly symmetric. So, when the arch is 18 m wide, it will be at exactly half the height of the arch. Therefore, the vertex will be at a height of 10 m above the base.
But hey, don't worry, even with a parabolic arch, there will never be a circus in town!
To find the height above the base at which the parabolic arch is 18 m wide, we need to determine the equation of the parabola.
The general equation of a parabola in vertex form is given by:
y = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola.
In this case, the vertex is at the top of the arch, so the y-coordinate of the vertex is the maximum height of the arch.
Also, the parabola has a width of 36 m at the base, which means that when y = 0 (at the base of the arch), the x-value is ±(36/2) = ±18.
We are given the height of the arch at the base, which is 20 m. So our vertex is at (±18, 20).
Substituting these values into the vertex form equation, we have:
0 = a(±18 - h)^2 + 20
Now we need one more point to find the value of a. We are given that the parabola is 18 m wide at a certain height above the base. Let's call this height y1.
So, at (±9, y1), the width is 18 m, which means x = ±9.
Now we can substitute these values into the equation to solve for a:
0 = a(±18 - h)^2 + 20
Since the width is 18 m at x = ±9, we have:
y1 = a(±9 - h)^2 + 20
Substituting in the known values:
0 = a(18 - h)^2 + 20
Now, let's solve for a.
First, subtract 20 from both sides:
-20 = a(18 - h)^2
Now divide by (18 - h)^2:
-20 / (18 - h)^2 = a
Now we have an equation for a in terms of h.
Next, we can substitute this value of a into the equation we found earlier for the vertex:
0 = (-20 / (18 - h)^2)(±18 - h)^2 + 20
Simplifying, we have:
0 = (-20 / (18 - h)^2)(324 - 36h + h^2) + 20
Now we can solve this equation to find the height above the base at which the parabolic arch is 18 m wide.
clearly the equation of the arch is
y = 20 - ax^2
y(18) = 0, so
a = 0.0617
So, y = 20-0.0617x^2
y(9) = y(-9) = 15