# parabolic arch

• Jan 7th 2007, 01:55 AM
archistrategos214
parabolic arch
Need help with a problem having to do with parabolas.
A tunnel has a height of 20m and a width of 36m at the base. If the vertex is at the top of the arch, at which point in above the base is it 18m wide?

Thanks!
• Jan 7th 2007, 02:37 AM
ticbol
Quote:

Originally Posted by archistrategos214
Need help with a problem having to do with parabolas.
A tunnel has a height of 20m and a width of 36m at the base. If the vertex is at the top of the arch, at which point in above the base is it 18m wide?

Thanks!

The standard equation of a vertical parabola that opens downward is
(y-k) = -a(x-h)^2
where
"a" is any real, non-zero number.
(h,k) is the vertex.

Vertex here is (0,0), so,
(y-0) = -a(x-0)^2
y = -ax^2 --------------(i)

At the two ends of the 36-m base, the points are (-18,-20) and (18,-20).
Plug the (18,-20) into (1),
-20 = -a(18^2)
a = -20/(-18^2) = 5/81
So the equation of the parabola is
y = -(5/81)x^2 -------------------------(ii)

If a "base" is 18 meters wide,
y = -(5/81)(9^2)
y = -5 m
Since the actual base, the 36-m wide base, is 20m below the vertex, then this 18-m wide "base" is [-5 -(-20)] = 15 meters above the actual base.
• Jan 7th 2007, 04:52 AM
Soroban
Hello, archistrategos214!

Did you make a sketch?

Quote:

A tunnel has a height of 20m and a width of 36m at the base.
If the vertex is at the top of the arch, at which point above the base is it 18m wide?

Code:

                20|                 * * *           *      |      *         *        |        *       *          |          *                   |     - * - - - - - + - - - - - * -     -18          |          18

The vertex of the parabola is on the y-axis.
. . The equation has the form: . $y \:=\:ax^2 + c$

Since $(0,20)$ is on the parabola: . $20 \:=\:a\!\cdot\!0^2 + c\quad\Rightarrow\quad c = 20$
. . The equation (so far) is: . $y \:=\:ax^2 + 20$

Since $(18,0)$ is on the parabola: . $0 \:=\:a\!\cdot\!18^2 + 20\quad\Rightarrow\quad a = -\frac{5}{81}$
. . The parabola is: . $y \:=\:-\frac{5}{81}x^2 + 20$

"Where is the parabola 18m wide?" $\rightarrow$ "Where is $x = 9$ ?"

. . When $x = 9\!:\;\;y \;=\;-\frac{5}{81}(9^2) + 20 \;=\;15$

The arch is 18m wide at a point 15m above the base.