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!
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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
(yk) = a(xh)^2
where
"a" is any real, nonzero number.
(h,k) is the vertex.
Vertex here is (0,0), so,
(y0) = a(x0)^2
y = ax^2 (i)
At the two ends of the 36m 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 36m wide base, is 20m below the vertex, then this 18m wide "base" is [5 (20)] = 15 meters above the actual base.
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 yaxis.
. . The equation has the form: .$\displaystyle y \:=\:ax^2 + c$
Since $\displaystyle (0,20)$ is on the parabola: .$\displaystyle 20 \:=\:a\!\cdot\!0^2 + c\quad\Rightarrow\quad c = 20$
. . The equation (so far) is: .$\displaystyle y \:=\:ax^2 + 20$
Since $\displaystyle (18,0)$ is on the parabola: .$\displaystyle 0 \:=\:a\!\cdot\!18^2 + 20\quad\Rightarrow\quad a = \frac{5}{81}$
. . The parabola is: .$\displaystyle y \:=\:\frac{5}{81}x^2 + 20$
"Where is the parabola 18m wide?" $\displaystyle \rightarrow$ "Where is $\displaystyle x = 9$ ?"
. . When $\displaystyle x = 9\!:\;\;y \;=\;\frac{5}{81}(9^2) + 20 \;=\;15$
The arch is 18m wide at a point 15m above the base.