SOLUTION: An entrance to a castle is in the form of parablolic arch 6m across at the base and 3m high in the center. What is the length of a beam across the entrance, parallel to the base an
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Question 1119014: An entrance to a castle is in the form of parablolic arch 6m across at the base and 3m high in the center. What is the length of a beam across the entrance, parallel to the base and 2m above it.
Found 2 solutions by ankor@dixie-net.com, greenestamps:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
n entrance to a castle is in the form of parablolic arch 6m across at the base and 3m high in the center.
What is the length of a beam across the entrance, parallel to the base and 2m above it.
:
Using the form ax^2 + bx + c = y, (c=0)
the parabola begins at origin 0,0 and ends at 6,0; max is 3,3
write an equation for each
3,3
9a + 3b = 3
:
6,0
36a + 6b = 0
multiply the first equation by 2, subtract from the above equation
36a + 6b = 0
18a + 6b = 6
---------------subtracting eliminates b, find a
18a = -6
a = -6/18
a = -.3333
find b using the first equation
9(-.333) + 3b = 3
-3 + 3b = 3
3b = 3 + 3
b = 6/3
b = 2
Graph the equation y = .33x^2 + 2x, green shows when y = 2
"What is the length of a beam across the entrance, parallel to the base and 2m above it."
y = 2
-.33x^2 + 2x = 2
-.33x^2 + 2x - 2 = 0
solve for x using the quadratic formula, a=-.33, b=2, c=-2
two solutions
x = 1.268
x = 4.732
Find the length of the beam
4.732 - 1.268 = 3.464 meters, length of a beam 2 m above the base
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
One way to set this up to make the calculations as simple as possible is to put the center of the arch at (0,3); then the bases of the arch are at (-3,0) and (3,0).
With the vertex of the parabola at (0,3), the equation of the parabola is of the form
for some constant a.
To find the value of a, use either of the other known point on the parabola:
So the equation of the parabola is
To find the length of the beam, set y=2 and solve for x. The length of the beam will be the difference between the two x values.
or
The length of the beam in meters is the difference between sqrt(3) and -sqrt(3), which is 2*sqrt(3), or 3.464 to 3 decimal places.
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