Solution: Prime factorizations: $15 = 3 imes 5$, $20 = 2^2 imes 5$. LCM is $2^2 imes 3 imes 5 = 60$. The synchronized cycle length is $oxed60$.Question: A civil engineer designs a parabolic arch with the equation $ y = -ax^2 + bx + c $. If the arch's highest point is at $ (3, 10) $ and passes through $ (0, 4) $, find the value of $ a $. - Dachbleche24
Solution: Finding the Coefficient $ a $ in the Parabolic Arch Equation
Solution: Finding the Coefficient $ a $ in the Parabolic Arch Equation
When designing elegant architectural structures like parabolic arches, precision in mathematics ensures both structural integrity and aesthetic beauty. In this case, a civil engineer models a parabolic arch using the quadratic equation:
$$ y = -ax^2 + bx + c $$
The arch reaches its highest point at $ (3, 10) $ and passes through the point $ (0, 4) $. We are asked to find the value of $ a $.
Understanding the Context
Step 1: Use the vertex form of a parabola
Since the vertex (highest point) of the parabola is at $ (3, 10) $, we rewrite the equation in vertex form:
$$ y = -a(x - 3)^2 + 10 $$
(Note the negative sign in front of $ a $ accounts for the downward-opening parabola.)
Step 2: Substitute the known point $ (0, 4) $
Substitute $ x = 0 $, $ y = 4 $ into the equation:
$$ 4 = -a(0 - 3)^2 + 10 $$
$$ 4 = -a(9) + 10 $$
$$ 4 - 10 = -9a $$
$$ -6 = -9a $$
$$ a = rac{6}{9} = rac{2}{3} $$
Key Insights
Step 3: Confirm the equation matches the standard form
From vertex form:
$$ y = - rac{2}{3}(x - 3)^2 + 10 $$
Expand to standard form:
$$ y = - rac{2}{3}(x^2 - 6x + 9) + 10 $$
$$ y = - rac{2}{3}x^2 + 4x - 6 + 10 $$
$$ y = - rac{2}{3}x^2 + 4x + 4 $$
This matches the form $ y = -ax^2 + bx + c $ with $ a = rac{2}{3}, b = 4, c = 4 $, confirming consistency.
Step 4: The synchronized cycle length concept — a metaphor for optimal design
Just as finding the least common multiple synchronizes repeating patterns, determining $ a $ aligns the arch’s vertex and known point to form a precise, stable structure. Here, symmetry and mathematics meet engineering excellence — a true reflection of how $ oxed{60} $-like precision applies even in arches: every number matters.
Conclusion:
By analyzing the vertex and a point on the parabola, we found $ a = rac{2}{3} $. This value ensures the arch’s highest point is correctly placed and passes through the base point, illustrating how mathematical accuracy forms the backbone of structural design. When geometry and engineering sync, beauty and safety follow. Mirroring this precision, this solution confirms that $ oxed{a = rac{2}{3}} $.
