SOLUTION: Hi, I was given a question that I need help in completing. The unit I have been completing in Maths lately involves calculus, the trapezoidal rule and the rule for catenary. In

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Hi, I was given a question that I need help in completing. The unit I have been completing in Maths lately involves calculus, the trapezoidal rule and the rule for catenary. In       Log On


   



Question 337592: Hi,
I was given a question that I need help in completing. The unit I have been completing in Maths lately involves calculus, the trapezoidal rule and the rule for catenary. In regards to the question i required assistance I had to trace a parabola using a necklace that formed a parabola, and then calculate the area under it between the x-axis, using the trapezoidal rule. The question that I am having trouble with is "Using an algebraic method, determine the equation of the parabola that you believe best models the necklace. How did you ensure this parabolic model was the best possible." I completed it using the turning point rule and using algebra on the parabola equation ax^2+bx+c, however my teacher said this method was too basic. What i need to do is continue to use similar methods that have alreay been used, namely the trapezoidal rule incorporating limits, etc. I am having difficulty completing this and any help would be greatly appreciated. Some coordinates that may prove useful include turning point at (0,0) and some other points include (18,4),(24,10) and if needed, (12,2).
Thank you,
Keiran

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You can solve for the approximation using the general equation of the parabola,
y=ax%5E2%2Bbx%2Bc and three points from the parabola.
Generate 3 equations with the three points (18,4),(24,10)(12,2)
(18,4)
4=a%2818%29%5E2%2B18b%2Bc
1.324a%2B18b%2Bc=4
.
.
.
(24,10)
10=a%2824%29%5E2%2Bb%2824%29%2Bc
2.576a%2B24b%2Bc=10
.
.
.
(12,2)
2=a%2812%29%5E2%2Bb%2812%29%2Bc
3.144a%2B12b%2Bc=2
Solve for a,b and c.
I used matrix inversion.
a=1%2F18
b=-4%2F3
c=10
.
.
.
highlight%28y=%281%2F18%29x%5E2-%284%2F3%29x%2B10%29
.
.
.