Parabolic Equation of an Arch
Date: 05/25/98 at 00:36:49 From: Derek Child Subject: parabolas For an assignment in my trig class, I was asked to find the equation of the parabola in the McDonald's golden arches. I have searched high and low all over the Internet. I don't want a full blown solution, just some hints as to how I can come up with a solution. I have tried taking a picture of the "M" and putting it on a x-y axis, but this doesn't work too well. Any help would be greatly appreciated. Thanks.
Date: 06/08/98 at 15:41:53 From: Doctor Jeremiah Subject: Re: parabolas Dear Derek: To define a parabola requires three (x,y) coordinates. Measure the height (I will use the symbol H for the value that you measure) and width at the widest part (I will use the symbol W for that value) of one of the parabolas in the "M." If you draw the parabola like this ... y | | | | | -W/2 | (0,0) W/2 --- -+- ----|--------+++++--------|-------- x | + | + | + | + | + | + H + | + | | | + | + | | | | -+- + -+- -H + (-W/2,-H) (W/2,-H) |----------W----------| ... then you have three points: (0,0), (W/2,-H), and (-W/2,-H). So all you have to do is solve the equation of the parabola with these three coordinates. The equation of a parabola is: y = ax^2 + bx + c We need to make three equations, because we want to solve for three variables: a, b and c. So we take our equation and we substitute in each of our coordinates for x and y, and we end up with three equations: 0 = a(0)^2 + b(0) + c -H = a(W/2)^2 + b(W/2) + c -H = a(-W/2)^2 + b(-W/2) + c The first obviously makes c = 0. To solve this problem, we need to find b. We have two equations left that just happen to be equal to the same thing (-H), so we can equate the equations and solve for b: a(W/2)^2 + b(W/2) + c = a(-W/2)^2 + b(-W/2) + c Now that we know a and b, we can substitute them into the second or third equation and solve for a. Now that we have a, b and c, we can change y = ax^2 + bx + c to the real equation by replacing the letters with the values of those variables. If you need more details, mail me back. -Doctor Jeremiah, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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