General Form Q 2 P N example essay topic

Dawn Lode wegen Ledocq Areas as Limits In 250 B.C., Archimedes derived the formula for the area of a circle. We were instructed to find the formula of the area of a circle using the same method that Archimedes did. Everyone should be familiar with the formula giving the area of a circle as a function of it's radius: A = pr 2. We were instructed to look at the combined areas of inscribed triangles, and increase the number of triangles until we get closer and closer to the actual area of a circle. In order to carry out the instructions above we must first determine the area of one of the inscribed congruent triangles in terms of the radius and some appropriate angle. In this figure, the angle q = p/4 radians.

If one is to put it into a general form q = 2 p / n, where 2 p is the distance around the circle and n is the number of triangles used. If we take one of these triangles out of the picture you get something like this: Using trigonometry you know that the sin (q/2) = b/2 r and the cos (q/2) = h / r. When you put this into a general formula for n, you then get sin (2 p/2 n) = b/2 r and cos (2 p/2 n) = h / r. When the two's cancel you then get your basic formulas and you can solve for b and h. sin (p / n ) = b/2 r 'a b = 2 rsin (p / n ) cos (p / n ) = h / r 'a h = rcos (p / n ) Once you have these formulas for b and h, you can then plug then into the formula, 1/2 nbh.

The limit of this formula is what will give you the area of your circle. We know this because if n = the number of triangles used and 1/2 bh is the formula of the area of the triangles. As stated before as the number of triangles gets infinitely large, the sum of the areas of these triangles will converge to the area of a circle. Therefore you can take this limit as n goes to infinity and hopefully derive the formula for the area of a circle.

Step 1: Substitution nbh = n (2 rsin (p / n ) ) (rcos (p / n ) ) 2 2 Step 2: Rearrangement / cancellation = (nsin (p / n ) ) (r 2 cos (p / n ) ) Step 3: Taking the Limit = Lim (nsin (p / n ) ) (r 2 cos (p / n ) ) n'a yen Step 4: Rearrangement = Lim (nsin (p / n ) ) Lim (r 2 cos (p / n ) ) n'a yen n'a yen Step 5: Taking the Limit of r 2 cos (p / n ) = Lim (r 2 cos (p / n ) ) n'a yen = r 2 Lim (cos (p / n ) ) n'a yen = r 2 cos (0) = (r 2) (1) = r 2 Step 6: Taking the Limit of nsin (p / n ) = Lim (nsin (p / n ) ) n'a yen = (yen) (sin (0) ) = 0 Therefore you must come up with some other way to get the limit of nsin (p / n ), as n'a yen, to p. You do this by using a trig identity that was learned earlier this semester. This identity is that: the Limit of (sint) /t = 1 as t goes to infinity. We do this by multiplying and dividing by p / n, which is equivalent to multiplying by 1. = Lim n (sin (p / n ) ) (p / n ) n'a yen (p / n ) When you do this the n's will cancel and you will get an identity where this limit will equal p. = Lim p sin (p / n ) n'a yen (p / n ) = p Lim sin (p / n ) = (p) (1) = p n'a yen (p / n ) Step 7: Combining the Limits Lims in = p Lim sin (p / n ) = (p) (1) = p n'a yen (p / n ) Lim cos = r 2 cos (0) = (r 2) (1) = r 2 pr 2 = Area of a circle Conclusion: As one can see this was not necessarily an easy process.

It takes time, patience, and knowledge. Although it is difficult to figure out, it is possible. It does help with the understanding of how all of these formulas, like area of a circle, came about.