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Hain · 02-13-2005, 07:41 PM

Anyways new weirdness of math.

I was bored out of my skull in calculus the other day and began experimenting with polar graphs and finding areas. Easy enough--generate infinitely small triangles and take the sum of them, Area = 1/2 * r^2 * dTheta. I got to this by saying that the sum of the area of arcs will be the area total. The area of a triangle is 1/2 r(theta) r(theta + change in theta) sin(change in theta)
So: The sum from a to b of the product of 1/2, r(theta), r(theta + change in theta), and sin of (change in theta) is area. Further you can take the limit of that expression for when the change in theta becomes zero.
I saw that the limit made the two radii approach each other, and that sin x as it approaches 0, approaches x.
So: the integral of the product of 1/2 and the square of r(theta) of dTheta from a to b was actually the area of polar curve! My teacher was amazed that I figured that out, kinda sad aint it for any calc nerds out there.

All in a day's work. Then I began working on polar arc lengths. Anyone that knows the method, you convert to parametric.
Let: x = r(theta) times cos(theta) and y = r(theta) times sin(theta).
To find a length you integrate the squareroot of the sum of the squares of r(theta) and r`(theta) by differential change in theta from a to b... that's simplified (maybe someone else wants to send a letter to TFP for a pretty print feature for math equations since copying from word doesn't work in here???). Well I wanted to do the same method.

Well law of Cosines: a = sqrt(b^2 + c^2 -2bc cos A)

By my logic, as the angle between approaches 0 then the two sides, b and c, become the radius taken at theta. Then that can be simplified to the integral of absolute value of r times root 2 times the squareroot of 1 minus cosine dTheta. I stumped even my teacher trying to find a different method from this point. Anyone want to take a stab at where I might have gone with this if I knew a little more about limits?

02-13-2005, 07:41 PM	#16 (permalink)
Hain has a plan Location: middle of Whywouldanyonebethere	Anyways new weirdness of math. I was bored out of my skull in calculus the other day and began experimenting with polar graphs and finding areas. Easy enough--generate infinitely small triangles and take the sum of them, Area = 1/2 * r^2 * dTheta. I got to this by saying that the sum of the area of arcs will be the area total. The area of a triangle is 1/2 r(theta) r(theta + change in theta) sin(change in theta) So: The sum from a to b of the product of 1/2, r(theta), r(theta + change in theta), and sin of (change in theta) is area. Further you can take the limit of that expression for when the change in theta becomes zero. I saw that the limit made the two radii approach each other, and that sin x as it approaches 0, approaches x. So: the integral of the product of 1/2 and the square of r(theta) of dTheta from a to b was actually the area of polar curve! My teacher was amazed that I figured that out, kinda sad aint it for any calc nerds out there. All in a day's work. Then I began working on polar arc lengths. Anyone that knows the method, you convert to parametric. Let: x = r(theta) times cos(theta) and y = r(theta) times sin(theta). To find a length you integrate the squareroot of the sum of the squares of r(theta) and r`(theta) by differential change in theta from a to b... that's simplified (maybe someone else wants to send a letter to TFP for a pretty print feature for math equations since copying from word doesn't work in here???). Well I wanted to do the same method. Well law of Cosines: a = sqrt(b^2 + c^2 -2bc cos A) By my logic, as the angle between approaches 0 then the two sides, b and c, become the radius taken at theta. Then that can be simplified to the integral of absolute value of r times root 2 times the squareroot of 1 minus cosine dTheta. I stumped even my teacher trying to find a different method from this point. Anyone want to take a stab at where I might have gone with this if I knew a little more about limits? __________________ ////\\// IRONSKY.NET \\//\\\\ Last edited by Hain; 02-15-2005 at 09:44 AM..