11-08-2005, 02:21 AM
Ok..I know you all have been holding your breaths wondering..is that really a true statement? Well, I present a proof. Any one who wants to read proof and/or proof read..notify me if I need corrections.
I will use "o" to mean theta and "a" to mean alpha, seeing as I dont have a greek keyboard.
let P have polar coordinates (r, theta) = (ao, o). Since x = r*cos(o) and y = r * sin(o) then we can substitute in with r = ao.
P = (ao*cos(o), ao*sin(o))
So the vector OP = (aocos(o), aosin(o)). But OP is perpindicular to the vector OT. So OT = (-aosin(o), aocos(o))
The Slope of OP is x/y = aosin(o)/aocos(o) = sin(o)/cos(o) cancelling out the ao's on top and bottom. Thus OT must have slope -cos(o)/sin(o)
(this is the negative reciprocal)
also, we now know the slop of the line k, so writing it in point slope form we can get y = mx = -cos(o)/sin(o) * x => sin(o)y + cos(o)x = 0
Set this up in parametric form with k = { x = -aosin(o)t, y = aocos(o)t}
Since the line i is tangent at the point P, where r = ao, the slope of i is the derivative at the point theta.
so take the derivative of x and y to get x' = acos(o) - aosin(o) and y' = asin(o) + aocos(o)
Put these in parametric form for the line i
i = { x = x_0 + x'(o)s, y = y_0 + y'(o)s} where x_0, y_0 are the points at P.
Set these equations for x(s) and x(t) equal AND y(s) and y(t) equal since were are looking for an intersection of two lines.
-aosin(o)t = aocos(o) + (acos(o)-ao(sin(o))s
AND
aocos(o)t = aosin(o) + (asin(o) + aocos(o))s
By inspection (or solving a system of 2 equations with 2 unknowns s and t, which archimedes wouldnt have known how to do), T = (ao^2sin(o), -ao^2cos(o)) so s = o and t = -o
Now to find the length of OT. This is the Square root of the squares of the components of T.
After much simplification this = a(o^2)
To find the arc length of PR, we take circumference times the ratio of the arc angle to 2pi.
PR = 2pi*r x o/2pi. Notice the 2pi's cancel so we get PR = ao * o = a(o^2) = OT. We have precisely proven that if T is the intersection of i and k, then arcPR = length of OT. Q.E.D.
I will use "o" to mean theta and "a" to mean alpha, seeing as I dont have a greek keyboard.
let P have polar coordinates (r, theta) = (ao, o). Since x = r*cos(o) and y = r * sin(o) then we can substitute in with r = ao.
P = (ao*cos(o), ao*sin(o))
So the vector OP = (aocos(o), aosin(o)). But OP is perpindicular to the vector OT. So OT = (-aosin(o), aocos(o))
The Slope of OP is x/y = aosin(o)/aocos(o) = sin(o)/cos(o) cancelling out the ao's on top and bottom. Thus OT must have slope -cos(o)/sin(o)
(this is the negative reciprocal)
also, we now know the slop of the line k, so writing it in point slope form we can get y = mx = -cos(o)/sin(o) * x => sin(o)y + cos(o)x = 0
Set this up in parametric form with k = { x = -aosin(o)t, y = aocos(o)t}
Since the line i is tangent at the point P, where r = ao, the slope of i is the derivative at the point theta.
so take the derivative of x and y to get x' = acos(o) - aosin(o) and y' = asin(o) + aocos(o)
Put these in parametric form for the line i
i = { x = x_0 + x'(o)s, y = y_0 + y'(o)s} where x_0, y_0 are the points at P.
Set these equations for x(s) and x(t) equal AND y(s) and y(t) equal since were are looking for an intersection of two lines.
-aosin(o)t = aocos(o) + (acos(o)-ao(sin(o))s
AND
aocos(o)t = aosin(o) + (asin(o) + aocos(o))s
By inspection (or solving a system of 2 equations with 2 unknowns s and t, which archimedes wouldnt have known how to do), T = (ao^2sin(o), -ao^2cos(o)) so s = o and t = -o
Now to find the length of OT. This is the Square root of the squares of the components of T.
After much simplification this = a(o^2)
To find the arc length of PR, we take circumference times the ratio of the arc angle to 2pi.
PR = 2pi*r x o/2pi. Notice the 2pi's cancel so we get PR = ao * o = a(o^2) = OT. We have precisely proven that if T is the intersection of i and k, then arcPR = length of OT. Q.E.D.
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