How do you prove that the circumference of a circle is #2pir#?

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Feb 28, 2015

If we imagine the circle centered in the origin with radius #r#, it has the equation:

#x^2+y^2=r^2#, (in the graph the radius is 2):

graph{x^2+y^2=4 [-10, 10, -5, 5]}

or #y=+-sqrt(r^2-x^2)#

And considering the fourth of circle in the first quadrant, we can obtain the lenght of a line with the integral:

#L=4int_0^rsqrt(1+(y')^2)dx#.

This integral is quite long, so we can parametrize the circle as usual:

#x=rcostheta#
#y=rsintheta#

and use this integral:

#L=int_a^bsqrt([x'(theta)]^2+[y'(theta)]^2)d theta#.

Since:

#x'=-rsintheta#
#y'=rcostheta#

So:

#L=4int_0^(pi/2)sqrt(r^2sin^2theta+r^2cos^2theta)d theta=#

#=4int_0^(pi/2)sqrt(r^2(sin^2theta+cos^2theta))d theta=#

#=4int_0^(pi/2)rd theta=4[rtheta]_0^(pi/2)=4rpi/2=2pir#.

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Roy E. Share
Jan 21, 2017

Answer:

I don't think you can prove it, because that is, or is equivalent, to the definition of #pi#.

Explanation:

Similarly I don't think you prove that #int_1^x 1/t dt=ln |x|#, because #ln x# is defined by precisely that integral. However, in both cases there are alternative definition in terms of limits of infinite series, but if you go down that path you just go round in circles: you can arbitrarily pick any one of the relationships as a definition and show that the others necessarily follow, and that they are all related to the same quantity, be it #pi# or #e#. Evaluating #pi# or #e# is a different matter entirely.

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Feb 27, 2015

Basically this is a definition thing.

#pi# is defined to be the ratio of the circumference of a circle over its diameter (or 2 times its radius).

This ratio is a constant since all circles are geometrically similar and linear proportions between any similar geometric figures are constant.

If you were looking for how the value of the ratio #pi# is calculated or how we know that the same ratio applies to the area of a circle, then those are different questions. (Ask if either of these are what you were looking for).

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