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

Question

Scarlett Allison · Accepted Answer

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#.

Paisley Johnson · Answer

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

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.

Christopher Agresta · Answer

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).

Adam Adkins · Answer

undefined The circumference of a circle can be proven to be $2\pi r$ using basic principles of geometry and trigonometry.

One method is to consider the definition of a radian, which is the angle subtended by an arc of a circle that has the same length as the radius. By definition, the circumference of a circle is $2\pi$ times the radius. Since the angle corresponding to the entire circumference is $2\pi$ radians, the ratio of the circumference to the radius is $2\pi:r$. Therefore, the circumference of a circle is $2\pi r$.

Another approach involves the concept of similar triangles. Consider a circle with radius $r$ and circumference $C$. If we draw a straight line from the center of the circle to any point on its circumference, we form a right triangle with the radius as one leg and the line segment from the center to the circumference as the hypotenuse. By the definition of a circle, this line segment is equal to the radius $r$. The other leg of the triangle is the height of the circle, which is also $r$ (since it's perpendicular to the base, which is the radius). Therefore, we have a right triangle with two legs of length $r$ each. Using the Pythagorean theorem, the hypotenuse (which is the circumference of the circle) is $C = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}$. Since $C = 2\pi r$, equating the expressions for $C$ gives $r\sqrt{2} = 2\pi r$, which implies $\sqrt{2} = 2\pi$. Squaring both sides yields $2 = 4\pi^2$, and dividing both sides by 2 gives $\pi^2 = \frac{1}{2}$, which is clearly false. Therefore, the assumption that $\sqrt{2} = 2\pi$ is incorrect. Hence, the circumference of a circle is indeed $2\pi r$.

Both methods demonstrate that the circumference of a circle is $2\pi r$.