# circumference of a cirlce

• Apr 5th 2013, 05:37 AM
Ant
circumference of a cirlce
I'm definitely doing something wrong here, just not sure what!

Let $\displaystyle \gamma$ be the anticlockwise oriented circle centered at $\displaystyle z_{o} \in \mathbb{C}$ of radius $\displaystyle r$.

Clearly (from the formula for the circumference of a circle) $\displaystyle | \gamma | = 2\pi r$.

Using formula though:

$\displaystyle | \gamma | = \int_a^b | \gamma'(t) | dt$.

Suppose we parameterize $\displaystyle \gamma$ as follows:

$\displaystyle \gamma(t)=z_{0} + re^{it} \ t \in [0,2\pi]$

Hence $\displaystyle \gamma'(t) = tre^{it}$

And so,

$\displaystyle | \gamma | = \int_a^b | \gamma'(t) | dt =\int_0^{2\pi} |tre^{it}| \ dt = r \int_0^{2\pi} t \ dt = r(\frac{4\pi^2}{2} - 0) = 2\pi^2 r \ne 2\pi r$

Can anyone tell me what I'm doing wrong? Thanks!
• Apr 5th 2013, 05:49 AM
xxp9
Re: circumference of a cirlce
the derivative is re^it*i
• Apr 5th 2013, 06:03 AM
Ruun
Re: circumference of a cirlce
Indeed, the parameter is $\displaystyle t$ so the derivative shall be with respect to $\displaystyle t$