# Logarithmic Spiral

• Aug 11th 2010, 10:51 AM
MechEng
Logarithmic Spiral
Good afternoon once again,

I hope it is acceptable keep posting these. It seems like I make some small mistake on each and every one of them, so the feedback is rather helpful.

Find the arc length of the logarithmic spiral $ae^{b\theta}$ for $0\leq\theta\leq\pi$.

I have the following:

r=ae^{b\theta}

\frac{r}{dr}=abe^{b\theta}

So...

$\displaystyle\int_{0}^{\pi}\sqrt{(ae^{b\theta})^2+ (abe^{b\theta})^2}d\theta$

$\displaystyle\int_{0}^{\pi}\sqrt{(a)^2(e^{b\theta} )^2+(ab)^2(e^{b\theta})^2}d\theta$

$\displaystyle\int_{0}^{\pi}a+ab\sqrt{(e^{b\theta}) ^2+(e^{b\theta})^2}d\theta$

$\displaystyle2a+b\int_{0}^{\pi}\sqrt{2(e^{b\theta} )^2}d\theta$

$\displaystyle2a+b\int_{0}^{\pi}e^{b\theta}\sqrt{2} d\theta$

$\displaystyle(2a+b)\sqrt{2}\int_{0}^{\pi}e^{b\thet a}d\theta$

$\displaystyle(2a+b)\sqrt{2}(\frac{e^{b\theta}}{b}| _{0}^{\pi})$

$\displaystyle(2a+b)\sqrt{2}(\frac{e^{b\pi}-e^{b0}}{b})$

$\displaystyle(2a+b)\sqrt{2}(\frac{e^{b\pi}-1}{b})$

How did I do this time around?
• Aug 11th 2010, 11:58 AM
Jester
The problem is in the third step

$\sqrt{a^2+(ab)^2} \ne a + ab$