Archimedes Spiral

• August 10th 2010, 01:19 PM
MechEng
Archimedes Spiral
Good afternoon,

I am working on finding the length of Archimedes Spiral $r=\theta$ for $0\leq\theta\leq2\pi$. I think I may be doing something wrong. This is what I'm getting:

$r=\theta=f(x)$

$f'(x)=1$

$\displaystyle\int_{0}^{2\pi}\sqrt{\theta^2-1^2}d\theta$

I have evaluated this to:

$\displaystyle\frac{\theta}{2}\sqrt{\theta^2-1}-\frac{ln(\theta+\sqrt{\theta^2-1}}{2}|_{0}^{2\pi}$

When I try to wrap this up by plugging in my terminals... I get a mess. Is this right?

$\displaystyle\pi\sqrt{(2\pi)^2-1}-\frac{ln(\2\pi+\sqrt{(2\pi)^2-1})}{2}$
• August 10th 2010, 02:02 PM
Quote:

Originally Posted by MechEng
Good afternoon,

I am working on finding the length of Archimedes Spiral

$r=\theta$

for... $0\leq\theta\leq2\pi$.

I think I may be doing something wrong. This is what I'm getting:

$r=\theta$

$\displaystyle\huge\frac{dr}{d\theta}=1$

$\displaystyle\int_{0}^{2\pi}\sqrt{r^2+\left(\frac{ dr}{d\theta}\right)^2}\ d\theta$

Shouldn't you have been working from the above ?

I have evaluated this to:

$\displaystyle\frac{\theta}{2}\sqrt{\theta^2-1}-\frac{ln(\theta+\sqrt{\theta^2-1}}{2}|_{0}^{2\pi}$

When I try to wrap this up by plugging in my terminals... I get a mess. Is this right?

$\displaystyle\pi\sqrt{(2\pi)^2-1}-\frac{ln(\2\pi+\sqrt{(2\pi)^2-1})}{2}$

$\displaystyle\huge\int_{0}^{2{\pi}}\sqrt{r^2+1^2}\ d\theta=\int_{0}^{2{\pi}}\sqrt{1+\theta^2}\ d\theta$

Working through the integration....

if you draw a right-angled triangle, perpendicular sides x and 1,
then the hypotenuse length is

$\sqrt{1+x^2}$

Hence.... $cos\theta=\frac{1}{\sqrt{1+x^2}}\ \Rightarrow\ \sqrt{1+x^2}=sec\theta$

$x=tan\theta$

$dx=sec^2\theta\ d\theta$

$\int{\sqrt{1+x^2}}dx=\int{\sqrt{1+tan^2\theta}sec^ 2\theta\ d\theta$

$=\int{sec\theta\ sec^2\theta\ d\theta=\int{sec^3\theta}d\theta$

Integration by parts...

$u=sec\theta,\ dv=sec^2\theta\ d\theta$

$\Rightarrow\ du=sec\theta\ tan\theta,\ v=tan\theta$

$uv-\int{v}du=sec\theta\ tan\theta-\int{sec\theta\ tan^2\theta}d\theta=sec\theta\ tan\theta-\int{sec\theta\left(sec^2\theta-1\right)d\theta$

$=sec\theta\ tan\theta-\int{sec^3\theta}d\theta+\int{sec\theta}d\theta$

$2\int{sec^3\theta}d\theta=sec\theta\ tan\theta+\int{sec\theta}d\theta$

$=sec\theta\ tan\theta+ln|sec\theta+tan\theta|$

$\int{sec^3\theta}=\frac{1}{2}\left(sec\theta\ tan\theta+ln|sec\theta+tan\theta|\right)$

$=\frac{1}{2}\left(x\sqrt{1+x^2}+ln|\sqrt{1+x^2}+x| \right)$

Rewrite using $\theta$ instead of x and evaluate using the limits.
• August 11th 2010, 06:46 AM
MechEng
...yep. I'm not sure how I wound up with

$\displaystyle\int_{0}^{2\pi}\sqrt{r^2-\left(\frac{dr}{d\theta}\right)^2}\ d\theta$

VS.

$\displaystyle\int_{0}^{2\pi}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\ d\theta$

I think I may have been looking at two different formulas at once, as they are all listed in the chapter summary. Since area formulas are genereally subtracting something from something else, that is the only explanation I can produce.

I thank you once again, Sir.
• August 11th 2010, 07:15 AM
MechEng
Anyway... Finishing this one up...

For $0\leq\theta\leq2\pi$...

$\displaystyle\frac{\left(2\pi\sqrt{1+2\pi^2}+ln|\s qrt{1+2\pi^2}+2\pi|\right)}{2}-\frac{\left(0\sqrt{1+0^2}+ln|\sqrt{1+0^2}+0|\right )}{2}$

$\displaystyle\frac{\left(2\pi\sqrt{1+2\pi^2}+ln|\s qrt{1+2\pi^2}+2\pi|\right)}{2}$

Is this a reasonable answer for this problem?
• August 11th 2010, 12:34 PM
Quote:

Originally Posted by MechEng
Anyway... Finishing this one up...

For $0\leq\theta\leq2\pi$...

$\displaystyle\frac{\left(2\pi\sqrt{1+(2\pi)^2}+ln| \sqrt{1+(2\pi)^2}+2\pi|\right)}{2}-\frac{\left(0\sqrt{1+0^2}+ln|\sqrt{1+0^2}+0|\right )}{2}$

$\displaystyle\frac{\left(2\pi\sqrt{1+4\pi^2}+ln|\s qrt{1+4\pi^2}+2\pi|\right)}{2}$

Is this a reasonable answer for this problem?

Looks good to me.
Be careful when squaring, parentheses helps.
Lastly, you may evaluate the final expression.