# Logarithmic spiral

• Jan 12th 2009, 07:07 AM
jpi
Logarithmic spiral
Hi
I have some questions regarding a puzzle.

http://img.geocaching.com/cache/fb69...d5b5f32eea.jpg

This is a logarithmic spiral, and i know that from "Startpunkt" to "A" the direct distance is 86,23. From "Startpunkt" to "B" the direct distance is 75,41.
If i start at "Startpunkt" and measure the spiral where will the point be when the distance is 3000 ?

I have found the definition r = ae^(b*theta), but i don't know how to use it. How do i find the two constants a and b ?

And how do i find the length ?

Regards
Jacob
• Jan 12th 2009, 07:39 AM
HallsofIvy
Quote:

Originally Posted by jpi
Hi
I have some questions regarding a puzzle.

http://img.geocaching.com/cache/fb69...d5b5f32eea.jpg

This is a logarithmic spiral, and i know that from "Startpunkt" to "A" the direct distance is 86,23. From "Startpunkt" to "B" the direct distance is 75,41.
If i start at "Startpunkt" and measure the spiral where will the point be when the distance is 3000 ?

I have found the definition r = ae^(b*theta), but i don't know how to use it. How do i find the two constants a and b ?

For $\theta= 0$,at startpunkt, r= a and when $\theta= \pi$, at A, $r= ae^(b\pi)$. And we are told that the distance between the points, which is the difference between those two, is 86,23: $a- ae^{b\pi}= 86,23$. When $\theta= 2\pi$, point B, [tex]r= ae^{2b\pi}. The distance between startpunkt and B, we are told, is 7541: ae^{2b\pi}- a= 7541. Solve those two equations for a and b.

Quote:

And how do i find the length ?

Regards
Jacob
As for the arclength, the general formula, given that x(t) and y(t) are functions of the parameter t, is
$\int \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt$

Here, since this is given in terms of r and $\theta$ so you need $x= rcos(\theta)$ and $y= rsin(\theta)$ as well as the fact that $r= ae^{b\theta}$. That is, $x= ae^{b\theta}cos(\theta)$ and $y= ae^{b\theta}sin(\theta)$.

The derivatives messy but the formula reduces nicely (since $sin^2(\theta)+ cos^2(\theta)= 1$) to
$\int a\sqrt{b^2+ 1}e^{b\theta} d\theta$
which is easy to integrate:
$\frac{a}{b}\sqrt{b^2+1}e^{b\theta}+ C$
to find the arclength between $\theta_1$ and the startpunkt, evaluate that at the two points and subtract:
$\frac{a}{b}\sqrt{b^2+1}\left(e^{b\theta}- 1\right)$
set that equal to 3000 and solve for $\theta$.
• Jan 12th 2009, 10:34 AM
jpi
Quote:

Originally Posted by HallsofIvy
For $\theta= 0$,at startpunkt, r= a and when $\theta= \pi$, at A, $r= ae^(b\pi)$. And we are told that the distance between the points, which is the difference between those two, is 86,23: $a- ae^{b\pi}= 86,23$.

Shouldn't it be $a+ ae^{b\pi}= 86,23$ ?
r = a is the piece from (0,0) to startpunkt and $r = ae^{b\pi}$ is the part from (0,0) to A. Or am i wrong here ?
• Jan 13th 2009, 03:20 AM
HallsofIvy
Quote:

Originally Posted by jpi
Shouldn't it be $a+ ae^{b\pi}= 86,23$ ?
r = a is the piece from (0,0) to startpunkt and $r = ae^{b\pi}$ is the part from (0,0) to A. Or am i wrong here ?

Yes, you are right: one point is left of (0,0), the other to the right. I though I had corrected that by editting!
• Jan 13th 2009, 11:36 AM
jpi
I have now solved the puzzle, and got the answer correct !!!

Thanks a lot. It is higly appreciated (Nod)

/Jacob