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Logarithmic spiral |
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| Jul26-12, 04:06 AM | #1 |
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Logarithmic spiral
1. The problem statement, all variables and given/known data
a logarithmic spiral is given. The center lies on the x-axis. after a turn of 180 degrees counterclockwise I am 86.23m away from the starting point, after 360 degrees I'm 75.41m away from the start (radius, not along the spiral). Where am I after I walked exactly 3km along the spiral trajectory? 2. Relevant equations equation: r=ae^b*theta starting point corresponds to theta=0 3. The attempt at a solution Um.. that's my problem. I never really learnt how to solve logarithmic calculations. Btw, this is not a homework, but part of a puzzle. I got the numbers from previous steps and found the equation online. I understand that r=distance from origin, theta = angle with x-axis, a and b some constants. Right. But I don't see any way of working with this this equation or any of the expressions connected to it I could need a little push in the right direction.Thanks a lot, Martine |
| Jul26-12, 06:57 AM | #2 |
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Recognitions:
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Hint1: "b" is negative in this case. After you solve for the parameters "a" and "b" you then have to work out how to find the length of a path along the spiral. Hint2: [itex] dl = \left( \sqrt{r^2 + \left( dr/d\theta \right)^2} \right) d\theta [/itex] |
| Jul26-12, 07:17 AM | #3 |
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You say "180 degrees" and "360 degrees" but radian measure should be use here.
when [itex]\theta= \pi[/itex], r= 86.23 so [itex]86.23= ae^{b\pi}[/itex]. When [itex]\theta= 2\pi[/itex], r= 75.42 so [itex]75.41= ae^{2b\pi}[/itex]. As uart says, dividing one equation by the other will cancel the 'a's leaving [tex]\frac{75.41}{86.23}= \frac{e^{2b\pi}}{e^{b\pi}}= e^{b\pi}[/tex] that is easy to solve for b. Then use that value of b with either of the first equations to find a. For the final problem you will need to integrate the "differential of arc length" that uart gave from 0 to [itex]\theta[/itex] and set it equal to 3000 to solve for [itex]\theta[/itex]. |
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