Mathematics in Nature: Nautilus shell
The Nautilus Spiral is given by the polar coordinates
r
=
e
θ
. Its shape resembles that of the Nautilus Shell, hence its name.
Which of the following statements about this curve is false ?
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2 solutions
It says distance along the curve, not distance from one point to the other.
The distance along the curve for a polar function r ( θ ) is given by
d s = r 2 + ( d θ d r ) 2
In this case, r = e θ and d θ d r = e θ , so our integral looks like this:
s = ∫ − ∞ 0 2 e 2 θ d θ
s = 2 ∫ − ∞ 0 e θ d θ
s = 2
The integral is convergent.
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Great job with the integration.
Actually, there is a much easier way to prove that the distance is finite, without performing the integration. All the options revolve around the self-similar nature of the Nautilus curve.
Hint: Cut the curve with the half line θ = 0 . The corresponding lengths form a (fill in the blank) and hence the sum of lengths is finite.
as ds is very small can we assume ds=r*d(theta) and then perform intregration?
But it asks for the distance measured a l o n g the curve and the curve makes infinite turns around the origin till θ = 0 . So, the length of the curve till θ = 0 should be infinite.
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Not true. If you make an infinite number of infinitely small loops, you can't say anything about the length of the curve. Though you're right that Paramjit Singh's solutions is wrong -- I posted the correct reasoning below.
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But the smallest distance that r can obtain is 0; the spiral is not necessarily a fractal; where θ is equal to 0, r will be 1. Even if there were to be infinitely small loops, there would have to be an r value of an extremely small value that is a ratio of e; since r=e^θ
Agree with Michael, take for example, 1+1/2+1/4+1/8+....... here the series has infinite terms but the sum is finite...:)
Since the distance from the center of the spiral to the outer shell is r, and the equation for finding r is r=e^θ, this choice stated that θ=0; r=e^0=1, and clearly, where r=1 there must be a corresponding point on the curve.
Simple math, r=eƟ, here Ɵ=0 so distance r=1
Origin for this curve means the case of Θ= -∞ . So, for the calculation of distance along curve from Θ= -∞ to Θ= 0, if the curve is integrated with dΘ element from Θ= -∞ to Θ= -∞ it will produce the result 1(one), not infinity. so the answer will be D
The curve is self similar, because the whole curve resembles a part of it.
The half-line θ = α is cut into segments of { r = e { 2 k π + θ } } k = 0 ∞ for any θ , which is a geometric progression.
The curve loops around the origin infinitely many times from θ = − ∞ to θ = 0 . Basically this can also be deduced from it's self similarity.
The distance of origin from θ = 0 is not infinite, but convergent, namely s = 2 , as shown by Michael Tong below. [I had misread this part]