The logarithmic spiral

The logarithmic spiral has some very interesting properties and Bernoulli was especially fascinated by it.I’ll prove it’s most important property(the angle between the curve and the radius at every angle is constant) and proceed with an example.

In polar co-ordinates,the equation of the spiral is given by:

r(\theta)=ae^{k\theta} where a,k are constants and a>0

Now,to prove that any line from the origin which intersects the curve does so by making a constant angle(say \phi) with the curve(direction of tangent line),we consider the derivatives of the parameter equations which correspond to r(\theta)

log_full-1

\Rightarrow x'{\theta}=kae^{kt\theta}cos(\theta)-ae^{k\theta}sin(\theta) and

y'(\theta)=kae^{k\theta}sin(\theta)+ae^{k\theta}cos(\theta)

So, \vec{r'}=(x'(\theta),y'(\theta))

Now,we consider the following equation for the tangential angle:

cos(\phi)=\frac{\vec{r}.\vec{r'}}{|\vec{r}||\vec{r'}|}

\vec{r}.\vec{r'}=(ka^{2}e^{2k\theta}-a^{2}e^{2k\theta}sin(\theta)cos(\theta))+(ka^{2}e^{2k\theta}+a^{2}e^{2k\theta}sin(\theta)cos(\theta))=k(a^{2}e^{2k\theta})=kr^{2}

After some calculation it is seen that |\vec{r}|=r and |\vec{r'}|=r\sqrt{k^{2}+1}

cos( \phi )=\frac{k}{\sqrt{k^{2}+1}}

The expression on the right is clearly a constant and is dependent only on k  !!

This means that the angle \phi=arcos(\frac{k}{\sqrt{k^{2}+1}} ) is constant and as a result,the astounding property of logarithmic spirals is proved.

In fact it can even be concluded that given two different logarithmic spirals with the same tangential angle(ie \phi_1=\phi_2),the spirals will only differ in scale(a_1,a_2).This means that the spirals will be in a sense,similar.

As an exercise,try deriving an expression for the length of the spiral in a definite interval,you’ll get a NEAT formula.


I’ll now give an example of a special type of spiral.I think I saw this example in one of Strang’s book though I’m not exactly sure about that.All talk aside,I was amazed at the example when I first saw it and still remember it.Truth be told,it isn’t something that extraordinary.

Finally,putting an end to my gibberish,here is the example:

r(\theta)=e^{-\theta}

Capture

We know that ds=\sqrt{e^{-2\theta}} d\theta .It can be noticed that the curve spirals to zero as \theta \to \infty.In fact,the total finite length(say l) of the spiral can be also be calculated.

l=\int\limits_{0}^{\infty} \sqrt{2e^{-2\theta}} d\theta=+\sqrt(2)e^{0}=\sqrt{2}


That’s about it for this week’s post.I might post something related to irrationality next time.

 

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