Think above picture in polar coordinates. Q is the angle, and radius 'r' is a function of 'Q'. Now in my problem, radius r takes it's minimum value on Q = 0 and as Q getting bigger, radius r is getting biggr as well. But, I'm trying to arrange it's inrease for a predetermined condition. To do that, I have to determine the function of radius r over Q. The condition is like this:
You see in above picture, There are 3 radiuses. First one is r(0) which is a constant. Second one is r(Q1) and third one is r(Q2). Now, what I need is, when angle Q1 goes to zero (Q1 -> 0), r(Q1) must be equal to a length that
$$ \lim_{Q1 \to 0} \dfrac{(r(Q1)-\dfrac{r(0)}{\cos(Q1)})\times \cos(Q1)}{r(Q1)\times \sin(Q1)} = \tan(Q\Delta) $$
just like for $ Q2-Q1 = Qx $
$$ \lim_{Qx \to 0}\, \dfrac{(r(Q2)-\dfrac{r(Q1)}{\cos(Qx)})\times \cos(Qx)}{r(Q2)\times \sin(Qx)} = \tan(Q\Delta) $$
$ Q\Delta $ here is an initial condition, a constant.
I hope I could explain the idea. The result ought to be something like this:
So obviously the function of r must be a differential equation. But I could not figure out the equation so could you please help me with that? It also would be a great example for engineering because in class we always begin from the equation and finish with the graph, but in this, we extract the equation from a given condition.
The geometric construction gives the radius progression as $$ r(Q_{k+1})\cos(Q_{k+1}-Q_k)-r(Q_k) = r(Q_{k+1})· \sin(Q_{k+1}-Q_k)\tan(Q\Delta) \\~\\\iff\\~\\ r(Q_{k+1})= \frac{r(Q_k)}{\cos(Q_{k+1}-Q_k)-\sin(Q_{k+1}-Q_k)\tan(Q\Delta)} $$ Assuming the angles are equally spaced, $Q_k=kA$ with some small $A$, then $$ \cos(A)-\sin(A)\tan(Q\Delta)=1-2\sin(A/2)(\sin(A/2)+\cos(A/2)\tan(Q\Delta)) $$ which approximately reduces to $$ =1-A\tan(Q\Delta)+O(A^2)=e^{-A\tan(Q\Delta)+O(A^2)} $$ giving $$ r(Q_k)=r(0)e^{(kA)(\tan(Q\Delta)+O(A))} $$ so that in the limit $A\to 0$ you get the logarithmic spiral $$ r(Q)=r(0)e^{Q·\tan(Q\Delta)} $$