# Thread: Differential equation--Moth problem

1. ## Differential equation--Moth problem

One theory about the behaviour of moths states that they navigate at night by keeping fixed angle between their velocity vector and the direction of the Moon [or some bright star]. A certain moth flies near to a candle and mistakes it for the Moon. What will happen to the moth?
Hints: in polar coordinates (r,θ ), the formula for the angle ω between the radius vector and the velocity vector is given by

Use the formula to solve for r as a function of .

I do not know how to set up the necessary equations. I tried sketching a diagram,but I cannot seem to find a relation between ω and θ necessary to solve the formula given. Is my diagram correct? Thank you for your help.

2. ## Re: Differential equation--Moth problem

Originally Posted by cyt91
One theory about the behaviour of moths states that they navigate at night by keeping fixed angle between their velocity vector and the direction of the Moon [or some bright star]. A certain moth flies near to a candle and mistakes it for the Moon. What will happen to the moth?
Hints: in polar coordinates (r,θ ), the formula for the angle ω between the radius vector and the velocity vector is given by

Use the formula to solve for r as a function of .

I do not know how to set up the necessary equations. I tried sketching a diagram,but I cannot seem to find a relation between ω and θ necessary to solve the formula given. Is my diagram correct? Thank you for your help.

First take the position of the candle as the origin.

CB

3. ## Re: Differential equation--Moth problem

What CB suggested is excellent!... to proceeding we suppose that the speed of the moth is in modulus a constant v, so that only its direction can be changed, and we indicate with $\alpha$ the 'fixed angle' defined in the original post. Setting the position of the moth as a complex number...

$z(t)= x(t)+ i\ y(t) = r(t)\ e^{i \theta(t)}$ (1)

... the complex equation describing the flight of the moth is...

$z^{'}= (r^{'} + i\ r\ \theta^{'})\ e^{i\ \theta}= v\ e^{i\ (\theta- \frac{\pi}{2} + \alpha)} \rightarrow -r\ \theta^{'} +i\ r^{'} = v\ e^{i\ \alpha}$ (2)

The (2) is equivalent to a system of two DE in two variables...

$r^{2}\ \theta^{'\ 2} + r^{'\ 2}= v^{2}$

$\frac{r^{'}}{r\ \theta^{'}}= - \tan \alpha$ (3)

... and a sucessive post is dedicated to investigate about the solution of (3)...

Kind regards

$\chi$ $\sigma$

4. ## Re: Differential equation--Moth problem

For semplicity sake we set v=1 so that the system of DE becomes...

$r\ \theta^{'}=- \cos \alpha$

$r^{'}=\sin \alpha$ (1)

Of course the solution second DE is immediate...

$r(t)= t\ \sin \alpha + c_{1}$ (2)

... and that means that...

a) for $0< \alpha< \frac{\pi}{2}$ the distance of the moth from the candle will increase without limits...

b) for $\alpha=0$ the distance of the moth from the candle remains constant and the motion is 'uniform circular'...

c) for $-\frac {\pi}{2}<\alpha<0$ the distance of the moth from the candle vanishes at the time $- \frac{c_{1}}{\sin \alpha}$ and the moth is 'kaput'...

Now if we substitute (2) in the first DE we obtain...

$\theta^{'}= - \frac{\cos \alpha}{t\ \sin \alpha + c_{1}}$ (3)

.... and the solution is...

$\theta(t)= \frac{1}{\tan \alpha}\ \ln \frac{1}{t\ \sin \alpha\ + c_{1}} + c_{2}$ (4)

Kind regards

$\chi$ $\sigma$

5. ## Re: Differential equation--Moth problem

Being in the trade I know that we do not normally go for an explicit solution, but observe that the range decreases at a constant rate, and that the angle rate goes to infinity as range goes to zero (as is obvious from chisigma's equations). Also since such systems are usually (body) rate limited you can calculate the range at which the lead-pursuit model breaks down (when you should end up circling the flame).

CB

6. ## Re: Differential equation--Moth problem

Originally Posted by CaptainBlack
Being in the trade I know that we do not normally go for an explicit solution, but observe that the range decreases at a constant rate, and that the angle rate goes to infinity as range goes to zero (as is obvious from chisigma's equations). Also since such systems are usually (body) rate limited you can calculate the range at which the lead-pursuit model breaks down (when you should end up circling the flame).

CB
I suppose You intend the case c) , when is $- \frac{\pi}{2}<\alpha<0$. If the hypothesis of speed constant in modulus is true, then in such a situation when the decreasing of the distance of the moth from the flame produces a great increasing of the angular speed so that it is not surprising that...

$\lim_{t \rightarrow t_{0}} \theta (t)= - \infty\ ,\ t_{0}= - \frac{c_{1}}{\sin \alpha}$

Of course different hypothesis about the speed produce a different equation and a different scenario...

Kind regards

$\chi$ $\sigma$

7. ## Re: Differential equation--Moth problem

Originally Posted by chisigma
I suppose You intend the case c) , when is $- \frac{\pi}{2}<\alpha<0$. If the hypothesis of speed constant in modulus is true, then in such a situation when the decreasing of the distance of the moth from the flame produces a great increasing of the angular speed so that it is not surprising that...

$\lim_{t \rightarrow t_{0}} \theta (t)= - \infty\ ,\ t_{0}= - \frac{c_{1}}{\sin \alpha}$

Of course different hypothesis about the speed produce a different equation and a different scenario...

Kind regards

$\chi$ $\sigma$
I think either you are using a different convention from me about which angle is $\alpha$ or you have the trig functions reversed. I think $\dot{r}=-\cos(\alpha)$, where $\alpha$ is the angle from the moth's velocity vector to the line-of-sight to the candle (when $\alpha$ is zero the moth flies directly towards the candle).

The cases with $-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ are all closing, and changing the sign of $\alpha$ gives the mirror image trajectory. The remaining cases are all opening or constant range trajectories.

Now because we are talking about moths I have assumed we are only interested in closing trajectories, as otherwise we would not be aware of the behavior.

CB