##### Part I:

A fox chases a rabbit. Both run at the same speed $ v$ . At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $ \alpha $ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is $ l$ .

When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation?

##### Part II:

Similarly think about the same situation, except now let the rabbit always move in the straight line of its initial direction in above part of the question.

When and where does the fox catch the rabbit (if it does)? If it never does, what is their eventual separation?

## Solutions

#### Part I

The relative speed of the fox and the rabbit, along the line connecting them, is always $ v_{\text{rel}}= v- v \cos \alpha$ . Therefore, the total time needed to decrease their separation from $ l$ to zero is $ T=\dfrac{l}{v-v \cos \alpha} =\dfrac{l}{v(1-\cos \alpha)} \ \ldots (1)$ which is valid unless $ \alpha=0$ , in which case the fox never catches the rabbit.

The location of their meeting is a little trickier to obtain. We have two methods to do :

##### SLICK METHOD

Imagine that the rabbit chases another rabbit, which chases another rabbit, etc. Each animal runs at an angle $ \alpha$ relative to the direction directly away from the animal chasing it. The initial positions of all the animals lie on a circle, which is easily seen to have radius $ R=\dfrac{l/2}{\sin (\alpha/2)} \ \ldots (2)$ .

The center of the circle is the point, O, which is the vertex of the isosceles triangle with vertex angle $ \alpha$ , and with the initial fox and rabbit positions as the othe two vertices. By symmetry, the positions of the animals at all times must lie on a circle with center O. Therefore, O, is the desired point where they meet. The animals simply spiral into O.

#### Remark

An equivalent solution is the the following:

At all times, the rabbit’s velocity vector is obtained by rotating the fox’s velocity vector by angle $ \alpha$ . The meeting point O, is therefore the vertex of the above mentioned isosceles triangle,

##### MESSIER METHOD

The speed of the rabbit in the direction orthogonal to the line connecting the two animals in $ v \sin \alpha$ . Therefore, during a time $ dt$ , the direction of the fox’s motion changes by an angle $ d\theta =\dfrac {v \sin \alpha}{l_t} dt$ , where $ l_t$ is the separation at time $ t $ . Hence the change in the fox’s velocity has magnitude $ |d\overrightarrow{v}|=v d\theta =v (v \sin \alpha dt/l_t)$ . The vector $ d\overrightarrow{v}$ is orthogonal to $ \overrightarrow{v}$ , therefore, to get the $ x$ -component of $ d\overrightarrow{v}$ , we need to multiply $ |d\overrightarrow{v}|$ by $ v_y/v$ . Similar reasoning holds for $ y$ -component of $ d\overrightarrow{v}$ , so we arrive at the two equations $ \dot{v_x}= \frac{vv_y \sin \alpha}{l_t} \ \ldots (3)$ $ \dot{v_y}=- \frac{vv_x \sin \alpha}{l_t} \ \ldots (4)$

Now, we know that $ l_t =\{ l-v(1-\cos \alpha) t \}$ . Multiplying the above equations (3) and (4) by $ l_t$ , and integrating from the initial to final times, yields $ v_{x,0}l+v(1-\cos \alpha)X=v \sin \alpha \, Y \ \ldots (5)$ $ v_{y,0}l+v(1-\cos \alpha)Y=-v \sin \alpha \, X \ \ldots (6)$

where (X,Y) is the total displacement vector and $ (v_{x,0},v_{y,0})$ is the initial velocity vector. Putting all the X and Y terms on the right sides, and squaring and adding the equations, we get $ l^2v^2=(X^2+Y^2)(v^2 \sin^2 \alpha +v^2{(1-\cos \alpha)}^2). \ \ldots (7)$ Therefore , the net displacement is

$ R=\sqrt{X^2+Y^2}=\dfrac{l}{\sqrt{2(1-\cos \alpha)}}=\dfrac{l/2}{\sin (\alpha/2)} \ \ldots (8)$

To find the exact location, we can, without loss of generality, set $ v_{x,0} =0$ , in which case we find $ Y/X=(1-\cos \alpha)/\sin \alpha =\tan \alpha/2$ . This agrees with the result of the first solution. $ \Box$

#### Part II:

##### SLICK METHOD

Let $ A(t)$ and $ B(t)$ be the positions of the fox and the rabbit respectively. Let $ C(t)$ be the foot of the perpendicular dropped from $ A$ to the line of the rabbit’s path. Let $ \alpha_t$ be the angle, dependent to the time, at which the rabbit moves relative to the direction directly away from the fox (so at $ t=0, \ \alpha_0=\alpha$ and at $ t=\infty , \ \alpha_{\infty}=0$ ).

The speed at which the distance AB decreases is equal to $ v-v \cos \alpha_t$ . Therefore, the sum of the distances AB and CB doesn’t change. Initially, the sum is $ l+l \cos \alpha$ and in the end , it is $ 2d$ where $ d$ is desired eventual separation. Therefore, the desired eventual separation

$ d=\dfrac{l(1+\cos \alpha)}{2} \ \ldots (9)$

##### STRAIGHT FORWARD METHOD

Let $ \alpha_t$ be defined as in the first solution, and let $ l_t$ be the separation at time $ t$ . The speed of the rabbit in the direction orthogonal to the line connecting the two animals is $ v \sin \alpha_t$ . The separation is $ l_t$ , so the angle $ \alpha_t$ changes at a rate $ \dot{\alpha_{t}}= – \dfrac{v \sin \alpha_t}{l_t} \ \ldots (10)$ . And $ l_t$ changes at a rate $ \dot{l_t}=-v(1-\cos \alpha_t) \ \ldots (11)$ .

Taking the quotient of the above two equations, separating variables, gives a differential equation $ \dfrac{\dot{l_t}}{l_t} = \dfrac {\dot{\alpha_t}(1-\cos \alpha_t)}{\sin \alpha_t} \ \ldots (12)$ which on solving gives $ \ln (l_t) = -\ln {(1+\cos \alpha_t)} + \ln (k) \ \ldots (13)$ . Where $ k$ is the constant of integration. Which gives $ k=l_t (1+\cos \alpha_t) \ \ldots (14)$ . Applying initial conditions $ k_0 = l_0 (1+\cos \alpha_0)= l(1+\cos \alpha) \ \ldots (15)$ . Therefore from (14), we get $ l(1+\cos \alpha) = l_t(1+\cos \alpha_t)$ , or $ l_t= \dfrac{l(1+\cos \alpha)}{1+\cos \alpha_t}$ .

Setting $ t= \infty$ and using $ \alpha_{\infty} =0$ , gives the final result $ l_{\infty} = \dfrac{l(1+\cos \alpha)}{2}$ . $ \Box$ .

#### Remark:

The solution of Part II is valid for all $ \alpha$ except $ \alpha= \pi$ . If $ \alpha = \pi$ , the rabbit run directly towards the fox and they will meet halfway in time $ l/2v$ .