# Hungry lioness

1. Aug 6, 2010

### Petr Mugver

A hungry lioness runs after a gazelle. $$v_{L}>v_{G}$$ are the constant magnitudes of the velocities of the two animals. The lioness always directs it's velocity towards the gazelle, the latter always runs in a straight line. Calculate when will the lioness eat the gazelle, as a function of the initial positions of the two animals in the savannah (the horizontal xy plane) and of the direction of the gazelle's speed.

2. Aug 6, 2010

### arildno

Sorry, I can only solve a similar problem where a cat chases a mouse.

You have to help me out here:

SHOW ME YOUR HOMEWORK SO FAR!!!! :grumpy:

3. Aug 6, 2010

### Petr Mugver

...it's about 10 years I don't have to do homework, luckily...:rofl:

Anyway...
...in the rest frame of the gazelle, or the mouse, or the thief...
...in plane polar coordinates $$r,\alpha$$...
...I found...

$$\dot r=-v_L-v_G\cos\alpha$$

$$r\dot\alpha=-v_G\sin\alpha$$

Theese should be the equations of motion.
The first one can be replaced by

$$v_G\ddot\alpha\sin\alpha+v_L\dot\alpha^2=0$$

But I can't go any further!

4. Aug 6, 2010

### arildno

Now, I am unsure as to why polar coordinates would be your salvation here.

I suggest that you first try the following special case:

The direction the gazelle will run in is orthogonal to the line connecting the initial positions of the gazelle and lioness.

Try and set up a set up equations of motion in the ground's rest frame; let for example the gazelle run up along the y-axis.

5. Aug 6, 2010

### Petr Mugver

In the rest frame it should be $$(x_G,y_G)=(0,v_Gt)$$ and

$$\dot x=\frac{-vx}{[x^2+(v_Gt-y)^2]^{1/2}}$$

$$\dot y=\frac{v(v_Gt-y)}{[x^2+(v_Gt-y)^2]^{1/2}}$$

(I suppressed the index L for readability).
I thought this looked awful, that's why I used polar coordinates. But maybe you have got the clue here?

EDIT:

Oh yeah and you can write

$$\frac{\dot x}{x}=\frac{\dot y}{y-v_Gt}$$

that looks better, but I still can't solve!

Last edited: Aug 6, 2010
6. Aug 6, 2010

7. Aug 6, 2010

### arildno

1. This can be written as:
$$x\frac{\dot{y}}{\dot{x}}=(y-v_{G}t)(1)$$

2. Now we have:
$$\frac{\dot{y}}{\dot{x}}=\frac{dy}{dx}(2)$$

Now, we need to eliminate "t" from (1)!
Remember that:
$$\int_{0}^{t}\sqrt{\dot{x}^{2}+\dot{y}^{2}}dt=v_{L}t$$
The integral, can, however, be rewritten, in terms of an arbitrary x as:
$$t=\frac{1}{v_{L}}\int_{x_{0}}^{x}\sqrt{1+(\frac{dy}{dx})^{2}}dx$$
whereby we get:
$$\frac{dt}{dx}=\frac{\sqrt{1+(\frac{dy}{dx})^{2}}}{v_{L}}$$

Now, let us differentiate (1) with respect to x:
$$\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}-\frac{v_{G}}{v_{L}}\sqrt{1+(\frac{dy}{dx})^{2}}$$
which is, at first a separable equation for the variable Y=dy/dx:
$$x\frac{dY}{dx}=-\frac{v_{G}}{v_{L}}\sqrt{1+Y^{2}}$$

8. Aug 6, 2010