DE problem: Dog chasing a rabbit

[camilus]

Main Question or Discussion Point

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Can anyone help me with this? Thanks.

 

[HallsofIvy]

Help you do what? You don't appear to have made any effort to do anything yourself.

Here's a starter: Since dog and rabbit run at the same speed, assume a time scale so that is 1. At time t, the rabbit is at (0, t). If the dog is at (x,y), it runs directly at the rabbit so it runs on a curve whose tangent line is the line from (x,y) to (0, t): the slope is (y-t)/x.

dy/dx= (y-t)/x. That has the problem that it still involves "t". Try to eliminate t by by using the fact that length of the vector $x'(t)\vec{i}+ y'(t)\vec{j}$ is 1.

 

[camilus]

I made a small effort and got stuck lol. I wanted to write the DE in linear form but in order to isolate the dy/dx, i would have to square both sides to get it out of the square root. But that means the left side would be x^2 times (d^2y/dx^2)^2, and the second derivative squared messes up the linearity. If I could get rid of it, like if it was x^2 times y-double-prime, i could bring the right over and solve the homogenous equation easily.

but the y-double-prime squared got me stuck!!

 

[HallsofIvy]

Try this. Multiply
$$\frac{dy}{dx}= \frac{y-vt}{x}$$
through by x to get
$$x\frac{dy}{dx}= y- vt$$

Now differentiate both sides of that, with respect to x, to get
$$x\frac{d^2y}{dx^2}+ \frac{dy}{dx}= \frac{dy}{dx}- v\frac{dt}{dx}$$
$$x\frac{d^2y}{dx^2}= -v\frac{dt}{dx}$$

The fact that the "speed", or length of the velocity vector, is 1 means
$$\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}= 1$$
so
$$dx^2+ dy^2= dt^2$$
or
$$1+ \left(\frac{dy}{dx}\right)^2= \left(\frac{dt}{dx}\right)^2$$
$$\frac{dt}{dx}= \sqrt{1+ \left(\frac{dy}{dx}\right)^2}$$
so the differential equation becomes
$$x\frac{d^2y}{dx^2}= -\sqrt{1+\left(\frac{dy}{dx}\right)^2}$$

Oops! I think I just did part a for you!

That's a complicated equation but notice that "y" itself does not appear in it. Let u= dy/dx and you have a first order, separable, differential equation for u.

 

[Defennder]

There's something I don't follow here. Why is the magnitude of the velocity vector 1? Shouldn't that apply only to v/|v|, which is T, the unit tangent vector?

 

[HallsofIvy]

The problem says dog and rabbit run with the same speed and that is a constant. Since one of the conditions is that the dog starts at (L, 0), we are given some distance scale but we are free to choose the time scale so that speed is 1 "distance unit/time unit".

If you prefer, you could call the speed "v" but it will cancel out.

 

[Defennder]

Oh okay thanks. Got it.

 

[RuDyTaBoOtIe]

Hi, i have a similar question that deals with the above problem! I already solved the problem but there is a question that corresponds with it and i am not sure how to do it...

The question i am having problem with is:

a) Suppose that the dog in problem 7 runs twice as fast as the rabbit. find a differential equation for the path of the dog. then solve it to find the point where the dog catches the rabbit.

b) Suppose the dog runs half as fast as the rabbit. how close does the dog get to the rabbit? what are their positions when they are closest?

what i have done so far is use the same process in question 7 by implementing the "twice as fast" part, but i am not sure if this is right...

 

[RuDyTaBoOtIe]

nevermind, i got it

 

[sennyk]

I don't see anywhere in that problem in which it states that the speed of the rabbit, or the dog, is constant. As a matter of fact, there is no negative in front of the radical either. I think you're missing something here.

 

[sennyk]

If you look at the DE that was derived, you notice that it says that f(x) is concave down. It is obviously concave up.

 

[sennyk]

Those darn negative signs.

Halls,

It took me a while, but I finally figured out what is missing. Although, v could be variable, you are correct in saying that it will "cancel" out. Here's the issue with the negative in the final DE.

$$\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}= v$$
$$\left|\frac{dx}{dt}\right|\sqrt{\left(\frac{dy}{dx}\right)^2+1}=v$$

Since dx/dt is negative, the following is true.

$$-\frac{dx}{dt}\sqrt{\left(\frac{dy}{dx}\right)^2+1}=v$$

 

[Defennder]

If we don't assume the speed is constant, but varies as a function of t, then we would have $$-x\frac{d^2y}{dx^2} = v \frac{dt}{dx} + t \frac{dv}{dx}$$ Any way to simplify this to the required DE?

 

[sennyk]

Yeah. "v" is not dependent upon x, so dv/dx is zero. "v" is only dependent upon time.

 

[Defennder]

v is a function of t, and x is also a function of t. Doesn't this mean that v can be regarded as a function of x?

 

[sennyk]

In a pure math sense that is correct. Since the DE is given in the problem statement, one would have to assume that the problem writer implies that the speed of the rabbit is not influenced by the x position of the dog; otherwise, another DE would be given. And another constraint would have to be given. You could chide the problem writer for his/her "ambiguity".

 

[D H]

There is no ambiguity here. The dog's speed at any time is equal to that of the rabbit.

A few auxiliary variables makes this problem a lot easier. Let

• $r(t)$ be the rabbit's position on the y-axis as a function of time. Vectorially, $\mathbf r(t) = r(t) \hat {\mathbf y}$
• $\mathbf d(t)$ be the dog's position as a function of time: $\mathbf d(t)=x(t)\hat {\mathbf x} + y(t)\hat {\mathbf y}$
• $\mathbf l(t)$ be the displacement vector from the dog to the rabbit: $\mathbf l(t) = \mathbf r(t) - \mathbf d(t)$
• $u(t)$ be the y-component of $\mathbf l(t)$: $u(t) = r(t) -y(t)$
• $l(t)$ be the distance between dog and rabbit: $l(t) = ||\mathbf l(t)|| = \sqrt{x^2(t)+u^2(t)}$

Per the problem statement, the dog's velocity vector at some time t is
$$\dot {\mathbf d}(t) = \dot r(t) \frac {\mathbf l(t)}{l(t)} = \frac{\dot r(t)}{l(t)} (-x(t)\hat {\mathbf x} + u(t)\hat {\mathbf y})$$
Thus
$$\begin{aligned} \dot x(t) &= -\dot r(t)\,\frac{x(t)}{l(t)} \\ \dot y(t) &= \dot r(t)\frac{u(t)}{l(t)} \\ \dot u(t) &= \dot r(t) - \dot y(t) = \dot r(t)\left(1-\frac{u(t)}{l(t)}\right) \end{aligned}$$

Applying the chain rule,
$$\frac {dy}{dx} = \frac {\dot y(t)}{\dot x(t)} = -\,\frac {u(t)}{x(t)}$$

The chain rule yields the second derivative of y wrt x,
$$\frac {d^2y}{dx^2} = \frac 1 {\dot x(t)} \frac d{dt}\left(\frac {dy}{dx}\right) = \frac 1 {\dot x(t)} \frac {\dot x(t) u(t) - \dot u(t) x(t)}{x^2(t)} = \frac{l(t)}{x^2(t)}$$

Note that
$$1 + \left(\frac {dy}{dx}\right)^2 = 1 + \frac {u^2(t)}{x^2(t)} = \frac {l^2(t)}{x^2(t)}$$

Thus
$$x(t) \frac {d^2y}{dx^2} = \sqrt{1 + \left(\frac {dy}{dx}\right)^2}$$

 

[Defennder]

That was brilliant, thanks.

 

[tdude]

Could someone also derive the equation in relation to y?

I'm having huge problems with a question of the sort. Hopefully with the steps laid out it would become clear. Thanks!

 

[corey2014]

So how would I do this if the rabbit has 1/2 the velocity? how would i set up an equation for how X, and Y will change?

 

[HallsofIvy]

v is a function of t, and x is also a function of t. Doesn't this mean that v can be regarded as a function of x?

Well, yes. A constant function for this problem.

 

[ppoundstone]

Here's another approach that doesn't assume that the dog and rabbit run at constant speeds (but their speeds are the same).

Suppose that after some time t>0 the dog has arrived at the point (x,y) and the rabbit has arrived at the point (0,r). Then the distance the dog has travelled is given by
$\int^{L}_{x}$$\sqrt{1+(dy/du)^2}du$ and the distance the rabbit has travelled is r. Since they are traveling at the same speed we then have that

r=$\int^{L}_{x}$$\sqrt{1+(dy/du)^2}du$.

Now, since the dog is heading straight towards the rabbit, we have that dy/dx = (y-r)/x, and so r=y-x(dy/dx), implying that

y-x(dy/dx)=$\int^{L}_{x}$$\sqrt{1+(dy/du)^2}du$.

Differentiating both sides with respect to x gives the differential equation in part a.

 

[HomogenousCow]

R(t) denotes the position of the rabbit. It works even if either speeds are not constant.
That minus sign is because I forgot to throw a plus/minus in front of the square root.

 

[HomogenousCow]

oooh..the next part is nice.
Interesting problem.