http://i159.photobucket.com/albums/t121/camilus23/SCAN0873.jpg

Can anyone help me with this? Thanks.

Which part?

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(\frac{d^2y}{dx^2})^2 = 1+(\frac{dy}{dx})^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!!

I'm thinking Laplace Transforms to solve this baby, I'm gonna try it now.

Nevermind on the Laplace transforms, I only know how to solve DE's with them when the DE has constant coefficients. L{y"}=s^2Y(s)-sy(0)-y'(0), but L{xy"}=?????

Question

Show that the dog's path is the graph of the function y = f(x) where y satisfies the differential equation:

xy'' = sqrt [1 + (y')^2]

Solution

At time t, let (x,y) represent the point the dog is at. Let s be the speed. So at time t, the rabbit is (0, st).

The arc length travelled by the dog is
st = int (from x to P) sqrt(1+(y')^2) dx
= - int (from P to x) sqrt(1+(y')^2) dx
(This is the formula for arc length.)

In order for the dog to always run straight towards the rabbit,
y' = (y-st)/(x-0)
xy' = y - st
xy' = y + int (from P to x) sqrt(1+(y')^2) dx
xy'' + y' = y' + sqrt(1+(y')^2)
xy'' = sqrt(1+(y')^2)

Question

Show that the dog's path is the graph of the function y = f(x) where y satisfies the differential equation:

xy'' = sqrt [1 + (y')^2]

Solution

At time t, let (x,y) represent the point the dog is at. Let s be the speed. So at time t, the rabbit is (0, st).

The arc length travelled by the dog is
st = int (from x to P) sqrt(1+(y')^2) dx
= - int (from P to x) sqrt(1+(y')^2) dx
(This is the formula for arc length.)

In order for the dog to always run straight towards the rabbit,
y' = (y-st)/(x-0)
xy' = y - st
xy' = y + int (from P to x) sqrt(1+(y')^2) dx (1)
xy'' + y' = y' + sqrt(1+(y')^2) (2)
xy'' = sqrt(1+(y')^2)

okay I follow you till step (1), but how'd you get from (1) to (2)?

I approached it in a different (less elegant?) way:

Show that the dog's path is the graph of the function y = f(x) where y satisfies the differential equation: x\frac{d^2y}{dx^2} = sqrt {1 + (\frac{dy}{dx})^2}

Consider the dog at an arbitrary position (x,y), and the rabbit at (0,r).
If the dog moves toward the rabbit, then the gradient of the dog's path will be:

f'(x) = \frac{y-r}{x}
Draw a diagram to check this.

Now, consider an instant later when the dog is now at (x + \Delta x, y + \Delta y), and the rabbit is at (0,r + \Delta r)
The gradient of the dog's path is now:

f'(x + \Delta x) = \frac{y + \Delta y - (r + \Delta r)}{x + \Delta x}

Now, you can use those two expressions in the definition of a derivative to get an expression for the second derivative:

f''(x) = \lim_{\Delta x \to 0} \frac {f'(x + \Delta x) - f'(x)}{\Delta x}

But first, you want to get \Delta y and \Delta r in terms of \Delta x. You can do this by noting that:

\frac{\Delta y}{\Delta x} = f'(x) (for small \Delta x), and
\Delta x^2 + \Delta y^2 = \Delta r^2 (because the dog is moving at the same speed as the rabbit)

Now it's just algebra. You start with something pretty ugly (no way I'm going to TeX it, plus this forces you to work through it yourself :)), but it pretty easily reduces to:

f''(x) = \frac sqrt {x^2 + (y-r)^2}{x^2}
Now, rearrange that equation to look like this:

xf''(x) = sqrt {1 + (\frac{y-r}{x})^2}
...and remembering from early on that:

f'(x) = \frac{y-r}{x}
We finish with:

xf''(x) = sqrt {1 + f'(x)^2}

Which is what we want. Hooray!

Wow. What an ugly post. I hope it makes sense!

I knew I could continue to be lazy and rely on you not to be Pete, but still the OP wanted to know if dog catches rabit and if so where?

Let the L=1 ie. dog at (1,0) initially.

Can I continue to be my normal lazy self?

PS - Why did Zeno not state it this way?

The last part of the question is easy, because the the dog and the rabbit are moving at the same speed.
The dog will soon end up on the y-axis behind the rabbit (close enough, anyway. Mathematically, the y-axis is an asymptote to the dog's path.)
Since the dog isn't going faster than the rabbit, it will never catch it unless it starts some way ahead of the rabbit to begin with. It might be interesting to figure out how far behind the rabbit the dog will be when it gets to th x-axis (strictly speaking, what value does the dog-rabbit distance approach as t goes to infinity).

I don't know how to solve the D.E. so I don't think I could answer that question, and I can't help with the second part of the original post.

Parametrically, with \kappa \equiv \frac s L,

x = L\,\text{sech}(\kappa t)
y = st - L\,\tanh(\kappa t)

The resultant curve is appropriately called a hundkurve, aka a tractrix. The hound never catches the rabbit because the distance between the hound and the rabbit is always L.

Proof of the above is left as an exercise to the reader.

The hound never catches the rabbit because the distance between the hound and the rabbit is always L.

Proof of the above is left as an exercise to the reader.

Hmm... I don't think so. I think the distance between the hound and the rabbit will decrease, asymptotically approaching some value between L and zero.

Hmm... I don't think so. I think the distance between the hound and the rabbit will decrease, asymptotically approaching some value between L and zero.Because of my respect for D.H. I am scared to do so, but I agree with you Pete in both your posts (11&9). The dog is only L away at the start and surely decreases the separation with his first step towards the rabbit. D. H. must be wrong.

I think relocation of the dog at T= 0 is an interesting extension of the problem.

For example, if dog were initially at (L, L) and as smart as some of the missiles I have helped evaluate, he could catch the rabbit at (0, L) if he ran straight towards that point. However, this dumb dog, always heading towards the rabbit, will never catch it if starting from the (L, L) start point.

Clearly, at least to me, if the dumb dog's initial position is, for example, (L, 3L) he gets to eat the rabbit, even though dumb. So the question becomes what is the least value of x for which the "dumb dog" still gets his rabbit meal starting from (L, xL)? And where is the capture point (0, y)? Obviously*, y = the length along the dog's curved path to the point (0,y).

Any guesses as to which is greater (or could they be equal?) x or y? I.e. is y > x, or x > y or x = y ?

Over to you, as lazy as ever, Billy T.
------------
*They start together and finish traveling at the same time so with same speed, the distances traveled are identical.

In my prior post I said:
"The dog is only L away at the start and surely decreases the separation with his first step towards the rabbit."

I now note that the separation continues to decrease even after the rabbit passes thru the point (0, 0.5L) but does approach an non-zero limit.

Proof:

Assume an extra dumb dog that runs along the x-axis to where he saw first saw the rabbit. Until he is at (0.5L, 0) and rabbit at (0, 0.5L) their separation has steadly decreasing to: L(root2)/2 =~0.707L but then this "x-axis loving" dumb dog lets it increase as he continues on to the origin along the x-axis where their separation is once again L.

The dumb dog that hates the x-axis and immediatley leaves it will surely get the separation reduced below 0.7L; I bet something like 0.5L is their long term separation, not L as D. H. asserted.

Hmm... I don't think so. I think the distance between the hound and the rabbit will decrease, asymptotically approaching some value between L and zero.From my post 13 proof, your limits are reduced to "between 0.7L and 0."

The dumb dog that hates the x-axis and immediatley leaves it will surely get the separation reduced below 0.7L; I bet something like 0.5L is their long term separation, not L as D. H. asserted.
You are correct, I was incorrect. The dog does not follow a tractrix. The asymptotic limit of the distance between the dog and rabbit is exactly L/2.

Defining some auxiliary variables,
\vec r(t)= rabbit's position as a function of time: \vec r(t)=st\hat y\quad
\vec d(t)= dog's position as a function of time: \vec d(t)=x(t)\hat x + y(t)\hat y\quad
u(t)= y-component of \vec r(t) - \vec d(t): u(t) = st-y(t)\quad
l(t)= distance between dog and rabbit: l(t) = ||\vec r(t) -\vec d(t)|| = \sqrt{x^2(t)+u^2(t)}


From the problem statement, dog's velocity vector is

\dot{\vec d}(t) = s\frac{\vec r(t) - \vec d(t)}{||\vec r(t) - \vec d(t)||}
= \frac s {l(t)} (-x(t)\hat x + u(t)\hat y)

Thus

\dot x = -\,\frac s l x
\dot y = \frac s l u
\dot u = s - \dot y = s - \frac s l u = \frac s l (l-u)

Since l^2=x^2+u^2,

l\dot l = x\dot x + u\dot u = su-\,\frac s l (x^2 + u^2) = su - sl = -s(l-u) = -l \dot u

Solving for \dot l yields \dot l = -\dot u. In other words, l(t)+u(t) is a constant. The initial conditions are l(0)=L and u(0)=0, so l(t)+u(t)\equiv L. As the dog approaches the y-axis, x\to0 and l\to u. Since l(t)+u(t) \equiv L, the asymptotic behavior is that l\to L/2.

Nice work, D H! I doff my hat to you.

u(t)= y-component of \vec r(t) - \vec d(t): u(t) = st-y(t)\quad
l(t)= distance between dog and rabbit: l(t) = ||\vec r(t) -\vec d(t)|| = \sqrt{x^2(t)+u^2(t)}


...

Solving for \dot l yields \dot l = -\dot u.
In other words, l(t)+u(t) is a constant.
...

I found that last result fascinating. It seems like it should mean something.

After thinking about constructing conic sections with a pencil and string, I figured out that the dog's path in the rabbit's reference frame is section of a parabola... but I don't know if that's very interesting, or just trivia.

I found a good discussion of these Pursuit curves (http://mathworld.wolfram.com/PursuitCurve.html) at Mathworld.

I've also just figured out that the hound will never catch the rabbit under any initial conditions, except for x=0, y>=0 (ie already on the rabbit's path).

This is the same stuff that lets a gladiator (or a Christian) keep away from a lion in a circular arena. The lion starts in the centre, and the gladiator at a point on the circumference.

Nice work, D H! I doff my hat to you. ...I've also just figured out that the hound will never catch the rabbit under any initial conditions, except for x=0, y>=0 (ie already on the rabbit's path).I again agree with you Pete (on both the above).

I was wrong to guess that there was an x greater than unity where the dog starting from (L, xL) could catch the rabbit.

The larger x is, the less progress the dog makes to the "y-axis" (line x = 0)with his initial steps. As time progresses the dog will still not reach that y-axis line -"close but no cigar" (nor rabbit). Eventually both dog and rabbit will have the same y value, Y, and dog with very tiny but positive x coordinate. This is just the original problem with dog at (0, Ln) all over again but with a much smaller value Ln < < L, and a new x-axis shifted up by Y from the original y = 0 x-azis.

(Except for the "quantum dog" of course. :D )

This is the same stuff that lets a gladiator (or a Christian) keep away from a lion in a circular arena. The lion starts in the centre, and the gladiator at a point on the circumference.

This case is about the specific situation where the dog and rabbit have the same speed. If the dog (lion) is faster than the rabbit (christian)... there will be blood!

I think it is safe to say (I don't know though how to express it mathematically), that for any
given initial position of the dog (except x=0 and y=0), the dog will never reach coordinate of y=0
(at any given time t), because the gradient of the dog's path curve is always negative.