Interesting one - A Cat and A mouse

[ujjagrawal]

There is a square ABCD with a Cat at A and a mouse at B. The mouse starts walking towards C, while the Cat walks directly towards the mouse. If the Cat walks n times as fast as the mouse, and catches the mouse at C, what is the value of n?

Additionally can you also find what is the equation of the Cat's path?

[hhh3]

wel.. this is too mathematical for me.. any way i'm gonna tak a stab n the dark...

the value of n is 2/pi ??? by any chance?? hi hi...

[bonanova]

The cat clearly travels faster than the mouse.

The mouse's path is a side of the square.

The cat's path length is longer than a diagonal but shorter than two sides.

So n lies between 1.414 and 2.

The slope seems to be somewhat proportional to the path length.

So I'm guessing a catenary. In which case,

n = 1.6161 times as fast as the mouse.

If the cat starts at the origin and the mouse goes from [1,0] to [1,1]

and if the curve is a catenary, the cat's path is given by

y = [cosh(nx)-1]/n

[ndk]

My guess would be that the cat must follow an arc covering a quarter of the circumference of the circle with the radius equal to the length of one side of the square ABCD. Therefore speed would be 1.5708 times that of the mouse.

[ujjagrawal]

No proper solution... apart from few guesses... Still waiting for anyone to solve this one ?

[phil1882]

so clearly the mouse will run in a straight line.

lets say the square is 1 meter and the mouse runs at the speed of 1 meter per sec.

mouse = 1 m/s

the position of the mouse after t seconds has passed would be

mouse = t meters

now the cat starts at 1 meter straight away from the mouse, perpendicular to the direction the mouse is running.

we know the cat catches the mouse after 1 sec.

this suggests to me that after a 1/2 second, the cat is half way to catching the mouse. after 7/8 a second, the cat is 7/8 of the way to catching the mouse. and so on.

so based on this, its sounds like the equation for the motion of the cat is

cat = n*t meters

so velocity of the cat is sqrt(2).

[fabpig]

I agree with Bonanova. I think the apex of the catenary lies at a point on the BD diagonal and the intersection of a line drawn between the midpoints of AB and BC

btw. I don't think negative reputation points are intended for wrong answers: more for rude or insulting statements.

[curr3nt]

n = pi * (2(AB² + BC²))^(1/2) / 4BC

It looks like the cat travels along a quarter of the circumference of a circle/oval. So if AB = 1 and BC = 1 the cat travels pi/2 to the mouse's 1.

[Yoruichi-san]

Defining my coordinate system (and orientating the cat the mouse) so that mouse starts at (s,s), wgere s=length of one side, cat starts at (0,s). Also defining mouse speed = v, and X(t) to be my function of the x position of the cat at time t. At time t, the mouse's distance from its starting point is vt. The cat's distance from it's starting point is X, the velocity of the cat is nv. Hence the distance b/w the cat and the mouse will be sqrt((s-X)²-v²t²).

Then the speed of the cat, in the x-direction, dX/dt = nv²t/sqrt((s-X)²-v²t²)

To be honest, it's been a while since diffEQ, so I don't recognize the equation or way to solve it off the top of my head. Bonanova and fabpig are probably right about it being a catenary. But solve this differential equation, and you will have the function (you can put it in terms of y, if you want, instead of t, by using the relation b/w the vt and y). You know it will take t_f=s/v for the mouse to reach it's destination, so plug that in, put the other side equal to s (the total X-length the cat travels in that time), and solve for n.

[curr3nt]

In order to draw it out I had to make the cat move towards where the mouse was going to be. Since if the cat only moves to where the mouse was it couldn't catch it.

Here is where the cat moves in 1 - 4 and 6 steps.

This is what lead me to the oval/circle path.

[Yoruichi-san]

Defining my coordinate system (and orientating the cat the mouse) so that mouse starts at (s,s), wgere s=length of one side, cat starts at (0,s). Also defining mouse speed = v, and X(t) to be my function of the x position of the cat at time t. At time t, the mouse's distance from its starting point is vt. The cat's distance from it's starting point is X, the velocity of the cat is nv. Hence the distance b/w the cat and the mouse will be sqrt((s-X)²-v²t²).

Then the speed of the cat, in the x-direction, dX/dt = nv(s-X)/sqrt((s-X)²-v²t²)

To be honest, it's been a while since diffEQ, so I don't recognize the equation or way to solve it off the top of my head. Bonanova and fabpig are probably right about it being a catenary. But solve this differential equation, and you will have the function (you can put it in terms of y, if you want, instead of t, by using the relation b/w the vt and y). You know it will take t_f=s/v for the mouse to reach it's destination, so plug that in, put the other side equal to s (the total X-length the cat travels in that time), and solve for n.

Oops, making a correction. I want the x-component, not the y-component. Also, I realized this whole thing might be easier if I put in terms of theta(t), instead of X(t). Will try when I have time.

[mewminator]

Maybe

If the cat takes a perfect parabola in approaching the mouse, assume DC y-axis and DA x-axis, the equation would be y= -x^2 + BC, the shape of the parabola will be a perfect quarter circle.

Assume BC=1m

Then the parabola= 2pi/4 =pi/2= 1.57m then the cat is 1.57 times faster

Edited July 4, 2012 by mewminator

[fabpig]

OK, my thinking is this. I'm assuming the cat moves at constant speed and the square is of unit length.

If the path of the cat was a perfect quarter circle, its halfway point would be where the DB diagonal intersected the arc ("X"). For the cat to be looking at the mouse, the tangent at this point must intersect line BC at point Y. YB has to be 0.5 (halfway point for the mouse) for this to be true.

From Pythagorus, we can see XB is ~0.41 [sqrt(2) - radius 1] XY must be the same (tangent at X intersects AB and BC at 45deg). Therefore length YB is sqrt((XY^2) + (XB^2)) =~ 0.58 which is clearly past halfway for the mouse.

So, to extend my thinking, halfway for the cat must be somewhere on the BD diagonal - and he must be looking at the midpoint of BC (new "Y")(halfway mouse). (This bit is more intuitive) - as the cat turns 90deg in total, at halfway he must've turned 45deg. YXB form an isosceles right triangle with YB length 0.5. Again from pythagarus, XB = sqrt((0.5^2)/2) =~ 0.353. so point X is 1.057 from D which is > radius 1.

Or is that a load of old tosh?

[EventHorizon]

(Hint: don't look for the curve, just examine the distance between them and how it changes with time)

Let the mouse start at (0,w) moving right with speed v.

Let (x,y) be the position of the cat that starts at the origin and moves at speed s.

Let z be the angle from the cat to the mouse (0 being straight right as usual).

Let D be the distance between the cat and the mouse.

dx/dt = s*cos z (eq1)

dD/dt = v*cos z - s (eq2)

rearranging (eq1) gives cos z = (1/s) dx/dt

plugging into (eq2) gives dD/dt = (v/s) dx/dt - s (eq3)

Integrating (eq3) gives D = (v/s) * x - st + c.

At time t=0, D=w and x=0. So w = 0-0+c, so c=w.

This means the equation for the distance between the cat and the mouse is...

D = (v/s) * x - st + w.

We need D=0 when x=vt=w.

0 = (v/s) * w - s*(w/v) + w

0 = (v/s) - (s/v) + 1 (divide by w)

0 = (s/v)^2 - (s/v) -1 (multiply by -s/v, and rearrange)

(s/v) = (-(-1) +/- sqrt(1-4(1)(-1)))/2(1) (quadratic equation)

(s/v) = (1 +/- sqrt(5))/2

Since (1-sqrt(5))/2 is negative.... the ratio is (1+sqrt(5)) / 2 = 1.618033989...

Looks like the cat is golden.

I'll try (again) to find the equation for the path in a bit.

Edit: I thought I'd include the last equation I got, but kept failing to verify. So I don't know it if is correct.

if you have the mouse at (1,0) moving up at a speed of 1 and the cat at the origin moving at speed s (now found to be the golden ratio) this is what I got but couldn't as of yet verify... so it could be way off. If I can verify it, I'll post how I got it.

y = (e^(x/s)+e^(-x/s)-2)/(e^(1/s)+e^(-1/s)-2)

equivalently,

y = (cosh(x/s)-1)/(cosh(1/s)-1)

Which is pretty close to what bonanova guessed in post 3.

[bonanova]

I put the cat at the origin and the mouse at (1,0) headed toward (1,1) in 1000 steps.

I headed the cat toward the midpoint of the mouse's step intervals.

I adjusted the cat's speed iteratively until the cat arrived at (1,1) after 1000 steps.

I got the golden ratio to eight or ten places and transcribed it erroneously in my post.

The number I posted was related somehow to the catenary, and strikingly close.

Not sure About the equation difference with EventHorizon's, but I think we've established

the curve as a catenary. I'm traveling but when I get home I'll post the curve I calculated,

along with our two equations. should be interesting to compare the three trajectories.

[EventHorizon]

if you have the mouse at (1,0) moving up at a speed of 1 and the cat at the origin moving at speed s (now found to be the golden ratio) this is what I got but couldn't as of yet verify... so it could be way off. If I can verify it, I'll post how I got it.

y = (e^(x/s)+e^(-x/s)-2)/(e^(1/s)+e^(-1/s)-2)

equivalently,

y = (cosh(x/s)-1)/(cosh(1/s)-1)

Which is pretty close to what bonanova guessed in post 3.

I found a mistake in the derivation, so my equation is wrong. I accidently omitted something which made it much easier to solve, but incorrect.

[Yoruichi-san]

Defining my coordinate system (and orientating the cat the mouse) so that mouse starts at (s,s), wgere s=length of one side, cat starts at (0,s). Also defining mouse speed = v, and X(t) to be my function of the x position of the cat at time t. At time t, the mouse's distance from its starting point is vt. The cat's distance from it's starting point is X, the velocity of the cat is nv. Hence the distance b/w the cat and the mouse will be sqrt((s-X)²+v²t²).

Then the speed of the cat, in the x-direction, dX/dt = nv(s-X)/sqrt((s-X)²+v²t²)

To be honest, it's been a while since diffEQ, so I don't recognize the equation or way to solve it off the top of my head. Bonanova and fabpig are probably right about it being a catenary. But solve this differential equation, and you will have the function (you can put it in terms of y, if you want, instead of t, by using the relation b/w the vt and y). You know it will take t_f=s/v for the mouse to reach it's destination, so plug that in, put the other side equal to s (the total X-length the cat travels in that time), and solve for n.

Further correction. Somehow I always inadvertently put a - instead of a +...

Haven't had a chance to further work on a solution, though. I'm thinking u-sub or trial and error (guess the general form of the solution and then solve for constants).

[EventHorizon]

Let the mouse start at (0,0) running up.

The cat starts at (1,0).

Let g be the golden ratio.

The equation for the path is as follows:

y = (1/2g) * x^g - (1/(4-2g)) * x^2-g + 1

Derivation:

y' = (t-y)/(0-x)

-xy' = t-y

y-xy' = t

Since t is equal to the arc length divided by the speed of the cat...

y-xy' = (1/g) * definite integral from x to 1 of (sqrt(1+y'²))

gy-gxy' = definite integral from x to 1 of (sqrt(1+y'²))

Take the derivative of both sides (Notice the definite integral above is some constant minus the value of the integral at x, so we'll need to negate it).

gy'-gy'-gxy'' = -sqrt(1+y'²)

0 = sqrt(1+y'²) - gxy''

0 = (1/gx) - (y''/sqrt(1+y'²))

Now integrate...

c = (ln x)/g - sinh^-1(y')

sinh^-1(y') = -c + (ln x)/g

y' = sinh(-c + (ln x)/g)

y' = (e^-cx^1/g - e^cx^-1/g)/2

y' = (ax^1/g - (1/a)x^-1/g)/2 (substituted in a for e^-c)

use y'(1) = 0 to find out a.

0 = (a - (1/a)) / 2, so a = 1.

y' = .5 * (x^1/g - x^-1/g)

Integrate again...

y = .5 * ( (1/(1+1/g)) * x^1+1/g - (1/(1-1/g)) * x^1-1/g ) + c

Find the value for the introduced constant using y(1) = 0.

0 = .5 * ( (1/(1+1/g)) - (1/(1-1/g)) ) + c

0 = .5 * ( (1/(1+g-1)) - (1/(1-(g-1))) ) + c

0 = .5 * ( (1/g) - (1/(2-g)) ) + c

0 = .5 * ( g-1 - g² ) + c

0 = .5 * ( g-1 - (1+g) ) + c

0 = .5 * (-2) + c

c = 1

I substituted in prettier values (1/g = g-1) and distributed the 1/2 through

y = (1/2g) * x^g - (1/(4-2g)) * x^2-g + 1

done.

[fabpig]

I'm out of my depth here, but you guys introducing the Golden Ratio to things made me do a bit more digging (for my own peace of mind) and I came up with this.....

http://en.wikipedia....i/Golden_spiral You've probably encountered it before, but it's new to me. And the more I think about it, the more this seems likely to be the cat's path. So I'm changing my vote from 'catenary' to 'Golden Spiral' !

Does that tie in with your thinking?

[bonanova]

Let the mouse start at (0,0) running up.

The cat starts at (1,0).

Let g be the golden ratio.

The equation for the path is as follows:

y = (1/2g) * x^g - (1/(4-2g)) * x^2-g + 1

Derivation:

y' = (t-y)/(0-x)

-xy' = t-y

y-xy' = t

Since t is equal to the arc length divided by the speed of the cat...

y-xy' = (1/g) * definite integral from x to 1 of (sqrt(1+y'²))

gy-gxy' = definite integral from x to 1 of (sqrt(1+y'²))

Take the derivative of both sides (Notice the definite integral above is some constant minus the value of the integral at x, so we'll need to negate it).

gy'-gy'-gxy'' = -sqrt(1+y'²)

0 = sqrt(1+y'²) - gxy''

0 = (1/gx) - (y''/sqrt(1+y'²))

Now integrate...

0 = d(1/gx)/dx - d(y''/sqrt(1+y'²))/dx

c = (ln x)/g - sinh^-1(y')

sinh^-1(y') = -c + (ln x)/g

y' = sinh(-c + (ln x)/g)

y' = (e^-cx^1/g - e^cx^-1/g)/2

y' = (ax^1/g - (1/a)x^-1/g)/2 (substituted in a for e^-c)

use y'(1) = 0 to find out a.

0 = (a - (1/a)) / 2, so a = 1.

y' = .5 * (x^1/g - x^-1/g)

Integrate again...

y = .5 * ( (1/(1+1/g)) * x^1+1/g - (1/(1-1/g)) * x^1-1/g ) + c

Find the value for the introduced constant using y(1) = 0.

0 = .5 * ( (1/(1+1/g)) - (1/(1-1/g)) ) + c

0 = .5 * ( (1/(1+g-1)) - (1/(1-(g-1))) ) + c

0 = .5 * ( (1/g) - (1/(2-g)) ) + c

0 = .5 * ( g-1 - g² ) + c

0 = .5 * ( g-1 - (1+g) ) + c

0 = .5 * (-2) + c

c = 1

I substituted in prettier values (1/g = g-1) and distributed the 1/2 through

y = (1/2g) * x^g - (1/(4-2g)) * x^2-g + 1

done.

That seems to tie a bow on it.

Below is a plot of

A your catenary equation

B my catenary equation

C your final equation quoted above [plotted vs 1-x to make curves congruent]

D my simulation of a thousand points using n = 1.618033989 to make cat and mouse arrive at [1,1] simultaneously.

The light blue arrow is the mouse's path, at unit speed.

C and D are indistinguishable, lending credence to both and casting doubt on the catenary hypothesis.

[bonanova]

I'm out of my depth here, but you guys introducing the Golden Ratio to things made me do a bit more digging (for my own peace of mind) and I came up with this.....

http://en.wikipedia....i/Golden_spiral You've probably encountered it before, but it's new to me. And the more I think about it, the more this seems likely to be the cat's path. So I'm changing my vote from 'catenary' to 'Golden Spiral' !

Does that tie in with your thinking?

Could very well be.

The above curve probably debunks the catenary hypothesis.

Care to cast the golden spiral inside the unit square? I'll add to the plot.