help!! (the cat, the mouse, and the hole)

K

korean

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Jun 30, 2005
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maybe i should post here since my problem is an advanced problem....sorry if im wrong ill delete this if i need to or if an admin says so cuz i posted in the diffy Q section beforehand.

problem---> the ratio between the running rates (speeds) of a cat and mouse is 5/3. it takes the cat 25 jumps to reach the wall. the distance between the cat and the mouse is (a) 8 or (b) 10 or (c) 12 jumps of the cat. there is a hole at the base of the wall and only the mouse can go through. determine whether the mouse can escape under (a), or (b), or (c).

and so here is somewhat of what i came up with but not finished or i dont see the cannection of what i did.

my stuff----> first of all i started with a general diffy equation like so....

S = meaning intergrate symbol
a = some constant
b = some constant
A = some constant
B = some constant
K = 5/3

so i start of with this....

y(0) = a; x(0)=b; K=5/3 then i make a general equation.

S dy/dt = K S dx/dt gives me

y(t) = Kx(t) + A

rewrite it as using parameters....

y(0)=Kx(0)+A gives me a = Kb+A gives me A = a - Kb

so... now going back to y(t) replacing where needed.

y(t) = Kx(t) + a - Kx(0)
y(t) = K(x(t) - x(0)) + a

replacing parameters were needed i get..

y(t) = K(x(t)-b) + a

y(t) = (5/3)(x(t)-b) + a


***so now i think i got me general solution somewhere, somehow....now all i need help is interpreting the problem and plugging in my variables where needed....basically applicating what i know. lol....i know the theories but i suck at applications.****

thx to everyone that helps me out
 
Your derivation looks ok, kpama. Just be careful not to re-label too many variables.

If you can understand the situation, then you should be fine to derive something similar.

Let's say the cat is initially 25 jumps from the wall. Call the distance (in jumps) from the cat to the wall C. So: C(0) = 25.

The mouse is in front of the cat. Let the distance from the mouse to the wall be M. So: M(0) = 25 minus the initial distance between the two (which we are to choose from 8, 10, or 12 jumps).

Draw a diagram for yourself. We are assuming the animals have negligible length.

The rate of change of C (the speed of the cat), dC/dt, is indeed equal to 5/3 times the rate of change of M (the speed of the mouse).

The question is asking: when M(t) = 0 (the mouse is at the wall), what is C(t) (how far is the cat from the wall)? If the mouse beats the cat, C(t) will still be positive. The cat hits the wall when C(t) = 0. If C(t) is negative, then it has either gobbled up the mouse or smashed through the wall, perhaps both.
 
so ok i did it but i have one question....a little confused but i think i know the right answer i just wanna double check on this.

well i have figured it out and one of the problems comes out to be 0...like you said if C(t)=0 this the cat hits the wall right?? well that is what i came up with on one of the problems and well my question is.....does the mouse make it in the hole or what???? i think it would but when i told my friend (who is in the class) about this he debated me on it and now im confused? he said it could go either way...either the mouse is eatin' or he will make it. at first i thought he was pulling my leg but he made some really good points on it.....i know you all think im stupid if i fall for this one right?? but please i cant seem to make up my mind.

oh and to UNCO thx for helping me out!!!
 
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