DE application with possible Euler steps

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Homework Statement


A mouse starts at the origin and runs up the y-axis with a speed a. At the same time, a cat running with speed b, starts at the point (c,0) and pursues the mouse.

i. What is the path of the cat?
ii. Assume a<b and solve for y(x). How far does the mouse run before being caught by the cat?
iii. Assume a=b and solve for y(x). How close does the cat come to the mouse?

Homework Equations


Circled in red are the steps in the problem that I'm comfortable with (refering to the picture below). The rest of the math was from a friend who conceded that he wasn't sure if he was on the right track. If he was on the right track then I would benefit from somebody explaining why that's the right track and possibly hinting what to do next.

Relevant equations is likely where I'm stuck. I don't know how to approach the cat's movement towards the mouse. I understand that the mouse's position changes the cat's movement, but I don't see how to show that relationship using equations.

The Attempt at a Solution


One classmate suggested that we use Euler's Method, which I haven't used yet, but it doesn't make much sense to me since the cat could start from anywhere on (c,0). All we've done so far in class are first order DEs and Wronskian DEs.

Intuitively I want to say the answer to iii. is what ever the distance is when both the cat and mouse start. Is there a clean way to prove that? (My prof. likes proofs)I haven't taken Linear Algebra, but if you throw any terminology at me from LA I will do my best to find out what you mean.
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Let ##x(t),y(t)## be the coordinates of the cat's position at time ##t## and use overdots to indicate differentiation wrt time. The mouse's position at time ##t## will be ##(0,at)##.

We can write two equations that we can try to solve to find out two unknown functions ##t\mapsto x(t)## and ##t\mapsto y(t)##.

The first equation relates ##\dot x(t)## and ##\dot y(t)## to ##b##. What is that equation?

The second equation makes the direction of the cat's velocity point towards the mouse. At time ##t## the mouse is at ##(0,at)## and the cat is at ##(x(t),y(t))##. Use those to write a formula for the gradient of the line that connects the cat to the mouse at that instant. That is also the direction in which the cat is running at that instant. So write a formula, in terms of ##\dot x## and ##\dot y##, for the gradient of the cat's direction at that instant. Equate the two formulas and you have a second equation.

I can't resist adding that the question is poorly framed. It does not state the crucial assumption that the cat's velocity vector always points towards where the mouse. I'm pretty sure that's what they want. But as stated the problem doesn't rule out the possibility that a clever cat will run in a straight line in such a direction that he will reach a point on the y-axis at the same time as the mouse does - ie he aims above the mouse, rather than at it. That's the quickest way to catch the mouse, and the maths is much, much easier!
 
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The cats position will be c +/- the x component of bt and 0 + the y component of bt. and the mouse will be xm(t)=0 and ym(t)=bt.

xc (t )= c+bx t
yc (t )= by t
I realize i can't just call them b sub x and y but I'm not sure how to show the relationship of the change, will it be something like e raised to the power of 1-b and b?
 
SeattleDrew said:
I realize i can't just call them b sub x and y
In my post the components of the cat's velocity at time ##t## are denoted as ##\dot x(t)## and ##\dot y(t)##.

Using those, try to write the first equation I refer to above.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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