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DE problem: Dog chasing a rabbit 
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#19
Oct2710, 10:40 PM

P: 4

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! 


#20
Apr1612, 07:45 PM

P: 22

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?



#21
Apr1712, 08:52 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682




#22
May114, 12:54 AM

P: 1

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 [itex]\int^{L}_{x}[/itex][itex]\sqrt{1+(dy/du)^2}du[/itex] and the distance the rabbit has travelled is r. Since they are traveling at the same speed we then have that r=[itex]\int^{L}_{x}[/itex][itex]\sqrt{1+(dy/du)^2}du[/itex]. Now, since the dog is heading straight towards the rabbit, we have that dy/dx = (yr)/x, and so r=yx(dy/dx), implying that yx(dy/dx)=[itex]\int^{L}_{x}[/itex][itex]\sqrt{1+(dy/du)^2}du[/itex]. Differentiating both sides with respect to x gives the differential equation in part a. 


#23
May114, 02:17 AM

P: 366

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. 


#24
May114, 02:34 AM

P: 366

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


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