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Suppose at $t=0$, a rabbit initially at the origin, runs along the x-axis with constant speed $v$. At this same time, a dog, initially 20 meters north of the origin, also runs with constant speed $v$. Assume dog's instantaneous velocity is always directed toward the rabbit's instantaneous velocity. $\mathbf{show \; that \; the \; dog \; never \; gets \; closer \; than \; 10 \; meters \; from \; the \; rabbit}$.

Attempt:

Here is the graph of the situation

enter image description here

Basically, after some time, suppose dog is at poinst $(x,y)$ and let $y = f(x)$ be the path of the dog. since both rabbit and dog has same speed, then they have travelled same distance $\alpha$, which is arclengt of curve $y = f(x)$. thus

$$ \alpha = \int\limits_0^x \sqrt{ 1 + (y')^2} \implies \frac{ d \alpha }{d x } = \sqrt{ 1 + (y')^2 }$$

now since dogs velocity is always directed towards the rabbit, we can find

$$ y' = \frac{ y }{x - \alpha } $$

I think from here I can obtain a DE to find $y=f(x)$, but it seems to be complicated... Am I on the right track?

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ILoveMath
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  • Your question says that the dog's instantaneous velocity is always pointed toward the rabbit's instantaneous velocity. Do you mean to say that the dog's instantaneous velocity is always pointed toward the rabbit? – Hrhm Oct 11 '16 at 12:40

1 Answers1

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Have a look at this paper on pursuit curves. The specific case you describe, where the dog's speed is the same as the rabbit, is solved on page $6$.

Jens
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