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I've been stuck on this problem of Modified pursuit curve, in which the dog chases the cat with a constant acceleration $a$, starting from rest. The cat moves horizontally with a uniform speed of $v_0$, and the dog wants to catch the cat at all points of time. The dog is at a distance of $h$ from the cat initially.

What I'm confused about:

What would be optimal for the dog, acceleration vector always towards the cat, or something else?

What I did: I assumed acceleration vector always towards the cat, and let the dog catch the cat in time $T$. And this is what I'm getting for the equations: $$v_0 T = \int_{0}^{T}{(v \cos{\alpha}) dt}$$ $$h = \int_{0}^{T}{(v \sin{\alpha}) dt} = \int_{0}^{T}{(v_0 \cos{\alpha} - v)}dt$$ I am a beginner in calculus. So can someone correct me/help me with the equations?

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The dog should want to travel the least distance, so should compute the point it will catch the cat if it travels on a straight line towards that point. I assume the cat starts at the origin and travels along the $+x$ axis, while the dog starts at $(0,h)$ and can travel anywhere that is within the allowed acceleration.

The cat's position at time $t$ is $(v_0t,0)$. In time $t$ the dog can travel $\frac 12at^2$. For the intercept, we must have $\frac 12at^2=\sqrt{(v_0t)^2+h^2}$ because the dog can cover the distance on the left and needs to cover the distance on the right. Solve for $t$, then have the dog follow a straight line to the intercept point. No integral is required.

If the dog always heads at the cat, let $(x(t),y(t))$ be the position of the dog at time $t$. The cat is at $(v_0t,0)$. The dog's speed is $v=at$. The slope of the dog's trajectory is $\frac {dy}{dx}=\frac y{v_0t-x}$. We have $$\frac {dy}{dt}=\frac {dy}{dx}\frac {dx}{dt}=\frac y{v_0t-x}\frac {dx}{dt}\\ at=\sqrt{\left(\frac{dy}{dt}\right)^2+\left(\frac {dx}{dt}\right)^2}\\ at=\frac {dx}{dt}\sqrt {1+\left(\frac y{v_0t-x}\right)^2}$$ but I don't know how to go further

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  • $\begingroup$ Because the dog can calculate the point to head for. It should then accelerate in a straight line towards that point and meet the cat there. If you substitute $z=t^2$ in the equation I gave and square it, you get a quadratic in $z$ which you can feed to the quadratic formula and find the time of the meeting,$t_i$. The dog then heads for $(v_0t_i,0)$ and meets the cat there. $\endgroup$
    – Ross Millikan
    Commented Feb 16, 2019 at 4:42
  • $\begingroup$ I am assuming the cat is required to run along the $+x$ axis at constant velocity. If we know that we can do the calculation I suggest. The problem specifies that. If the cat is free to move differently, the dog's strategy needs to be different. Given that we know how the cat will move, a straight line lets the dog go a shorter distance, which will result in a shorter time to catch the cat. People in sports do that all the time-run to where the opponent will be, not where s/he is. $\endgroup$
    – Ross Millikan
    Commented Feb 16, 2019 at 5:03
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    $\begingroup$ If the dog is running toward the cat you get differential equations for the dog's position as a function of $t$. You calculate the speed at any given time, resolve the vector into $x$ and $y$ components, and that gives you $\frac {dx}{dt}$ and $\frac {dy}{dt}$. The pursuit curve comes up if the dog is too dumb to realize what the cat will do or is worried the cat will change strategy. It is a different and much harder problem. $\endgroup$
    – Ross Millikan
    Commented Feb 16, 2019 at 6:01
  • $\begingroup$ If the dog is dumb, are the differential equations I wrote in the original post correct? And can you please solve them? $\endgroup$
    – kushal
    Commented Feb 16, 2019 at 6:09
  • $\begingroup$ I don't think they are sufficient. You need something to define $\alpha$ as a function of $t$. Please show your work, as it is easier to check than to do. $\endgroup$
    – Ross Millikan
    Commented Feb 16, 2019 at 15:16

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