There is no ambiguity here. The dog's speed at any time is equal to that of the rabbit.
A few auxiliary variables makes this problem a lot easier. Let
- r(t) be the rabbit's position on the y-axis as a function of time. Vectorially, \mathbf r(t) = r(t) \hat {\mathbf y}
- \mathbf d(t) be the dog's position as a function of time: \mathbf d(t)=x(t)\hat {\mathbf x} + y(t)\hat {\mathbf y}
- \mathbf l(t) be the displacement vector from the dog to the rabbit: \mathbf l(t) = \mathbf r(t) - \mathbf d(t)
- u(t) be the y-component of \mathbf l(t): u(t) = r(t) -y(t)
- l(t) be the distance between dog and rabbit: l(t) = ||\mathbf l(t)|| = \sqrt{x^2(t)+u^2(t)}
Per the problem statement, the dog's velocity vector at some time
t is
\dot {\mathbf d}(t) = \dot r(t) \frac {\mathbf l(t)}{l(t)} = \frac{\dot r(t)}{l(t)} (-x(t)\hat {\mathbf x} + u(t)\hat {\mathbf y})
Thus
\begin{aligned}<br />
\dot x(t) &= -\dot r(t)\,\frac{x(t)}{l(t)} \\<br />
\dot y(t) &= \dot r(t)\frac{u(t)}{l(t)} \\<br />
\dot u(t) &= \dot r(t) - \dot y(t) = \dot r(t)\left(1-\frac{u(t)}{l(t)}\right)<br />
\end{aligned}
Applying the chain rule,
\frac {dy}{dx} = \frac {\dot y(t)}{\dot x(t)} = -\,\frac {u(t)}{x(t)}
The chain rule yields the second derivative of
y wrt
x,
\frac {d^2y}{dx^2} = \frac 1 {\dot x(t)} \frac d{dt}\left(\frac {dy}{dx}\right)<br />
= \frac 1 {\dot x(t)} \frac {\dot x(t) u(t) - \dot u(t) x(t)}{x^2(t)}<br />
= \frac{l(t)}{x^2(t)}<br />
Note that
1 + \left(\frac {dy}{dx}\right)^2 = 1 + \frac {u^2(t)}{x^2(t)} = \frac {l^2(t)}{x^2(t)}
Thus
x(t) \frac {d^2y}{dx^2} = \sqrt{1 + \left(\frac {dy}{dx}\right)^2}
