I have the following (apparently simple) problem:
A point $P$ is moving downwards with constant speed $|v|=1$ along a vertical line of equation $x = 5\pi$. At time $t=0$, $P$ has cartesian coordinates $(5\pi, 25)$ so obviously at time $t$: $P(5\pi, 25-t)$.
A point $M$ needs to reach point $P$, starting from the origin $O$; at $t=0$, $M(0,0)$. $M$ is constantly adjusting its trajectory so that it is always on the shortest path from $O$ (not $M$) to $P$. So, if $x(t)$ and $y(t)$ are the parametric coordinates of $M$ as functions of time, $y(t) = \frac{(25-t)}{5\pi} x(t)$
$M$ is moving with constant speed $V$ of norm $3$: $\left(\frac{dy(t)}{dt}\right)^2 + \left(\frac{dx(t)}{dt}\right)^2 = 9$.
The question is: when will $M$ reach $P$?
I did not manage to calculate $x(t)$ or $y(t)$.
From your problem, we have the following \begin{align} \frac{dy}{dx} &= \frac{y'(t)}{x'(t)}\\ 1 + \left(\frac{dy}{dx}\right)^2 &= \left(\frac{3}{x'(t)}\right)^2 \quad\Longrightarrow \quad \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \frac{3}{x'(t)}. \end{align} Also \begin{equation} y = \frac{25-t}{5\pi}x \quad \Longrightarrow \quad y'(t) = -\frac{1}{5\pi}x(t) + \frac{25-t}{5\pi} x'(t). \end{equation} Thereofore \begin{align} \frac{dy}{dx} - \frac{y}{x} = -\frac{x}{5\pi} \frac{1}{x'(t)} = -\frac{x}{15\pi}\sqrt{1 + \left(\frac{dy}{dx}\right)^2}. \end{align} Denote $y_x' = \frac{dy}{dx}$ we deduce $$ y'_x - \frac{y}{x} = -\frac{x}{15\pi}\sqrt{1 + \left(y'_x\right)^2}$$ In other words, $$\left(\frac{y}{x}\right)'_x = -\frac{1}{15\pi}\sqrt{1+(y'_x)^2}$$