Path of particle in velocity field

The path of a particle is found by integrating $dx/dt = u$, $dy/dt = v$, $dz/dt = w$ with the initial position at the given time, then eliminating $t$.

Given: $u = 5x$, $v = 15y + 11$, $w = 19t^2$; at $t = 3$ s the particle is at $(4, 6, 2)$.

x-motion

$$\frac{dx}{x} = 5\,dt \;\Rightarrow\; x = 4\,e^{5(t-3)}$$

y-motion

$$\begin{aligned} \frac{dy}{15y + 11} &= dt\\ \ln(15y + 11) &= 15t + C\\ y + \tfrac{11}{15} &= \left(6 + \tfrac{11}{15}\right)e^{15(t-3)} = 6.733\,e^{15(t-3)} \end{aligned}$$

z-motion

$$z = 2 + \frac{19}{3}\left(t^3 - 27\right)$$

Parametric path (parameter $t$):

$$x = 4e^{5(t-3)},\quad y = 6.733e^{15(t-3)} - 0.733,\quad z = 2 + 6.333(t^3 - 27)$$

Eliminating t: from $x$, $e^{5(t-3)} = x/4$ and $t = 3 + \tfrac{1}{5}\ln(x/4)$. Hence

$$y = 6.733\left(\frac{x}{4}\right)^{3} - 0.733 = 0.1052\,x^3 - 0.733$$ $$z = 2 + 6.333\left[\left(3 + \tfrac{1}{5}\ln\tfrac{x}{4}\right)^3 - 27\right]$$

Answer: the path is the curve $y = 0.1052x^3 - 0.733$ in the xy-plane, with $z = 2 + 6.333\,(t^3 - 27)$ and $t = 3 + 0.2\ln(x/4)$. It passes through $(4, 6, 2)$; the particle moves away from the origin as $t$ increases.