To determine if the curve $c(t) = \left( e^{4t}, \ln(|t|), \frac{1}{t} \right)$ is a flow line of the vector field $F(x, y, z) = (4x, z, -z^2)$, we need to show that $\frac{dr}{dt} = F(r(t))$. Here, $r(t) = c(t)$, and we’ll differentiate $c(t)$ with respect to $t$ and compare it to $F(c(t))$.

1. Differentiate $c(t)$: $c(t) = \left( e^{4t}, \ln(|t|), \frac{1}{t} \right)$ We’ll find $\frac{dc}{dt}$ by differentiating each component of $c(t)$ with respect to $t$:

   • For the $x$-component: $x(t) = e^{4t}$. $\frac{dx}{dt} = \frac{d}{dt} e^{4t} = 4e^{4t}$

   • For the $y$-component: $y(t) = \ln(|t|)$. $\frac{dy}{dt} = \frac{d}{dt} \ln(|t|) = \frac{1}{t}$

   • For the $z$-component: $z(t) = \frac{1}{t}$. $\frac{dz}{dt} = \frac{d}{dt} \left( \frac{1}{t} \right) = -\frac{1}{t^2}$

   Therefore, the derivative $\frac{dc}{dt}$ is: $\frac{dc}{dt} = \left( 4e^{4t}, \frac{1}{t}, -\frac{1}{t^2} \right)$

2. Evaluate $F(c(t))$: Now, we substitute $x = e^{4t}$, $y = \ln(|t|)$, and $z = \frac{1}{t}$ into $F(x, y, z)$: $F(x, y, z) = (4x, z, -z^2)$ Substituting, we get: $F(c(t)) = \left( 4 \cdot e^{4t}, \frac{1}{t}, -\left( \frac{1}{t} \right)^2 \right)$ Simplifying the last component: $F(c(t)) = \left( 4e^{4t}, \frac{1}{t}, -\frac{1}{t^2} \right)$

3. Compare $\frac{dc}{dt}$ and $F(c(t))$: We have: $\frac{dc}{dt} = \left( 4e^{4t}, \frac{1}{t}, -\frac{1}{t^2} \right)$ and $F(c(t)) = \left( 4e^{4t}, \frac{1}{t}, -\frac{1}{t^2} \right)$

   Since $\frac{dc}{dt} = F(c(t))$, this confirms that $c(t)$ is indeed a flow line of the vector field $F(x, y, z)$.

Would you like further clarification on this result or additional details?

Here are some related questions for further exploration:

1. How would the flow line change if the vector field $F(x, y, z)$ were different?
2. What is the physical interpretation of flow lines in fluid dynamics?
3. Can flow lines intersect in a vector field? Why or why not?
4. How would you find the flow line through a different initial point in the same vector field?
5. What are streamlines, and how are they related to flow lines?

Tip: Always check if the vector field’s dimensions and the curve’s dimensions match to ensure the possibility of them being flow lines.