Vector Fields & Field Lines - Calculus 3

1. A vector field $\vec{F}(x,y)$ is defined such that its field lines satisfy the differential equation $\frac{dy}{dx} = \frac{x+y}{x-y}$. Which point represents a location where field lines intersect (a singular point of the field)?
   Answer: $(0,0)$ Field lines can only intersect at points where the vector field is zero (singular points). The differential equation $\frac{dy}{dx} = \frac{x+y}{x-y}$ represents $\frac{Q(x,y)}{P(x,y)}$ where $\vec{F} = \langle P, Q \rangle$. We can take $P(x,y) = x-y$ and $Q(x,y) = x+y$. The field is zero when both components are zero: $x-y = 0$ and $x+y = 0$. From the first equation $x = y$, and substituting into the second gives $x + x = 0$, so $x = 0$ and $y = 0$. Therefore, the singular point is at $(0,0)$. At all other given points, the field has a well-defined non-zero direction, so field lines cannot intersect there. Choice A gives an undefined slope (denominator zero but numerator non-zero), indicating a vertical field line, not an intersection.
2. The vector field $\vec{F}(x,y) = \langle 2x, -y \rangle$ has field lines that can be described by the family of curves $x^ay^b = C$ for constants $a$, $b$, and $C$. What are the values of $a$ and $b$?
   Answer: $a = 1, b = 2$ When you encounter vector fields and field lines, remember that field lines are curves whose tangent vectors at every point match the direction of the vector field. This means the slope of a field line at any point $(x,y)$ equals the ratio of the vector field components: $\frac{dy}{dx} = \frac{F_y}{F_x}$. For $\vec{F}(x,y) = \langle 2x, -y \rangle$, the field lines satisfy $\frac{dy}{dx} = \frac{-y}{2x}$. To solve this separable differential equation, separate variables: $\frac{dy}{y} = \frac{-dx}{2x}$. Integrating both sides gives $\ln|y| = -\frac{1}{2}\ln|x| + C_1$. Using properties of logarithms, this becomes $\ln|y| + \frac{1}{2}\ln|x| = C_1$, or $\ln|x^{1/2}y| = C_1$. Exponentiating yields $x^{1/2}y = C$, which we can write as $x^1y^2 = C^2$ (absorbing the constant into a new constant). Therefore, $a = 1$ and $b = 2$. Choice A gives $a = 1, b = -2$, which would produce $xy^{-2} = C$, corresponding to a different differential equation. Choice B gives $a = 2, b = 1$, yielding $x^2y = C$, which doesn't match our solution. Choice D gives $a = -1, b = 2$, producing $x^{-1}y^2 = C$, also incorrect. Only choice C matches our derived relationship. Study tip: Always verify field line equations by checking that their derivatives match the vector field's direction ratio $F_y/F_x$.
3. Consider the vector field $\vec{F}(x,y) = \langle y\cos(xy), x\cos(xy) \rangle$. Near the origin, the field lines most closely resemble which of the following patterns?
   Answer: Lines with slope 1 passing through the origin To analyze the field near the origin, we use the linear approximation. Since $\cos(xy) \approx 1$ for small $xy$, we have $\vec{F}(x,y) \approx \langle y, x \rangle$ near the origin. The slope of field lines is $\frac{dy}{dx} = \frac{x}{y} \approx \frac{x}{y}$ near the origin. This gives $y\,dy = x\,dx$, which integrates to $\frac{y^2}{2} = \frac{x^2}{2} + C$, or $y^2 - x^2 = \text{constant}$. These are hyperbolas, but very close to the origin, they approximate to lines $y \approx \pm x$, which have slope $\pm 1$. Since the field $\langle y, x \rangle$ points in the first and third quadrants for the line $y = x$, the field lines have slope 1. Choice A would require $\vec{F} \perp \vec{r}$, choice B would require $\vec{F} \parallel \vec{r}$, and choice C would require constant $y$-direction.
4. The vector field $\vec{F}(x,y) = \langle f(x,y), g(x,y) \rangle$ has the property that all field lines are parallel to the line $y = 2x + 1$. Which of the following must be true about the components of $\vec{F}$?
   Answer: $g(x,y) = 2f(x,y)$ for all points $(x,y)$ If field lines are parallel to $y = 2x + 1$, then the direction vectors of the field must be parallel to the direction vector of this line, which is $\langle 1, 2 \rangle$. This means $\vec{F}(x,y) = k(x,y)\langle 1, 2 \rangle$ for some scalar function $k(x,y)$. Therefore $f(x,y) = k(x,y)$ and $g(x,y) = 2k(x,y)$, giving $g(x,y) = 2f(x,y)$. Choice B reverses the relationship, choice C incorrectly adds the constant from the line equation, and choice D misinterprets the parallel condition as a dot product constraint.
5. Consider the vector field $\vec{F}(x,y) = \langle -y, x \rangle$. A particle starts at point $(2,0)$ and follows a field line for time $t = \frac{\pi}{2}$. If the particle moves with speed equal to the magnitude of the field at each point, what is the particle's final position?
   Answer: $(0, 2)$ The vector field $\vec{F}(x,y) = \langle -y, x \rangle$ has field lines that are circles centered at the origin. Following a field line means solving $\frac{d\vec{r}}{dt} = \vec{F}(\vec{r})$, which gives $\frac{dx}{dt} = -y$ and $\frac{dy}{dt} = x$. This system has solutions $x(t) = r\cos(t + \phi)$ and $y(t) = r\sin(t + \phi)$. Starting at $(2,0)$ means $r = 2$ and $\phi = 0$, so $x(t) = 2\cos(t)$ and $y(t) = 2\sin(t)$. At $t = \frac{\pi}{2}$, we get $(0, 2)$. Choice B results from incorrect sign in the field equations, choice C from using $t = \pi$, and choice D assumes no movement.

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