벡터장과 흐름선

H

d. $-\text{F}+\text{G}$

a. $\text{F}+\text{G}$

Correct answers should look similar to

Correct answers should look similar to

Correct answers should look similar to

Correct answers should look similar to

The arrows near the point $A$ are pointing to the right, indicating that $P>0$, and upward, indicating that $Q>0$. Moving from left to right near $A$, the vertical component of the arrows is decreasing, indicating that $\frac{\partial Q}{\partial x}<0$. Moving vertically upwards near $A$, the vertical component of the arrows is increasing, indicating that $\frac{\partial Q}{\partial y}>0$.

Beginning as in the text, we note

$\mathbf{v}(x,y)\cdot\hat{\boldsymbol{\imath}}=x^2 \begin{cases} >0 & x \neq 0\\ =0 & x=0 \end{cases}$ and $\mathbf{v}(x,y)\cdot\hat{\boldsymbol{\jmath}}=y \begin{cases} >0 & y > 0\\ =0 & y=0\\ <0 & y<0 \end{cases}$ .

That leads to the following picture:

This gives us a general idea to start with. Refining, we notice that when $x^2>|y|$, then the vector $\mathbf{v}(x,y)$ will be more horizontal than vertical. As we move away from the $y$-axis in a horizontal line, the difference between $x^2$ and $|y|$ grows, so the vectors get more and more horizontal. However, for a fixed value of $x$, vectors farther from the axis will be more vertical than vectors closer to it.

Although ultimately we'll sketch only unit-length vectors, we can still find the direction of $\mathbf{v}(x,y)$ by finding its $x$- and $y$ components.

Note $\mathbf{v}(x,y)\cdot \hat{\boldsymbol{\imath}}$ is the distance from $(x,y)$ to the origin, while $\mathbf{v}(x,y)\cdot\hat{\boldsymbol{\jmath}}$ is the distance from $(x,y)$ to the point $(1,1)$. Both these numbers are always nonnegative. This leads to the following sketch:

When $(x,y)$ is far from the origin, its distance from $(0,0)$ is almost the same as its distance from $(1,0)$. So, we expect $\mathbf{v}(x,y)$ to be approximately a scalar multiple of $(1,1)$.

At $(0,0)$, $v(0,0)\cdot \hat{\boldsymbol{\imath}}=0$, so our vector is horizontal; similarly, $v(1,1)\cdot\hat{\boldsymbol{\jmath}}=0$ so this vector is horizontal. Vectors very near to $(0,0)$ are nearly horizontal, while vectors near to $(1,1)$ are nearly vertical.

For the direction field, we normalize our vectors to have unit length.

The field $\mathbf{v}(x,y)$ is the sum, scaled by 1/3, of the unit vector pointing away from the origin and the unit vector pointing away from $(1,0)$. This tells us about a few regions:

• Along the $x$ axis between $(0,0)$ and $(1,0)$, the vectors away from these points are pointing in opposite directions (and have the same length), so they cancel each other out. That is, $v(x,0)=0$ for all $x \in (0,1)$.
• $v(0,0)$ and $v(1,0)$ are not defined.
• Along the $x$-axis outside of $[0,1]$, the vector pointing away from the point $(0,0)$ is the same as the vector pointing away from the point $(1,0)$. So, $v(x,0)=(-2/3,0)$ for $x<0$ and $v(x,0)=(2/3,0)$ for $x>1$.

• As the distance from $(x,y)$ to the origin grows, the vector pointing away from $(0,0)$ looks more and more like the vector pointing away from $(1,0)$. So, our vectors far away from the origin look like vectors of length about 2/3, pointing away from the origin.

A particle of unit mass at position $(x,y)$ has distance $D_1=\sqrt{x^2+y^2}$ from the 5kg mass, so that mass exerts a force of magnitude $\frac{G(5)}{x^2+y^2}$ on the particle. This force has direction $(-x,-y)$. So, the force exerted by the 5kg mass is $\mathbf{f}_1(x,y)=\frac{-5G}{(x^2+y^2)^{3/2}}(x,y)$.

Similarly, the 3 kg mass at $(2,3)$ exerts a force of $\mathbf{f}_2(x,y)=\frac{3G}{((x-2)^2+(y-3)^2)^{3/2}}(2-x,3-y)$; and the 7 kg mass at $(4,0)$ exerts a force of $\mathbf{f}_3(x,y)=\frac{7G}{((x-4)^2+y^2)^{3/2}}(4-x,-y)$.

The net force on a unit mass is therefore

$$\begin{aligned} \mathbf{f}(x,y)&=\mathbf{f}_1(x,y)+\mathbf{f}_2(x,y)+\mathbf{f}_3(x,y)\\ &=\frac{-5G(x,y)}{(x^2+y^2)^{3/2}}+\frac{3G(2-x,3-y)}{((x-2)^2+(y-3)^2)^{3/2}}+\frac{7G(4-x,-y)}{((x-4)^2+y^2)^{3/2}} \end{aligned}$$

a. Consider a point $P$ on the pole that is a distance $p$ away from the bottom end. Use this point to make a smaller right triangle, as in the picture below.

Using similar triangles:

$$\begin{aligned} h&=\frac{p}{2}H & b&=\frac{p}{2}\sqrt{4-H^2} \\ \text{If $P$ is at position $(x,y)$, then:} \\ y&=h=\frac{p}{2}H & x&=\sqrt{4-H^2}-b=\left(1-\frac{p}{2}\right)\sqrt{4-H^2}\\ \frac{\mathrm{d}y}{\mathrm{d}t}&=\frac{p}{2}\frac{\mathrm{d}H}{\mathrm{d}t}=-\frac{p}{4} &\frac{\mathrm{d}x}{\mathrm{d}t}&=\left(1-\frac{p}{2}\right)\frac{-H}{\sqrt{4-H^2}}\frac{\mathrm{d}H}{\mathrm{d}t}=\left(1-\frac{p}{2}\right)\frac{H}{2\sqrt{4-H^2}} \\ \text{When $H=1$:} \\ \left.\frac{\mathrm{d}y}{\mathrm{d}t}\right|_{H=1}&=-\frac{p}{4} &\left.\frac{\mathrm{d}x}{\mathrm{d}t}\right|_{H=1}&=\left(1-\frac{p}{2}\right)\frac{1}{2\sqrt{3}} \end{aligned}$$

Therefore, $$\mathbf{v}(p)=\left.\left( \frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t}\right)\right|_{H=1}=\left( \left(1-\frac{p}{2}\right)\frac{1}{2\sqrt{3}}~,~-\frac{p}{4} \right)$$

For our model, we set the domain of this function to be $[0,2]$. b. Let's start by seeing what we can salvage from our work on part a. As in part a., consider a point $P$ on one of the poles, $p$ metres from the bottom end.

Let $P$ have position $(x,y,z)$. Noting that $\frac{\mathrm{d}H}{\mathrm{d}t}$ is now positive, not negative, if we stick to this two-dimensional slice, $$\mathbf{V}(p)=\left( \left(1-\frac{p}{2}\right)\frac{-1}{2\sqrt{3}}~,~\frac{p}{4} \right)$$ where the second coordinate is $z$ and the first coordinate refers to the (horizontal) line in the direction of the vector $(x,y,0)$.

So, we know $\displaystyle\left.\frac{\mathrm{d}z}{\mathrm{d}t}\right|_{H=1}=\frac{p}{4}$, and we know $\displaystyle\left.\left(\frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t}\right)\right|_{H=1}=(x,y)c$ for some negative constant $c$ with $|(x,y)c|=\left(1-\frac{p}{2}\right)\frac{1}{2\sqrt{3}}$. Since we have the direction and the magnitude of the vector, we can find the vector: $$\left.\left( \frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t}\right)\right|_{H=1}=(x,y)c=-\frac{\left(1-\frac{p}{2}\right)}{2\sqrt 3\sqrt{x^2+y^2}}(x,y)$$

We want our equation to be in terms of $x$, $y$, and $z$, so we need to get rid of $p$. Using similar triangles, $\frac{p}{2}=\frac{\sqrt{4-H^2}-\sqrt{x^2+y^2}}{\sqrt{4-H^2}}$. When $H=1$, then $1-\frac{p}{2}=\frac{\sqrt{x^2+y^2}}{\sqrt3}$. So:

$$\left.\left( \frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t}\right)\right|_{H=1} =-\frac{1}{6}(x,y)$$

Finally: $$\mathbf{V}(x,y,z)=\left.\left( \frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t},\frac{\mathrm{d}z}{\mathrm{d}t}\right)\right|_{H=1}=\left( -\frac{1}{6}x~,~ -\frac{1}{6}y~,~ \frac{1}{2}z\right)$$

Not all values of $(x,y,z)$ are on the frame. But, for those values of $(x,y,z)$ that are on the frame, this equation holds.

(a) At every point of the positive $y$-axis, the velocity vector $\mathbf{v}(0,y)$ points straight down. So a rubber ducky placed in the water at $(0,2)$ just floats straight down the positive $y$-axis towards the origin.

(b) At every point of the positive $x$-axis, the velocity vector $\mathbf{v}(x,0)$ points straight to the right. So a rubber ducky placed in the water at $(1,0)$ just floats rightward along the positive $x$-axis.

(c) At every point of the first quadrant away from the axes, the velocity vector $\mathbf{v}(x,y)$ points downwards and towards the right. So a rubber ducky placed in the water at $(1,2)$ always floats down and to the right. The closer the ducky gets to the $x$–axis the more rightward its motion becomes.

The field lines obey

$$\begin{aligned} &\frac{\mathrm{d}x}{x}=\frac{\mathrm{d}y}{3y}\qquad\text{ if $x,y\ne 0$}\\ &\implies 3\ln |x|=\ln |y|+C\\ &\implies |x|^3=e^C|y| \\ &\implies y=\pm e^{-C}x^3 \\ &\implies y=C'x^3 \end{aligned}$$

with $C'$ a nonzero constant. $x=0$ and $y=0$ are also field lines, since on the $y$-axis $\mathbf{F}\parallel\hat{\boldsymbol{\jmath}}$ and on the $x$-axis $\mathbf{F}\parallel\hat{\boldsymbol{\imath}}$.