벡터 미적분 전체 문제

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 $\mathbf{F}\parallel\hat{\boldsymbol{\imath}}$.

Since $\mathbf{F}$ is conservative, and $A$ and $B$ start and end at the same points, by path-independence $\int_B \mathbf{F}\cdot\mathrm{d}\mathbf{r}=\int_A \mathbf{F}\cdot\mathrm{d}\mathbf{r}=5.$

(a) $P$ is the $x$-component of $\mathbf{F}$. As we travel vertically upward through $A$, that $x$-component decreases. Hence $P_y<0$ at $A$.

(b) $Q$ is the $y$-component of $\mathbf{F}$. As we travel horizontally to the right through $A$, that $y$-component increases. Hence $Q_x>0$ at $A$.

(c) ${ \boldsymbol{\nabla} }\times\mathbf{F} = (Q_x-P_y)\hat{\mathbf{k}}$ and $Q_x-P_y>0$ at $A$, so that the curl of $\mathbf{F}$ at $A$ is in the direction of $+\hat{\mathbf{k}}$.

(d) Along the curve $\mathcal{C}_1$ the magnitude of the angle between $\mathbf{F}$ and $\mathrm{d}\mathbf{r}$ is less than $90^\circ$, so that $\mathbf{F}\cdot\mathrm{d}\mathbf{r}>0$ and $\int_{\mathcal{C}_1}\mathbf{F}\cdot\mathrm{d}\mathbf{r}>0$.

(e) Along the curve $\mathcal{C}_2$ the magnitude of the angle between $\mathbf{F}$ and $\mathrm{d}\mathbf{r}$ is greater than $90^\circ$, so that $\mathbf{F}\cdot\mathrm{d}\mathbf{r}<0$ and $\int_{\mathcal{C}_2}\mathbf{F}\cdot\mathrm{d}\mathbf{r}<0$.

(f) If $\mathbf{F}$ were conservative, we would have $\int_{\mathcal{C}_1}\mathbf{F}\cdot\mathrm{d}\mathbf{r} =\int_{\mathcal{C}_2}\mathbf{F}\cdot\mathrm{d}\mathbf{r}$. As these two integrals have opposite signs $\mathbf{F}$ is not conservative. (Since $\mathbf{F}$ is not conservative, it is not the gradient of some function. At $A$, $P_x>0$ and $Q_y>0$. So $\mathbf{F}$ is not divergence free and is not the curl of a vector potential.)