벡터와 공간 전체 문제

(a) The graph is $z=\sin x$ with $(x,y)$ running over $0\le x\le 2\pi$, $0\le y\le 1$. For each fixed $y_0$ between $0$ and $1$, the intersection of this graph with the vertical plane $y=y_0$ is the same sin graph $z=\sin x$ with $x$ running from $0$ to $2\pi$. So the whole graph is just a bunch of 2-d sin graphs stacked side-by-side. This gives the graph on the left below.

(b) The graph is $z=\sqrt{x^2+y^2}$. For each fixed $z_0\ge 0$, the intersection of this graph with the horizontal plane $z=z_0$ is the circle $\sqrt{x^2+y^2}=z_0$. This circle is centred on the $z$-axis and has radius $z_0$. So the graph is the upper half of a cone. It is the sketch on the right above.

(c) The graph is $z=|x|+|y|$. For each fixed $z_0\ge 0$, the intersection of this graph with the horizontal plane $z=z_0$ is the square $|x|+|y|=z_0$. The side of the square with $x,y\ge 0$ is the straight line $x+y=z_0$. The side of the square with $x\ge 0$ and $y\le 0$ is the straight line $x-y=z_0$ and so on. The four corners of the square are $(\pm z_0,0,z_0)$ and $(0, \pm z_0,z_0)$. So the graph is a stack of squares. It is an upside down four-sided pyramid. The part of the pyramid in the first octant (that is, $x,y,z\ge 0$) is the sketch below.

(a) For each fixed $z_0$, the $z=z_0$ cross-section (parallel to the $xy$-plane) of this surface is an ellipse centered on the origin with one semiaxis of length 2 along the $x$-axis and one semiaxis of length 4 along the $y$-axis. So this is an elliptic cylinder parallel to the $z$-axis. Here is a sketch of the part of the surface above the $xy$–plane.

(b) This is a plane through $(4,0,0)$, $(0,4,0)$ and $(0,0,2)$. Here is a sketch of the part of the plane in the first octant.

(c) For each fixed $x_0$, the $x=x_0$ cross-section parallel to the $yz$-plane is an ellipse with semiaxes $3\sqrt{1+\frac{x_0^2}{16}}$ parallel to the $y$-axis and $2\sqrt{1+\frac{x_0^2}{16}}$ parallel to the $z$-axis. As you move out along the $x$-axis, away from $x=0$, the ellipses grow at a rate proportional to $\sqrt{1+\frac{x^2}{16}}$, which for large $x$ is approximately $\frac{|x|}{4}$. This is called a hyperboloid of one sheet. Its

(d) For each fixed $y_0$, the $y=x_0$ cross-section (parallel to the $xz$-plane) is a circle of radius $|y|$ centred on the $y$-axis. When $y_0=0$ the radius is $0$. As you move further from the $xz$-plane, in either direction, i.e. as $|y_0|$ increases, the radius grows linearly. The full surface consists of a bunch of these circles stacked sideways. This is a circular cone centred on the $y$-axis.

(e) This is an ellipsoid centered on the origin with semiaxes $3$, $\sqrt{12}=2\sqrt{3}$ and $3$ along the $x$, $y$ and $z$-axes, respectively.

(f) Completing three squares, we have that $x^2+y^2+z^2+4x-by+9z-b=0$ if and only if $(x+2)^2+\big(y-\frac{b}{2}\big)^2+\big(z+\frac{9}{2}\big)^2 =b+4+\frac{b^2}{4}+\frac{81}{4}$. This is a sphere of radius $r_b=\frac{1}{2}\sqrt{b^2+4b+97}$ centered on $\frac{1}{2}(-4,b,-9)$.

(g) There are no points on the surface with $x<0$. For each fixed $x_0>0$ the cross-section $x=x_0$ parallel to the $yz$-plane is an ellipse centred on the $x$–axis with semiaxes $\sqrt{x_0}$ in the $y$-axis direction and $\frac{3}{2}\sqrt{x_0}$ in the $z$–axis direction. As you increase $x_0$, i.e. move out along the $x$-axis, the ellipses grow at a rate proportional to $\sqrt{x_0}$. This is an elliptic paraboloid with axis the $x$-axis.

(h) This is called a parabolic cylinder. For any fixed $y_0$, the $y=y_0$ cross-section (parallel to the $xz$-plane) is the upward opening parabola $z=x^2$ which has vertex on the $y$-axis.

Since the level curves are circles centred at the origin (in the $xy$-plane), when $z$ is a constant, the equation will have the form $x^2+y^2=c$ for some constant. That is, our equation looks like $$x^2+y^2=g(z),$$ where $g(z)$ is a function depending only on $z$.

Because our cross-sections are so nicely symmetric, we know the intersection of the figure with the left side of the $yz$-plane as well: $z=3(-y-1)=-3(y+1)$ (when $z\ge0$) and $z=-3(-y-1)=3(y+1)$ (when $z<0$). Below is the intersection of our surface with the $yz$ plane.

Setting $x=0$, our equation becomes $y^2=g(z)$. Looking at the right side of the $yz$ plane, this should lead to: $\left.\begin{cases} z=3(y-1) &\text{if } z\geq 0,\ y\ge 1\\ z=-3(y-1) &\text{if } z< 0,\ y\ge 1 \end{cases}\right\}$. That is:

$$\begin{aligned} |z|&=3(y-1)\\ \frac{|z|}{3}+1&=y\\ \left(\frac{|z|}{3}+1\right)^2&=y^2 &(*) \end{aligned}$$

A quick check: when we squared both sides of the equation in $(*)$, we added another solution, $\frac{|z|}{3}+1=-y$. Let's make sure we haven't diverged from our diagram.

$$\begin{aligned} &&& \left(\frac{|z|}{3}+1\right)^2=y^2\\ &\Leftrightarrow&& \underbrace{\frac{|z|}{3}+1}_{\text{positive}}=\pm y\\ &\Leftrightarrow&& \begin{cases} \frac{|z|}{3}+1 =y & y>0\\[0.05in] \frac{|z|}{3}+1 =-y & y<0 \end{cases}\\[0.1in] & \Leftrightarrow&& \begin{cases} \frac{|z|}{3}+1 =y & y\ge1\\[0.05in] \frac{|z|}{3}+1 =-y & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} |z| =3(y-1) & y\ge1\\[0.05in] |z|=-3(y+1) & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} z =\pm \underbrace{3(y-1)}_{\text{positive}} & y\ge1\\ z=\pm \underbrace{3(y+1)}_{\text{negative}} & y\le-1 \end{cases}\\[0.1in] &\Leftrightarrow&& \begin{cases} z =3(y-1) & y\ge1,\ z\ge0\\ z =-3(y-1) & y\ge1,\ z\le0\\ z=-3(y+1) & y\le-1,\ z \ge 0\\ z=3(y+1) & y\le-1,\ z\le 0 \end{cases} \end{aligned}$$

This matches our diagram eactly. So, all together, the equation of the surface is $$x^2+y^2=\left( \frac{|z|}{3}+1\right)^2$$

The curve is $y=y(x)={x^3}/{3}$. Since $y'(x) = x^2$ and $y''(x) = 2x$, the curvature is

$$\begin{aligned} \kappa (x) &=\frac{\big|\frac{\mathrm{d^2}y}{\mathrm{d}{x}^2}(x)\big|} {\Big[1+\big(\frac{\mathrm{d}y}{\mathrm{d}x}(x)\big)^2\Big]^{3/2}} =\frac{\big|2x\big|} {\big[1+x^4\big]^{3/2}} \end{aligned}$$

We'd like to find the critical points of $\kappa (x)$, but differentiating it looks messy. Since $\kappa (x)$ has only nonnegative values, its maxima correspond the maxima of the function $\kappa ^2(x)$. So, we find the critical points of $\kappa ^2(x)$ instead, to save ourselves some computational toil.

$$\begin{aligned} 0&=\frac{\mathrm{d}}{\mathrm{d}x}\kappa (x)^2 =\frac{\mathrm{d}}{\mathrm{d}x}\frac{4x^2}{{(1+x^4)}^3} =\frac{8x}{{(1+x^4)}^3} - 3\frac{16x^5}{{(1+x^4)}^4} =\frac{8x(1+x^4)-3\times 16x^5}{{(1+x^4)}^4} \\ &=\frac{8x(1-5x^4)}{{(1+x^4)}^4} \end{aligned}$$

Note that $\kappa (0)=0$ and $\kappa (x)\rightarrow 0$ as $x\rightarrow\pm \infty$. So the maximum occurs when $x=\pm 1/\sqrt[4]{5}$.

$\hat{\mathbf{T}}$ is tangent to the curve, while $\hat{\mathbf{N}}$ is perpendicular to it.

Using the right-hand rule and $\hat{\mathbf{B}}=\hat{\mathbf{T}} \times \hat{\mathbf{N}}$, $\hat{\mathbf{B}}$ points out of the page (towards the reader).

To see this, point the fingers of your right hand in the direction of $\hat{\mathbf{T}}$, and curl them inwards until they are in the direction of $\hat{\mathbf{N}}$. To do this, your thumb must be pointing towards you, not away from you. Your thumb shows the direction of $\hat{\mathbf{T}} \times \hat{\mathbf{N}}$.

We don't have enough information to gauge the size of the vectors, but we can figure out their direction. Gravity pulls straight down, so the vector $-mg\hat{\boldsymbol{\jmath}}$ points straight down. The normal force will be normal to the curve.

Solution 1:

For large, negative values of $x$, the wire is closer and closer to a vertical line. If the bead were sliding down a vertical wire, it could do so without even touching the wire, so the force exerted on the bead would be zero. As $x$ approaches 0 from the left, the wire approximates a horizontal line. If the bead were sitting on a horizontal line, the wire would be pushing up to counter gravity. So, we imagine the magnitude of the force exerted by the wire might increase as $x$ increases. That is, $\frac{\mathrm{d}W}{\mathrm{d}x}>0$.

Solution 2:

The net force exerted on the bead is

$$\begin{aligned} F=m\mathbf{a}&= W\hat{\mathbf{N}} -mg\hat{\boldsymbol{\jmath}} \\ \text{We dot both sides with $\hat{\mathbf{N}}$.} \\ W\hat{\mathbf{N}} \cdot \hat{\mathbf{N}} - mg\hat{\boldsymbol{\jmath}}\cdot\hat{\mathbf{N}}&=m\mathbf{a}\cdot\hat{\mathbf{N}} \\ \text{Using the equation $\mathbf{a}(t) = \frac{\mathrm{d}^2s}{\mathrm{d}{t}^2}\hat{\mathbf{T}} + \kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\hat{\mathbf{N}}$,} \\ W-mg\hat{\boldsymbol{\jmath}}\cdot\hat{\mathbf{N}}&=m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\\ W&=mg\hat{\mathbf{n}}\cdot\hat{\mathbf{N}}+m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2\\ &=mg\cos\theta+m\kappa \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^2 \end{aligned}$$

where $\theta$ is the angle between $\hat{\boldsymbol{\jmath}}$ and $\hat{\mathbf{N}}$.

As $x$ moves from a highly negative number to zero, $\theta$ moves from nearly $\pi/2$ to nearly $0$. Therefore $\cos \theta$ increases from nearly zero to nearly one. Then $mg\cos\theta$ is increasing.

Furthermore, as $x$ increases, we see from the picture that the curvature $\kappa$ increases, and speed $\frac{\mathrm{d}s}{\mathrm{d}t}$ increases as well (kinetic energy is increasing as potential energy decreases).

So, $\frac{\mathrm{d}W}{\mathrm{d}x}>0$.