벡터와 공간 전체 문제

Solution 1: That's too easy. We just guess. The plane $x+y+z=1$ contains all three given points.

Solution 2: The plane does not pass through the origin. (You can see this by just making a quick sketch.) So the plane has an equation of the form $ax+by+cz=1$.

• For $(1,0,0)$ to be on the plane we need that

$$a(1) +b(0) +c(0) =1 \implies a=1$$

• For $(0,1,0)$ to be on the plane we need that

$$a(0) +b(1) +c(0) =1 \implies b=1$$

• For $(0,0,1)$ to be on the plane we need that

$$a(0) +b(0) +c(1) =1 \implies c=1$$

So the plane is $x+y+z=1$.

Solution 3: Both the head and the tail of the vector from $(1,0,0)$ to $(0,1,0)$, namely $\left< -1,1,0\right>$, lie in the plane. Similarly, both the head and the tail of the vector from $(1,0,0)$ to $(0,0,1)$, namely $\left< -1,0,1\right>$, lie in the plane. So the vector

$$\begin{aligned} \left< -1,1,0\right> \times \left< -1,0,1\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix}\right] =\left< 1 , 1 ,1 \right> \end{aligned}$$

is a normal vector for the plane. As $(1,0,0)$ is a point in the plane,

$$\begin{aligned} \left< 1 , 1 ,1 \right> \cdot \left< x-1\,,\,y-0\,,\, z-0 \right> = 0\qquad \text{or}\quad x+y+z=1 \end{aligned}$$

is an equation for the plane.

The vector from $(1,2,3)$ to $(2,3,4)$, namely $\left< 1,1,1\right>$ is parallel to the vector from $(1,2,3)$ to $(3,4,5)$, namely $\left< 2,2,2\right>$. So the three given points are collinear. Precisely, all three points $(1,2,3),\ (2,3,4)$ and $(3,4,5)$ are on the line $\left< x-1,y-2,z-3\right>=t\left< 1,1,1\right>$. There are many planes through that line.

Note $12\left< \frac13,-\frac12,\frac16 \right>=\left< 4,-6,2 \right>$. So, we actually only know one normal direction to the line we're supposed to be describing. That means there are actually infinitely many lines satisfying the given conditions.

There are infinitely many correct answers. One is

$$\begin{aligned} L_1:\ \left< x\,,\,y\,,\,z-1\right> &= t\left< 1,0,0\right> & L_2:\ \left< x\,,\,y\,,\,z-2\right> &= t\left< 0,1,0\right> \\ L_3:\ \left< x\,,\,y\,,\,z-3\right> &= t\left< 1,1,0\right> & L_4:\ \left< x\,,\,y\,,\,z-4\right> &= t\left< 1,-1,0\right> \end{aligned}$$

No two of the lines are parallel because

• $L_1$ and $L_2$ are not parallel because $\left< 1,0,0\right>\times \left< 0,1,0\right>=\left< 0,0,1\right>\ne\mathbf{0}$.
• $L_1$ and $L_3$ are not parallel because $\left< 1,0,0\right>\times \left< 1,1,0\right>=\left< 0,0,1\right>\ne\mathbf{0}$.
• $L_1$ and $L_4$ are not parallel because $\left< 1,0,0\right>\times \left< 1,-1,0\right>=\left< 0,0,-1\right>\ne\mathbf{0}$.
• $L_2$ and $L_3$ are not parallel because $\left< 0,1,0\right>\times \left< 1,1,0\right>=\left< 0,0,-1\right>\ne\mathbf{0}$.
• $L_2$ and $L_4$ are not parallel because $\left< 0,1,0\right>\times \left< 1,-1,0\right>=\left< 0,0,-1\right>\ne\mathbf{0}$.
• $L_3$ and $L_4$ are not parallel because $\left< 1,1,0\right>\times \left< 1,-1,0\right>=\left< 0,0,-2\right>\ne\mathbf{0}$.

No two of the lines intersect because

• every point on $L_1$ has $z=1$ and
• every point on $L_2$ has $z=2$ and
• every point on $L_3$ has $z=3$ and
• every point on $L_4$ has $z=4$.

(a) $\varphi =\frac{\pi}{3}$ is a surface of constant (spherical coordinate) $\varphi$. So it is a cone with vertex at the origin. We can express $\varphi=\frac{\pi}{3}$ in cartesian coordinates by observing that $0\le\varphi\le\frac{\pi}{2}$ so that $z\ge 0$, and

$$\begin{aligned} \varphi=\frac{\pi}{3} \iff \tan\varphi =\frac{\sqrt{3}}{2} \iff \rho\sin\varphi =\frac{\sqrt{3}}{2}\rho\cos\varphi \iff \sqrt{x^2+y^2}=\frac{\sqrt{3}}{2} z \end{aligned}$$

So the picture that corresponds to (a) is (C).

(b) As $r$ and $\theta$ are cylindrical coordinates

$$\begin{aligned} r=2\cos\theta \iff r^2=2 r\cos\theta \iff x^2+y^2 = 2x \iff (x-1)^2+y^2=1 \end{aligned}$$

There is no $z$ appearing in $(x-1)^2+y^2=1$. So every constant $z$ cross–section of $(x-1)^2+y^2=1$ is a (horizontal) circle of radius $1$ centred on the line $x=1$, $y=0$. It is a cylinder of radius $1$ centred on the line $x=1$, $y=0$. So the picture that corresponds to (b) is (F).

(c) Each constant $z$ cross–section of $x^2+y^2=z^2+1$ is a (horizontal) circle centred on the $z$–axis. The radius of the circle is $1$ when $z=0$ and grows as $z$ moves away from $z=0$. So $x^2+y^2=z^2+1$ consists of a bunch of (horizontal) circles stacked on top of each other, with the radius increasing with $|z|$. It is a hyperboloid of one sheet. The picture that corresponds to (c) is (D).

(d) Every point of $y=x^2+z^2$ has $y\ge 0$. Only (B) has that property. We can also observe that every constant $y$ cross–section is a circle centred on $x=z=0$. The radius of the circle is zero when $y=0$ and increases as $y$ increases. The surface $y=x^2+z^2$ is a paraboloid. The picture that corresponds to (d) is (B).

(e) As $\rho$ and $\varphi$ are spherical coordinates

$$\begin{aligned} \rho=2\cos\varphi \iff \rho^2=2\rho\cos\varphi \iff x^2+y^2+z^2=2z \iff x^2+y^2 + (z-1)^2=1 \end{aligned}$$

This is the sphere of radius $1$ centred on $(0,0,1)$. The picture that corresponds to (e) is (A).

(f) The only possibility left is that the picture that corresponds to (f) is (E).

Each solution below consists of three sketchs.

• In the first sketch, we just redraw the given level curves with the $x$- and $y$-axes reoriented so that the sketch looks like we are high on the $z$-axis looking down at the $xy$-plane.
• In the second sketch, we lift up each level curve $f(x,y)=C$ and draw it in the horizontal plane $z=C$. That is we draw

$$\begin{aligned} &\big\{ \ (x,y,z)\ \big|\ f(x,y)=C,\ z=C,\ x\ge 0, y\ge 0\ \big\} \\ &=\big\{ \ (x,y,z)\ \big|\ z=f(x,y),\ z=C,\ x\ge 0, y\ge 0\ \big\} \end{aligned}$$

• Finally, in the third sketch, we draw the part of graph $z=f(x,y)$ in the first octant, just by “filling in the gaps in the second sketch”.

(a)

(b)

We first add into the sketch of the graph the horizontal planes $z=C$, for $C=3$, $2$, $1$, $0.5$, $0.25$.

To reduce clutter, for each $C$, we have drawn in only

• the (gray) intersection of the horizontal plane $z=C$ with the $yz$–plane, i.e. with the vertical plane $x=0$, and
• the (blue) intersection of the horizontal plane $z=C$ with the graph $z=f(x,y)$.

We have also omitted the label for the plane $z=0.25$.

The intersection of the plane $z=C$ with the graph $z=f(x,y)$ is line

$$\big\{ \ (x,y,z)\ \big|\ z=f(x,y),\ z=C\ \big\} = \big\{ \ (x,y,z)\ \big|\ f(x,y)=C,\ z=C\ \big\}$$

Drawing this line (which is parallel to the $x$-axis) in the $xy$-plane, rather than in the plane $z=C$, gives a level curve. Doing this for each of $C=3$, $2$, $1$, $0.5$, $0.25$ gives five level curves.

The arclength from $P$ to $P$ will be 0, so $P$ is the point where $s=0$. That is, $\mathbf{r}(0)$, or $\left( \sin(1/2),\cos(1/2),\sqrt{3}/2\right)$.

• Solution 1: We consider the situation geometrically. If we plot $\mathbf{R}$ in space (of the relevant dimension), regardless of its parametrization, the derivative at a point will give a vector tangent to $\mathbf{R}$, in the direction the curve moves when the parameter is increasing. Since $\mathbf{a}(t_0)$ and $\mathbf{b}(s_0)$ describe the same spot on the curve, $\mathbf{a}'(t_0)$ and $\mathbf{b}'(s_0)$ will be parallel (Since we specified the derivatives are nonzero, there's no messiness about vectors being parallel to a zero vector.) — they're both tangent to the same piece of curve. Furthermore, as $t$ increases, so does $s$, so the direction of increasing $t$ is the same as the direction of increasing $s$. Therefore, A. holds.

Now we consider the magnitudes of the vectors, to rule out E. Recall $|\mathbf{a}'(t)|$ is the speed at which the curve changes relative to $t$; this could be any (nonnegative) number. By the same token, $|\mathbf{b}'(s)|=1$. So, $\mathbf{b}'(s_0)$ is a unit vector, while $\mathbf{a}'(t_0)$ may or may not be. Then the two vectors are not necessarily equal (although they could be).

So, the best answer is A.

• Solution 2: The chain rule gives us a relationship between $\mathbf{b}'(s)$ and $\mathbf{a}'(t)$.

$$\begin{aligned} \frac{\mathrm{d}\mathbf{b}}{\mathrm{d}s}&=\frac{\mathrm{d}}{\mathrm{d}s}[\mathbf{a}(t(s))]=\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}\,\frac{\mathrm{d}t}{\mathrm{d}s} \end{aligned}$$

So, the vectors $\frac{\mathrm{d}\mathbf{b}}{\mathrm{d}s}$ and $\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}$ differ only by the scalar function $\frac{\mathrm{d}t}{\mathrm{d}s}$. So, at any point along the curve, these vectors are parallel.

Furthermore, we know that $t$ and $s$ are positively correlated: as $t$ increases, so does $s$, because we're covering more arclength. So, $\frac{\mathrm{d}t}{\mathrm{d}s}$ is nonnegative. Furthermore, since the derivatives are nonzero, $\frac{\mathrm{d}t}{\mathrm{d}s}$ is nonzero. So, $\mathbf{b}'(s_0)$ and $\mathbf{a}'(t_0)$ are positive scalar multiples of each other. That is, they are parallel, and pointing in the same direction. However, unless $\frac{\mathrm{d}t}{\mathrm{d}s}=1$ (that is, $t(s)=s+C$ for some constant $C$), the vectors do not have the same magnitude, and hence are not equal.

So, A is the best solution.

$\hat{\mathbf{T}}$ 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 $\hat{\mathbf{T}} \times \hat{\mathbf{N}}$.

In this equation, $s$ stands for arclength.

When we take a very small interval from $t$ to $t+h$, the change in arclength $s(t+h)-s(t)$ is approximately $|\mathbf{r}(t+h)-\mathbf{r}(t)|$, because our curve is approximated by a straight line. So, $\frac{s(t+h)-s(t)}{h} \approx \frac{|\mathbf{r}(t+h)-\mathbf{r}(t)|}{h}$, leading to $\frac{\mathrm{d}s}{\mathrm{d}t}=|\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}|=|\mathbf{v}(t)|$.

The magnitude of velocity is speed; in this text we generally call this $v$. That is, $v=|\mathbf{v}(t)|$. This leads to the potentially confusing (but standard) convention that $s$ stands for arclength, while $v$ stands for speed.