벡터와 공간 전체 문제

(a) $x=y$ is a straight line and passes through the points $(0,0)$ and $(1,1)$. So it is the straight line through the origin that makes an angle $45^\circ$ with the $x$– and $y$–axes. It is sketched in the figure on the left below.

(b) $x+y=1$ is the straight line through the points $(1,0)$ and $(0,1)$. It is sketched in the figure on the right above.

(c) $x^2+y^2$ is the square of the distance from $(0,0)$ to $(x,y)$. So $x^2+y^2=4$ is the circle with centre $(0,0)$ and radius 2. It is sketched in the figure on the left below.

(d) The equation $x^2+y^2=2y$ is equivalent to $x^2+(y-1)^2=1$. As $x^2+(y-1)^2$ is the square of the distance from $(0,1)$ to $(x,y)$, $x^2+(y-1)^2=1$ is the circle with centre $(0,1)$ and radius 1. It is sketched in the figure on the right above.

(e) As in part (d),

$$x^2+y^2<2y \iff x^2+y^2-2y<0 \iff x^2+y^2-2y+1<1 \iff x^2+(y-1)^2<1$$

As $x^2+(y-1)^2$ is the square of the distance from $(0,1)$ to $(x,y)$, $x^2+(y-1)^2<1$ is the set of points whose distance from $(0,1)$ is strictly less than $1$. That is, it is the set of points strictly inside the circle with centre $(0,1)$ and radius 1. That set is the shaded region (not including the dashed circle) in the sketch below.

(a) For each fixed $y_0$, $z=x,\ y=y_0$ is a straight line that lies in the plane, $y=y_0$ (which is parallel to the plane containing the $x$ and $z$ axes and is a distance $y_0$ from it). This line passes through $x=z=0$ and makes an angle $45^\circ$ with the $xy$–plane. Such a line (with $y_0=0$) is sketched in the figure below. The set $z=x$ is the union of all the lines $z=x,\ y=y_0$ with all values of $y_0$. As $y_0$ varies $z=x,\ y=y_0$ sweeps out the plane which contains the $y$–axis and which makes an angle $45^\circ$ with the $xy$–plane. Here is a sketch of the part of the plane that is in the first octant.

(b) $x+y+z=1$ is the plane through the points $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$. Here is a sketch of the part of the plane that is in the first octant.

(c) $x^2+y^2+z^2$ is the square of the distance from $(0,0,0)$ to $(x,y,z)$. So $x^2+y^2+z^2=4$ is the set of points whose distance from $(0,0,0)$ is $2$. It is the sphere with centre $(0,0,0)$ and radius 2. Here is a sketch of the part of the sphere that is in the first octant.

(d) $x^2+y^2+z^2=4$, $z=1$ or equivalently $x^2+y^2=3$, $z=1$, is the intersection of the plane $z=1$ with the sphere of centre $(0,0,0)$ and radius 2. It is a circle in the plane $z=1$ that has centre $(0,0,1)$ and radius $\sqrt{3}$. The part of the circle in the first octant is the heavy quarter circle in the sketch

(e) For each fixed $z_0$, $x^2+y^2=4$, $z=z_0$ is a circle in the plane $z=z_0$ with centre $(0,0,z_0)$ and radius $2$. So $x^2+y^2=4$ is the union of $x^2+y^2=4,\ z=z_0$ for all possible values of $z_0$. It is a vertical stack of horizontal circles. It is the cylinder of radius $2$ centered on the $z$–axis. Here is a sketch of the part of the cylinder that is in the first octant.

(f) For each fixed $z_0\ge 0$, the curve $z = x^2 + y^2,\ z=z_0$ is the circle in the plane $z=z_0$ with centre $(0,0,z_0)$ and radius $\sqrt{z_0}$. As $z = x^2 + y^2$ is the union of $z = x^2 + y^2,\ z=z_0$ for all possible values of $z_0\ge 0$, it is a vertical stack of horizontal circles. The intersection of the surface with the $yz$–plane is the parabola $z=y^2$. Here is a sketch of the part of the paraboloid that is in the first octant.

$\mathbf{a}+\mathbf{b}=\left< 3,1\right>$, $\mathbf{a}+2\mathbf{b}=\left< 4,2\right>$, $2\mathbf{a}-\mathbf{b}=\left< 3,-1\right>$

(a) Each point on the $y$-axis is of the form $(0,y,0)$. Such a point is on the plane $P$ if

$$3(0)+\frac{1}{2}y+0=4 \iff y=8$$

So the intersection of $P$ with the $y$-axis is the single point $(0,8,0)$. (b) Each point on the $z$-axis is of the form $(0,0,z)$. Such a point is on the plane $P$ if

$$3(0)+\frac{1}{2}(0)+z=4 \iff z=4$$

So the intersection of $P$ with the $z$-axis is the single point $(0,0,4)$. (c) The intersection of the plane $P$ with the $yz$-plane is a line. We have shown in parts (a) and (b) that the points $(0,8,0)$ and $(0,0,4)$ are on that line. Here is a sketch of the part of that line that is in the first octant.

(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.

$\hat{\mathbf{T}}$ 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, $\frac{\mathrm{d}W}{\mathrm{d}x}>0$.