벡터와 공간 전체 문제

radius of curvature is $1/45$.

$x^2+(y-1/2)^2 = 1/4$, or $\vec c(t) = \left\langle 1/2\cos t, 1/2\sin t+ 1/2\right\rangle$

$(x-8/3)^2+y^2 = 1/9$, or $\vec c(t) = \left\langle \frac13\cos t+\frac83, \frac13\sin t\right\rangle$

$x^2+(y+8)^2 = 81$, or $\vec c(t) = \left\langle 9\cos t, 9\sin t-8\right\rangle$

$(x-1/2)^2+(y-1/2)^2 = 1/2$, or $\vec c(t) = \left\langle \frac{\sqrt{2}}{2}\cos t+\frac12, \frac{\sqrt{2}}{2}\sin t+\frac12\right\rangle$

(a) The point $(x,y,z)$ satisfies $x^2 +y^2+z^2= 2x-4y+4$ if and only if it satisfies $x^2-2x +y^2+4y+z^2= 4$, or equivalently $(x-1)^2 +(y+2)^2+z^2=9$. Since $\sqrt{(x-1)^2 +(y+2)^2+z^2}$ is the distance from $(1, -2, 0)$ to $(x,y,z)$, our point satisfies the given equation if and only if its distance from $(1,-2,0)$ is three. So the set is the sphere of radius 3 centered on $(1,-2,0)$.

(b) As in part (a), $x^2 +y^2+z^2< 2x-4y+4$ if and only if $(x-1)^2 +(y+2)^2+z^2<9$. Hence our point satifies the given inequality if and only if its distance from $(1,-2,0)$ is strictly smaller than three. The set is the interior of the sphere of radius 3 centered on $(1,-2,0)$.

(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.

(a) The $z$ coordinate of any point is the signed distance from the point to the $xy$-plane. So the distance from $(2,1,3)$ to the $xy$-plane is $|3|=3$.

(b) The $y$ coordinate of any point is the signed distance from the point to the $xz$-plane. So the distance from $(2,1,3)$ to the $xz$-plane is $|1|=1$.

(c) The distance from $(2,1,3)$ to $(x,0,0)$ is

$$\sqrt{(2-x)^2 + (1-0)^2 + (3-0)^2} = \sqrt{(x-2)^2 + 10}$$

(d) Since $(x-2)^2\ge 0$, the distance $\sqrt{(x-2)^2 + 10}$ is minimized when $x=2$. Alternatively,

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \sqrt{(x-2)^2 + 10} =\frac{x-2}{\sqrt{(x-2)^2 + 10}} =0 &\iff x=2 \end{aligned}$$

So the point on the $x$-axis that is closest to $A$ is $(2,0,0)$.

(e) As $(2,0,0)$ is the point on the $x$-axis that is nearest $(2,1,3)$, the distance from $A$ to the $x$-axis is

$$\sqrt{(2-2)^2 + (1-0)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{10}$$

Call the centre of the circumscribing circle $(\bar x,\bar y)$. This centre must be equidistant from the three vertices. So

$$\bar x^2+\bar y^2=(\bar x-a)^2+\bar y^2=(\bar x-b)^2+(\bar y-c)^2$$

or, subtracting $\bar x^2+\bar y^2$ from the three equal expressions,

$$0=a^2-2a\bar x=b^2-2b\bar x+c^2-2c\bar y$$

which implies

$$\bar x=\frac{a}{2}\qquad\qquad \bar y =\frac{b^2+c^2-2b\bar x}{2c}=\frac{b^2+c^2-ab}{2c}$$

The radius is the distance from the vertex $(0,0)$ to the centre $(\bar x,\bar y)$, which is $\sqrt{\big(\frac{a}{2}\big)^2+\big(\frac{b^2+c^2-ab}{2c}\big)^2}$.

The distance from $P$ to the point $(0,0,1)$ is $\sqrt{x^2+y^2+(z-1)^2}$. The distance from $P$ to the specified plane is $|z+1|$. Hence the equation of the surface is

$$x^2+y^2+(z-1)^2=(z+1)^2\text{ or } x^2+y^2=4z$$

All points on this surface have $z\ge 0$. The set of points on the surface that have any fixed value, $z_0\ge 0$, of $z$ consists of a circle that is centred on the $z$–axis, is parallel to the $xy$-plane and has radius $2\sqrt{z_0}$. The surface consists of a stack of these circles, starting with a point at the origin and with radius increasing vertically. The surface is a paraboloid and is sketched below.

Let $(x,y,z)$ be a point in $P$. The distances from $(x,y,z)$ to $(3,-2,3)$ and to $(3/2,1,0)$ are

$$\sqrt{(x-3)^2+(y+2)^2+(z-3)^2}\quad\text{ and }\quad \sqrt{(x-3/2)^2+(y-1)^2+z^2}$$

respectively. To be in $P$, $(x,y,z)$ must obey

$$\begin{aligned} \sqrt{(x-3)^2+(y+2)^2+(z-3)^2}&=2\sqrt{(x-3/2)^2+(y-1)^2+z^2} \\ (x-3)^2+(y+2)^2+(z-3)^2&=4(x-3/2)^2+4(y-1)^2+4z^2 \\ x^2-6x+9+y^2+4y+4+z^2-6z+9&=4x^2-12x+9+4y^2-8y+4+4z^2 \\ 3x^2-6 x+3y^2-12y+3z^2+6z-9&=0 \\ x^2-2 x+y^2-4y+z^2+2z-3&=0 \\ (x-1)^2+(y-2)^2+(z+1)^2&=9 \end{aligned}$$

This is a sphere of radius 3 centered on $(1,2,-1)$.