15. 곡선의 매개화Parametrizations of Curves

Solution. (1) For $x=\cos t$, $y=\sin t$ $(0 \lt t \lt 2\pi)$, we have $x^2+y^2=\cos^2t+\sin^2t=1$, so the path traced by $P$ lies on the unit circle $x^2+y^2=1$. The point $(1,0)$, corresponding to $t=0,2\pi$, is technically excluded, but as $t\to0^+$ and $t\to2\pi^-$ the point approaches $(1,0)$ arbitrarily closely, so the trace is essentially the entire unit circle. As $t$ increases, $P$ starts near $(1,0)$, goes counterclockwise once around the circle, and returns near $(1,0)$.

(2) For $x=\cos2t$, $y=\sin2t$ $(0 \lt t \lt 2\pi)$, again $x^2+y^2=\cos^22t+\sin^22t=1$, so the trace is the same unit circle as in (1). Setting $\varphi=2t$, as $t$ ranges over $(0,2\pi)$, $\varphi$ ranges over $(0,4\pi)$, so $P$ goes counterclockwise around the unit circle twice. That is, the trace (as a set of points) is the same as in (1), but the angular speed is doubled.

Solution. (1) For $\displaystyle x=t+\frac1t$, $\displaystyle y=t-\frac1t$ $(t\gt0)$, $$x^2-y^2=\left(t+\frac1t\right)^2-\left(t-\frac1t\right)^2=\left(t^2+2+\frac1{t^2}\right)-\left(t^2-2+\frac1{t^2}\right)=4,$$ so $P$ lies on the hyperbola $x^2-y^2=4$. By AM–GM, $x=t+\dfrac1t\ge2$ for $t\gt0$, with equality at $t=1$, so the trace is the right branch ($x\ge2$). Also $\dfrac{dy}{dt}=1+\dfrac1{t^2}\gt0$, so $y=t-\dfrac1t$ is increasing in $t$, with $y\to-\infty$ as $t\to0^+$ and $y\to\infty$ as $t\to\infty$. Hence as $t$ increases, $P$ moves up the right branch from below, through the vertex $(2,0)$ (at $t=1$), to above.

(2) For $x=\sqrt{4+t^2}$, $y=t$ ($t$ any real number), $x^2=4+t^2=4+y^2$, so $x^2-y^2=4$, and $x=\sqrt{4+t^2}\ge2$ (equality at $t=0$), so the trace is again the right branch. Since $y=t$ increases outright, as $t$ increases from $-\infty$ to $\infty$, $P$ moves up the same right branch as in (1), from below through the vertex $(2,0)$ to above.

(3) For $x=2\sec t$, $y=2\tan t$ $(-\pi/2\lt t\lt\pi/2)$, $$\frac{x^2}{4}-\frac{y^2}{4}=\sec^2t-\tan^2t=1,$$ so $x^2-y^2=4$. On this interval $\sec t\ge1$ (equality at $t=0$), so $x\ge2$, and $\tan t$ increases from $-\infty$ to $\infty$, so $y=2\tan t$ also increases. As in (1),(2), $P$ moves up the right branch.

(4) For $x=2\cosh t$, $y=2\sinh t$ ($t$ any real number), by the hyperbolic identity $\cosh^2t-\sinh^2t=1$, $$\frac{x^2}{4}-\frac{y^2}{4}=\cosh^2t-\sinh^2t=1,$$ so $x^2-y^2=4$. Since $\cosh t\ge1$, $x\ge2$, and since $\sinh t$ is increasing, $y=2\sinh t$ is also increasing, so again $P$ moves up the right branch.

Thus (1)–(4) all trace the same curve — the right branch ($x\ge2$) of the hyperbola $x^2-y^2=4$ — moving upward (from $y\to-\infty$ to $y\to\infty$) as the parameter increases.

Solution. For $x=\sec^2t-1$, $y=\tan t$ $(-\pi/2\lt t\lt\pi/2)$, using the identity $\sec^2t=1+\tan^2t$, $$x=\sec^2t-1=\tan^2t=y^2,$$ so the trace lies on the parabola $x=y^2$. On $-\pi/2\lt t\lt\pi/2$, $y=\tan t$ takes every real value (as $t\to-\pi/2^+$, $y\to-\infty$; as $t\to\pi/2^-$, $y\to\infty$), so the entire parabola $x=y^2\ge0$ is traced. Since $\tan t$ is increasing, $y$ increases as $t$ increases, so $P$ starts from below ($y\to-\infty$), passes through the vertex $(0,0)$ (at $t=0$), and moves to above ($y\to\infty$).

Solution. Let $O$ be the origin. The segment $\overline{AB}$ passes through $A(4,0)$ and $B(0,2)$, so its equation is $$\frac x4+\frac y2=1.$$ If $P(x,y)$ lies in the direction $\theta=\angle POA$ from the origin at distance $r=\overline{OP}$, then $x=r\cos\theta$, $y=r\sin\theta$. Substituting into the line's equation, $$\frac{r\cos\theta}4+\frac{r\sin\theta}2=1 \iff r\left(\frac{\cos\theta}4+\frac{\sin\theta}2\right)=1 \iff r=\frac{4}{\cos\theta+2\sin\theta}.$$ Hence $$x=\frac{4\cos\theta}{\cos\theta+2\sin\theta},\qquad y=\frac{4\sin\theta}{\cos\theta+2\sin\theta}\qquad\left(0\le\theta\le\frac\pi2\right)$$ is a parametrization of $\overline{AB}$. Indeed, $\theta=0$ gives $(x,y)=(4,0)=A$ and $\theta=\pi/2$ gives $(x,y)=(0,2)=B$, so as $\theta$ increases from $0$ to $\pi/2$, $P$ moves from $A$ to $B$.

Solution. (1) Let $(x,y)=(-2+s\cos\theta,\ s\sin\theta)$ $(s\ge0)$ be a point on the ray from $B(-2,0)$ in direction $\theta$. For this point to lie on $C:x^2+y^2=1$, $$(-2+s\cos\theta)^2+(s\sin\theta)^2=1 \iff 4-4s\cos\theta+s^2=1 \iff s^2-4\cos\theta\,s+3=0.$$ Solving for $s$ gives $s=2\cos\theta\pm\sqrt{4\cos^2\theta-3}$. At $\theta=0$ we must get $P=A(1,0)$; taking $s=2\cos0+\sqrt{4-3}=3$ gives $(x,y)=(-2+3,0)=(1,0)=A$, so the $+$ sign is the correct choice: $$s(\theta)=2\cos\theta+\sqrt{4\cos^2\theta-3}.$$ Since $P$ moves in the first quadrant from $A(1,0)$ to the point $(0,1)$ where the circle meets the $y$-axis, we find $\theta$ at $(0,1)$: from $\overrightarrow{BP}=(2,1)$, $\overrightarrow{BA}=(3,0)$, we get $\tan\theta=\dfrac12$. Hence a parametrization is $$x=-2+s(\theta)\cos\theta,\qquad y=s(\theta)\sin\theta,\qquad s(\theta)=2\cos\theta+\sqrt{4\cos^2\theta-3},$$ with $0\le\theta\le\tan^{-1}\dfrac12$.

(2) The line through $P(x,y)$ and $A(1,0)$ is $y=m(x-1)$. Substituting into $x^2+y^2=1$, $$x^2+m^2(x-1)^2=1 \iff (1+m^2)x^2-2m^2x+(m^2-1)=0.$$ Since $x=1$ is one root (the point $A$), using the product of roots $\dfrac{m^2-1}{1+m^2}$, the other root is $$x=\frac{m^2-1}{1+m^2},$$ and $$y=m(x-1)=m\left(\frac{m^2-1}{1+m^2}-1\right)=\frac{-2m}{1+m^2}.$$ Hence a parametrization of $C$ (excluding the point $A$) is $$x=\frac{m^2-1}{m^2+1},\qquad y=\frac{-2m}{m^2+1}\qquad(m\in\mathbb{R}),$$ with $A(1,0)$ corresponding to the limit $m\to\pm\infty$.

Solution. (1) For $x\neq0$, set $t=\dfrac yx$, so $y=tx$; substituting into $y^2=x^3$, $$t^2x^2=x^3 \iff x^2(x-t^2)=0,$$ so for $x\neq0$, $x=t^2$ and $y=tx=t^3$. The case $x=0$ ($y=0$, the origin) is also included at $t=0$, where $x=t^2=0$, $y=t^3=0$. Hence a parametrization is $$x=t^2,\qquad y=t^3\qquad(t\in\mathbb{R}).$$

(2) For $y^2=x^2+x^3=x^2(1+x)$, dividing both sides by $x^2$ (for $x\neq0$) and setting $t=\dfrac yx$, $$t^2=1+x \iff x=t^2-1,\qquad y=tx=t(t^2-1)=t^3-t.$$ The origin $(0,0)$ is also included, at $x=t^2-1=0$, i.e. $t=\pm1$, where $y=t^3-t=0$. Hence a parametrization is $$x=t^2-1,\qquad y=t^3-t\qquad(t\in\mathbb{R}).$$

Solution. Let $C(\theta)$ denote the center of the circle. Since the circle rolls without slipping, while it rotates through angle $\theta$ the center moves the same distance as the arc length $a\theta$ traced along the $x$-axis. Hence $$C(\theta)=(a\theta,\ a).$$

Initially ($\theta=0$), $P$ was at the bottom of the circle, i.e. displaced $(0,-a)$ from the center. As the circle rolls in the positive $x$-direction it rotates clockwise, so the position vector of $P$ relative to the center also rotates clockwise through angle $\theta$. A clockwise rotation by $\theta$ is the map $$(u,v)\ \longmapsto\ (u\cos\theta+v\sin\theta,\ -u\sin\theta+v\cos\theta),$$ so applying it to the initial vector $(0,-a)$ gives $$\big(0\cdot\cos\theta+(-a)\sin\theta,\ -0\cdot\sin\theta+(-a)\cos\theta\big)=(-a\sin\theta,\ -a\cos\theta).$$ Hence the position of $P$ is $$P(\theta)=C(\theta)+(-a\sin\theta,-a\cos\theta)=(a\theta-a\sin\theta,\ a-a\cos\theta),$$ that is, $$x=a\theta-a\sin\theta,\qquad y=a-a\cos\theta.$$ This curve is called a cycloid.

Solution. We argue as in Example 03. Since the circle has radius $a$, when it rolls through angle $\theta$ the center moves a distance $a\theta$, so $$C(\theta)=(a\theta,\ a).$$ Initially ($\theta=0$), $P$ is displaced $(0,\,(a-b)-a)=(0,-b)$ from the center $(0,a)$, i.e. a distance $b$ directly below the center. As the circle rolls in the positive $x$-direction it rotates clockwise, so the position vector of $P$ relative to the center also rotates clockwise through angle $\theta$. Applying the rotation of Example 03 to the initial vector $(0,-b)$, $$\big(0\cdot\cos\theta+(-b)\sin\theta,\ -0\cdot\sin\theta+(-b)\cos\theta\big)=(-b\sin\theta,\ -b\cos\theta).$$ Hence $$P(\theta)=C(\theta)+(-b\sin\theta,-b\cos\theta)=(a\theta-b\sin\theta,\ a-b\cos\theta),$$ that is, $$x=a\theta-b\sin\theta,\qquad y=a-b\cos\theta.$$ This curve is called a trochoid. (When $b=a$ it becomes a cycloid; when $b\lt a$, $P$ lies inside the circle and the trace is flatter.)

Note. With center $(0, R-r)$ the point $P(R,0)$ cannot lie on $C_2$; we correct the center to $(R-r, 0)$.

Proof. (1) Let $O_2(\theta)$ be the center of $C_2$ and $T(\theta)$ the point of tangency between $C_1$ and $C_2$. Since $O_2$ revolves about the origin counterclockwise through angle $\theta$, $$O_2(\theta)=\big((R-r)\cos\theta,\ (R-r)\sin\theta\big),$$ and $T(\theta)$ lies on the line through the origin and $O_2(\theta)$, at distance $R$ from the origin, so $$T(\theta)=(R\cos\theta,\ R\sin\theta).$$ (At $\theta=0$, $O_2(0)=(R-r,0)$ and $T(0)=(R,0)=P$, matching the initial condition.)

Since $C_2$ rolls without slipping, the arc length $R\theta$ traced on $C_1$ from the initial contact $T(0)=P$ to $T(\theta)$ must equal the arc length traced on $C_2$ from $P$ to the current contact point, which means the disk $C_2$ has rotated through angle $\dfrac{R\theta}{r}$ relative to the line $OO_2$. For rolling on the inside, this spin is superposed on the revolution in the opposite (clockwise) sense, so the direction angle of the vector $O_2P$ is that of $O_2T(\theta)$, namely $\theta$, rotated a further $\dfrac{R\theta}{r}$ clockwise: $$\theta-\frac{R\theta}{r}=-\frac{(R-r)\theta}{r}.$$ Hence $$P(\theta)=O_2(\theta)+r\left(\cos\!\left(-\frac{(R-r)\theta}{r}\right),\ \sin\!\left(-\frac{(R-r)\theta}{r}\right)\right),$$ and since cosine is even and sine is odd, $$x=(R-r)\cos\theta+r\cos\!\left(\frac{(R-r)\theta}{r}\right),\qquad y=(R-r)\sin\theta-r\sin\!\left(\frac{(R-r)\theta}{r}\right).$$ This curve is called a hypocycloid.

(2) When $r=\dfrac R4$, $R-r=\dfrac{3R}4$ and $\dfrac{R-r}{r}=3$, so substituting into (1), $$x=\frac{3R}4\cos\theta+\frac R4\cos3\theta,\qquad y=\frac{3R}4\sin\theta-\frac R4\sin3\theta.$$ Using the triple-angle formulas $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\sin3\theta=3\sin\theta-4\sin^3\theta$, $$x=\frac{3R}4\cos\theta+\frac R4(4\cos^3\theta-3\cos\theta)=\frac{3R}4\cos\theta+R\cos^3\theta-\frac{3R}4\cos\theta=R\cos^3\theta,$$ $$y=\frac{3R}4\sin\theta-\frac R4(3\sin\theta-4\sin^3\theta)=\frac{3R}4\sin\theta-\frac{3R}4\sin\theta+R\sin^3\theta=R\sin^3\theta.$$ That is, $x=R\cos^3\theta$, $y=R\sin^3\theta$, so $$x^{2/3}+y^{2/3}=R^{2/3}\cos^2\theta+R^{2/3}\sin^2\theta=R^{2/3}.$$ This is the equation of an astroid, so when $r=R/4$ the path is an astroid.

Note. With center $(0, R+r)$ the point $P(R,0)$ cannot lie on $C_2$; we correct the center to $(R+r, 0)$.

Solution. (With the corrected center $(R+r,0)$, we argue as in Exercise 03-2, except now $C_2$ rolls along the outside of $C_1$.)

Let $O_2(\theta)$ be the center of $C_2$ and $T(\theta)$ the point of tangency. Then $$O_2(\theta)=\big((R+r)\cos\theta,\ (R+r)\sin\theta\big),$$ and $T(\theta)$ lies on the segment from the origin to $O_2(\theta)$, at distance $R$ from the origin, so $$T(\theta)=(R\cos\theta,\ R\sin\theta).$$ (At $\theta=0$, $O_2(0)=(R+r,0)$ and $T(0)=(R,0)=P$, matching the initial condition.)

Since $C_2$ rolls without slipping, the arc $R\theta$ on $C_1$ (from the initial contact $P$ to $T(\theta)$) equals the arc on $C_2$ (from $P$ to the current contact), which means the disk $C_2$ has rotated through angle $\dfrac{R\theta}{r}$ relative to the line $OO_2$. For rolling on the outside, this spin is superposed on the revolution in the same (counterclockwise) sense, so the direction angle of $O_2P$ is the opposite of $O_2T(\theta)$ — i.e. the direction angle $\theta+\pi$ pointing toward the origin — rotated a further $\dfrac{R\theta}{r}$ counterclockwise: $$\theta+\pi+\frac{R\theta}{r}=\pi+\frac{(R+r)\theta}{r}.$$ Hence $$P(\theta)=O_2(\theta)+r\left(\cos\!\left(\pi+\frac{(R+r)\theta}{r}\right),\ \sin\!\left(\pi+\frac{(R+r)\theta}{r}\right)\right)=O_2(\theta)-r\left(\cos\frac{(R+r)\theta}{r},\ \sin\frac{(R+r)\theta}{r}\right),$$ so $$x=(R+r)\cos\theta-r\cos\!\left(\frac{(R+r)\theta}{r}\right),\qquad y=(R+r)\sin\theta-r\sin\!\left(\frac{(R+r)\theta}{r}\right).$$ This curve is called an epicycloid.

Solution. For $x=a\theta-a\sin\theta$, $y=a-a\cos\theta$ $(0\le\theta\le2\pi)$, $$\frac{dx}{d\theta}=a(1-\cos\theta),\qquad \frac{dy}{d\theta}=a\sin\theta.$$

(1) $$\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2=a^2(1-\cos\theta)^2+a^2\sin^2\theta=a^2(2-2\cos\theta)=4a^2\sin^2\frac\theta2$$ (by the half-angle identity $1-\cos\theta=2\sin^2(\theta/2)$), and since $\sin(\theta/2)\ge0$ on $0\le\theta\le2\pi$, $\dfrac{ds}{d\theta}=2a\sin\dfrac\theta2$. Hence the length is $$L=\int_0^{2\pi}2a\sin\frac\theta2\,d\theta=2a\left[-2\cos\frac\theta2\right]_0^{2\pi}=2a\big(-2\cos\pi+2\cos0\big)=2a(2+2)=8a.$$

The area under the curve is $$A=\int_0^{2\pi a}y\,dx=\int_0^{2\pi}(a-a\cos\theta)\cdot a(1-\cos\theta)\,d\theta=a^2\int_0^{2\pi}(1-\cos\theta)^2\,d\theta,$$ and since $(1-\cos\theta)^2=1-2\cos\theta+\cos^2\theta$ with $\displaystyle\int_0^{2\pi}1\,d\theta=2\pi$, $\displaystyle\int_0^{2\pi}\cos\theta\,d\theta=0$, $\displaystyle\int_0^{2\pi}\cos^2\theta\,d\theta=\pi$, $$A=a^2(2\pi-0+\pi)=3\pi a^2.$$

(2) The volume of the solid obtained by revolving the region under the cycloid about the $x$-axis is $$V=\int_0^{2\pi a}\pi y^2\,dx=\pi\int_0^{2\pi}a^2(1-\cos\theta)^2\cdot a(1-\cos\theta)\,d\theta=\pi a^3\int_0^{2\pi}(1-\cos\theta)^3\,d\theta.$$ Since $(1-\cos\theta)^3=1-3\cos\theta+3\cos^2\theta-\cos^3\theta$ and $\displaystyle\int_0^{2\pi}\cos^3\theta\,d\theta=0$ (integral over a full period), $$\int_0^{2\pi}(1-\cos\theta)^3\,d\theta=2\pi-0+3\pi-0=5\pi.$$ Hence $$V=\pi a^3\cdot5\pi=5\pi^2a^3.$$

Solution. The curve $x=\cos\theta$, $y=1+\sin\theta$ $(0\le\theta\le2\pi)$ is a circle of radius $1$ centered at $(0,1)$, and since $y=1+\sin\theta\ge0$, the entire curve lies at or above the $x$-axis (tangent to it at one point). The surface-area formula is $$S=\int2\pi y\,ds,\qquad ds=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}\,d\theta.$$ Since $\dfrac{dx}{d\theta}=-\sin\theta$ and $\dfrac{dy}{d\theta}=\cos\theta$, $$\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2=\sin^2\theta+\cos^2\theta=1,\qquad ds=d\theta.$$ Hence $$S=\int_0^{2\pi}2\pi(1+\sin\theta)\,d\theta=2\pi\Big[\theta-\cos\theta\Big]_0^{2\pi}=2\pi\big[(2\pi-1)-(0-1)\big]=2\pi\cdot2\pi=4\pi^2.$$

Solution. This curve is the limaçon given in polar form by $r=1+2\sin\theta$, which on $0\le\theta\le2\pi$ forms a large outer loop and a small inner loop about the origin. Using double-angle identities, $$x=(1+2\sin\theta)\cos\theta=\cos\theta+\sin2\theta,\qquad y=(1+2\sin\theta)\sin\theta=\sin\theta+1-\cos2\theta,$$ so $$\frac{dx}{d\theta}=-\sin\theta+2\cos2\theta,\qquad \frac{dy}{d\theta}=\cos\theta+2\sin2\theta.$$

The point corresponding to $\theta=\pi/2$ has $r=1+2\sin(\pi/2)=3$, so $(x,y)=\big(3\cos(\pi/2),3\sin(\pi/2)\big)=(0,3)$.

At $\theta=\pi/2$, $$\frac{dx}{d\theta}=-\sin\frac\pi2+2\cos\pi=-1+2(-1)=-3,\qquad \frac{dy}{d\theta}=\cos\frac\pi2+2\sin\pi=0+0=0,$$ so $$\left.\frac{dy}{dx}\right|_{\theta=\pi/2}=\frac{dy/d\theta}{dx/d\theta}=\frac0{-3}=0.$$ That is, the tangent line at $(0,3)$ has slope $0$, so its equation is $$y=3.$$