매개변수와 극좌표 전체 문제

The left hand sketch below contains the points, $(x_1,y_1)$, $(x_3,y_3)$, $(x_5,y_5)$, that are on the axes. The right hand sketch below contains the points, $(x_2,y_2)$, $(x_4,y_4)$, that are not on the axes.

Recall that the polar coordinates $r$, $\theta$ are related to the cartesian coordinates $x$, $y$, by $x=r\cos\theta$, $y=r\sin\theta$. So $r=\sqrt{x^2+y^2}$ and $\tan\theta=\frac{y}{x}$ (assuming that $x\ne 0$) and

$$\begin{aligned} (x_1,y_1) &= (3,0) &&\implies &r_1=3,\ \tan\theta_1=0 &&\implies \theta_1&=0 \text{ as $(x_1,y_1)$ is on the positive $x$-axis} \\ (x_2,y_2) &= (1,1) &&\implies &r_2=\sqrt{2},\ \tan\theta_2=1 &&\implies \theta_2&=\frac{\pi}{4} \text{ as $(x_2,y_2)$ is in the first octant} \\ (x_3,y_3) &= (0,1) &&\implies &r_3=1,\ \cos\theta_3=0 &&\implies \theta_3&=\frac{\pi}{2} \text{ as $(x_3,y_3)$ is on the positive $y$-axis} \\ (x_4,y_4) &= (-1,1) &&\implies &r_4=\sqrt{2},\ \tan\theta_4=-1 &&\implies \theta_4&=\frac{3\pi}{4} \text{ as $(x_4,y_4)$ is in the third octant} \\ (x_5,y_5) &= (-2,0) &&\implies &r_5=2,\ \tan\theta_5=0 &&\implies \theta_5&=\pi \text{ as $(x_5,y_5)$ is on the negative $x$-axis} \end{aligned}$$

(a) The lengths are

$$\begin{aligned} |\hat{\mathbf{e}}_r(\theta)| &= \sqrt{\cos^2\theta+\sin^2\theta} = 1 \\ |\hat{\mathbf{e}}_\theta(\theta)| &= \sqrt{(-\sin\theta)^2+\cos^2\theta} = 1 \end{aligned}$$

As

$$\hat{\mathbf{e}}_r(\theta) \cdot \hat{\mathbf{e}}_\theta(\theta) = (\cos\theta)(-\sin\theta) +(\sin\theta)(\cos\theta)=0$$

the two vectors are perpendicular and the angle between them is $\frac{\pi}{2}$. The cross product is

$$\begin{aligned} \hat{\mathbf{e}}_r(\theta) \times \hat{\mathbf{e}}_\theta(\theta) &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \end{matrix}\right] =\hat{\mathbf{k}} \end{aligned}$$

(b) Note that for $\theta$ determined by $x=r\cos\theta$, $y=r\sin\theta$,

• the vector $\hat{\mathbf{e}}_r(\theta)$ is a unit vector in the same direction as the vector from $(0,0)$ to $(x,y)$ and
• the vector $\hat{\mathbf{e}}_\theta(\theta)$ is a unit vector that is perpendicular to $\hat{\mathbf{e}}_r(\theta)$.
• The $y$-component of $\hat{\mathbf{e}}_\theta(\theta)$ has the same sign as the $x$-component of $\hat{\mathbf{e}}_r(\theta)$. The $x$-component of $\hat{\mathbf{e}}_\theta(\theta)$ has opposite sign to that of the $y$-component of $\hat{\mathbf{e}}_r(\theta)$.

Here is a sketch of $(x_i,y_i)$, $\hat{\mathbf{e}}_r(\theta_i)$, $\hat{\mathbf{e}}_\theta(\theta_i)$ for $i =1,3,5$ (the points on the axes)

and here is a sketch (to a different scale) of $(x_i,y_i)$, $\hat{\mathbf{e}}_r(\theta_i)$, $\hat{\mathbf{e}}_\theta(\theta_i)$ for $i =2,4$ (the points off the axes).

(a) Since $-1\le \sin(4\theta)\le 1$, the coordinate $r=2+\sin(4\theta)$ oscillates between $r=1$ and $r=3$ as $\theta$ runs from $0$ to $2\pi$. The maximum value $r=3$ is achieved when $\sin(4\theta)=1$, i.e when $4\theta=\frac{\pi}{2} +2n\pi$, i.e. when $\theta=\frac{\pi}{8}+\frac{n\pi}{2}$. That matches figure (E).

(b) Since $-1\le \sin(4\theta)\le 1$, the coordinate $r=1+2\sin(4\theta)$ takes its maximum value $r=3$ when $\sin(4\theta)=1$, i.e. when $\theta=\frac{\pi}{8}+\frac{n\pi}{2}$, just as the case with $(a)$. But now $r$ can also take the value $0$. That matches figure (B).

(c) $r=1$ is completely indepedent of $\theta$. All points on the curve $r=1$ are a distance $1$ from the origin. That is, $r=1$ is the circle of radius $1$ centred on the origin. That's figure (F).

(d) In this case, $\theta$ is subject to the restriction $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$, like figure (C). Figure (C) looks like a circle. We can verify that $r=2\cos(\theta)$ is indeed a circle by converting to Cartesian coordinates. We can convert the right hand side to exactly $2x=2r\cos(\theta)$ by multiplying the whole equation by $r$.

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

So $r=2\cos(\theta)$ is the circle of radius $1$ centred on $x=1$, $y=0$, which indeed matches figure (C).

(e) When $\theta=0$, $r=e^{\theta/10}+e^{-\theta/10}=2$. As

$$\frac{\mathrm{d}}{\mathrm{d}\theta}\big(e^{\theta/10}+e^{-\theta/10}\big) =\frac{1}{10}\big(e^{\theta/10}-e^{-\theta/10}\big) >0\qquad\text{for all }\theta > 0$$

$r=e^{\theta/10}+e^{-\theta/10}$ increases as $\theta$ increases for all $\theta\ge 0$. Furthermore the rate of increase gets bigger and bigger as $\theta$ gets bigger and bigger. So $r$ starts at $r=2$ when $\theta=0$ and increases faster and faster as $\theta$ increases. That matches figure (A).

(f) When $\theta=0$, $r=\theta=0$. As

$$\frac{\mathrm{d}}{\mathrm{d}\theta} \theta =1 \qquad\text{for all }\theta$$

$r=\theta$ increases as $\theta$ increases for all $\theta\ge 0$. Furthermore the rate of increase is independent of $\theta$. So $r$ starts at $r=0$ when $\theta=0$ and increases at a constant rate as $\theta$ increases. That matches figure (D).