매개변수와 극좌표 전체 문제

$P(3/2,\pi/3)$, $P(3/2,-\pi/3)$

For all points, $r=1$; $\theta = \pi/12,\ 5\pi/12,\ 7\pi/12,\ 11\pi/12,\ 13\pi/12,\ 17\pi/12,\ 19\pi/12,\ 23\pi/12$.

$P(0,0)=P(0,3\pi/2)$, $P(1+\sqrt{2}/2,3\pi/4)$, $P(1-\sqrt{2}/2,7\pi/4)$

(a) $\frac{dy}{dx} = -\cot\theta$ (b) tangent line: $y=-(x-\sqrt{2}/2)+\sqrt{2}/2$; normal line: $y=x$

(a) $\frac{dy}{dx} = 1/2(\tan \theta-\cot\theta)$ (b) tangent line: $y=1/2$; normal line: $x=1/2$

(a) $\frac{dy}{dx} = \frac{\cos\theta(1+2\sin\theta)}{\cos^2\theta-\sin\theta(1+\sin\theta)}$ (b) tangent line: $x=3\sqrt{3}/4$; normal line: $y=3/4$

(a) $\frac{dy}{dx} = \frac{3 \sin ^2(t)+(1-3 \cos (t)) \cos (t)}{3 \sin (t) \cos (t)-\sin (t) (1-3 \cos (t))}$ (b) tangent line: $y=\frac{1}{1+3\sqrt{2}}(x+(1/\sqrt{2}+3/2))+1/\sqrt{2}+3/2 \approx y=0.19(x+2.21)+2.21$; normal line: $y=-(1+3\sqrt{2})(x+(1/\sqrt{2}+3/2))+1/\sqrt{2}+3/2$

(a) $\frac{dy}{dx} = \frac{\theta\cos\theta+\sin\theta}{\cos\theta-\theta\sin\theta}$ (b) tangent line: $y=-2/\pi x+\pi/2$; normal line: $y=\pi/2x+\pi/2$

(a) $\frac{dy}{dx} = \frac{\cos\theta\cos(3\theta)-3\sin\theta\sin(3\theta)}{-\cos(3\theta)\sin\theta-3\cos\theta\sin(3\theta)}$ (b) tangent line: $y=x/\sqrt{3}$; normal line: $y=-\sqrt{3}x$

(a) $\frac{dy}{dx} = \frac{4 \sin (\theta) \cos (4 \theta)+\sin (4 \theta) \cos (\theta)}{4 \cos (\theta) \cos (4 \theta)-\sin (\theta) \sin (4 \theta)}$ (b) tangent line: $y=5\sqrt{3}(x+\sqrt{3}/4)-3/4$; normal line: $y=-1/5\sqrt{3}(x+\sqrt{3}/4)-3/4$

(a) $\frac{dy}{dx} = 1$ (b) tangent line: $y=x+1$; normal line: $y=-x-1$

$4\pi$

$4\pi$

area = $\sqrt{2}\pi$

$8$

$L\approx 2.2592$; (actual value $L=2.22748$)

$x\,'(\theta) = f\,'(\theta)\cos\theta -f(\theta)\sin\theta$, $y\,'(\theta) = f\,'(\theta)\sin\theta + f(\theta)\cos\theta$. Square each and add; applying the Pythagorean Theorem twice achieves the result.

Note that the distance from the point $\big(r\cos\theta\,,\,r\sin\theta\big)$ to the origin is

$$\sqrt{r^2\cos^2\theta + r^2\sin^2\theta} =\sqrt{r^2} =|r|$$

Thus $r$ can be either the distance to the origin or minus the distance to the origin.

(a) The distance from $(-2,0)$ to the origin is $2$. So either $r=2$ or $r=-2$.

• If $r=2$, then $\theta$ must obey

$$\begin{aligned} (-2,0) = \big(2\cos\theta\,,\,2\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=-1 \\ &\iff \theta= n\pi,\ n\text{ integer },\ \cos\theta=-1 \\ &\iff \theta= n\pi,\ n\text{ odd integer } \end{aligned}$$

• If $r=-2$, then $\theta$ must obey

$$\begin{aligned} (-2,0) = \big(-2\cos\theta\,,\,-2\sin\theta\big) &\iff \sin\theta=0,\ \cos\theta=1 \\ &\iff \theta= n\pi,\ n\text{ integer },\ \cos\theta=1 \\ &\iff \theta= n\pi,\ n\text{ even integer } \end{aligned}$$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\pi$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\pi$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=0$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=0$, $r<0$.

(b) The distance from $(1,1)$ to the origin is $\sqrt{2}$. So either $r=\sqrt{2}$ or $r=-\sqrt{2}$.

• If $r=\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (1,1) = \big(\sqrt{2}\,\cos\theta\,,\,\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta={1}/{\sqrt{2}} \\ &\iff \theta= {\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

• If $r=-\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (1,1) = \big(-\sqrt{2}\,\cos\theta\,,\,-\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=-{1}/{\sqrt{2}} \\ &\iff \theta= {5\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r<0$.

(c) The distance from $(-1,-1)$ to the origin is $\sqrt{2}$. So either $r=\sqrt{2}$ or $r=-\sqrt{2}$.

• If $r=\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (-1,-1) = \big(\sqrt{2}\,\cos\theta\,,\,\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta=-{1}/{\sqrt{2}} \\ &\iff \theta= {5\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

• If $r=-\sqrt{2}$, then $\theta$ must obey

$$\begin{aligned} (-1,-1) = \big(-\sqrt{2}\,\cos\theta\,,\,-\sqrt{2}\,\sin\theta\big) &\iff \sin\theta=\cos\theta={1}/{\sqrt{2}} \\ &\iff \theta= {\pi}/{4} + 2n\pi,\ n\text{ integer } \end{aligned}$$

In the figure on the left below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\frac{5\pi}{4}$, $r<0$. In the figure on the right below, the blue half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r>0$ and the pink half-line is the set of all points with polar coordinates $\theta=\frac{\pi}{4}$, $r<0$.