## Exercises4.3Polar Coordinates

Use polar coordinates to solve the following problems.

###### 1.

Express the polar equation $r=\cos 2\theta$ in rectangular coordinates.

Hint
Multiply by $r^2$ and use the fact that $\cos 2\theta =cos ^2\theta -\sin ^2\theta\text{.}$
$(x^2+y^2)^3=(y^2-x^2)\text{.}$

Sketch polar graphs of:

###### 2.

$r=1+\sin \theta\text{.}$

###### 3.

$r=\cos 3\theta\text{.}$

For the each of the following circles find a polar equation, i.e. an equation in $r$ and $\theta\text{:}$

###### 4.

$x^2+y^2=4$

$r = 2\text{.}$
###### 5.

$(x-1)^2+y^2=1$

$r = 2\cos \theta\text{.}$
###### 6.

$x^2+(y-0.5)^2=0.25$

$r = \sin \theta\text{.}$
###### 7.

Find the maximum height above the $x$-axis of the cardioid $r=1+\cos \theta\text{.}$

$\ds y = \frac{3\sqrt{3}}{4}\text{.}$
Solution

On the given cardioid, $x = (1 + \cos \theta ) \cos \theta$ and $y = (1 + \cos \theta ) \sin \theta\text{.}$ The question is to find the maximum value of $y\text{.}$ Note that $y > 0$ is equivalent to $\sin \theta > 0\text{.}$ From $\ds \frac{dy}{d\theta } = 2\cos ^2 \theta + \cos \theta -1$ we get that the critical numbers of the function $y = y(\theta )$ are the values of $\theta$ for which $\ds \cos \theta = \frac{-1 \pm 3}{4}\text{.}$ It follows that the critical numbers are the values of $\theta$ for which $\ds \cos \theta = -1$ or $\ds \cos \theta = \frac{1}{2}\text{.}$ Since $y_{\mbox{max} } > 0$ it follows that $\ds \sin \theta = \sqrt{1-\left( \frac{1}{2}\right) ^2}=\frac{\sqrt{3}}{2}$ and the maximum height equals $\ds y = \frac{3\sqrt{3}}{4}\text{.}$

###### 8.

Sketch the graph of the curve whose equation in polar coordinates is $r=1-2\cos\theta\text{,}$ $0\leq \theta \lt 2\pi\text{.}$

###### 9.

Sketch the graph of the curve whose equation in polar coordinates is $r=3\cos 3\theta\text{.}$

###### 10.

Sketch the curve whose polar equation is $r=-1+\cos \theta\text{,}$ indicating any symmetries. Mark on your sketch the polar coordinates of all points where the curve intersects the polar axis.

Sketch a polar coordinate plot of:

###### 11.

$\ds r=\frac{1}{2}+\sin \theta$

###### 12.

$r=2\cos 3\theta$

###### 13.

$r^2=-4\sin 2\theta$

###### 14.

$r=2\sin \theta$

###### 15.

$r=2\cos \theta$

###### 16.

$r=4+7\cos \theta$

###### 17.

Consider the curve given by the polar equation $r=1-\cos \theta\text{,}$ for $0\leq \theta \lt 2\pi\text{.}$

1. Given a point $P$ on this curve with polar coordinates $(r,\theta)\text{,}$ represent its Cartesian coordinates $(x,y)$ in terms of $\theta\text{.}$

2. Find the slope of the tangent line to the curve where $\ds \theta = \frac{\pi }{2}\text{.}$

3. Find the points on this curve where the tangent line is horizontal or vertical.

1. $(x,y)=((1-\cos\theta)\cdot \cos\theta, (1-\cos\theta)\cdot \sin\theta)\text{.}$

2. $\ds \left.\frac{dy}{dx}\right|_{\theta=\frac{\pi}{2}}=-1\text{.}$

3. Horizontal tangent lines at $\ds \left(-\frac{3}{4},\frac{3\sqrt{3}}{4}\right), \ \left(-\frac{3}{4},-\frac{3\sqrt{3}}{4}\right)\text{;}$ Vertical tangent line at $\ds (-2,0), \left(\frac{1}{4},\frac{\sqrt{3}}{4}\right),\ \left(\frac{1}{4},-\frac{\sqrt{3}}{4}\right)\text{.}$

###### 18.

Consider the curve given by the polar equation $r=\cos (2\theta)\text{,}$ for $0\leq \theta \lt 2\pi\text{.}$

1. Find $\ds \frac{dy}{dx}$ in terms of $\theta\text{.}$

2. Find the Cartesian coordinates for the point on the curve corresponding to $\ds \theta = \frac{\pi }{8}\text{.}$

3. Find the tangent line to the curve at the point corresponding to $\ds \theta = \frac{\pi }{8}\text{.}$

4. Sketch this curve for $\displaystyle 0\leq \theta \leq \frac{\pi}{4}$ and label the point from part (b) on your curve.

1. $\ds \frac{dy}{dx}=\frac{-2\sin 2\theta\sin\theta+\cos\theta\cos2\theta}{-2\sin 2\theta\cos\theta-\sin \theta\cos 2\theta}\text{.}$

2. $\ds \left(\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{2}}},\frac{1}{2}\sqrt{1-\frac{1}{\sqrt{2}}}\right)\text{.}$

3. $y- \frac{1}{2}\sqrt{1-\frac{1}{\sqrt{2}}}= \frac{2\sqrt{2-\sqrt{2}}-\sqrt{2+\sqrt{2}}}{2\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}}\cdot\left(x-\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{2}}}\right)\text{.}$

###### 19.

Consider the curve given by the polar equation $r=4\cos (3\theta)\text{,}$ for $0\leq \theta \lt 2\pi\text{.}$

1. Find the Cartesian coordinates for the point on the curve corresponding to $\ds \theta = \frac{\pi }{3}\text{.}$

2. One of graphs in the Figure below is the graph of $r=4\cos(3\theta)\text{.}$ Indicate which one by circling it.

3. Find the slope of the tangent line to the curve where $\ds \theta = \frac{\pi }{3}\text{.}$

1. $(-2,-2\sqrt{3})\text{.}$

2. Right.

###### 20.

Consider the curve given by the polar equation $r=4\sin (3\theta)\text{,}$ for $0\leq \theta \lt 2\pi\text{.}$

1. Find the Cartesian coordinates for the point on the curve corresponding to $\ds \theta = \frac{\pi }{6}\text{.}$

2. One of graphs in the Figure below is the graph of $r=4\sin(3\theta)\text{.}$ Indicate which one by circling it.

3. Find the slope of the tangent line to the curve where $\ds \theta = \frac{\pi }{3}\text{.}$

1. $(2\sqrt{3},2)\text{.}$

2. Middle.

3. $\ds \sqrt{3}\text{.}$

###### 21.

Consider the curve given by the polar equation $r=1+3\cos(2\theta)\text{,}$ for $0\leq \theta \lt 2\pi\text{.}$

1. Find the Cartesian coordinates for the point on the curve corresponding to $\ds \theta = \frac{\pi }{6}\text{.}$

2. One of graphs in the Figure below is the graph of $r=1+3\cos(2\theta)\text{.}$ Indicate which one by putting a checkmark in the box below the graph you chose.

3. Find the slope of the tangent line to the curve where $\ds \theta = \frac{\pi }{6}\text{.}$

1. $\left(\frac{5\sqrt{3}}{4},\frac{5}{4}\right)\text{.}$

2. Right. Observe that $r(0)=r(\pi)=4$ and $\ds r\left(\frac{\pi}{2}\right)= r\left(\frac{3\pi}{2}\right)=1\text{.}$

3. $\ds \frac{\sqrt{3}}{23}\text{.}$

###### 22.

Consider the curve given by the polar equation $r=1-2\sin \theta\text{,}$ for $0\leq \theta \lt 2\pi\text{.}$

1. Find the Cartesian coordinates for the point on the curve corresponding to $\ds \theta = \frac{3\pi }{2}\text{.}$

2. The curve intersects the $x$-axis at two points other than the pole. Find polar coordinates for these other points.

3. On the Figure below, identify the graphs that correspond to the following two polar curves.

\begin{equation*} \begin{array}{cc} \fbox { } \ r=1-2\sin \theta \amp \fbox { } \ r=1+2\sin \theta \end{array} \end{equation*}
1. $(0,-3)\text{.}$

2. $(\pm 1,0)\text{.}$ Solve $y=(1-2\sin \theta)\sin \theta =0\text{.}$

3. The middle graph corresponds to $r=1+\sin 2\theta$ and the right graph corresponds to $r=1-2\sin \theta\text{.}$

###### 23.

Consider the curve $C$ given by the polar equation $r=1+2\cos \theta\text{,}$ for $0\leq \theta \lt 2\pi\text{.}$

1. Find the Cartesian coordinates for the point on the curve corresponding to $\ds \theta = \frac{\pi }{3}\text{.}$

2. Find the slope of the tangent line where $\ds \theta = \frac{\pi }{3}\text{.}$

3. On the Figure below, identify the graph of $C\text{.}$

1. $(1,\sqrt{3})\text{.}$

2. $\ds \frac{1}{3\sqrt{3}}\text{.}$

3. B.

###### 24.
1. Sketch a polar coordinate plot of

\begin{equation*} r=1+2\sin 3\theta, \ 0\leq \theta \leq 2\pi\text{.} \end{equation*}
2. How many points lie in the intersection of the two polar graphs

\begin{equation*} r=1+2\sin 3\theta, \ 0\leq \theta \leq 2\pi \end{equation*}

and

\begin{equation*} r=1? \end{equation*}
3. Algebraically find all values of $\theta$ that

\begin{equation*} 1=1+2\sin 3\theta, \ 0\leq \theta \leq 2\pi\text{.} \end{equation*}
4. Explain in a sentence or two why the answer to part (b) differs from (or is the same as) the number of solutions you found in part (c).

1. 9.

2. $\theta = 0\text{,}$ $\ds \frac{\pi }{3}\text{,}$ $\ds \frac{2\pi }{3}\text{,}$ $\pi\text{,}$ $\ds \frac{4\pi }{3}\text{,}$ $\ds \frac{5\pi }{3}\text{,}$ $2\pi\text{.}$

3. The remaining points of intersection are obtained by solving $-1 = 1 + 2 \sin 3\theta\text{.}$

###### 25.

Consider the following curve $C$ given in polar coordinates as

\begin{equation*} r(\theta )=1+\sin \theta +e^{\sin \theta }, \ 0\leq \theta \leq 2\pi\text{.} \end{equation*}
1. Calculate the value of $r(\theta )$ for $\ds \theta =0, \frac{\pi }{2}, \frac{3\pi }{2}\text{.}$

2. Sketch a graph of $C\text{.}$

3. What is the minimum distance from a point on the curve $C$ to the origin? (i.e. determine the minimum of $|r(\theta )|=r(\theta )=1+\sin \theta +e^{\sin \theta }$ for $\theta \in [0,2\pi ]$).

1. $r(0) = 2\text{,}$ $\ds r \left( \frac{\pi }{2}\right) = 2+e\text{,}$ $\ds r \left( \frac{3\pi }{2}\right) = e^{-1}\text{.}$

2. $e^{-1}\text{.}$

Solution

(c) From $\ds \frac{dr}{d\theta }= \cos \theta (1 + e^{\sin \theta }) = 0$ we conclude that the critical numbers are $\ds \frac{\pi }{2}$ and $\ds \frac{3\pi }{2}\text{.}$ By the Extreme Value Theorem, the minimum distance equals $e^{-1}\text{.}$

###### 26.
1. Give polar coordinates for each of the points $A\text{,}$ $B\text{,}$ $C$ and $D$ on the Figure below.

2. On the Figure below identify the graphs that correspond to the following three polar curves.

\begin{equation*} \fbox { } \ r=1-2\sin \theta \ \ \fbox { } \ r^2\theta =1 \ \ \fbox { } \ r=\frac{1}{1-2\sin \theta} \end{equation*}
1. $\ds A=\left( r=\sqrt{2},\theta= \frac{\pi }{4}\right)\text{,}$ $\ds B=\left( 4,\frac{5\pi }{3}\right)\text{,}$ $\ds C=\left( 2,\frac{7\pi }{6}\right)\text{,}$ $\ds D=\left( 2\sqrt{2}-1,\frac{3\pi }{4}\right)\text{.}$

2. A, B, D.

###### 27.
1. Sketch the curve defined by $r=1+2\sin \theta\text{.}$

2. For what values of $\theta\text{,}$ $\theta \in [-\pi ,\pi )\text{,}$ is the radius $r$ positive?

3. For what values of $\theta\text{,}$ $\theta \in [-\pi ,\pi )\text{,}$ is the radius $r$ maximum and for what values is it minimum?

1. $\ds \theta \in \left[ -\pi, -\frac{5\pi }{6}\right) \cup \left( -\frac{\pi }{6},\pi \right)\text{.}$ Solve $\ds \sin \theta > -\frac{1}{2}\text{.}$

2. To find critical numbers solve $\ds \frac{dr}{d\theta }=2\cos \theta =0$ in $[-\pi ,\pi )\text{.}$ It follows that $\ds \theta =-\frac{\pi }{2}$ and $\ds \theta =\frac{\pi }{2}$ are critical numbers. Compare $r(-\pi )=r(\pi )=1\text{,}$ $\ds r\left( -\frac{\pi }{2}\right) =-1\text{,}$ and $\ds r\left( \frac{\pi }{2}\right) =3$ to answer the question.

###### 28.
1. Sketch the graph described in polar coordinates by the equation $r=\theta$ where $-\pi \leq \theta \leq 3\pi\text{.}$

2. Find the slope of this curve when $\ds \theta =\frac{5\pi }{2}\text{.}$ Simplify your answer for full credit.

3. Express the polar equation $r=\theta$ in cartesian coordinates, as an equation in $x$ and $y\text{.}$

1. $-\frac{2}{5\pi }\text{.}$

2. $\ds \sqrt{x^2+y^2}=\arctan \frac{y}{x}\text{.}$

Solution

(b) The slope is given by $\ds \left. \frac{dy}{dx}\right| _{\theta =\frac{5\pi }{2}}\text{.}$ From $x=r\cos \theta =\theta \cos \theta$ and $y=\theta \sin \theta$ it follows that $\ds \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{\sin \theta +\theta \cos \theta }{\cos \theta -\theta \sin \theta}\text{.}$ Thus $\ds \left. \frac{dy}{dx}\right| _{\theta =\frac{5\pi }{2}}=-\frac{2}{5\pi }\text{.}$

###### 29.

1. Let $C$ denote the graph of the polar equation $r=5\sin \theta\text{.}$ Find the rectangular coordinates of the point on $C$ corresponding to $\ds \theta =\frac{3\pi }{2}\text{.}$

2. Write a rectangular equation (i.e. using the variables $x$ and $y$) for $C\text{.}$ (in other words, convert the equation for $C$ into rectangular coordinates.)

3. Rewrite the equation of $C$ in parametric form, i.e. express both $x$ and $y$ as functions of $\theta\text{.}$

4. Find an expression for $\ds \frac{dy}{dx}$ in terms of $\theta\text{.}$

5. Find the equation of the tangent line to $C$ at the point corresponding to $\ds \theta =\frac{\pi }{6}\text{.}$

1. $(0,5)\text{.}$

2. $x^2+y^2=5y\text{.}$

3. $x=5\sin \theta \cos \theta\text{,}$ $y=5\sin ^2\theta\text{.}$

4. $\ds \frac{dy}{dx}=\frac{2\sin \theta\cos \theta }{\cos ^2\theta - \sin ^2\theta}=\tan 2\theta\text{.}$

5. $\ds y -\frac{5}{4}=\sqrt{3}\left( x-\frac{5\sqrt{3}}{4}\right)\text{.}$

###### 30.

Find the slope of the tangent line to the polar curve $r=2$ at the points where it intersects the polar curve $r=4\cos \theta\text{.}$ (Hint: After you find the intersection points, convert one of the curves to a pair of parametric equations with $\theta$ as the perimeter.

$-\frac{\sqrt{3}}{3}\text{.}$
Solution

Solve $2=4\cos \theta$ to get that curve intersect at $( 1, \sqrt{3})\text{.}$ To find the slope we note that the circle $r=2$ is given by parametric equations $x=2\cos \theta$ and $y=2\sin \theta\text{.}$ It follows that $\ds \frac{dy}{dx}=\frac{2\cos \theta }{-2\sin \theta }=-\cot \theta\text{.}$ The slope of the tangent line at the intersection point equals $\ds \left. \frac{dy}{dx}\right| _{\theta = \frac{\pi }{3}}=-\frac{\sqrt{3}}{3}\text{.}$

###### 31.

A bee goes out from its hive in a spiral path given in polar coordinates by $r=be^{kt}$ and $\theta =ct\text{,}$ where $b\text{,}$ $k\text{,}$ and $c$ are positive constants. Show that the angle between the bee's velocity and acceleration remains constant as the bee moves outward.

Solution

See the Figure below for the graph of the case $b=1\text{,}$ $k=0.01\text{,}$ and $c=2\text{.}$ The position in $(x,y)-$plane of the bee at time $t$ is given by a vector function $\ds \vec{s}(t)=\langle be^{kt}\cos ct,be^{kt}\sin ct\rangle\text{.}$ Recall that the angle $\alpha$ between the velocity and acceleration is given by $\ds \cos\alpha=\frac{\vec{v}\cdot\vec{a}}{|\vec{v}||\vec{a}|}\text{,}$ where $\vec{v}(t)=\vec{s}^\prime(t)$ and $\vec{a}(t)=\vec{s}^{\prime\prime}(t)\text{.}$ One way to solve this problem is to consider that the bee moves in the complex plane. In that case its position is given by

\begin{equation*} F(t)=be^{kt}\cos ct+i\cdot be^{kt}\sin ct=be^{kt}(\cos ct +i\sin ct)=be^{(k+ic)t}\text{,} \end{equation*}

where $i$ is the imaginary unit. Observe that $\vec{v}(t)=\langle \mbox{Re} (F^\prime(t)),\mbox{Im} (F^\prime(t))\rangle$ and $\vec{a}(t)=\langle \mbox{Re} (F^{\prime\prime}(t)),\mbox{Im} (F^{\prime\prime}(t))\rangle\text{.}$ Next, observe that $F^\prime(t)=(k+ic)F(t)$ and $F^{\prime\prime}(t)=(k+ic)^2F(t)\text{.}$ From $F^{\prime\prime}(t)=(k+ic)F^\prime(t)$ it follows that $\mbox{Re} (F^{\prime\prime}(t))=k\cdot \mbox{Re} (F^\prime(t))-c\cdot \mbox{Im} (F^\prime(t))$ and $\mbox{Im} (F^{\prime\prime}(t))=k\cdot \mbox{Im} (F^\prime(t))+c\cdot \mbox{Re} (F^\prime(t))\text{.}$ Finally, $\vec{v}\cdot\vec{a}=\mbox{Re} (F^\prime(t))\cdot \mbox{Re} (F^{\prime\prime}(t))+\mbox{Im} (F^\prime(t))\cdot \mbox{Im} (F^{\prime\prime}(t))=k((\mbox{Re} (F^\prime(t)))^2+(\mbox{Im} (F^\prime(t)))^2)=k|F^\prime(t)|^2$ which immediately implies the required result.