CHAPTER 28
Polar Coordinates; Parametric Equations
Polar Coordinate System
A polar coordinate system specifies points in the plane in terms of directed distances r from a fixed point called the pole and angles $ $ measured from a fixed ray (with initial point the pole) called the polar axis. The polar axis is the positive half of a number line, drawn to the right of the pole. See Fig. 28-1.
For any point P, $ $ is an angle formed by the polar axis and the ray connecting the pole to P, and r is the distance measured along this ray from the pole to P. For any ordered pair $ (r,) $ , if r is positive, take $ $ as an angle with vertex the pole and initial side the polar axis, and measure r units along the terminal side of $ $ . If r is negative, measure $ |r| $ units along the ray directed opposite to the terminal side of $ $ . Any pair with r = 0 represents the pole. In this manner, every ordered pair $ (r,) $ is represented by a unique point.
EXAMPLE 28.1 Graph the points specified by $ (3,/3) $ and $ (-3,/3) $ .
Polar Coordinates of a Point Are not Unique
The polar coordinates of a point are not unique, however. Given point P, there is an infinite set of polar coordinates corresponding to P, since there are an infinite number of angles with terminal sides passing through P.
EXAMPLE 28.2 List four alternative sets of polar coordinates corresponding to the point $ P(3, /3) $ .
Adding any multiple of $ 2$ yields an angle coterminal with a given angle; hence $ (3, 7/3) $ and $ (3, 13/3) $ are two possible alternative polar coordinates. Since $ + /3 = 4/3 $ has terminal side the ray opposite to $ /3 $ , the coordinates $ (-3, 4/3) $ and $ (-3, 10/3) $ are further alternative polar coordinates for P.
Polar and Cartesian Coordinates
If a polar coordinate system is superimposed upon a Cartesian coordinate system, as in Fig. 28-3, the transformation relationships below hold between the two sets of coordinates.
If P has polar coordinates $ (r, ) $ and
Cartesian coordinates $ (x, y) $ , then
\[ x=r\cos\theta\quad y=r\sin\theta \]
\[ r^{2}=x^{2}+y^{2}\quad\tan\theta=\frac{y}{x}\quad(x\neq0) \]
EXAMPLE 28.3 Convert $ (6, 2/3) $ to Cartesian coordinates.
Since \(r = 6\) and \(\theta = 2\pi/3\), applying the transformation relationships yields
\[ x=r\cos\theta=6\cos2\pi/3=-3\qquad y=r\sin\theta=6\sin2\pi/3=3\sqrt{3} \]
Thus the Cartesian coordinates are $ (-3, 3) $ .
EXAMPLE 28.4 Convert $ (-5, -5) $ to polar coordinates with r > 0 and $ 0 $ .
Since x = -5 and y = -5, applying the transformation relationships yields
\[ r^{2}=x^{2}+y^{2}=(-5)^{2}+(-5)^{2}=50\qquad\tan\theta=\frac{y}{x}=\frac{-5}{-5}=1 \]
Since r is required positive, $ r = = 5 $ . Since the point $ (-5, -5) $ is in quadrant III, $ = 5/4 $ . The polar coordinates that satisfy the given conditions are $ (5, 5/4) $ .
Equations in Polar Coordinates
Any equation in the variables r and $ $ may be interpreted as a polar coordinate equation. Often r is specified as a function of $ $ .
EXAMPLE 28.5 $ r= 1 $ and $ r^{2} = 2$ are examples of polar coordinate equations. $ r = 2$ and $ r = 3 - 3$ are examples of polar coordinate equations with r specified as a function of $ $ .
Parametric Equations
An equation for a curve may be given by specifying x and y separately as functions of a third variable, often t, called a parameter. These functions are called the parametric equations for the curve. Points on the curve may be found by assigning permissible values of t. Often, t may be eliminated algebraically, but any restrictions placed on t are needed to determine the portion of the curve that is specified by the parametric equations.
EXAMPLE 28.6 Graph the curve specified by the parametric equations x = 1 - t, $ y = 2t + 2 $ .
First note that t can be eliminated by solving the equation specifying x for t to obtain t = 1 - x, then substituting into the equation specifying y to obtain $ y = 2(1 - x) + 2 = 4 - 2x $ . Thus, for every value of t, the point $ (x, y) $ lies on the graph of
y = 4 - 2x. Moreover, since there are no restrictions on t and the functions $ x(t) $ and $ y(t) $ are one-to-one, it follows that x and y can take on any value and the graph is the entire line y = 4 - 2x. Form a table of values, then plot the points and connect them (Fig. 28-4).
First note that t can be eliminated by adding the equations specifying x and y to obtain $ x + y = 1 $ . However, both variables are restricted by these equations to the interval $ [0, 1] $ . In fact, since both x and y are periodic with period $ $ , the graph is the portion of the line $ x + y = 1 $ on the interval $ 0 x $ , and is traced out repeatedly as t varies through all possible real values. Form a table of values, then plot the points and connect them (Fig. 28-5).
| t | 0 | $ /4 \(</td><td>\) /2 \(</td><td>\) 3/4 \(</td><td>\) $ | |||
| x | 1 | 1/2 | 0 | 1/2 | 1 |
| y | 0 | 1/2 | 1 | 1/2 | 0 |
Polar Coordinates and Parametric Equations
According to the transformation relationships, the Cartesian coordinates of a point are given in terms of its polar coordinates by the equations $ x = r $ and $ y = r $ . Hence any polar coordinate equation specifying $ r = f() $ can be regarded as giving parametric equations for x and y of the form $ x = f() $ , $ y = f() $ , with $ $ as the parameter.
EXAMPLE 28.8 Write the parametric equations for x and y specified by $ r = 1 + $ .
\[ x=(1+\sin\theta)\cos\theta\qquad y=(1+\sin\theta)\sin\theta \]
SOLVED PROBLEMS
28.1. Graph the points with the following polar coordinates:
\[ \begin{array}{r}{(a)A(4,\pi/6),B(6,-\pi/4);(b)C(-2,5\pi/3),D(-5,\pi).}\end{array} \]
$ (4, /6) $ is located 4 units along the ray at an angle of $ /6 $ to the polar axis. $ (6, -/4) $ is located 6 units along the ray at an angle of $ -/4 $ to the polar axis (Fig. 28-6).
$ (-2, 5/3) $ is located $ |-2| = 2 $ units along the ray directed opposite to $ 5/3 $ . $ (-5, ) $ is located $ |-5| = 5 $ units along the ray directed opposite to $ $ (Fig. 28-7).
28.2. Give all possible polar coordinates that can describe a point $ P(r,) $ .
Since any angle of measure $ + 2$ , where n is an integer, is coterminal with $ $ , any point with coordinates $ (r, + 2n) $ is coincident with $ (r, ) $ . Also, since the ray forming an angle of $ + $ with the polar axis is directed opposite to the ray forming angle $ $ , the coordinates $ (r, ) $ and $ (-r, + ) $ name the same point. Finally, any point with coordinates $ (-r, + + 2n) $ is coincident with $ (-r, + ) $ . Summarizing, the coordinates $ (r, + 2n) $ and $ (-r, + (2n + 1)) $ describe the same point as $ (r, ) $ .
28.3. Establish the transformation relationships between polar and Cartesian coordinates.
See Fig. 28-3. Let P be a point with Cartesian coordinates $ (x,y) $ and polar coordinates $ (r,) $ . Since $ P(x,y) $ is a point on the terminal side of an angle $ $ , and r is the distance of P from the origin, it follows that
\[ \tan\theta=\frac{y}{x}\left(x\neq0\right)\quad\cos\theta=\frac{x}{r}\quad\sin\theta=\frac{y}{r} \]
Hence $ x = r $ and $ y = r $ . From these relationships, it follows that
\[ x^{2}+y^{2}=r^{2}\cos^{2}\theta+r^{2}\sin^{2}\theta=r^{2}(\cos^{2}\theta+\sin^{2}\theta)=r^{2} \]
The last relationship also follows immediately from the distance formula.
28.4. Convert to Cartesian coordinates: (a) $ (4,) $ ; (b) $ (-5,-) $ .
- Since $ r = 4 $ and $ = 4/3 $ , it follows from the transformation relationships that
\[ x=r\cos\theta=4\sqrt{3}\cos(4\pi/3)=4\sqrt{3}(-1/2)=-2\sqrt{3} \]
and
\[ y=r\sin\theta=4\sqrt{3}\sin(4\pi/3)=4\sqrt{3}(-\sqrt{3}/2)=-6. \]
The Cartesian coordinates are $ (-2, -6) $ .
- Since r = -5 and $ = -/2 $ , it follows from the transformation relationships that
\[ x=r\cos\theta=-5\cos(-\pi/2)=0\qquad and\qquad y=r\sin\theta=-5\sin(-\pi/2)=5. \]
The Cartesian coordinates are (0,5).
28.5. Convert $ (-8, 8) $ to polar coordinates with r > 0 and $ 0 $ .
Since $ x = -8 $ and $ y = 8 $ , applying the transformation relationships yields
\[ r^{2}=x^{2}+y^{2}=(-8\sqrt{2})^{2}+(8\sqrt{2})^{2}=256\qquad and\qquad\tan\theta=\frac{\mathbf{Y}}{x}=\frac{8\sqrt{2}}{-8\sqrt{2}}=-1 \]
Since r is required positive, $ r = = 16 $ . Since the point $ (-8, 8) $ is in quadrant II, $ = 3/4 $ . The polar coordinates that satisfy the given conditions are $ (16, 3/4) $ .
28.6. Transform the following polar coordinate equations to Cartesian coordinates:
r = 4; (b) $ r = 4$ ; (c) $ r^{2}= 4 $ .
Since the polar coordinate equation specifies all points that are at a distance of 4 units from the origin, this is the equation of a circle with radius 4 and center at the origin. Hence the Cartesian coordinate equation is $ x^{2} + y^{2} = 16 $ .
Multiply both sides by r to obtain an equation that is easier to work with: $ r^{2} = 4r$ . This operation adds the pole to the graph (r = 0). But the pole was already part of the graph (choose $ = /2 $ ), so nothing has been changed. Now apply the transformation relationships $ r^{2} = x^{2} + y^{2} $ and $ x = r$ to obtain $ x^{2} + y^{2} = 4x $ . This equation can be rewritten as $ (x - 2)^{2} + y^{2} = 4 $ ; thus, it is the equation of a circle with center at (2,0) and radius 2.
Rewrite $ r^{2}= 4 $ as follows:
\[ \begin{aligned}r^{2}(2\sin\theta\cos\theta)&=4\quad&Double angle identity\\2r\cos\theta r\sin\theta&=4\quad&Algebra\\2xy&=4\quad&Transformation relationships\\xy&=2\quad&Algebra\end{aligned} \]
28.7. Transform the following Cartesian coordinate equations to polar coordinates:
$ x + y = 3 $ ; (b) $ x^{2} + y^{2} = 3y $ ; (c) $ y^{2} = 4x $ .
Apply the transformation equations $ x = r $ and $ y = r $ to obtain $ r + r = 3 $ .
Apply the transformation equations $ x^{2} + y^{2} = r^{2} $ and $ y = r$ to obtain $ r^{2} = 3r$ . This can be further simplified as follows:
\[ \begin{aligned}r^{2}-3r\sin\theta&=0\\r(r-3\sin\theta)&=0\\r=0\quad or\quad r-3\sin\theta&=0\\r&=3\sin\theta\end{aligned} \]
The graph of \(r=0\) consists only of the pole. Since the pole is included in the graph of \(r=3\sin\theta\) (choose \(\theta=0\)), it is sufficient to consider only \(r=3\sin\theta\) as the transformed equation.
- Apply the transformation equations $ x = r $ and $ y = r $ to obtain $ r^{2} ^{2} = 4 r $ . Proceeding as in part (b), this can be simplified to $ r ^{2} = 4 $ , which can be further rewritten as follows:
\[ \begin{aligned}r&=\frac{4\cos\theta}{\sin^{2}\theta}\\&=4\frac{\cos\theta}{\sin\theta}\frac{1}{\sin\theta}\\&=4\cot\theta\csc\theta\end{aligned} \]
28.8. Sketch a graph of $ r = 1 + $
Before making a table of values it is helpful to consider the general behavior of the function $ r() = 1 + $ . From knowledge of the behavior of the cosine function:
| As $ $ increases | $ \(</td><td>\) 1 + $ | |
| from 0 to $ /2 $ | decreases from 1 to 0 | decreases from 2 to 1 |
| from $ /2 $ to $ $ | decreases from 0 to -1 | decreases from 1 to 0 |
| from $ $ to $ 3/2 $ | increases from -1 to 0 | increases from 0 to 1 |
| from $ 3/2 $ to $ 2$ | increases from 0 to 1 | increases from 1 to 2 |
Since the cosine function is periodic with period $ 2$ , this shows the behavior of $ 1 + $ for all $ $ . Now form a table of values and sketch the graph (see Fig. 28-8).
| $ \(</td><td>0</td><td>\) /4 \(</td><td>\) /2 \(</td><td>\) 3/4 \(</td><td>\) \(</td></tr><tr><td>r</td><td>2</td><td>1.7</td><td>1</td><td>0.3</td><td>0</td></tr><tr><td>\) \(</td><td></td><td>\) 5/4 \(</td><td>\) 3/2 \(</td><td>\) 7/4 \(</td><td>\) 2$ | |||||
| r | 0.3 | 1 | 1.7 | 2 |
The curve is known as a cardioid because of its heart shape.
28.9. Sketch a graph of $ r = $
Before making a table of values, it is helpful to consider the general behavior of the function $ r() = $ . From knowledge of the behavior of the cosine function:
| As 2 $ $ increases | $ $ increases | $ $ |
| from 0 to $ /2 $ | from 0 to $ /4 $ | decreases from 1 to 0 |
| from $ /2 $ to $ $ | from $ /4 $ to $ /2 $ | decreases from 0 to -1 |
| from $ $ to $ 3/2 $ | from $ /2 $ to $ 3/4 $ | increases from -1 to 0 |
| from $ 3/2 $ to $ 2$ | from $ 3/4 $ to $ $ | increases from 0 to 1 |
| from $ 2$ to $ 5/2 $ | from $ $ to $ 5/4 $ | decreases from 1 to 0 |
| from $ 5/2 $ to $ 3$ | from $ 5/4 $ to $ 3/2 $ | decreases from 0 to -1 |
| from $ 3$ to $ 7/2 $ | from $ 3/2 $ to $ 7/4 $ | increases from -1 to 0 |
| from $ 7/2 $ to $ 4$ | from $ 7/4 $ to $ 2$ | increases from 0 to 1 |
Since the cosine function is periodic with period $ 2$ , this shows the behavior of $ $ for all $ $ . Now form a table of values and sketch the graph (Fig. 28-9).
| $ \(</td><td>0</td><td>\) /8 \(</td><td>\) /4 \(</td><td>\) 3/8 \(</td><td>\) /2 \(</td></tr><tr><td>r</td><td>1</td><td>0.7</td><td>0</td><td>-0.7</td><td>-1</td></tr><tr><td>\) \(</td><td></td><td>\) 5/8 \(</td><td>\) 3/4 \(</td><td>\) 7/8 \(</td><td>\) \(</td></tr><tr><td>r</td><td></td><td>-0.7</td><td>0</td><td>0.7</td><td>1</td></tr><tr><td>\) \(</td><td></td><td>\) 9/8 \(</td><td>\) 5/4 \(</td><td>\) 11/8 \(</td><td>\) 3/2 \(</td></tr><tr><td>r</td><td></td><td>0.7</td><td>0</td><td>-0.7</td><td>-1</td></tr><tr><td>\) \(</td><td></td><td>\) 13/8 \(</td><td>\) 7/4 \(</td><td>\) 15/8 \(</td><td>\) 2$ | |||||
| r | -0.7 | 0 | 0.7 | 1 |
The curve is known as a four-leaved rose.
28.10. Graph the curve specified by the parametric equations $ x = t^{2} $ , $ y = t^{2} + 2 $ .
First note that t can be eliminated by substituting x for $ t^{2} $ to obtain $ y = x + 2 $ . However, both variables are restricted by these equations so that $ x $ and therefore $ y $ . Hence the graph is the portion of the line $ y = x + 2 $ in the first quadrant, but is traced out twice, once for negative t and once for positive t. Form a table of values, then plot the points and connect them (see Fig. 28-10).
| t | -2 | -1 | 0 | 1 | 2 |
| x | 4 | 1 | 0 | 1 | 4 |
| y | 6 | 3 | 2 | 3 | 6 |
28.11. Graph the curve specified by the parametric equations $ x = 2 t $ , $ y = 2 t $ .
First note that t can be eliminated by squaring the equations specifying x and y and adding to obtain $ x^{2} + y^{2} = 4 $ . Thus the graph consists of the circle with center the origin and radius 2, and is traced out once each time t increases by an amount $ 2$ . Form a table of values, then plot the points and draw the circle (Fig. 28-11).
| t | 0 | $ /4 \(</td><td>\) /2 \(</td><td>\) 3/4 \(</td><td>\) \(</td></tr><tr><td>x</td><td>2</td><td>\) \(</td><td>0</td><td>\) - \(</td><td>-2</td></tr><tr><td>y</td><td>0</td><td>\) \(</td><td>2</td><td>\) \(</td><td>0</td></tr><tr><td>t</td><td></td><td>\) 5/4 \(</td><td>\) 3/2 \(</td><td>\) 7/4 \(</td><td>\) 2\(</td></tr><tr><td>x</td><td></td><td>\) - \(</td><td>0</td><td>\) \(</td><td>2</td></tr><tr><td>y</td><td></td><td>\) - \(</td><td>-2</td><td>\) - $ | 0 |
28.12. When a wheel of radius a rolls without slipping on a horizontal surface, the curve traced out by a point on the rim of the wheel is called a cycloid. (a) Show that the parametric equations of a cycloid can be written as
\[ x=a(\phi-\sin\phi) \]
\[ y=a(1-\cos\phi) \]
Sketch a graph of a cycloid for a = 1.
Draw a figure (see Fig. 28-12). The parameter $ $ is the angle through which the wheel has rotated.
The coordinates of P, the point on the rim, are $ (x, y) $ . Because the wheel rotates without slipping, the length of arc $ $ is equal to the length of line segment $ $ . Hence $ x = - = a- a$ and $ y = = - = a - a$ .
- In this case $ x = - $ , $ y = 1 - $ . Form a table of values and connect the points. The curve (Fig. 28-13) is shown for $ 0 $ ; for other values of $ $ the arch shape is repeated, since y is a periodic function of $ $ .
| φ | 0 | π/4 | π/2 | 3π/4 | π |
| x | 0 | 0.08 | 0.57 | 1.65 | π |
| y | 0 | 0.29 | 1 | 1.71 | 2 |
| φ | 5π/4 | 3π/2 | 7π/4 | 2π | |
| x | 4.63 | 5.71 | 6.20 | 2π | |
| y | 1.71 | 1 | 0.29 | 0 |
SUPPLEMENTARY PROBLEMS
28.13. Convert to Cartesian coordinates: $ (5,0) $ , $ (5,) $ , $ (6,-/3) $ , $ (-2,3/4) $ , $ (-20,-5/2) $ .
Ans. $ (5,0) $ , $ (-5,0) $ , $ (3,-3) $ , $ (2,-2) $ , $ (0,20) $
28.14. Convert to polar coordinates with r > 0 and 0 ≤ θ ≤ 2π: (0,2), (0,-3), (−4,4), (6, −6√3).
Ans. $ (2, /2), (3, 3/2), (4, 3/4), (12, 5/3) $
28.15. Transform the following polar coordinate equations to Cartesian coordinates:
- $ r = 3 $ ; (b) $ = /4 $ ; (c) $ r = 2 $ ; (d) $ r = 1 + $ .
\[ x^{2}+y^{2}=3y;\quad y=x;\quad x^{4}+x^{2}y^{2}=4y^{2};\quad x^{4}+y^{4}-2x^{3}-2xy^{2}+2x^{2}y^{2}-y^{2}=0 \]
28.16. Transform the following Cartesian coordinate equations to polar coordinates:
\[ y=5;(b)xy=4;(c)x^{2}+y^{2}=16;(d)x^{2}-y^{2}=16. \]
Ans. (a) $ r = 5 $ ; (b) $ r^{2} = 4 $ ; (c) r = 4; (d) $ r^{2} = 16 $
28.17. Sketch a graph of the following polar coordinate equations:
- $ r = $ ( $ 0 $ ); (b) $ r = 1 + 2$
Ans. (a) Fig. 28-14; (b) Fig. 28-15
28.18. Eliminate the parameter t and state any restrictions on the variables in the resulting equation:
\[ \left(\mathrm{a}\right)x=3t,y=2t-5;\left(\mathrm{b}\right)x=\sqrt[{}]{t-1},y=t-2;\left(\mathrm{c}\right)x=e^{t},y=e^{-t}. \]
Ans. (a) 2x - 3y = 15; (b) $ y = x^{2} - 1 $ , $ x $ ; (c) xy = 1, x, y > 0
28.19. A projectile is fired at an angle of inclination $ $ ( $ 0 < < /2 $ ) at an initial speed of $ v_{0} $ . Parametric equations for its path can be shown to be $ x = v_{0}t$ , $ y = v_{0}t- (gt^{2})/2 $ (t represents time).
Eliminate the parameter t and find the value of t when the projectile hits the ground.
Sketch the path of the projectile for the case α = π/6, ν0 = 32 ft/sec, g = 32 ft/sec2.
Ans. (a) $ y = x - (gx^{2} ^{2} ) / (2v_{0}^{2}) $ ; y = 0 when $ t = $ ; (b) Fig. 28-16
