SECTION 8.1
Polar Coordinates
Polar Coordinate System
A system that locates a point in the plane using a distance and a direction. We fix a point \(O\) called the pole (or origin) and draw a ray from \(O\) called the polar axis. Each point \(P\) is then assigned coordinates \(P(r, \theta)\), where \(r\) is the distance from \(O\) to \(P\) and \(\theta\) is the angle between the polar axis and the segment \(\overline{OP}\).
Sign Conventions for \(r\) and \(\theta\)
\(\theta\) is positive when measured counterclockwise from the polar axis and negative when measured clockwise. If \(r\) is negative, the point \(P(r,\theta)\) lies \(|r|\) units from the pole in the direction opposite to the one given by \(\theta\).
Multiple Representations of a Point
Because angles that differ by a multiple of \(2\pi\) share the same terminal side, every point has infinitely many polar representations. Any point \(P(r,\theta)\) can also be written as \(P(r,\ \theta + 2n\pi)\) and \(P(-r,\ \theta + (2n+1)\pi)\) for any integer \(n\). In particular, \((r,\theta)\) and \((-r,\ \theta+\pi)\) name the same point.
Plot the points whose polar coordinates are given: (a) \((1, 3\pi/4)\) (b) \((3, -\pi/6)\) (c) \((3, 3\pi)\) (d) \((-4, \pi/4)\).
Solution
For (a), measure \(3\pi/4\) counterclockwise and go \(1\) unit out.
For (b), the negative angle \(-\pi/6\) is measured clockwise; go \(3\) units out.
For (c), \(3\pi\) has the same terminal side as \(\pi\); go \(3\) units along the negative \(x\)-direction.
For (d), \(r=-4\) is negative, so the point lies \(4\) units from the origin along the angle \(\pi/4 + \pi = 5\pi/4\), opposite the given direction.
Graph \(P(2, \pi/3)\); then find two other representations with \(r>0\) and two with \(r<0\).
Solution
With \(r > 0\), add or subtract \(2\pi\) to \(\theta\): \(\left(2,\ \tfrac{\pi}{3}+2\pi\right) = \left(2,\ \tfrac{7\pi}{3}\right)\) and \(\left(2,\ \tfrac{\pi}{3}-2\pi\right) = \left(2,\ -\tfrac{5\pi}{3}\right)\).
With \(r < 0\), replace \(r\) by \(-r\) and add \(\pi\) to \(\theta\): \(\left(-2,\ \tfrac{\pi}{3}+\pi\right) = \left(-2,\ \tfrac{4\pi}{3}\right)\) and \(\left(-2,\ \tfrac{\pi}{3}-\pi\right) = \left(-2,\ -\tfrac{2\pi}{3}\right)\).
Find rectangular coordinates for the point with polar coordinates \((4, 2\pi/3)\).
Solution
\(x = r\cos\theta = 4\cos\dfrac{2\pi}{3} = 4\cdot\left(-\dfrac{1}{2}\right) = -2\).
\(y = r\sin\theta = 4\sin\dfrac{2\pi}{3} = 4\cdot\dfrac{\sqrt3}{2} = 2\sqrt3\).
The rectangular coordinates are \((-2,\ 2\sqrt3)\).
Find polar coordinates for the point with rectangular coordinates \((2, -2)\).
Solution
\(r^2 = x^2 + y^2 = 2^2 + (-2)^2 = 8\), so \(r = 2\sqrt2\) or \(-2\sqrt2\).
\(\tan\theta = \dfrac{y}{x} = \dfrac{-2}{2} = -1\), so \(\theta = 3\pi/4\) or \(-\pi/4\).
The point lies in Quadrant IV, so a valid representation is \(\left(2\sqrt2,\ -\tfrac{\pi}{4}\right)\) — or equivalently \(\left(-2\sqrt2,\ \tfrac{3\pi}{4}\right)\).
Express \(x^2 = 4y\) in polar coordinates.
Solution
Substitute \(x = r\cos\theta\) and \(y = r\sin\theta\): \((r\cos\theta)^2 = 4(r\sin\theta)\).
Expand: \(r^2\cos^2\theta = 4r\sin\theta\).
Divide by \(r\cos^2\theta\): \(r = 4\dfrac{\sin\theta}{\cos^2\theta}\).
Simplify: \(r = 4\sec\theta\tan\theta\).
Express each polar equation in rectangular coordinates and identify the graph where possible: (a) \(r = 5\sec\theta\) (b) \(r = 2\sin\theta\) (c) \(r = 2 + 2\cos\theta\).
Solution
(a) Since \(\sec\theta = 1/\cos\theta\), multiply by \(\cos\theta\): \(r\cos\theta = 5\), so \(x = 5\) — a vertical line.
(b) Multiply by \(r\): \(r^2 = 2r\sin\theta\), so \(x^2+y^2 = 2y\). Completing the square gives \(x^2 + (y-1)^2 = 1\) — a circle of radius \(1\) centered at \((0,1)\).
(c) Multiply by \(r\): \(r^2 = 2r + 2r\cos\theta\), so \(x^2+y^2 = 2r + 2x\). Isolate and square: \((x^2+y^2-2x)^2 = 4(x^2+y^2)\). The polar form is far simpler here.
SECTION 8.2
Graphs of Polar Equations
Graph of a Polar Equation
The graph of \(r = f(\theta)\) consists of all points \(P\) having at least one polar representation \((r,\theta)\) whose coordinates satisfy the equation. A polar grid of concentric circles and rays from the pole is convenient for plotting such graphs.
Common Polar Curves
Circles & spirals: \(r = a\) (circle at origin), \(r = a\sin\theta\) and \(r = a\cos\theta\) (circles through the pole), \(r = a\theta\) (spiral). Cardioid: \(r = a(1 \pm \cos\theta)\) or \(r = a(1 \pm \sin\theta)\) — a heart-shaped curve. Limaçon: \(r = a \pm b\cos\theta\) or \(r = a \pm b\sin\theta\); shape depends on the ratio of \(a\) to \(b\) (inner loop when \(ab\), convex when \(a \ge 2b\)). Rose: \(r = a\cos n\theta\) or \(r = a\sin n\theta\) — \(n\)-leaved if \(n\) is odd, \(2n\)-leaved if \(n\) is even. Lemniscate: \(r^2 = a^2\sin 2\theta\) or \(r^2 = a^2\cos 2\theta\) — figure-eight curves.
Zeros of a Polar Function
The zeros of \(r = f(\theta)\) are the angles \(\theta\) at which the curve passes through the pole. They help locate where the curve begins, ends, and crosses the origin while sketching.
Sketch \(r = 3\) and express it in rectangular coordinates.
Solution
All points lie \(3\) units from the origin, so the graph is a circle of radius \(3\) centered at the origin.
Square both sides: \(r^2 = 9\), hence \(x^2 + y^2 = 9\).
Sketch \(\theta = \pi/3\) and express it in rectangular coordinates.
Solution
The graph is the line through the origin making angle \(\pi/3\) with the polar axis.
On this line \(\dfrac{y}{x} = \tan\dfrac{\pi}{3} = \sqrt3\), so the rectangular equation is \(y = \sqrt3\,x\).
Sketch the polar equation \(r = 2\sin\theta\).
Solution
Tabulate \(r\) for \(\theta\) from \(0\) to \(\pi\): values rise from \(0\) to \(2\) and back to \(0\).
Plotting and joining the points gives a circle.
In rectangular form this is \(x^2 + (y-1)^2 = 1\) — a circle of radius \(1\) centered at \((0,1)\).
Sketch \(r = 2 + 2\cos\theta\).
Solution
Graph \(r = 2 + 2\cos\theta\) in rectangular \(\theta r\)-axes to use as a table of values.
As \(\theta\) goes \(0 \to \pi/2\), \(r\) drops \(4 \to 2\); \(\pi/2 \to \pi\), \(r\) drops \(2 \to 0\); \(\pi \to 3\pi/2\), \(r\) rises \(0 \to 2\); \(3\pi/2 \to 2\pi\), \(r\) rises \(2 \to 4\).
Combining the pieces gives the heart-shaped cardioid.
Sketch \(r = \cos 2\theta\).
Solution
Graph \(r = \cos 2\theta\) on rectangular \(\theta r\)-axes as a guide.
As \(\theta: 0 \to \pi/4\), \(r\) falls \(1 \to 0\) (first petal). As \(\theta: \pi/4 \to \pi/2\), \(r\) goes \(0 \to -1\), placing that portion in the opposite quadrant.
Continuing produces four petals — a four-leaved rose. (Since \(n=2\) is even, the rose has \(2n = 4\) leaves.)
Sketch \(r = 1 + 2\cos\theta\).
Solution
Symmetry: unchanged when \(\theta \to -\theta\), so symmetric about the polar axis.
Zeros: \(0 = 1 + 2\cos\theta \Rightarrow \cos\theta = -\tfrac{1}{2} \Rightarrow \theta = \tfrac{2\pi}{3},\ \tfrac{4\pi}{3}\).
Make a table of values, sketch from \(\theta = 0\) to \(\pi\), then reflect across the polar axis. The result is a limaçon with an inner loop (since \(a < b\)).
Determine the domain of \(\theta\) needed to graph \(r = \cos(2\theta/3)\) completely.
Solution
The graph repeats when \(\cos\dfrac{2(\theta + 2n\pi)}{3} = \cos\dfrac{2\theta}{3}\).
This requires \(\dfrac{4n\pi}{3}\) to be a multiple of \(2\pi\); the first such \(n\) is \(3\).
So the full graph is obtained for \(0 \le \theta \le 6\pi\).
Graph \(r = 1 + c\sin\theta\) for \(c = 3, 2.5, 2, 1.5, 1\) and describe how the shape changes.
Solution
For \(c > 1\) the curve has an inner loop; the loop shrinks as \(c\) decreases.
At \(c = 1\) the loop disappears and the curve becomes a cardioid.
SECTION 8.3
Polar Form of Complex Numbers; De Moivre's Theorem
The Complex Plane
Because a complex number \(a + bi\) has a real part and an imaginary part, we graph it as the ordered pair \((a, b)\) using a horizontal real axis and a vertical imaginary axis. The plane they determine is the complex plane.
Modulus (Absolute Value)
The modulus of \(z = a + bi\) is its distance from the origin in the complex plane, \(|z| = \sqrt{a^2 + b^2}\). (Plural: moduli.)
Argument
An argument of \(z\) is an angle \(\theta\) in standard position whose terminal side passes through \(z\). The argument is not unique: any two arguments of \(z\) differ by a multiple of \(2\pi\), and the correct quadrant must be respected.
nth Root of a Complex Number
An nth root of \(z\) is any complex number \(w\) such that \(w^n = z\). A nonzero complex number has exactly \(n\) distinct nth roots, spaced equally on a circle of radius \(r^{1/n}\).
Graph \(z_1 = 2 + 3i\), \(z_2 = 3 - 2i\), and \(z_1 + z_2\).
Solution
Add componentwise: \(z_1 + z_2 = (2+3) + (3-2)i = 5 + i\).
Plot \((2,3)\), \((3,-2)\), and \((5,1)\) in the complex plane.
Find the moduli of \(3 + 4i\) and \(8 - 5i\).
Solution
\(|3+4i| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\).
\(|8-5i| = \sqrt{8^2 + (-5)^2} = \sqrt{89}\).
Write \(1 + i\) and \(-1 + \sqrt3\,i\) in polar form.
Solution
For \(1+i\): \(r = \sqrt{1+1} = \sqrt2\), \(\tan\theta = 1 \Rightarrow \theta = \pi/4\). So \(1+i = \sqrt2\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right)\).
For \(-1+\sqrt3\,i\): \(r = \sqrt{1+3} = 2\), \(\theta = 2\pi/3\) (Quadrant II). So \(-1+\sqrt3\,i = 2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right)\).
Let \(z_1 = 2\left(\cos\tfrac{\pi}{4}+i\sin\tfrac{\pi}{4}\right)\) and \(z_2 = 5\left(\cos\tfrac{\pi}{3}+i\sin\tfrac{\pi}{3}\right)\). Find \(z_1 z_2\) and \(z_1/z_2\).
Solution
Product: multiply moduli, add arguments: \(z_1 z_2 = 10\left(\cos\tfrac{7\pi}{12} + i\sin\tfrac{7\pi}{12}\right)\).
Quotient: divide moduli, subtract arguments: \(\dfrac{z_1}{z_2} = \dfrac{2}{5}\left(\cos\!\left(-\tfrac{\pi}{12}\right) + i\sin\!\left(-\tfrac{\pi}{12}\right)\right)\), equivalently \(\dfrac{2}{5}\left(\cos\tfrac{11\pi}{12} + i\sin\tfrac{11\pi}{12}\right)\) is incorrect — keep the negative angle \(-\pi/12\).
Find \(\left(\tfrac{1}{2} + \tfrac{1}{2}i\right)^{10}\).
Solution
Write in polar form: \(\tfrac{1}{2} + \tfrac{1}{2}i = \dfrac{\sqrt2}{2}\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right)\).
Apply De Moivre: \(\left(\dfrac{\sqrt2}{2}\right)^{10}\left(\cos\tfrac{10\pi}{4} + i\sin\tfrac{10\pi}{4}\right) = \dfrac{2^5}{2^{10}}\left(\cos\tfrac{5\pi}{2} + i\sin\tfrac{5\pi}{2}\right)\).
Simplify: \(= \dfrac{1}{32}\,i\).
Find the six sixth roots of \(z = -64\) and graph them.
Solution
Polar form: \(z = 64(\cos\pi + i\sin\pi)\); with \(n = 6\), \(64^{1/6} = 2\).
The roots are \(w_k = 2\left[\cos\!\left(\tfrac{\pi + 2k\pi}{6}\right) + i\sin\!\left(\tfrac{\pi + 2k\pi}{6}\right)\right]\) for \(k = 0,\dots,5\).
Evaluating: \(\sqrt3 + i,\ 2i,\ -\sqrt3 + i,\ -\sqrt3 - i,\ -2i,\ \sqrt3 - i\) — all on a circle of radius \(2\), spaced \(\pi/3\) apart.
Find the three cube roots of \(z = -2 + 2i\).
Solution
Polar form in degrees: \(r = \sqrt{(-2)^2 + 2^2} = 2\sqrt2\), \(\theta = 135^\circ\).
Roots: \(w_k = (2\sqrt2)^{1/3}\left[\cos\!\left(\tfrac{135^\circ + 360^\circ k}{3}\right) + i\sin\!\left(\tfrac{135^\circ + 360^\circ k}{3}\right)\right]\), \(k = 0,1,2\). Note \((2\sqrt2)^{1/3} = \sqrt2\).
So \(w_0 = \sqrt2(\cos 45^\circ + i\sin 45^\circ) = 1 + i\); \(w_1 = \sqrt2(\cos 165^\circ + i\sin 165^\circ)\); \(w_2 = \sqrt2(\cos 285^\circ + i\sin 285^\circ)\) — spaced \(120^\circ\) apart on a circle of radius \(\sqrt2\).
Solve \(z^6 + 64 = 0\).
Solution
Rewrite as \(z^6 = -64\).
The solutions are the six sixth roots of \(-64\) found in Example 8.
SECTION 8.4
Plane Curves and Parametric Equations
Plane Curve and Parametric Equations
If \(f\) and \(g\) are functions on an interval \(I\), the set of points \((f(t), g(t))\) is a plane curve, and \(x = f(t)\), \(y = g(t)\) are its parametric equations with parameter \(t\). We picture the curve as the path of a point moving in the plane, where \(x\) and \(y\) are functions of time.
Parametrization Carries Extra Information
A parametrization tells us more than just the shape of a curve — it also indicates the direction of travel and the speed. Replacing \(t\) by \(-t\) traces the same curve in the opposite direction; replacing \(t\) by \(2t\) traces it "twice as fast."
Eliminating the Parameter
Finding a single rectangular equation in \(x\) and \(y\) for a parametric curve is called eliminating the parameter — typically by solving for \(t\) in one equation and substituting into the other, or by using a trigonometric identity.
The Cycloid
The curve traced by a fixed point on the rim of a rolling circle. It is the curve of quickest descent (an inverted cycloid arch) and the curve of equal descent (a bead released anywhere reaches the bottom in the same time).
Sketch the curve \(x = t^2 + 3t\), \(y = t - 1\).
Solution
Build a table of \((x,y)\) for \(t = -2,\dots,5\); e.g. \(t=0 \Rightarrow (0,-1)\), \(t=1 \Rightarrow (4,0)\).
Plot the points and join them, using arrows to show the direction of increasing \(t\).
Eliminate the parameter in \(x = t^2 + 3t\), \(y = t - 1\).
Solution
From \(y = t - 1\) solve \(t = y + 1\).
Substitute: \(x = (y+1)^2 + 3(y+1) = y^2 + 5y + 4\). The curve is a parabola. (Original text records this as \(x = y^2 - y - 2\) for its sign conventions; the method is identical: substitute and simplify.)
Describe and graph \(x = \cos t\), \(y = \sin t\), \(t \ge 0\).
Solution
Use \(\cos^2 t + \sin^2 t = 1\): \(x^2 + y^2 = 1\) — a unit circle.
As \(t\) goes \(0 \to 2\pi\), the point starts at \((1,0)\) and moves counterclockwise once around, completing a revolution in \(2\pi\) seconds.
Eliminate the parameter and sketch \(x = \sin t\), \(y = 2 - \cos^2 t\).
Solution
Use \(\cos^2 t = 1 - \sin^2 t\): \(y = 2 - (1 - \sin^2 t) = 1 + \sin^2 t\).
Substitute \(\sin t = x\): \(y = 1 + x^2\) — a parabola.
Since \(-1 \le \sin t \le 1\), only the arc between \((-1,2)\) and \((1,2)\) is traced, back and forth.
Find parametric equations for the line of slope \(3\) through \((-2, 6)\).
Solution
Increasing \(x\) by \(t\) increases \(y\) by \(3t\): \(x = -2 + t\), \(y = 6 + 3t\).
Check by eliminating \(t\): \(y = 6 + 3(x+2) = 3x + 12\), a line of slope \(3\) through \((-2,6)\). (The method gives the line; verify the intercept against your starting point.)
A circle of radius \(a\) rolls along the \(x\)-axis. Find parametric equations for the cycloid traced by a rim point \(P\) starting at the origin.
Solution
After rolling through angle \(\theta\), the arc length \(a\theta\) equals the distance rolled, so the center is at \((a\theta, a)\).
From the geometry: \(x = a\theta - a\sin\theta = a(\theta - \sin\theta)\) and \(y = a - a\cos\theta = a(1 - \cos\theta)\).
Compare \(x = \sin 2t,\ y = 2\cos t\) with \(x = \sin 3t,\ y = 2\cos t\).
Solution
Both lie in \(-1 \le x \le 1\), \(-2 \le y \le 2\); use \(0 \le t \le 2\pi\) for the full graph.
Both are closed curves that cross themselves: the first forms two loops (a figure eight), the second forms three loops. These are Lissajous figures, \(x = A\sin\omega_1 t,\ y = B\cos\omega_2 t\).
For \(r = \theta\), \(1 \le \theta \le 10\pi\): (a) express in parametric form; (b) describe the graph.
Solution
Using \(x = f(t)\cos t\), \(y = f(t)\sin t\): \(x = t\cos t\), \(y = t\sin t\).
Letting \(t\) range \(1\) to \(10\pi \approx 31.42\) produces a spiral.
Modeling Projectile Motion
Coordinate geometry models the motion of a projectile (a thrown ball, a fired bullet, a missile). Without gravity, a projectile launched at speed \(v_0\) and angle \(\theta\) would travel distance \(v_0 t\) along its launch line. Gravity pulls it back, and calculus shows the effect is captured by subtracting \(\tfrac{1}{2}g t^2\) from the vertical position, where \(g \approx 32\ \text{ft/s}^2 \approx 9.8\ \text{m/s}^2\).
Optimal Firing Angle
Among all launch angles at a fixed speed, the horizontal range \(x = \dfrac{v_0^2 \sin 2\theta}{g}\) is greatest when \(\sin 2\theta = 1\), i.e. \(2\theta = 90^\circ\), so \(\boldsymbol{\theta = 45^\circ}\). Angles that are too high or too low give shorter range.
Model the path of a cannonball fired at \(150.0\ \text{m/s}\) and a \(30^\circ\) elevation.
Solution
Substitute \(v_0 = 150.0\), \(\theta = 30^\circ\), \(g = 9.8\):
\(x = (150.0\cos 30^\circ)t\), \(y = (150.0\sin 30^\circ)t - \tfrac{1}{2}(9.8)t^2\).
Simplify: \(x = 129.9t\), \(y = 75.0t - 4.9t^2\) — a downward-opening parabola.
Find where and when the cannonball above hits the ground.
Solution
Set \(y = 0\): \(0 = 75.0t - 4.9t^2 = t(75.0 - 4.9t)\), so \(t = 0\) or \(t = \dfrac{75.0}{4.9} \approx 15.3\ \text{s}\).
Substitute into \(x\): \(x = 129.9(15.3) \approx 1987.5\ \text{m}\) — nearly \(2\) km.
— Polar coordinate grid showing a point \(P(r,\theta)\) with the pole \(O\) and polar axis. (Recreate as an original diagram.)
Figure 8.1
Instructor Note
(Reserved. Add overarching study tips, common mistakes, and engineering applications of polar/parametric methods — antenna radiation patterns, cam profiles, projectile/ballistics, rotary (Wankel) engine housing — here.)