Chapter 7: Vectors & Polar Coordinates
Vectors and polar coordinates are two of the most powerful tools in precalculus mathematics. Vectors give us a way to represent quantities that have both magnitude and direction—forces, velocities, displacements—and to compute with them algebraically. Polar coordinates offer an alternative to the familiar Cartesian \((x, y)\) system, describing points by their distance from the origin and the angle they make with the positive \(x\)-axis. Together, these topics prepare you for calculus, physics, and engineering by providing geometric intuition and computational techniques that go far beyond what rectangular coordinates alone can offer.
Table of Contents
7.1 Vector Basics: Magnitude, Direction, and Components
A scalar is a quantity described by a single number—temperature, mass, or speed. Many physical quantities, however, require both a size and a direction. A 60 mph wind blowing northeast carries very different information from a 60 mph wind blowing south. Vectors capture exactly this kind of information.
Definition: Vector
A vector is a quantity that has both magnitude (length) and direction. We denote a vector as \(\vec{v}\), \(\mathbf{v}\), or in component form \(\langle a, b \rangle\). The vector from point \(A\) to point \(B\) is written \(\overrightarrow{AB}\).
Definition: Component Form and Magnitude
If a vector \(\vec{v}\) has its initial point at the origin and terminal point at \((a, b)\), then its component form is \(\vec{v} = \langle a, b \rangle\). More generally, the vector from \(A(x_1, y_1)\) to \(B(x_2, y_2)\) is:
\[\overrightarrow{AB} = \langle x_2 - x_1,\; y_2 - y_1 \rangle\]The magnitude (or norm) of \(\vec{v} = \langle a, b \rangle\) is:
\[\|\vec{v}\| = \sqrt{a^2 + b^2}\]Two vectors are equal if and only if they have the same magnitude and the same direction. This means \(\langle a, b \rangle = \langle c, d \rangle\) if and only if \(a = c\) and \(b = d\). A vector can be freely translated in the plane without changing its identity—only the components matter, not where the vector is drawn.
Example 7.1.1
Problem: Find the component form and magnitude of the vector from \(P(3, -1)\) to \(Q(-2, 4)\).
Solution:
\[\overrightarrow{PQ} = \langle -2 - 3,\; 4 - (-1) \rangle = \langle -5, 5 \rangle\] \[\|\overrightarrow{PQ}\| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\]Example 7.1.2
Problem: A vector \(\vec{w}\) has magnitude 10 and makes an angle of \(120^\circ\) with the positive \(x\)-axis. Find its component form.
Solution:
Using the relationship between magnitude, angle, and components:
\[\vec{w} = \langle \|\vec{w}\|\cos\theta,\; \|\vec{w}\|\sin\theta \rangle = \langle 10\cos 120^\circ,\; 10\sin 120^\circ \rangle\] \[= \left\langle 10\left(-\frac{1}{2}\right),\; 10\left(\frac{\sqrt{3}}{2}\right) \right\rangle = \langle -5,\; 5\sqrt{3} \rangle\]7.2 Vector Operations
Vectors can be added, subtracted, and scaled. These operations correspond to natural geometric actions: placing vectors tip-to-tail for addition, stretching or compressing for scalar multiplication, and reversing direction for negation.
Definition: Vector Addition and Scalar Multiplication
Let \(\vec{u} = \langle u_1, u_2 \rangle\) and \(\vec{v} = \langle v_1, v_2 \rangle\) be vectors, and let \(c\) be a scalar.
- Addition: \(\vec{u} + \vec{v} = \langle u_1 + v_1,\; u_2 + v_2 \rangle\)
- Subtraction: \(\vec{u} - \vec{v} = \langle u_1 - v_1,\; u_2 - v_2 \rangle\)
- Scalar multiplication: \(c\vec{u} = \langle cu_1,\; cu_2 \rangle\)
Geometrically, \(\vec{u} + \vec{v}\) is formed by placing the tail of \(\vec{v}\) at the head of \(\vec{u}\) (the parallelogram law or tip-to-tail rule). Scalar multiplication by \(c > 0\) stretches the vector by factor \(c\); if \(c < 0\), it also reverses the direction.
Theorem: Properties of Vector Operations
For vectors \(\vec{u}\), \(\vec{v}\), \(\vec{w}\) and scalars \(c\), \(d\):
- \(\vec{u} + \vec{v} = \vec{v} + \vec{u}\) (commutativity)
- \((\vec{u} + \vec{v}) + \vec{w} = \vec{u} + (\vec{v} + \vec{w})\) (associativity)
- \(\vec{u} + \vec{0} = \vec{u}\) (additive identity)
- \(\vec{u} + (-\vec{u}) = \vec{0}\) (additive inverse)
- \(c(\vec{u} + \vec{v}) = c\vec{u} + c\vec{v}\) (distributivity over vector addition)
- \((c + d)\vec{u} = c\vec{u} + d\vec{u}\) (distributivity over scalar addition)
- \(c(d\vec{u}) = (cd)\vec{u}\) (associativity of scalar multiplication)
- \(1\vec{u} = \vec{u}\) and \(\|c\vec{u}\| = |c|\;\|\vec{u}\|\)
Example 7.2.1
Problem: Let \(\vec{u} = \langle 3, -2 \rangle\) and \(\vec{v} = \langle -1, 5 \rangle\). Compute \(2\vec{u} - 3\vec{v}\) and \(\|2\vec{u} - 3\vec{v}\|\).
Solution:
\[2\vec{u} = \langle 6, -4 \rangle, \qquad 3\vec{v} = \langle -3, 15 \rangle\] \[2\vec{u} - 3\vec{v} = \langle 6 - (-3),\; -4 - 15 \rangle = \langle 9, -19 \rangle\] \[\|2\vec{u} - 3\vec{v}\| = \sqrt{9^2 + (-19)^2} = \sqrt{81 + 361} = \sqrt{442} \approx 21.02\]Example 7.2.2
Problem: Two forces act on an object: \(\vec{F}_1 = \langle 40, 30 \rangle\) N and \(\vec{F}_2 = \langle -10, 50 \rangle\) N. Find the resultant force and its magnitude.
Solution:
\[\vec{F}_R = \vec{F}_1 + \vec{F}_2 = \langle 40 + (-10),\; 30 + 50 \rangle = \langle 30, 80 \rangle \text{ N}\] \[\|\vec{F}_R\| = \sqrt{30^2 + 80^2} = \sqrt{900 + 6400} = \sqrt{7300} = 10\sqrt{73} \approx 85.44 \text{ N}\]7.3 The Dot Product and Its Applications
The dot product is the single most important operation on vectors beyond addition and scalar multiplication. It takes two vectors and returns a scalar, and it encodes geometric information about the angle between the vectors.
Definition: Dot Product
The dot product (or scalar product) of \(\vec{u} = \langle u_1, u_2 \rangle\) and \(\vec{v} = \langle v_1, v_2 \rangle\) is:
\[\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2\]The result is a scalar, not a vector.
Theorem: Geometric Interpretation of the Dot Product
If \(\theta\) is the angle between nonzero vectors \(\vec{u}\) and \(\vec{v}\) (where \(0 \le \theta \le \pi\)), then:
\[\vec{u} \cdot \vec{v} = \|\vec{u}\|\;\|\vec{v}\|\cos\theta\]Equivalently, the angle between two vectors is:
\[\cos\theta = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\|\;\|\vec{v}\|}\]Key consequences:
- If \(\vec{u} \cdot \vec{v} > 0\), the angle is acute (\(\theta < 90^\circ\))
- If \(\vec{u} \cdot \vec{v} = 0\), the vectors are orthogonal (perpendicular, \(\theta = 90^\circ\))
- If \(\vec{u} \cdot \vec{v} < 0\), the angle is obtuse (\(\theta > 90^\circ\))
Example 7.3.1
Problem: Find the angle between \(\vec{u} = \langle 3, 4 \rangle\) and \(\vec{v} = \langle -4, 3 \rangle\).
Solution:
\[\vec{u} \cdot \vec{v} = (3)(-4) + (4)(3) = -12 + 12 = 0\]Since \(\vec{u} \cdot \vec{v} = 0\), the vectors are orthogonal. The angle between them is \(\theta = 90^\circ\).
Example 7.3.2
Problem: Find the angle between \(\vec{a} = \langle 2, 1 \rangle\) and \(\vec{b} = \langle -1, 3 \rangle\).
Solution:
\[\vec{a} \cdot \vec{b} = (2)(-1) + (1)(3) = -2 + 3 = 1\] \[\|\vec{a}\| = \sqrt{4 + 1} = \sqrt{5}, \qquad \|\vec{b}\| = \sqrt{1 + 9} = \sqrt{10}\] \[\cos\theta = \frac{1}{\sqrt{5}\;\sqrt{10}} = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}} = \frac{\sqrt{2}}{10}\] \[\theta = \cos^{-1}\!\left(\frac{\sqrt{2}}{10}\right) \approx 81.87^\circ\]Vector Projections
The projection of one vector onto another is fundamental in physics (computing work) and geometry (decomposing forces). It answers the question: how much of \(\vec{u}\) points in the direction of \(\vec{v}\)?
Theorem: Scalar and Vector Projection
The scalar projection (or component) of \(\vec{u}\) onto \(\vec{v}\) is:
\[\text{comp}_{\vec{v}}\,\vec{u} = \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|}\]The vector projection of \(\vec{u}\) onto \(\vec{v}\) is:
\[\text{proj}_{\vec{v}}\,\vec{u} = \left(\frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}\right)\vec{v} = \left(\frac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}}\right)\vec{v}\]Example 7.3.3
Problem: Find the vector projection of \(\vec{u} = \langle 6, 2 \rangle\) onto \(\vec{v} = \langle 3, 4 \rangle\).
Solution:
\[\vec{u} \cdot \vec{v} = (6)(3) + (2)(4) = 18 + 8 = 26\] \[\|\vec{v}\|^2 = 3^2 + 4^2 = 25\] \[\text{proj}_{\vec{v}}\,\vec{u} = \frac{26}{25}\langle 3, 4 \rangle = \left\langle \frac{78}{25},\; \frac{104}{25} \right\rangle\]The scalar projection is \(\dfrac{26}{5} = 5.2\).
7.4 Unit Vectors and Direction Angles
A unit vector is a vector of length 1. Unit vectors are used to describe direction independently of magnitude, and they form the building blocks for expressing any vector in terms of standard basis vectors.
Definition: Unit Vector
A unit vector is any vector with magnitude 1. The unit vector in the direction of a nonzero vector \(\vec{v}\) is:
\[\hat{v} = \frac{\vec{v}}{\|\vec{v}\|}\]The standard unit vectors are \(\mathbf{i} = \langle 1, 0 \rangle\) and \(\mathbf{j} = \langle 0, 1 \rangle\). Any vector \(\vec{v} = \langle a, b \rangle\) can be written as \(\vec{v} = a\,\mathbf{i} + b\,\mathbf{j}\).
Definition: Direction Angle
The direction angle \(\theta\) of a nonzero vector \(\vec{v} = \langle a, b \rangle\) is the angle measured counterclockwise from the positive \(x\)-axis to \(\vec{v}\):
\[\tan\theta = \frac{b}{a}\]Equivalently, \(\vec{v} = \|\vec{v}\|\langle \cos\theta, \sin\theta \rangle\). When computing \(\theta\), use the signs of \(a\) and \(b\) to determine the correct quadrant (i.e., use \(\theta = \text{atan2}(b, a)\)).
Example 7.4.1
Problem: Find the unit vector in the direction of \(\vec{v} = \langle -3, 4 \rangle\) and express \(\vec{v}\) in terms of its magnitude and direction angle.
Solution:
\[\|\vec{v}\| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5\] \[\hat{v} = \frac{1}{5}\langle -3, 4 \rangle = \left\langle -\frac{3}{5},\; \frac{4}{5} \right\rangle\]The direction angle: since \(a = -3 < 0\) and \(b = 4 > 0\), the vector is in Quadrant II.
\[\theta = 180^\circ - \arctan\!\left(\frac{4}{3}\right) \approx 180^\circ - 53.13^\circ = 126.87^\circ\]So \(\vec{v} = 5\langle \cos 126.87^\circ,\; \sin 126.87^\circ \rangle\).
Example 7.4.2
Problem: A boat heads due north at 12 km/h in a river whose current flows east at 5 km/h. Find the resultant speed and direction of the boat.
Solution:
Let north be the positive \(y\)-direction and east be positive \(x\). The boat's velocity is \(\vec{v}_b = \langle 0, 12 \rangle\) and the current is \(\vec{v}_c = \langle 5, 0 \rangle\).
\[\vec{v}_R = \langle 0 + 5,\; 12 + 0 \rangle = \langle 5, 12 \rangle\] \[\|\vec{v}_R\| = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ km/h}\] \[\theta = \arctan\!\left(\frac{12}{5}\right) \approx 67.38^\circ\]The boat actually travels at 13 km/h in a direction \(67.38^\circ\) from east (or equivalently, \(22.62^\circ\) east of north).
7.5 The Polar Coordinate System
In the Cartesian (rectangular) coordinate system, we locate a point by measuring horizontal and vertical distances from the origin. The polar coordinate system takes a fundamentally different approach: we specify a distance from the origin and an angle from the positive \(x\)-axis. This system is especially natural for describing curves with rotational symmetry, spirals, and periodic phenomena.
Definition: Polar Coordinates
In the polar coordinate system, a point \(P\) in the plane is represented by the ordered pair \((r, \theta)\), where:
- \(r\) is the directed distance from the origin (the pole) to \(P\). If \(r > 0\), the point is in the direction of \(\theta\); if \(r < 0\), the point is in the opposite direction.
- \(\theta\) is the angle measured counterclockwise from the positive \(x\)-axis (the polar axis) to the line segment from the origin to \(P\).
Unlike Cartesian coordinates, polar representations are not unique. The same point has infinitely many representations: \((r, \theta)\), \((r, \theta + 2\pi)\), \((-r, \theta + \pi)\), and so on.
The pole (origin) has the special property that \((0, \theta)\) represents the same point for every angle \(\theta\).
7.6 Converting Between Rectangular and Polar Coordinates
Conversion Formulas
Polar to Rectangular:
\[x = r\cos\theta, \qquad y = r\sin\theta\]Rectangular to Polar:
\[r = \sqrt{x^2 + y^2}, \qquad \tan\theta = \frac{y}{x}\]When converting to polar, choose \(\theta\) so that the point \((r, \theta)\) lies in the correct quadrant. It is often safest to use \(\theta = \text{atan2}(y, x)\).
Example 7.6.1
Problem: Convert the polar point \(\left(4, \dfrac{5\pi}{6}\right)\) to rectangular coordinates.
Solution:
\[x = 4\cos\frac{5\pi}{6} = 4\left(-\frac{\sqrt{3}}{2}\right) = -2\sqrt{3}\] \[y = 4\sin\frac{5\pi}{6} = 4\left(\frac{1}{2}\right) = 2\]The rectangular coordinates are \(\left(-2\sqrt{3},\; 2\right)\).
Example 7.6.2
Problem: Convert the rectangular point \((-3, -3)\) to polar coordinates with \(r > 0\) and \(0 \le \theta < 2\pi\).
Solution:
\[r = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = 3\sqrt{2}\] \[\tan\theta = \frac{-3}{-3} = 1\]Since the point is in Quadrant III, \(\theta = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}\).
The polar coordinates are \(\left(3\sqrt{2},\; \dfrac{5\pi}{4}\right)\).
Example 7.6.3
Problem: Convert the rectangular equation \(x^2 + y^2 - 6x = 0\) to a polar equation.
Solution:
Substitute \(x = r\cos\theta\), \(y = r\sin\theta\), and \(x^2 + y^2 = r^2\):
\[r^2 - 6r\cos\theta = 0\] \[r(r - 6\cos\theta) = 0\]Since \(r = 0\) is just the pole (already included when \(\theta = \pi/2\)), the polar equation is \(r = 6\cos\theta\). This is a circle of diameter 6 centered at \((3, 0)\).
7.7 Polar Equations and Graphs
The beauty of polar coordinates lies in the elegant curves they produce. Equations that would be enormously complicated in rectangular form often have simple, memorable polar forms. This section surveys the most important families of polar curves.
Circles in Polar Coordinates
The simplest polar curves are circles:
- \(r = a\): circle of radius \(|a|\) centered at the origin
- \(r = a\cos\theta\): circle of diameter \(|a|\) centered at \((a/2, 0)\)
- \(r = a\sin\theta\): circle of diameter \(|a|\) centered at \((0, a/2)\)
Cardioids and Limaçons
Definition: Limaçon
A limaçon has the form \(r = a + b\cos\theta\) or \(r = a + b\sin\theta\) where \(a > 0\) and \(b > 0\). The shape depends on the ratio \(a/b\):
- \(a/b < 1\): inner loop limaçon
- \(a/b = 1\): cardioid (heart-shaped curve passing through the origin)
- \(1 < a/b < 2\): dimpled limaçon
- \(a/b \ge 2\): convex limaçon
Rose Curves
Definition: Rose Curves
A rose curve has the form \(r = a\cos(n\theta)\) or \(r = a\sin(n\theta)\) where \(a \neq 0\) and \(n\) is a positive integer.
- If \(n\) is odd, the rose has \(n\) petals.
- If \(n\) is even, the rose has \(2n\) petals.
Each petal has length \(|a|\) (the maximum value of \(|r|\)).
Lemniscates
Definition: Lemniscate
A lemniscate is a figure-eight shaped curve with polar equation:
\[r^2 = a^2\cos(2\theta) \qquad \text{or} \qquad r^2 = a^2\sin(2\theta)\]The curve exists only where the right side is non-negative. The lemniscate \(r^2 = a^2\cos(2\theta)\) is symmetric about the polar axis, while \(r^2 = a^2\sin(2\theta)\) is symmetric about the line \(\theta = \pi/4\).
Example 7.7.1
Problem: Describe and sketch the curve \(r = 2 + 3\cos\theta\).
Solution:
This is a limaçon with \(a = 2\) and \(b = 3\). Since \(a/b = 2/3 < 1\), this is an inner loop limaçon.
- At \(\theta = 0\): \(r = 2 + 3 = 5\) (rightmost point)
- At \(\theta = \pi\): \(r = 2 - 3 = -1\) (inner loop)
- At \(\theta = \pi/2\): \(r = 2 + 0 = 2\)
- Inner loop when \(r = 0\): \(2 + 3\cos\theta = 0 \Rightarrow \cos\theta = -2/3 \Rightarrow \theta \approx 131.8^\circ\) and \(\theta \approx 228.2^\circ\)
The inner loop forms between these two angles where \(r < 0\).
Example 7.7.2
Problem: How many petals does \(r = 3\sin(4\theta)\) have, and over what interval of \(\theta\) is one petal traced?
Solution:
Since \(n = 4\) is even, the rose has \(2(4) = 8\) petals. One petal is traced where \(\sin(4\theta) \ge 0\) over a half-period. One such petal occurs for \(0 \le 4\theta \le \pi\), i.e., \(0 \le \theta \le \pi/4\).
Interactive Polar Graph Explorer
Use the interactive calculator below to explore rose curves and limaçons. Adjust the sliders for \(a\), \(b\), and \(n\) to see how the parameters affect the shape of each curve.
Try these experiments:
- Set \(n = 3\) for a 3-petal rose, then change to \(n = 4\) for an 8-petal rose
- Set \(c = d\) to see a cardioid; set \(c < d\) for an inner-loop limaçon
- Increase \(a\) to see how petal length changes
7.8 Practice Problems
Test your understanding of vectors and polar coordinates with these problems. Attempt each one before revealing the solution.
Problem 1
Given \(\vec{u} = \langle 5, -12 \rangle\), find \(\|\vec{u}\|\) and the unit vector in the direction of \(\vec{u}\).
Show Solution
\(\|\vec{u}\| = \sqrt{25 + 144} = \sqrt{169} = 13\).
\(\hat{u} = \dfrac{1}{13}\langle 5, -12 \rangle = \left\langle \dfrac{5}{13},\; -\dfrac{12}{13} \right\rangle\).
Problem 2
Let \(\vec{a} = \langle 1, 7 \rangle\) and \(\vec{b} = \langle -3, 2 \rangle\). Find \(\vec{a} \cdot \vec{b}\) and the angle between the vectors.
Show Solution
\(\vec{a} \cdot \vec{b} = (1)(-3) + (7)(2) = -3 + 14 = 11\).
\(\|\vec{a}\| = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}\), \(\|\vec{b}\| = \sqrt{9 + 4} = \sqrt{13}\).
\(\cos\theta = \dfrac{11}{5\sqrt{2}\;\sqrt{13}} = \dfrac{11}{5\sqrt{26}} = \dfrac{11\sqrt{26}}{130}\).
\(\theta = \cos^{-1}\!\left(\dfrac{11\sqrt{26}}{130}\right) \approx 64.65^\circ\).
Problem 3
Find the vector projection of \(\vec{u} = \langle 4, 3 \rangle\) onto \(\vec{v} = \langle 1, 1 \rangle\).
Show Solution
\(\vec{u} \cdot \vec{v} = 4 + 3 = 7\).
\(\|\vec{v}\|^2 = 1 + 1 = 2\).
\(\text{proj}_{\vec{v}}\,\vec{u} = \dfrac{7}{2}\langle 1, 1 \rangle = \left\langle \dfrac{7}{2},\; \dfrac{7}{2} \right\rangle\).
Problem 4
An airplane flies at 250 km/h on a bearing of \(N\,30^\circ\,W\) (i.e., \(30^\circ\) west of north). A wind blows from the east at 40 km/h. Find the ground speed and actual direction of the airplane.
Show Solution
Taking east as positive \(x\) and north as positive \(y\), the airplane's airspeed vector is:
\(\vec{v}_a = 250\langle -\sin 30^\circ,\; \cos 30^\circ \rangle = 250\left\langle -\dfrac{1}{2},\; \dfrac{\sqrt{3}}{2} \right\rangle = \langle -125,\; 125\sqrt{3} \rangle\).
Wind blows from the east (toward the west): \(\vec{v}_w = \langle -40, 0 \rangle\).
\(\vec{v}_R = \langle -125 - 40,\; 125\sqrt{3} \rangle = \langle -165,\; 125\sqrt{3} \rangle\).
\(\|\vec{v}_R\| = \sqrt{165^2 + (125\sqrt{3})^2} = \sqrt{27225 + 46875} = \sqrt{74100} \approx 272.2\) km/h.
The direction angle from east: \(\theta = \arctan\!\left(\dfrac{125\sqrt{3}}{165}\right) \approx 52.8^\circ\) in the second quadrant, so the bearing is approximately \(N\,37.2^\circ\,W\).
Problem 5
Determine whether the vectors \(\vec{u} = \langle 6, -2 \rangle\) and \(\vec{v} = \langle 1, 3 \rangle\) are orthogonal, parallel, or neither.
Show Solution
\(\vec{u} \cdot \vec{v} = (6)(1) + (-2)(3) = 6 - 6 = 0\).
Since \(\vec{u} \cdot \vec{v} = 0\), the vectors are orthogonal.
Problem 6
Convert the polar point \(\left(6, \dfrac{7\pi}{4}\right)\) to rectangular coordinates.
Show Solution
\(x = 6\cos\dfrac{7\pi}{4} = 6 \cdot \dfrac{\sqrt{2}}{2} = 3\sqrt{2}\).
\(y = 6\sin\dfrac{7\pi}{4} = 6 \cdot \left(-\dfrac{\sqrt{2}}{2}\right) = -3\sqrt{2}\).
The rectangular coordinates are \(\left(3\sqrt{2},\; -3\sqrt{2}\right)\).
Problem 7
Convert the rectangular equation \(x^2 + y^2 + 4y = 0\) to a polar equation.
Show Solution
Substitute \(x^2 + y^2 = r^2\) and \(y = r\sin\theta\):
\(r^2 + 4r\sin\theta = 0\).
\(r(r + 4\sin\theta) = 0\).
Since \(r = 0\) is the pole, the polar equation is \(r = -4\sin\theta\).
This is a circle of diameter 4 centered at \((0, -2)\).
Problem 8
Classify the curve \(r = 3 + 3\cos\theta\) and identify key features (intercepts, symmetry).
Show Solution
This is a limaçon with \(a = 3\) and \(b = 3\). Since \(a/b = 1\), it is a cardioid.
- At \(\theta = 0\): \(r = 6\) (rightmost point)
- At \(\theta = \pi/2\): \(r = 3\)
- At \(\theta = \pi\): \(r = 0\) (passes through the pole)
- Symmetric about the polar axis (since \(\cos\theta = \cos(-\theta)\))
Problem 9
How many petals does \(r = 5\cos(5\theta)\) have? Determine the interval of \(\theta\) for the petal along the positive \(x\)-axis.
Show Solution
Since \(n = 5\) is odd, the rose has \(5\) petals.
The petal along the positive \(x\)-axis occurs where \(\cos(5\theta) \ge 0\) and the curve is near \(\theta = 0\). We need \(-\pi/2 \le 5\theta \le \pi/2\), i.e., \(-\pi/10 \le \theta \le \pi/10\).
Problem 10
A force \(\vec{F} = \langle 8, 6 \rangle\) acts on an object that moves from point \(A(1, 2)\) to point \(B(5, 7)\). Find the work done by the force.
Show Solution
The displacement vector is \(\overrightarrow{AB} = \langle 5 - 1,\; 7 - 2 \rangle = \langle 4, 5 \rangle\).
Work is the dot product of force and displacement:
\(W = \vec{F} \cdot \overrightarrow{AB} = (8)(4) + (6)(5) = 32 + 30 = 62\) units of work.
Problem 11
Convert the rectangular point \((0, -5)\) to polar coordinates with \(r > 0\) and \(0 \le \theta < 2\pi\).
Show Solution
\(r = \sqrt{0^2 + (-5)^2} = 5\).
The point is on the negative \(y\)-axis, so \(\theta = \dfrac{3\pi}{2}\).
The polar coordinates are \(\left(5,\; \dfrac{3\pi}{2}\right)\).
Problem 12
For what values of \(a\) and \(b\) does the limaçon \(r = a + b\cos\theta\) (\(a, b > 0\)) have an inner loop? Determine the curve type when \(a = 1\) and \(b = 4\), and find the values of \(\theta\) where \(r = 0\).
Show Solution
An inner loop occurs when \(a < b\) (i.e., \(a/b < 1\)).
With \(a = 1\), \(b = 4\): since \(1 < 4\), this is an inner loop limaçon.
Setting \(r = 0\): \(1 + 4\cos\theta = 0 \Rightarrow \cos\theta = -\dfrac{1}{4}\).
\(\theta = \cos^{-1}\!\left(-\dfrac{1}{4}\right) \approx 104.48^\circ\) and \(\theta \approx 255.52^\circ\).