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cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1119 Given the polar equation of a cardioid, sketch its graph. 1. Check equation for the three types of symmetry. 2. Find the zeros. Set r = 0. 3. Find the maximum value of the equation according to the maximum value of the trigonometric expression. 4. Make a table of values for r and θ. 5. Plot the points and sketch the graph. Example 10.27 Sketching the Graph of a Cardioid Sketch the graph of r = 2 + 2cos θ. Solution First, testing the equation for symmetry, we find that the graph of this equation will be symmetric about the polar axis. Next, we find the zeros and maximums. Setting r = 0, we have θ = π + 2kπ. The zero of the equation is located at (0, π). The graph passes through this point. The maximum value of r = 2 + 2cos θ occurs when cos θ is a maximum, which is when cos θ = 1 or when θ = 0. Substitute θ = 0 into the equation, and solve for r. r = 2 + 2cos(0) r = 2 + 2(1) = 4 The point (4, 0) is the maximum value on the graph. We found that the polar equation is symmetric with respect to the polar axis, but as it extends to all four quadrants, we need to plot values over the interval [0, π]. The upper portion of the graph is then reflected over the polar axis. Next, we make a table of values, as in Table 10.3, and then we plot the points and draw the graph. See Figure 10.74. θ r 0 π 4 π 2 2π 3 4 3.41 2 1 π 0 Table 10.3 1120 Chapter 10 Further Applications of Trigonometry Figure 10.74 Investigating Limaçons The word limaçon is Old French for “snail,” a name that describes the shape of the graph. As mentioned earlier, the cardioid is a member of the limaçon family, and we can see the similarities in the graphs. The other images in this category include the one-loop limaçon and the two-loop (or inner-loop) limaçon. One-loop
limaçons are sometimes referred to as dimpled b ≥ 2. limaçons when 1 < b < 2 and convex limaçons when a a Formulas for One-Loop Limaçons The formulas that produce the graph of a dimpled one-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ a b < 2. All four graphs are shown in Figure 10.75. where a > 0, b > 0, and 1< Figure 10.75 Dimpled limaçons This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1121 Given a polar equation for a one-loop limaçon, sketch the graph. 1. Test the equation for symmetry. Remember that failing a symmetry test does not mean that the shape will not exhibit symmetry. Often the symmetry may reveal itself when the points are plotted. 2. Find the zeros. 3. Find the maximum values according to the trigonometric expression. 4. Make a table. 5. Plot the points and sketch the graph. Example 10.28 Sketching the Graph of a One-Loop Limaçon Graph the equation r = 4 − 3sin θ. Solution First, testing the equation for symmetry, we find that it fails all three symmetry tests, meaning that the graph may or may not exhibit symmetry, so we cannot use the symmetry to help us graph it. However, this equation has a, yet it fails all the three symmetry tests. A graph that clearly displays symmetry with respect to the line θ = π 2 graphing calculator will immediately illustrate the graph’s reflective quality. Next, we find the zeros and maximum, and plot the reflecting points to verify any symmetry. Setting r = 0 results in θ being undefined. What does this mean? How could θ be undefined? The angle θ is undefined for any value of sin θ > 1. Therefore, θ is undefined because there is no value of θ for which sin θ > 1. Consequently, the graph does not pass through the pole. Perhaps the graph does cross the polar axis, but not at the pole. We can investigate other intercepts by calculating r when θ = 0. r(0) = 4 − 3sin(0 So, there is at least one polar axis intercept at (4, 0). Next
, as the maximum value of the sine function is 1 when θ =, we will substitute θ = π 2 into the equation π 2 and solve for r. Thus, r = 1. Make a table of the coordinates similar to Table 10.4 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π 4 2.5 1.4 1 1.4 2.5 4 5.5 6.6 7 6.6 5.5 4 Table 10.4 The graph is shown in Figure 10.76. 1122 Chapter 10 Further Applications of Trigonometry Figure 10.76 One-loop limaçon Analysis This is an example of a curve for which making a table of values is critical to producing an accurate graph. The symmetry tests fail; the zero is undefined. While it may be apparent that an equation involving sin θ is likely symmetric with respect to the line θ =, evaluating more points helps to verify that the graph is correct. π 2 10.18 Sketch the graph of r = 3 − 2cos θ. Another type of limaçon, the inner-loop limaçon, is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer(1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing. A century later, the father of mathematician Blaise Pascal, Étienne Pascal(1588-1651), rediscovered it. Formulas for Inner-Loop Limaçons The formulas that generate the inner-loop limaçons are given by r = a ± bcos θ and r = a ± bsin θ where a > 0, b > 0, and a < b. The graph of the inner-loop limaçon passes through the pole twice: once for the outer loop, and once for the inner loop. See Figure 10.77 for the graphs. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1123 Figure 10.77 Example 10.29 Sketching the Graph of an Inner-Loop Limaçon Sketch the graph of r = 2 + 5cos θ. Solution Testing for symmetry, we find that the graph
of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when r = 0, θ = 1.98. The maximum \|r\| is found when cos θ = 1 or when θ = 0. Thus, the maximum is found at the point (7, 0). Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge. See Table 10.5 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π 7 6.3 4.5 2 −0.5 −2.3 −3 −2.3 −0.5 2 4.5 6.3 7 Table 10.5 As expected, the values begin to repeat after θ = π. The graph is shown in Figure 10.78. 1124 Chapter 10 Further Applications of Trigonometry Figure 10.78 Inner-loop limaçon Investigating Lemniscates The lemniscate is a polar curve resembling the infinity symbol ∞ or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition. Formulas for Lemniscates The formulas that generate the graph of a lemniscate are given by r 2 = a2 cos 2θ and r 2 = a2 sin 2θ where a ≠ 0. The formula r 2 = a2 sin 2θ is symmetric with respect to the pole. The formula r 2 = a2 cos 2θ is symmetric with respect to the pole, the line θ =, and the polar axis. See Figure 10.79 for the graphs. π 2 Figure 10.79 Example 10.30 Sketching the Graph of a Lemniscate This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1125 Sketch the graph of r 2 = 4cos 2θ. Solution The equation exhibits symmetry with respect to the line θ =, the polar axis, and the pole. π 2 Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution u = 2θ. 0 = 4cos 2θ 0 = 4cos u 0 = cos u cos− = 2θ = �
� = Substitute 2θ back in for u. So, the point ⎛ ⎝0, ⎞ ⎠ π 4 is a zero of the equation. Now let’s find the maximum value. Since the maximum of cos u = 1 when u = 0, the maximum cos 2θ = 1 when 2θ = 0. Thus, r 2 = 4cos(0) r 2 = 4(1) = 4 r = ± 4 = 2 We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line θ =, and the π 2 polar axis, we only need to plot points in the first quadrant. Make a table similar to Table 10.6 Table 10.6 Plot the points on the graph, such as the one shown in Figure 10.80. Figure 10.80 Lemniscate 1126 Chapter 10 Further Applications of Trigonometry Analysis Making a substitution such as u = 2θ is a common practice in mathematics because it can make calculations simpler. However, we must not forget to replace the substitution term with the original term at the end, and then solve for the unknown. Some of the points on this graph may not show up using the Trace function on the TI-84 graphing calculator, and the calculator table may show an error for these same points of r. This is because there are no real square roots for these values of θ. In other words, the corresponding r-values of 4cos(2θ) are complex numbers because there is a negative number under the radical. Investigating Rose Curves The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern. Rose Curves The formulas that generate the graph of a rose curve are given by r = acos nθ and r = asin nθ where a ≠ 0. If n is even, the curve has 2n petals. If n is odd, the curve has n petals. See Figure 10.81. Figure 10.81 Example 10.31 Sketching the Graph of a Rose Curve (n Even) Sketch the graph of r = 2cos 4θ. Solution Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the
line θ = and the pole. π 2 Now we will find the zeros. First make the substitution u = 4θ. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1127 0 = 2cos 4θ 0 = cos 4θ 0 = cos u cos− = 4θ = The zero is θ =. The point ⎛ ⎝0, π 8 π 8 ⎞ ⎠ is on the curve. Next, we find the maximum \|r\|. We know that the maximum value of cos u = 1 when θ = 0. Thus, r = 2cos(4 ⋅ 0) r = 2cos(0) r = 2(1) = 2 The point (2, 0) is on the curve. The graph of the rose curve has unique properties, which are revealed in Table 10.7. π 8 π 4 3π 8 π 2 5π 8 3π 4 0 −2 0 2 0 −2 θ r 0 2 Table 10.7 As r = 0 when θ = π 8, it makes sense to divide values in the table by π 8 units. A definite pattern emerges. Look at the range of r-values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at r = 0, each petal extends out a distance of r = 2, and then turns back to zero 2n times for a total of eight petals. See the graph in Figure 10.82. Figure 10.82 Rose curve, n even Analysis 1128 Chapter 10 Further Applications of Trigonometry When these curves are drawn, it is best to plot the points in order, as in the Table 10.7. This allows us to see how the graph hits a maximum (the tip of a petal), loops back crossing the pole, hits the opposite maximum, and loops back to the pole. The action is continuous until all the petals are drawn. 10.19 Sketch the graph of r = 4sin(2θ). Example 10.32 Sketching the Graph of a Rose Curve (n Odd) Sketch the graph of r = 2sin(5θ). Solution The graph of the equation shows symmetry with respect to
the line θ =. Next, find the zeros and maximum. π 2 We will want to make the substitution u = 5θ. 0 = 2sin(5θ) 0 = sin u sin−1 0 = 0 u = 0 5θ = 0 θ = 0 The maximum value is calculated at the angle where sin θ is a maximum. Therefore, ⎛ r = 2sin ⎝5 ⋅ ⎞ ⎠ π 2 r = 2(1) = 2 Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for n odd yields the same number of petals as n, there will be five petals on the graph. See Figure 10.83. Create a table of values similar to Table 10.8. π 6 π 3 π 2 2π 3 5π 6 1 −1.73 2 −1.73 1 π 0 θ r 0 0 Table 10.8 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1129 Figure 10.83 Rose curve, n odd 10.20 Sketch the graph of r = 3cos(3θ). Investigating the Archimedes’ Spiral The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE - c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics. Archimedes’ Spiral The formula that generates the graph of the Archimedes’ spiral is given by r = θ for θ ≥ 0. As θ increases, r increases at a constant rate in an ever-widening, never-ending, spiraling path. See Figure 10.84. Figure 10.84 1130 Chapter 10 Further Applications of Trigonometry Given an Archimedes’ spiral over [0, 2π], sketch the graph. 1. Make a table of values for r and θ over the given domain. 2. Plot the points and sketch the graph. Example 10.33 Sketching the Graph of an Archimedes’ Spiral Sketch the graph of r = θ over [0, 2π]. Solution As r is equal to θ, the plot of the Archimedes’ spiral begins
at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted. Create a table such as Table 10.9. θ r π 4 π 2 π 3π 2 7π 4 2π 0.785 1.57 3.14 4.71 5.50 6.28 Table 10.9 Notice that the r-values are just the decimal form of the angle measured in radians. We can see them on a graph in Figure 10.85. Figure 10.85 Archimedes’ spiral Analysis The domain of this polar curve is [0, 2π]. In general, however, the domain of this function is (−∞, ∞). Graphing the equation of the Archimedes’ spiral is rather simple, although the image makes it seem like it would be complex. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1131 10.21 Sketch the graph of r = − θ over the interval [0, 4π]. Summary of Curves We have explored a number of seemingly complex polar curves in this section. Figure 10.86 and Figure 10.87 summarize the graphs and equations for each of these curves. Figure 10.86 Figure 10.87 Access these online resources for additional instruction and practice with graphs of polar coordinates. • Graphing Polar Equations Part 1 (http://openstaxcollege.org/l/polargraph1) • Graphing Polar Equations Part 2 (http://openstaxcollege.org/l/polargraph2) • Animation: The Graphs of Polar Equations (http://openstaxcollege.org/l/polaranim) • Graphing Polar Equations on the TI-84 (http://openstaxcollege.org/l/polarTI84) 1132 Chapter 10 Further Applications of Trigonometry 10.4 EXERCISES Verbal 240. Describe the three types of symmetry in polar graphs, and compare them to the symmetry of the Cartesian plane. 241. Which of the three types of symmetries for polar graphs correspond to the symmetries with respect to the xaxis, y-axis, and origin? What are the steps to follow when graphing polar 242.
equations? Describe the shapes of the graphs of cardioids, 243. limaçons, and lemniscates. What part of the equation determines the shape of the 244. graph of a polar equation? Graphical 261. r = 3 + 2sin θ 262. r = 7 + 4sin θ 263. r = 4 + 3cos θ 264. r = 5 + 4cos θ 265. r = 10 + 9cos θ 266. r = 1 + 3sin θ 267. r = 2 + 5sin θ 268. r = 5 + 7sin θ 269. r = 2 + 4cos θ For the following exercises, test the equation for symmetry. 270. r = 5 + 6cos θ 245. r = 5cos 3θ 246. r = 3 − 3cos θ 247. r = 3 + 2sin θ 248. r = 3sin 2θ 249. r = 4 250. r = 2θ 251. r = 4cos θ 2 252. r = 2 θ 253. r = 3 1 − cos2 θ 254. r = 5sin 2θ For the following exercises, graph the polar equation. Identify the name of the shape. 255. r = 3cos θ 256. r = 4sin θ 257. r = 2 + 2cos θ 258. r = 2 − 2cos θ 259. r = 5 − 5sin θ 260. r = 3 + 3sin θ This content is available for free at https://cnx.org/content/col11758/1.5 271. r 2 = 36cos(2θ) 272. r 2 = 10cos(2θ) 273. r 2 = 4sin(2θ) 274. r 2 = 10sin(2θ) 275. r = 3sin(2θ) 276. r = 3cos(2θ) 277. r = 5sin(3θ) 278. r = 4sin(4θ) 279. r = 4sin(5θ) 280. r = −θ 281. r = 2θ 282. r = − 3θ Technology For the following exercises, use a graphing calculator to sketch the graph of the polar equation. 283. r = 1 θ 284. r = 1 θ Chapter 10 Further Applications of Trigonometry 1133 285.
r = 2sin θtan θ, a cissoid 286. r = 2 1 − sin2 θ, a hippopede On a graphing utility, graph each polar equation. 302. Explain the similarities and differences you observe in the graphs. r1 = 3θ r2 = 2θ r3 = θ Extensions For the following exercises, draw each polar equation on the same set of polar axes, and find the points of intersection. 303. r1 = 3 + 2sin θ, r2 = 2 304. r1 = 6 − 4cos θ, r2 = 4 305. r1 = 1 + sin θ, r2 = 3sin θ 306. r1 = 1 + cos θ, r2 = 3cos θ 307. r1 = cos(2θ), r2 = sin(2θ) 308. 309. 310. r1 = sin2 (2θ), r2 = 1 − cos(4θ) r1 = 3, r2 = 2sin(θ) r1 2 = sin θ, r2 2 = cos θ 311. r1 = 1 + cos θ, r2 = 1 − sin θ 287. r = 5 + cos(4θ) 288. r = 2 − sin(2θ) 289. r = θ 2 290. r = θ + 1 291. r = θsin θ 292. r = θcos θ For the following exercises, use a graphing utility to graph each pair of polar equations on a domain of [0, 4π] and then explain the differences shown in the graphs. 293. r = θ, r = − θ 294. r = θ, r = θ + sin θ 295. r = sin θ + θ, r = sin θ − θ 296. 297. 298. ⎛ r = 2sin ⎝ θ 2 ⎞ ⎛ ⎠, r = θsin ⎝ ⎞ ⎠ θ 2 r = sin⎛ ⎝cos(3θ)⎞ ⎠ r = sin(3θ) On a graphing utility, [0, 4π], [0, 8π], [0, 12π], and ⎡ effect of increasing the width of the domain.
⎣0, 16π⎤ ⎛ graph r = sin ⎝ 16 5 ⎦. Describe θ⎞ ⎠ on the 299. On a r = sin θ + graphing 3 ⎛ ⎛ ⎝sin ⎝ ⎞ θ⎞ ⎠ ⎠ 5 2 on [0, 4π]. utility, graph and sketch 300. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. r1 = 3sin(3θ) r2 = 2sin(3θ) r3 = sin(3θ) 301. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. r1 = 3 + 3cos θ r2 = 2 + 2cos θ r3 = 1 + cos θ 1134 Chapter 10 Further Applications of Trigonometry 10.5 \| Polar Form of Complex Numbers Learning Objectives In this section, you will: 10.5.1 Plot complex numbers in the complex plane. 10.5.2 Find the absolute value of a complex number. 10.5.3 Write complex numbers in polar form. 10.5.4 Convert a complex number from polar to rectangular form. 10.5.5 Find products of complex numbers in polar form. 10.5.6 Find quotients of complex numbers in polar form. 10.5.7 Find powers of complex numbers in polar form. 10.5.8 Find roots of complex numbers in polar form. “God made the integers; all else is the work of man.” This rather famous quote by nineteenth-century German mathematician Leopold Kronecker sets the stage for this section on the polar form of a complex number. Complex numbers were invented by people and represent over a thousand years of continuous investigation and struggle by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and others. Complex numbers answered questions that for centuries had puzzled the greatest minds in science. We first encountered complex numbers in Complex Numbers. In this section, we will focus on the mechanics of working with complex numbers: translation of complex numbers from polar form to rectangular form and vice versa, interpretation of complex numbers in the scheme of applications, and application of De Moivre’s Theorem. Plotting Complex Numbers in the Complex
Plane Plotting a complex number a + bi is similar to plotting a real number, except that the horizontal axis represents the real part of the number, a, and the vertical axis represents the imaginary part of the number, bi. Given a complex number a + bi, plot it in the complex plane. 1. Label the horizontal axis as the real axis and the vertical axis as the imaginary axis. 2. Plot the point in the complex plane by moving a units in the horizontal direction and b units in the vertical direction. Example 10.34 Plotting a Complex Number in the Complex Plane Plot the complex number 2 − 3i in the complex plane. Solution From the origin, move two units in the positive horizontal direction and three units in the negative vertical direction. See Figure 10.88. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1135 Figure 10.88 10.22 Plot the point 1 + 5i in the complex plane. Finding the Absolute Value of a Complex Number The first step toward working with a complex number in polar form is to find the absolute value. The absolute value of a complex number is the same as its magnitude, or \|z\|. It measures the distance from the origin to a point in the plane. For example, the graph of z = 2 + 4i, in Figure 10.89, shows \|z\|. Figure 10.89 Absolute Value of a Complex Number Given z = x + yi, a complex number, the absolute value of z is defined as It is the distance from the origin to the point (x, y). \|z\| = x2 + y2 1136 Chapter 10 Further Applications of Trigonometry Notice that the absolute value of a real number gives the distance of the number from 0, while the absolute value of a complex number gives the distance of the number from the origin, (0, 0). Example 10.35 Finding the Absolute Value of a Complex Number with a Radical Find the absolute value of z = 5 − i. Solution Using the formula, we have See Figure 10.90. 2 + (−1)2 \|z\| = x2 + y2 \|z\| = ⎛ ⎝ 5⎞ \|z\| = 5 + 1 \|z\| = 6 ⎠ Figure 10.90 10.23 Find the absolute value of the complex number z = 12 − 5i. Example 10.36
Finding the Absolute Value of a Complex Number Given z = 3 − 4i, find \|z\|. Solution Using the formula, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1137 \|z\| = x2 + y2 \|z\| = (3)2 + (−4)2 \|z\| = 9 + 16 \|z\| = 25 \|z\| = 5 The absolute value z is 5. See Figure 10.91. Figure 10.91 10.24 Given z = 1 − 7i, find \|z\|. Writing Complex Numbers in Polar Form The polar form of a complex number expresses a number in terms of an angle θ and its distance from the origin r. Given a complex number in rectangular form expressed as z = x + yi, we use the same conversion formulas as we do to write the number in trigonometric form: We review these relationships in Figure 10.92. x = rcos θ y = rsin θ r = x2 + y2 1138 Chapter 10 Further Applications of Trigonometry Figure 10.92 We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point (x, y). The modulus, then, is the same as r, the radius in polar form. We use θ to indicate the angle of direction (just as with polar coordinates). Substituting, we have z = x + yi z = rcos θ + (rsin θ)i z = r(cos θ + isin θ) Polar Form of a Complex Number Writing a complex number in polar form involves the following conversion formulas: Making a direct substitution, we have x = rcos θ y = rsin θ r = x2 + y2 z = x + yi z = (rcos θ) + i(rsin θ) z = r(cos θ + isin θ) where r is the modulus and θ is the argument. We often use the abbreviation rcis θ to represent r(cos θ + isin θ). Example 10.37 Expressing a Complex Number Using Polar Coordinates Express the complex number 4i using polar coordinates. Solution On the complex plane, the number z = 4i is the same as z = 0 + 4i. Writing it in
polar form, we have to calculate r first. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1139 r = x2 + y2 r = 02 + 42 r = 16 r = 4 Next, we look at x. If x = rcos θ, and x = 0, then θ = the complex number z = 0 + 4i can be written as z = 4 ⎛ ⎛ ⎝cos ⎝ π 2 ⎞ ⎛ ⎠ + isin ⎝ π 2 π. In polar coordinates, 2 ⎝ π ⎞ ⎞ ⎛ ⎠ or 4cis ⎠ 2 ⎞ ⎠. See Figure 10.93. Figure 10.93 10.25 Express z = 3i as r cis θ in polar form. Example 10.38 Finding the Polar Form of a Complex Number Find the polar form of − 4 + 4i. Solution First, find the value of r. r = x2 + y2 r = (−4)2 + r = 32 r = 4 2 ⎛ ⎝42⎞ ⎠ Find the angle θ using the formula: 1140 Chapter 10 Further Applications of Trigonometry x r cos θ = cos θ = −4 4 2 cos θ = − 1 2 θ = cos−1 ⎛ ⎝− 1 2 ⎞ ⎠ = 3π 4 ⎛ Thus, the solution is 4 2cis ⎝ ⎞ ⎠. 3π 4 10.26 Write z = 3 + i in polar form. Converting a Complex Number from Polar to Rectangular Form Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given z = r(cos θ + isin θ), first evaluate the trigonometric functions cos θ and sin θ. Then, multiply through by r. Example 10.39 Converting from Polar to Rectangular Form Convert the polar form of the given complex number to rectangular form: z = 12 ⎛ ⎛ ⎝cos ⎝ π 6 ⎛ ⎞ ⎠ + isin ⎝ ⎞ �
�� ⎠ ⎠ π 6 Solution We begin by evaluating the trigonometric expressions. ⎛ cos ⎝ π 6 ⎞ ⎠ = 3 2 ⎛ and sin ⎝ π 6 ⎞ ⎠ = 1 2 After substitution, the complex number is z = 12 ⎛ ⎝ 3 2 + 1 2 i⎞ ⎠ We apply the distributive property: z = 12 ⎛ ⎝ 3 2 = (12) 3 2 i⎞ ⎠ + 1 2 + (12)1 2 i The rectangular form of the given point in complex form is 6 3 + 6i. = 6 3 + 6i This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1141 Example 10.40 Finding the Rectangular Form of a Complex Number Find the rectangular form of the complex number given r = 13 and tan θ = 5 12. Solution If tan θ = 5, and tan θ = 12 y r. and sin θ = y x, we first determine r = x2 + y2 = 122 + 52 = 13. We then find cos θ = x r z = 13(cos θ + isin θ) 12 = 13 13 = 12 + 5i + 5 13 i⎞ ⎠ ⎛ ⎝ The rectangular form of the given number in complex form is 12 + 5i. 10.27 Convert the complex number to rectangular form: ⎛ ⎝cos11π 6 z = 4 + isin11π 6 ⎞ ⎠ Finding Products of Complex Numbers in Polar Form Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments. Products of Complex Numbers in Polar Form If z1 = r1(cos θ1 + isin θ1) and z2 = r2(cos θ2 + isin θ2), then the product of these numbers is given as:
⎡ ⎣cos⎛ z1 z2 = r1 r2 z1 z2 = r1 r2 cis⎛ ⎝θ1 + θ2 ⎝θ1 + θ2 ⎞ ⎠ ⎞ ⎠ + isin⎛ ⎝θ1 + θ2 ⎤ ⎞ ⎦ ⎠ Notice that the product calls for multiplying the moduli and adding the angles. Example 10.41 Finding the Product of Two Complex Numbers in Polar Form Find the product of z1 z2, given z1 = 4(cos(80°) + isin(80°)) and z2 = 2(cos(145°) + isin(145°)). Solution Follow the formula 1142 Chapter 10 Further Applications of Trigonometry z1 z2 = 4 ⋅ 2[cos(80° + 145°) + isin(80° + 145°)] z1 z2 = 8[cos(225°) + isin(225°)] ⎡ 5π ⎞ ⎛ ⎛ z1 z2 = 8 ⎠ + isin ⎣cos ⎝ ⎝ 4 ⎡ ⎤ ⎞ + i⎛ ⎣− 2 ⎝− 2 ⎦ ⎠ 2 2 z1 z2 = − 4 2 − 4i 2 z1 z2 = 8 5π 4 ⎤ ⎞ ⎦ ⎠ Finding Quotients of Complex Numbers in Polar Form The quotient of two complex numbers in polar form is the quotient of the two moduli and the difference of the two arguments. Quotients of Complex Numbers in Polar Form If z1 = r1(cos θ1 + isin θ1) and z2 = r2(cos θ2 + isin θ2), then the quotient of these numbers is z1 z2 z1 z2 = = r1 r2 r1 r2 ⎡ ⎣cos⎛ ⎝θ1 − θ2 ⎞ ⎠ + isin⎛ ⎝θ1 − θ2 ⎤ ⎞ ⎠ ⎦, z2 ≠ 0 cis⎛ ⎝θ1 − θ
2 ⎞ ⎠, z2 ≠ 0 Notice that the moduli are divided, and the angles are subtracted. Given two complex numbers in polar form, find the quotient. 1. Divide r1 r2. 2. Find θ1 − θ2. 3. Substitute the results into the formula: z = r(cos θ + isin θ). Replace r with r1 r2, and replace θ with θ1 − θ2. 4. Calculate the new trigonometric expressions and multiply through by r. Example 10.42 Finding the Quotient of Two Complex Numbers Find the quotient of z1 = 2(cos(213°) + isin(213°)) and z2 = 4(cos(33°) + isin(33°)). Solution Using the formula, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1143 [cos(213° − 33°) + isin(213° − 33°)] [cos(180°) + isin(180°)] z1 z2 z1 z2 z1 z2 z1 z2 z1 z2 = + 0i] + 0i 10.28 Find the product and the quotient of z1 = 2 3(cos(150°) + isin(150°)) and z2 = 2(cos(30°) + isin(30°)). Finding Powers of Complex Numbers in Polar Form Finding powers of complex numbers is greatly simplified using De Moivre’s Theorem. It states that, for a positive integer n, zn is found by raising the modulus to the nth power and multiplying the argument by n. It is the standard method used in modern mathematics. De Moivre’s Theorem If z = r(cos θ + isin θ) is a complex number, then zn zn where n is a positive integer. Example 10.43 = r n ⎡ = r n ⎣cos(nθ) + isin(nθ)⎤ cis(nθ) ⎦ Evaluating an Expression Using De Moivre’s Theorem Evaluate the expression (1 + i)5 using De Moivre’s Theorem. Solution Since De Moivre
’s Theorem applies to complex numbers written in polar form, we must first write (1 + i) in polar form. Let us find r. r = x2 + y2 r = (1)2 + (1)2 r = 2 Then we find θ. Using the formula tan θ = y x gives 1144 Chapter 10 Further Applications of Trigonometry tan θ = 1 1 tan θ = 1 π θ = 4 Use De Moivre’s Theorem to evaluate the expression. n = r n (a + bi) (1 + i)5 = ( 2)5 ⎡ [cos(nθ) + isin(nθ)] π ⎞ ⎛ ⎛ ⎠ + isin ⎣cos ⎝5 ⋅ ⎝5 ⋅ 4 ⎤ ⎡ 5π 5π ⎞ ⎞ ⎛ ⎛ (1 + i)5 = 4 2 ⎠ + isin ⎣cos ⎦ ⎠ ⎝ ⎝ 4 4 ⎡ ⎤ ⎞ + i⎛ ⎣− 2 ⎝− 2 (1 + i)5 = 4 2 ⎦ ⎠ 2 2 (1 + i)5 = − 4 − 4i ⎤ ⎞ ⎦ ⎠ π 4 Finding Roots of Complex Numbers in Polar Form To find the nth root of a complex number in polar form, we use the nth Root Theorem or De Moivre’s Theorem and raise the complex number to a power with a rational exponent. There are several ways to represent a formula for finding nth roots of complex numbers in polar form. The nth Root Theorem To find the nth root of a complex number in polar form, use the formula given as 1 n z 1 n ⎡ ⎛ ⎣cos ⎝ = r n + 2kπ θ n ⎞ ⎛ ⎠ + isin ⎝ n + 2kπ θ n ⎤ ⎞ ⎦ ⎠ where k = 0, 1, 2, 3,..., n − 1. We add 2kπ n to θ n in order to obtain the periodic roots. Example 10.44 Finding
the nth Root of a Complex Number Evaluate the cube roots of z = 8 ⎛ ⎛ ⎝cos ⎝ 2π 3 ⎛ ⎞ ⎠ + isin ⎝ ⎞ ⎞ ⎠. ⎠ 2π 3 Solution We have cos ⎜ ⎣ ⎝ 2π 3 3 + 2kπ 3 ⎛ ⎞ ⎜ ⎟ + isin ⎝ ⎠ 2π 3 3 + 2kπ cos ⎝ 2π 9 + 2kπ 3 ⎛ ⎞ ⎠ + isin ⎝ 2π 9 + 2kπ 3 ⎤ ⎞ ⎦ ⎠ There will be three roots: k = 0, 1, 2. When k = 0, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1145 1 3 = 2 z ⎛ ⎛ ⎝cos ⎝ 2π 9 ⎛ ⎞ ⎠ + isin ⎝ ⎞ ⎞ ⎠ ⎠ 2π 9 When k = 1, we have cos ⎝ 2π 9 + 6π 9 ⎛ ⎞ ⎠ + isin ⎝ 2π 9 + 6π 9 ⎤ ⎞ ⎦ ⎠ Add 2(1)π 3 to each angle. ⎛ ⎛ ⎝cos ⎝ 8π 9 ⎛ ⎞ ⎠ + isin ⎝ ⎞ ⎞ ⎠ ⎠ 8π 9 When k = 2, we have cos ⎝ 2π 9 + 12π 9 ⎛ ⎞ ⎠ + isin ⎝ 2π 9 + 12π 9 ⎤ ⎞ ⎦ ⎠ Add 2(2)π 3 to each angle. ⎛ ⎛ ⎝cos ⎝ 14π 9 ⎛ ⎞ ⎠ + isin ⎝ 14π 9 ⎞ ⎞ ⎠ ⎠ Remember to find the common denominator to simplify fractions in situations like this one. For k = 1, the angle simplification is
2(1)π 3 ⎞ ⎠ ⎛ ⎝ 3 3 + 2π 2(1)π = 2π 3 3 3 3 = 2π 9 = 8π 9 ⎞ ⎛ 1 ⎠ + ⎝ 3 + 6π 9 10.29 Find the four fourth roots of 16(cos(120°) + isin(120°)). Access these online resources for additional instruction and practice with polar forms of complex numbers. • The Product and Quotient of Complex Numbers in Trigonometric Form (http://openstaxcollege.org/l/prodquocomplex) • De Moivre’s Theorem (http://openstaxcollege.org/l/demoivre) 1146 Chapter 10 Further Applications of Trigonometry 10.5 EXERCISES Verbal 312. A complex number is a + bi. Explain each part. 332. z = 3cis(240°) 333. z = 2cis(100°) What does the absolute value of a complex number 313. represent? For the following exercises, find z1 z2 in polar form. 314. How is a complex number converted to polar form? How do we find the product of 315. numbers? two complex 316. What is De Moivre’s Theorem and what is it used for? Algebraic For the following exercises, find the absolute value of the given complex number. 317. 5 + 3i 318. −7 + i 319. −3 − 3i 320. 2 − 6i 321. 2i 322. 2.2 − 3.1i 334. 335. 336. 337. 338. 339. z1 = 2 3cis(116°); z2 = 2cis(82°) z1 = 2cis(205°); z2 = 2 2cis(118°) z1 = 3cis(120°); z2 = 1 4 cis(60°) ⎛ z1 = 3cis ⎝ π 4 ⎛ ⎞ ⎠; z2 = 5cis ⎝ ⎞ ⎠ π 6 ⎛ z1 = 5cis ⎝ 5π 8 ⎞ ⎛ ⎠; z2 = 15cis ⎝ ⎞ ⎠ π 12 ⎛ z1
= 4cis ⎝ π 2 ⎞ ⎛ ⎠; z2 = 2cis ⎝ ⎞ ⎠ π 4 For the following exercises, find z1 z2 in polar form. 340. z1 = 21cis(135°); z2 = 3cis(65°) 341. z1 = 2cis(90°); z2 = 2cis(60°) For the following exercises, write the complex number in polar form. 342. z1 = 15cis(120°); z2 = 3cis(40°) 323. 2 + 2i 324. 8 − 4i 325. − 1 2 − 1 2 i 326. 3 + i 327. 3i For the following exercises, convert the complex number from polar to rectangular form. 328. ⎛ z = 7cis ⎝ 329. ⎛ z = 2cis ⎝ ⎞ ⎠ π 6 ⎞ ⎠ π 3 330. ⎛ z = 4cis ⎝ ⎞ ⎠ 7π 6 331. z = 7cis(25°) This content is available for free at https://cnx.org/content/col11758/1.5 343. ⎛ z1 = 6cis ⎝ π 3 ⎞ ⎛ ⎠; z2 = 2cis ⎝ ⎞ ⎠ π 4 344. ⎛ z1 = 5 2cis(π); z2 = 2cis ⎝ ⎞ ⎠ 2π 3 345. ⎛ z1 = 2cis ⎝ 3π 5 ⎞ ⎛ ⎠; z2 = 3cis ⎝ ⎞ ⎠ π 4 For the following exercises, find the powers of each complex number in polar form. 346. 347. 348. 349. Find z3 when z = 5cis(45°). Find z4 when z = 2cis(70°). Find z2 when z = 3cis(120°). ⎛ Find z2 when z = 4cis ⎝ ⎞ ⎠. π 4 350. ⎛ Find z4 when z = cis ⎝ ⎞ ⎠. 3π
16 Chapter 10 Further Applications of Trigonometry 1147 351. ⎛ Find z3 when z = 3cis ⎝ ⎞ ⎠. 5π 3 Use the polar to rectangular feature on the graphing 371. calculator to change 2cis(45°) to rectangular form. For the following exercises, evaluate each root. 352. Evaluate the cube root of z when z = 27cis(240°). Use the polar to rectangular feature on the graphing 372. calculator to change 5cis(210°) to rectangular form. 353. Evaluate z = 16cis(100°). the square root of z when 354. ⎛ Evaluate the cube root of z when z = 32cis ⎝ ⎞ ⎠. 2π 3 355. Evaluate the square root of z when z = 32cis(π). 356. ⎛ Evaluate the cube root of z when z = 8cis ⎝ ⎞ ⎠. 7π 4 Graphical For the following exercises, plot the complex number in the complex plane. 357. 2 + 4i 358. −3 − 3i 359. 5 − 4i 360. −1 − 5i 361. 3 + 2i 362. 2i 363. −4 364. 6 − 2i 365. −2 + i 366. 1 − 4i Technology For the following exercises, find all answers rounded to the nearest hundredth. Use the rectangular to polar feature on the graphing 367. calculator to change 5 + 5i to polar form. Use the rectangular to polar feature on the graphing 368. calculator to change 3 − 2i to polar form. Use the rectangular to polar feature on the graphing 369. calculator to change −3 − 8i to polar form. Use the polar to rectangular feature on the graphing 370. calculator to change 4cis(120°) to rectangular form. 1148 Chapter 10 Further Applications of Trigonometry 10.6 \| Parametric Equations Learning Objectives In this section, you will: 10.6.1 Parameterize a curve. 10.6.2 Eliminate the parameter. 10.6.3 Find a rectangular equation for a curve defined parametrically. 10.6.4 Find parametric equations for curves defined by rectangular equations. Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 10
.94. At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon’s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time. Figure 10.94 In this section, we will consider sets of equations given by x(t) and y(t) where t is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values of t, the orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations: we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations. Parameterizing a Curve When an object moves along a curve—or curvilinear path—in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. However, both x and y vary over time and so are functions of time. For this reason, we add another variable, the parameter, upon which both x and y are dependent functions. In the example in the section opener, the parameter is time, t. The x position of the moon at time, t, is represented as the function x(t), and the y position of the moon at time, t, is represented as the function y(t). Together, x(t) and y(t) are called parametric equations, and generate an ordered pair ⎛ ⎠. Parametric equations primarily describe motion and direction. ⎝x(t), y(t)⎞ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1149 When we parameterize a curve, we are translating a single equation in two variables, such as x and y, into an
equivalent pair of equations in three variables, x, y, and t. One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time. When we graph parametric equations, we can observe the individual behaviors of x and of y. There are a number of shapes that cannot be represented in the form y = f (x), meaning that they are not functions. For example, consider the graph of a circle, given as r 2 = x2 + y2. Solving for y gives y = ± r 2 − x2, or two equations: y1 = r 2 − x2 and y2 = − r 2 − x2. If we graph y1 and y2 together, the graph will not pass the vertical line test, as shown in Figure 10.95. Thus, the equation for the graph of a circle is not a function. Figure 10.95 However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward. Parametric Equations Suppose t is a number on an interval, I. The set of ordered pairs, ⎛ a plane curve based on the parameter t. The equations x = f (t) and y = g(t) are the parametric equations. ⎠, where x = f (t) and y = g(t), forms ⎝x(t), y(t)⎞ Example 10.45 Parameterizing a Curve Parameterize the curve y = x2 − 1 letting x(t) = t. Graph both equations. Solution If x(t) = t, then to find y(t) we replace the variable x with the expression given in x(t). In other words, y(t) = t 2 − 1. Make a table of values similar to Table 10.10, and sketch the graph. 1150 Chapter 10 Further Applications of Trigonometry t x(t) y(t) −4 −4 y(−4) = (−4)2 − 1 = 15 −3 −3 y(−3) = (−3)2 − 1 = 8 −
2 −2 y(−2) = (−2)2 − 1 = 3 −1 −1 y(−1) = (−1)(0) = (0)2 − 1 = − 1 y(1) = (1)2 − 1 = 0 y(2) = (2)2 − 1 = 3 y(3) = (3)2 − 1 = 8 y(4) = (4)2 − 1 = 15 Table 10.10 See the graphs in Figure 10.96. It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated as t increases. Figure 10.96 (a) Parametric y(t) = t 2 − 1 (b) Rectangular y = x2 − 1 Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1151 The arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of y = x2 − 1. Construct 10.30 x(t) = t − 3, y(t) = 2t + 4; − 1 ≤ t ≤ 2. table of a values and plot the parametric equations: Example 10.46 Finding a Pair of Parametric Equations Find a pair of parametric equations that models the graph of y = 1 − x2, using the parameter x(t) = t. Plot some points and sketch the graph. Solution If x(t) = t and we substitute t for x into the y equation, then y(t) = 1 − t 2. Our pair of parametric equations is x(t) = t y(t) = 1 − t 2 To graph the equations, first we construct a table of values like that in Table 10.11. We can choose values around t = 0, from t = − 3 to t = 3. The values in the x(t) column will be the same as those in the t column because x(t) = t. Calculate values for the column y(t). 1152 Chapter 10 Further Applications of Trigonometry t x(t) = t y(t) = 1 − t 2 −3 −3 y(−3) = 1 − (−3)2 = − 8 −2 −2 y(−2) = 1 − (−2)2 = − 3 −1 −1 y
(−1) = 1 − (−1)(0) = 1 − 0 = 1 y(1) = 1 − (1)2 = 0 y(2) = 1 − (2)2 = − 3 y(3) = 1 − (3)2 = − 8 Table 10.11 The graph of y = 1 − t 2 is a parabola facing downward, as shown in Figure 10.97. We have mapped the curve over the interval [−3, 3], shown as a solid line with arrows indicating the orientation of the curve according to t. Orientation refers to the path traced along the curve in terms of increasing values of t. As this parabola is symmetric with respect to the line x = 0, the values of x are reflected across the y-axis. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1153 Figure 10.97 10.31 Parameterize the curve given by x = y3 − 2y. Example 10.47 Finding Parametric Equations That Model Given Criteria An object travels at a steady rate along a straight path (−5, 3) to (3, −1) in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object. Solution The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The x-value of the object starts at −5 meters and goes to 3 meters. This means the distance x has changed by 8 meters in 4 seconds, which is a rate of 8 m, or 2 m / s. We can write the x-coordinate as a linear function 4 s with respect to time as x(t) = 2t − 5. In the linear function template y = mx + b, 2t = mx and − 5 = b. Similarly, the y-value of the object starts at 3 and goes to −1, which is a change in the distance y of −4 meters in 4 seconds, which is a rate of −4 m, or − 1m / s. We can also write the y-coordinate as the linear function 4 s y(t) = − t + 3. Together, these are the parametric equations for the position of the object, where x and y are expressed in meters and t represents time
: 1154 Chapter 10 Further Applications of Trigonometry x(t) = 2t − 5 y(t) = − t + 3 Using these equations, we can build a table of values for t, x, and y (see Table 10.12). In this example, we limited values of t to non-negative numbers. In general, any value of t can be used. t 0 1 2 3 4 x(t) = 2t − 5 y(t) = − t + 3 x = 2(00) + 3 = 3 x = 2(11) + 3 = 2 x = 2(22) + 3 = 1 x = 2(3) − 5 = 1 y = − (3) + 3 = 0 x = 2(4) − 5 = 3 y = − (4) + 3 = − 1 Table 10.12 From this table, we can create three graphs, as shown in Figure 10.98. Figure 10.98 (a) A graph of x vs. t, representing the horizontal position over time. (b) A graph of y vs. t, representing the vertical position over time. (c) A graph of y vs. x, representing the position of the object in the plane at time t. Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1155 Again, we see that, in Figure 10.98(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows. Eliminating the Parameter In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as x and y. Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter t from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations. Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for
t. We substitute the resulting expression for t into the second equation. This gives one equation in x and y. Example 10.48 Eliminating the Parameter in Polynomials Given x(t) = t 2 + 1 and y(t) = 2 + t, eliminate the parameter, and write the parametric equations as a Cartesian equation. Solution We will begin with the equation for y because the linear equation is easier to solve for t. Next, substitute y − 2 for t in x(t). y − 2)2 + 1 x = y2 − 4y + 4 + 1 x = y2 − 4y + 5 x = y2 − 4y + 5 The Cartesian form is x = y2 − 4y + 5. Substitute the expression for t into x. Analysis This is an equation for a parabola in which, in rectangular terms, x is dependent on y. From the curve’s vertex at (1, 2), the graph sweeps out to the right. See Figure 10.99. In this section, we consider sets of equations given by the functions x(t) and y(t), where t is the independent variable of time. Notice, both x and y are functions of time; so in general y is not a function of x. 1156 Chapter 10 Further Applications of Trigonometry Figure 10.99 10.32 Given the equations below, eliminate the parameter and write as a rectangular equation for y as a function of x. Example 10.49 x(t) = 2t 2 + 6 y(t) = 5 − t Eliminating the Parameter in Exponential Equations Eliminate the parameter and write as a Cartesian equation: x(t) = e−t and y(t) = 3et, t > 0. Solution Isolate et. Substitute the expression into y(t). The Cartesian form is y = 3 x. Analysis x = e−t et = 1 x y = 3et ⎛ ⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1157 The graph of the parametric equation is shown in Figure 10.100(a). The domain is restricted to t > 0. The Cartesian equation, y = 3 x is shown in Figure 10.100(b) and has only one restriction on the domain, x ≠ 0.
Figure 10.100 Example 10.50 Eliminating the Parameter in Logarithmic Equations Eliminate the parameter and write as a Cartesian equation: x(t) = t + 2 and y(t) = log(t). Solution Solve the first equation for tx − 2)2 = t Square both sides. Then, substitute the expression for t into the y equation. y = log(t) y = log(x − 2)2 The Cartesian form is y = log(x − 2)2. Analysis 1158 Chapter 10 Further Applications of Trigonometry To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain on x = t + 2 to t > 0; we restrict the domain on x to x > 2. The domain for the parametric equation y = log(t) is restricted to t > 0; we limit the domain on y = log(x − 2)2 to x > 2. 10.33 Eliminate the parameter and write as a rectangular equation. x(t) = t 2 y(t) = ln t t > 0 Eliminating the Parameter from Trigonometric Equations Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem. First, we use the identities: Solving for cos t and sin t, we have Then, use the Pythagorean Theorem: Substituting gives Example 10.51 x(t) = acos t y(t) = bsin t x a = cos t y b = sin t cos2 t + sin2 t = 1 cos2 t + sin2 Eliminating the Parameter from a Pair of Trigonometric Parametric Equations Eliminate the parameter from the given pair of trigonometric equations where 0 ≤ t ≤ 2π and sketch the graph. x(t) = 4cos t y(t) = 3sin t Solution Solving for cos t and sin t, we have x = 4cos t x = cos t 4 y = 3sin t y = sin t 3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1159 Next, use the Pythagorean identity and make the substitutions. cos2 t + sin2 y2 9 + = 1
The graph for the equation is shown in Figure 10.101. x2 16 Figure 10.101 Analysis Applying the general equations for conic sections (introduced in Analytic Geometry, we can identify x2 16 t = y2 9 the coordinates are (0, 3). This shows the orientation of the curve with increasing values of t. = 1 as an ellipse centered at (0, 0). Notice that when t = 0 the coordinates are (4, 0), and when + π 2 Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: 10.34 x(t) = 2cos t and y(t) = 3sin t. Finding Cartesian Equations from Curves Defined Parametrically When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as x(t) = t. In this case, y(t) can be any expression. For example, consider the following pair of equations. x(t) = t y(t) = t 2 − 3 Rewriting this set of parametric equations is a matter of substituting x for t. Thus, the Cartesian equation is y = x2 − 3. Example 10.52 1160 Chapter 10 Further Applications of Trigonometry Finding a Cartesian Equation Using Alternate Methods Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations. x(t) = 3t − 2 y(t) = t + 1 Solution Method 1. First, let’s solve the x equation for t. Then we can substitute the result into the y equation. x = 3t − 2 x + 2 = 3t x + 2 3 = t Now substitute the expression for t into the y equation Method 2. Solve the y equation for t and substitute this expression in the x equation. Make the substitution and then solve for y(y − 1) − 2 x = 3y − 3 − 2 x = 3y − 5 x + 5 = 3y 10.35 Write the given parametric equations as a Cartesian equation: x(t) = t 3 and y(t) = t 6. Finding Param
etric Equations for Curves Defined by Rectangular Equations Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent x, and then substitute it into the y equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for x as the domain of the rectangular equation, then the graphs will be different. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1161 Example 10.53 Finding a Set of Parametric Equations for Curves Defined by Rectangular Equations Find a set of equivalent parametric equations for y = (x + 3)2 + 1. Solution An obvious choice would be to let x(t) = t. Then y(t) = (t + 3)2 + 1. But let’s try something more interesting. What if we let x = t + 3? Then we have y = (x + 3)2 + 1 y = ((t + 3) + 3)2 + 1 y = (t + 6)2 + 1 x(t) = t + 3 y(t) = (t + 6)2 + 1 The set of parametric equations is See Figure 10.102. Figure 10.102 Access these online resources for additional instruction and practice with parametric equations. • Introduction to Parametric Equations (http://openstaxcollege.org/l/introparametric) • Converting Parametric Equations to Rectangular Form (http://openstaxcollege.org/l/ convertpara) 1162 Chapter 10 Further Applications of Trigonometry 10.6 EXERCISES Verbal 373. What is a system of parametric equations? 389. x(t) = e2t ⎧ ⎨ y(t) = e6t ⎩ 374. Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter. 390. �
�� x(t) = t 5 ⎨ y(t) = t 10 ⎩ Explain how to eliminate a parameter given a set of 375. parametric equations. What is a benefit of writing a system of parametric 376. equations as a Cartesian equation? 377. What is a benefit of using parametric equations? Why are there many sets of parametric equations to 378. represent on Cartesian function? Algebraic For the following exercises, eliminate the parameter t to rewrite the parametric equation as a Cartesian equation. 379. x(t) = 5 − t ⎧ ⎨ y(t) = 8 − 2t ⎩ 380. x(t) = 6 − 3t ⎧ ⎨ y(t) = 10 − t ⎩ 381. x(t) = 2t + 1 ⎧ ⎨ y(t) = 3 t ⎩ 382. x(t) = 3t − 1 ⎧ ⎨ y(t) = 2t 2 ⎩ 383. x(t) = 2et ⎧ ⎨ y(t) = 1 − 5t ⎩ 384. x(t) = e−2t ⎧ ⎨ y(t) = 2e−t ⎩ 385. x(t) = 4log(t) ⎧ ⎨ y(t) = 3 + 2t ⎩ 386. x(t) = log(2t) ⎧ ⎨ y(t) = t − 1 ⎩ 387. 388. ⎧ x(t) = t 3 − t ⎨ y(t) = 2t ⎩ ⎧ x(t) = t − t 4 ⎨ y(t) = t + 2 ⎩ This content is available for free at https://cnx.org/content/col11758/1.5 391. x(t) = 4cos t ⎧ ⎨ y(t) = 5sin t ⎩ 392. x(t) = 3sin t ⎧ ⎨ y(t) = 6cos t ⎩ 393. ⎧ x(t) = 2cos2 t ⎨ y(t) = − sin t ⎩ 394. x(t) = cos
t + 4 ⎧ ⎨ y(t) = 2sin2 t ⎩ 395. x(t) = t − 1 ⎧ ⎨ y(t) = t 2 ⎩ 396. x(t) = − t ⎧ ⎨ y(t) = t 3 + 1 ⎩ 397. x(t) = 2t − 1 ⎧ ⎨ y(t) = t 3 − 2 ⎩ For the following exercises, rewrite the parametric equation as a Cartesian equation by building an x-y table. 398. x(t) = 2t − 1 ⎧ ⎨ y(t) = t + 4 ⎩ 399. x(t) = 4 − t ⎧ ⎨ y(t) = 3t + 2 ⎩ 400. x(t) = 2t − 1 ⎧ ⎨ y(t) = 5t ⎩ 401. x(t) = 4t − 1 ⎧ ⎨ y(t) = 4t + 2 ⎩ For the following exercises, parameterize (write parametric equations for) each Cartesian equation by setting x(t) = t or by setting y(t) = t. 402. y(x) = 3x2 + 3 Chapter 10 Further Applications of Trigonometry 1163 403. y(x) = 2sin x + 1 404. x(y) = 3log(y) + y 405. x(y) = y + 2y For the following exercises, parameterize (write parametric equations using by each Cartesian x(t) = acos t and y(t) = bsin t. Identify the curve. equation for) 406. 407. x2 4 + y2 9 = 1 x2 16 + y2 36 = 1 408. x2 + y2 = 16 409. x2 + y2 = 10 Parameterize the line from (3, 0) to (−2, −5) so the line is at (3, 0) at t = 0, and at (−2, −5) at 410. that t = 1. Parameterize the line from (−1, 0) to (3, −2) so the line is at (−1, 0) at t = 0, and at (3, −2) at 411. that
t = 1. Parameterize the line from (−1, 5) to (2, 3) so that 412. the line is at (−1, 5) at t = 0, and at (2, 3) at t = 1. Parameterize the line from (4, 1) to (6, −2) so that 413. the line is at (4, 1) at t = 0, and at (6, −2) at t = 1. Technology For the following exercises, use the table feature in the the graphs graphing calculator intersect. to determine whether 414. 415. x1(t) = 3t ⎧ ⎨ y1(t) = 2t − 1 ⎩ and x2(t) = t + 3 ⎧ ⎨ y2(t) = 4t − 4 ⎩ ⎧ x1(t) = t 2 ⎨ y1(t) = 2t − 1 ⎩ and x2(t) = − t + 6 ⎧ ⎨ y2(t) = t + 1 ⎩ For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. 416. ⎧ x1 (t) = 3t 2 − 3t + 7 ⎨ y1 (t) = 2t + 3 ⎩ x y t –1 0 1 417. ⎧ x1 (t) = t 2 − 4 ⎨ y1 (t) = 2t 418. ⎧ x1 (t) = t 4 ⎨ y1 (t1 0 1 2 Extensions Find two different sets of parametric equations for 419. y = (x + 1)2. Find two different sets of parametric equations for 420. y = 3x − 2. Find two different sets of parametric equations for 421. y = x2 − 4x + 4. 1164 Chapter 10 Further Applications of Trigonometry 10.7 \| Parametric Equations: Graphs Learning Objectives In this section you will: 10.7.1 Graph plane curves described by parametric equations by plotting points. 10.7.2 Graph parametric equations. It is the bottom of the ninth inning, with two outs and two men on base. The home team is losing by two runs. The batter swings and hits the baseball at 140 feet per second and
at an angle of approximately 45° to the horizontal. How far will the ball travel? Will it clear the fence for a game-winning home run? The outcome may depend partly on other factors (for example, the wind), but mathematicians can model the path of a projectile and predict approximately how far it will travel using parametric equations. In this section, we’ll discuss parametric equations and some common applications, such as projectile motion problems. Figure 10.103 Parametric equations can model the path of a projectile. (credit: Paul Kreher, Flickr) Graphing Parametric Equations by Plotting Points In lieu of a graphing calculator or a computer graphing program, plotting points to represent the graph of an equation is the standard method. As long as we are careful in calculating the values, point-plotting is highly dependable. Given a pair of parametric equations, sketch a graph by plotting points. 1. Construct a table with three columns: t, x(t), and y(t). 2. Evaluate x and y for values of t over the interval for which the functions are defined. 3. Plot the resulting pairs (x, y). Example 10.54 Sketching the Graph of a Pair of Parametric Equations by Plotting Points Sketch the graph of the parametric equations x(t) = t 2 + 1, y(t) = 2 + t. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1165 Construct a table of values for t, x(t), and y(t), as in Table 10.13, and plot the points in a plane. t −5 −4 −3 −2 −1 0 1 2 3 4 5 x(t) = t 2 + 1 y(t) = 2 + t 26 17 10 5 2 1 2 5 10 17 26 −3 −2 −1 0 1 2 3 4 5 6 7 Table 10.13 The graph is a parabola with vertex at the point (1, 2), opening to the right. See Figure 10.104. 1166 Chapter 10 Further Applications of Trigonometry Figure 10.104 Analysis As values for t progress in a positive direction from 0 to 5, the plotted points trace out the top half of the parabola. As values of t become negative, they trace out the lower half of the parabola. There are no
restrictions on the domain. The arrows indicate direction according to increasing values of t. The graph does not represent a function, as it will fail the vertical line test. The graph is drawn in two parts: the positive values for t, and the negative values for t. 10.36 Sketch the graph of the parametric equations x = t, y = 2t + 3, 0 ≤ t ≤ 3. Example 10.55 Sketching the Graph of Trigonometric Parametric Equations Construct a table of values for the given parametric equations and sketch the graph: x = 2cos t y = 4sin t Solution Construct a table like that in Table 10.14 using angle measure in radians as inputs for t, and evaluating x and y. Using angles with known sine and cosine values for t makes calculations easier. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1167 t x = 2cos t y = 4sin t 0 π 6 π 3 π 2 2π 3 5π 6 π 7π 6 4π 3 3π 2 5π 3 11π 6 2π x = 2cos(0) = 2 y = 4sin(0) = 0 ⎛ x = 2cos ⎝ ⎞ ⎠ = 3 π 6 ⎛ y = 4sin ⎝ ⎞ ⎠ = 2 π 6 ⎛ x = 2cos ⎝ ⎞ ⎠ = 1 π 3 ⎛ y = 4sin ⎝ ⎞ ⎠ = 2 3 π 3 ⎛ x = 2cos ⎝ ⎞ ⎠ = 0 π 2 ⎛ y = 4sin ⎝ ⎞ ⎠ = 4 π 2 ⎛ x = 2cos ⎝ ⎞ ⎠ = − 1 2π 3 ⎛ y = 4sin ⎝ ⎞ ⎠ = 2 3 2π 3 ⎛ x = 2cos ⎝ 5π 6 ⎞ ⎠ = − 3 ⎛ y = 4sin ⎝ ⎞ ⎠ = 2 5π 6 x = 2cos(π) = − 2 y = 4sin(π) = 0 ⎛ x = 2cos ⎝ 7π 6
⎞ ⎠ = − 3 ⎛ y = 4sin ⎝ ⎞ ⎠ = − 2 7π 6 ⎛ x = 2cos ⎝ ⎞ ⎠ = − 1 4π 3 ⎛ y = 4sin ⎝ 4π 3 ⎞ ⎠ = − 2 3 ⎛ x = 2cos ⎝ ⎞ ⎠ = 0 3π 2 ⎛ y = 4sin ⎝ ⎞ ⎠ = − 4 3π 2 ⎛ x = 2cos ⎝ ⎞ ⎠ = 1 5π 3 ⎛ y = 4sin ⎝ 5π 3 ⎞ ⎠ = − 2 3 ⎛ x = 2cos ⎝ 11π 6 ⎞ ⎠ = 3 ⎛ y = 4sin ⎝ 11π 6 ⎞ ⎠ = − 2 x = 2cos(2π) = 2 y = 4sin(2π) = 0 Table 10.14 Figure 10.105 shows the graph. 1168 Chapter 10 Further Applications of Trigonometry Figure 10.105 By the symmetry shown in the values of x and y, we see that the parametric equations represent an ellipse. The ellipse is mapped in a counterclockwise direction as shown by the arrows indicating increasing t values. Analysis We have seen that parametric equations can be graphed by plotting points. However, a graphing calculator will save some time and reveal nuances in a graph that may be too tedious to discover using only hand calculations. Make sure to change the mode on the calculator to parametric (PAR). To confirm, the Y = window should show X1T = Y1T = instead of Y1 =. 10.37 Graph the parametric equations: x = 5cos t, y = 3sin t. Example 10.56 Graphing Parametric Equations and Rectangular Form Together Graph the parametric equations x = 5cos t and y = 2sin t. First, construct the graph using data points generated from the parametric form. Then graph the rectangular form of the equation. Compare the two graphs. Solution Construct a table of values like that in Table 10.15. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1169
t x = 5cos t y = 2sin t 0 1 2 3 4 5 −1 −2 −3 −4 −5 x = 5cos(0) = 5 y = 2sin(0) = 0 x = 5cos(1) ≈ 2.7 y = 2sin(1) ≈ 1.7 x = 5cos(2) ≈ −2.1 y = 2sin(2) ≈ 1.8 x = 5cos(3) ≈ −4.95 y = 2sin(3) ≈ 0.28 x = 5cos(4) ≈ −3.3 y = 2sin(4) ≈ −1.5 x = 5cos(5) ≈ 1.4 y = 2sin(5) ≈ −1.9 x = 5cos(−1) ≈ 2.7 y = 2sin(−1) ≈ −1.7 x = 5cos(−2) ≈ −2.1 y = 2sin(−2) ≈ −1.8 x = 5cos(−3) ≈ −4.95 y = 2sin(−3) ≈ −0.28 x = 5cos(−4) ≈ −3.3 y = 2sin(−4) ≈ 1.5 x = 5cos(−5) ≈ 1.4 y = 2sin(−5) ≈ 1.9 Table 10.15 Plot the (x, y) values from the table. See Figure 10.106. 1170 Chapter 10 Further Applications of Trigonometry Figure 10.106 Next, translate the parametric equations to rectangular form. To do this, we solve for t in either x(t) or y(t), and then substitute the expression for t in the other equation. The result will be a function y(x) if solving for t as a function of x, or x(y) if solving for t as a function of y. x = 5cos t x = cos t 5 y = 2sin t y = sin t 2 Solve for cos t. Solve for sin t. Then, use the Pythagorean Theorem. cos2 t + sin2 y2 4 x2 25 + = 1 Analysis In Figure 10.107, the data from the parametric equations and the rectangular equation are plotted together. The parametric equations are plotted in blue; the graph for the
rectangular equation is drawn on top of the parametric in a dashed style colored red. Clearly, both forms produce the same graph. Figure 10.107 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1171 Example 10.57 Graphing Parametric Equations and Rectangular Equations on the Coordinate System Graph the parametric equations x = t + 1 and y = t, t ≥ 0, and the rectangular equivalent y = x − 1 on the same coordinate system. Solution Construct a table of values for the parametric equations, as we did in the previous example, and graph y = t, t ≥ 0 on the same grid, as in Figure 10.108. Figure 10.108 Analysis With the domain on t restricted, we only plot positive values of t. The parametric data is graphed in blue and the graph of the rectangular equation is dashed in red. Once again, we see that the two forms overlap. 10.38 Sketch the graph of the parametric equations x = 2cos θ and y = 4sin θ, along with the rectangular equation on the same grid. Applications of Parametric Equations Many of the advantages of parametric equations become obvious when applied to solving real-world problems. Although rectangular equations in x and y give an overall picture of an object's path, they do not reveal the position of an object at a specific time. Parametric equations, however, illustrate how the values of x and y change depending on t, as the location of a moving object at a particular time. A common application of parametric equations is solving problems involving projectile motion. In this type of motion, an object is propelled forward in an upward direction forming an angle of θ to the horizontal, with an initial speed of v0, and at a height h above the horizontal. The path of an object propelled at an inclination of θ to the horizontal, with initial speed v0, and at a height h above the horizontal, is given by x = (v0 cosθ)t y = − 1 2 gt 2 + (v0 sinθ)t + h 1172 Chapter 10 Further Applications of Trigonometry where g accounts for the effects of gravity and h is the initial height of the object. Depending on the units involved in the problem, use g = 32 ft / s2 or g = 9.8 m / s2. The equation for
x gives horizontal distance, and the equation for y gives the vertical distance. Given a projectile motion problem, use parametric equations to solve. 1. The horizontal distance is given by x = ⎛ ⎝v0 cos θ⎞ ⎠t. Substitute the initial speed of the object for v0. 2. The expression cos θ indicates the angle at which the object is propelled. Substitute that angle in degrees for cos θ. 3. The vertical distance is given by the formula y = − 1 2 the effect of gravity. Depending on units involved, use g = 32 ft/s2 or g = 9.8 m/s2. Again, substitute the initial speed for v0, and the height at which the object was propelled for h. ⎠t + h. The term − 1 2 gt 2 represents ⎝v0 sin θ⎞ gt 2 + ⎛ 4. Proceed by calculating each term to solve for t. Example 10.58 Finding the Parametric Equations to Describe the Motion of a Baseball Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of 45° to the horizontal, making contact 3 feet above the ground. a. Find the parametric equations to model the path of the baseball. b. Where is the ball after 2 seconds? c. How long is the ball in the air? d. Is it a home run? Solution a. Use the formulas to set up the equations. The horizontal position is found using the parametric equation for x. Thus, The vertical position is found using the parametric equation for y. Thus, x = (v0 cos θ)t x = (140cos(45°))t b. Substitute 2 into the equations to find the horizontal and vertical positions of the ball. y = − 16t 2 + (v0 sin θ)t + h y = − 16t 2 + (140sin(45°))t + 3 x = (140cos(45°))(2) x = 198 feet y = − 16(2)2 + (140sin(45°))(2) + 3 y = 137 feet After 2 seconds, the ball is 198 feet away from the batter’s box and 137 feet above the ground. This content is
available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1173 c. To calculate how long the ball is in the air, we have to find out when it will hit ground, or when y = 0. Thus, ⎝140sin(45∘)⎞ ⎠t + 3 y = − 16t 2 + ⎛ y = 0 t = 6.2173 Set y(t) = 0 and solve the quadratic. When t = 6.2173 seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.) d. We cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity’s sake, let’s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let’s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when x = 400 feet. So we will set x = 400, solve for t, and input t into y. ⎠t ⎝140cos(45°)⎞ ⎠t ⎝140cos(45°)⎞ x = ⎛ 400 = ⎛ t = 4.04 y = − 16(4.04)2 + ⎛ y = 141.8 ⎝140sin(45°)⎞ ⎠(4.04) + 3 The ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run. See Figure 10.109. Figure 10.109 Access the following online resource for additional instruction and practice with graphs of parametric equations. • Graphing Parametric Equations on the TI-84 (http://openstaxcollege.org/l/graphpara84) 1174 Chapter 10 Further Applications of Trigonometry 10.7 EXERCISES Verbal What are two methods used to graph parametric 422. equations? What is one difference in point-plotting parametric 423. equations compared to Cartesian equations? 424. Why are some graphs drawn with arrows? Name a few common types of graphs
of parametric 425. equations. t x y −3 −2 −1 0 1 2 Why are 426. understanding projectile motion? parametric graphs important in Graphical For the following exercises, graph each set of parametric equations by making a table of values. Include the orientation on the graph. 427. x(t) = t ⎧ ⎨ y(t 429. x(t) = 2 + t ⎧ ⎨ y(t) = 3 − 2t ⎩ −2 −1 0 1 2 3 t x y −3 −2 −1 0 1 2 3 428. x(t) = t − 1 ⎧ ⎨ y(t) = t 2 ⎩ 430. x(t) = − 2 − 2t ⎧ ⎨ y(t) = 3 + t ⎩ −3 −2 −1 0 1 t x y 431. ⎧ x(t) = t 3 ⎨ y(t) = t + 2 ⎩ −2 −1 0 1 2 t x y This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1175 432. ⎧ x(t) = t 2 ⎨ y(t) = t + 3 ⎩ −2 −1 0 1 2 t x y For the following exercises, sketch the curve and include the orientation. 444. x(t) = t − 1 ⎧ ⎨ y(t) = − t 2 ⎩ 445. ⎧ x(t) = t 3 ⎨ y(t) = t + 3 ⎩ 446. x(t) = 2cos t ⎧ ⎨ y(t) = − sin t ⎩ 447. x(t) = 7cos t ⎧ ⎨ y(t) = 7sin t ⎩ 448. x(t) = e2t ⎧ ⎨ y(t) = − e t ⎩ 433. x(t) = t ⎧ ⎨ y(t) = t ⎩ 434. x(t) = − t ⎧ ⎨ y(t) = t ⎩ 435. x(t) = 5 − \|t\| ⎧ �
�� y(t) = t + 2 ⎩ 436. x(t) = − t + 2 ⎧ ⎨ y(t) = 5 − \|t\| ⎩ 437. x(t) = 4sin t ⎧ ⎨ y(t) = 2cos t ⎩ 438. x(t) = 2sin t ⎧ ⎨ y(t) = 4cos t ⎩ 439. 440. ⎧ x(t) = 3cos2 t ⎨ y(t) = −3sin t ⎩ ⎧ x(t) = 3cos2 t ⎨ y(t) = −3sin2 t ⎩ 441. x(t) = sec t ⎧ ⎨ y(t) = tan t ⎩ 442. x(t) = sec t ⎧ ⎨ y(t) = tan2 t ⎩ 443. ⎧ x(t) = 1 e2t ⎨ y(t) = e− t ⎩ For the following exercises, graph the equation and include the orientation. Then, write the Cartesian equation. For the following exercises, graph the equation and include the orientation. 449. 450. 451. x = t 2, y = 3t, 0 ≤ t ≤ 5 x = 2t, y = t 2, y = 25 − t 2, 0 < t ≤ 5 452. x(t) = − t, y(t) = t, t ≥ 0 453. x = − 2cos t, y = 6 sin t, 0 ≤ t ≤ π 454. x = − sec t, y = tan t, − π 2 < t < π 2 For the following exercises, use the parametric equations for integers a and b: x(t) = acos((a + b)t) y(t) = acos((a − b)t) Graph on the domain [−π, 0], where a = 2 and 455. b = 1, and include the orientation. Graph on the domain [−π, 0], where a = 3 and 456. b = 2, and include the orientation. Graph on the domain [−π, 0], where a = 4 and 457. b = 3, and include the orientation. Graph on the
domain [−π, 0], where a = 5 and 458. b = 4, and include the orientation. If a is 1 more than b, describe the effect the values 459. of a and b have on the graph of the parametric equations. 460. Describe the graph if a = 100 and b = 99. 1176 Chapter 10 Further Applications of Trigonometry What happens if b is 1 more than a? Describe the 461. graph. 462. If the parametric equations x(t) = t 2 and y(t) = 6 − 3t have the graph of a horizontal parabola opening to the right, what would change the direction of the curve? For the following exercises, describe the graph of the set of parametric equations. 463. 464. 465. x(t) = − t 2 and y(t) is linear y(t) = t 2 and x(t) is linear y(t) = − t 2 and x(t) is linear Write the parametric equations of a circle with center 466. (0, 0), radius 5, and a counterclockwise orientation. 475. Write the parametric equations of an ellipse with 467. center (0, 0), major axis of length 10, minor axis of length 6, and a counterclockwise orientation. the window [−3, 3] by [−3, 3] on For the following exercises, use a graphing utility to graph on domain [0, 2π) for the following values of a and b, and include the orientation. the x(t) = sin(at) ⎧ ⎨ y(t) = sin(bt) ⎩ 468. a = 1, b = 2 469. a = 2, b = 1 470. a = 3, b = 3 471. a = 5, b = 5 472. a = 2, b = 5 473. a = 5, b = 2 Technology 476. by equations parametric For the following exercises, look at the graphs that were form created x(t) = acos(bt) ⎧ ⎨ y(t) = csin(dt) ⎩ graphing calculator to find the values of a, b, c, to achieve each graph.. Use the parametric mode on the and d the of 474. This content is available for free
at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1177 477. and then eliminate time to write height as a function of horizontal position. be 485. A skateboarder riding on a level surface at a constant speed of 9 ft/s throws a ball in the air, the height of which can equation y(t) = − 16t 2 + 10t + 5. Write parametric equations for the ball’s position, and then eliminate time to write height as a function of horizontal position. described the by For the following exercises, use this scenario: A dart is thrown upward with an initial velocity of 65 ft/s at an angle of elevation of 52°. Consider the position of the dart at any time t. Neglect air resistance. Find parametric equations that model the problem 486. situation. Find all possible values of x that 487. situation. represent the 488. When will the dart hit the ground? 489. Find the maximum height of the dart. 490. At what time will the dart reach maximum height? For the following exercises, look at the graphs of each of the four parametric equations. Although they look unusual and beautiful, they are so common that they have names, as indicated in each exercise. Use a graphing utility to graph each on the indicated domain. 491. x(t) = 14cos t − cos(14t) ⎧ An epicycloid: ⎨ y(t) = 14sin t + sin(14t) ⎩ on the domain [0, 2π]. 492. A hypocycloid: x(t) = 6sin t + 2sin(6t) ⎧ ⎨ y(t) = 6cos t − 2cos(6t) ⎩ on the domain [0, 2π]. For the following exercises, use a graphing utility to graph the given parametric equations. a. b. c. x(t) = cost − 1 ⎧ ⎨ y(t) = sint + t ⎩ x(t) = cost + t ⎧ ⎨ y(t) = sint − 1 ⎩ x(t) = t − sint ⎧ ⎨ y(t) = cost − 1 ⎩ Graph all three sets of parametric equations on the
478. domain [0, 2π]. Graph all three sets of parametric equations on the 479. domain [0, 4π]. Graph all three sets of parametric equations on the 480. domain ⎡ ⎣−4π, 6π⎤ ⎦. The graph of each set of parametric equations appears 481. to “creep” along one of the axes. What controls which axis the graph creeps along? 493. A hypotrochoid: x(t) = 2sin t + 5cos(6t) ⎧ ⎨ y(t) = 5cos t − 2sin(6t) ⎩ on the domain [0, 2π]. 494. x(t) = 5sin(2t)sint ⎧ A rose: ⎨ y(t) = 5sin(2t)cost ⎩ on the domain [0, 2π]. Explain the effect on the graph of the parametric 482. equation when we switched sin t and cos t. Explain the effect on the graph of the parametric 483. equation when we changed the domain. Extensions An object is thrown in the air with vertical velocity of 484. 20 ft/s and horizontal velocity of 15 ft/s. The object’s height can be described by the equation y(t) = − 16t 2 + 20t, while the object moves horizontally with constant velocity 15 ft/s. Write parametric equations for the object’s position, 1178 Chapter 10 Further Applications of Trigonometry 10.8 \| Vectors In this section you will: Learning Objectives 10.8.1 View vectors geometrically. 10.8.2 Find magnitude and direction. 10.8.3 Perform vector addition and scalar multiplication. 10.8.4 Find the component form of a vector. 10.8.5 Find the unit vector in the direction of v. 10.8.6 Perform operations with vectors in terms of i and j. 10.8.7 Find the dot product of two vectors. An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in Figure 10.110. What are the ground speed and actual bearing of the plane? Figure 10.110 Ground speed
refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s groundspeed and bearing, while investigating another approach to problems of this type. First, however, let’s examine the basics of vectors. A Geometric View of Vectors A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities: • Lower case, boldfaced type, with or without an arrow on top such as v, u, w, v→, u→, w→. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1179 • Given initial point P and terminal point Q, a vector can be represented as PQ →. The arrowhead on top is what indicates that it is not just a line, but a directed line segment. • Given an initial point of (0, 0) and terminal point (a, b), a vector may be represented as 〈 a, b 〉. This last symbol 〈 a, b 〉 has special significance. It is called the standard position. The position vector has an initial point (0, 0) and a terminal point 〈 a, b 〉. To change any vector into the position vector, we think about the change in the → is C(x1, y1) and the terminal x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector CD point is D(x2, y2), then the position vector is found by calculating → = 〈 x2 − x1, y2 − y1 〉 AB = 〈 a, b 〉 In Figure 10.111, we see the original vector CD → → and the position vector AB
. Figure 10.111 Properties of Vectors A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at (0, 0) and is identified by its terminal point 〈 a, b 〉. Example 10.59 Find the Position Vector Consider the vector whose initial point is P(2, 3) and terminal point is Q(6, 4). Find the position vector. Solution The position vector is found by subtracting one x-coordinate from the other x-coordinate, and one y-coordinate from the other y-coordinate. Thus v = 〈 6 − 2, 4 − 3 〉 = 〈 4, 1 〉 The position vector begins at (0, 0) and terminates at (4, 1). The graphs of both vectors are shown in Figure 10.112. 1180 Chapter 10 Further Applications of Trigonometry Figure 10.112 We see that the position vector is 〈 4, 1 〉. Example 10.60 Drawing a Vector with the Given Criteria and Its Equivalent Position Vector Find the position vector given that vector v has an initial point at (−3, 2) and a terminal point at (4, 5), then graph both vectors in the same plane. Solution The position vector is found using the following calculation: v = 〈 4 − ( − 3), 5 − 2 〉 = 〈 7, 3 〉 Thus, the position vector begins at (0, 0) and terminates at (7, 3). See Figure 10.113. Figure 10.113 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1181 10.39 Draw a vector v that connects from the origin to the point (3, 5). Finding Magnitude and Direction To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function. Magnitude and Direction of a Vector the magnitude is found by \|v\| = a2 + b2. The direction is equal to the angle Given a position vector v = �
�� a, b 〉, formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by tan θ = ⎞ ⎠, as illustrated in Figure 10.114. ⎠ ⇒ θ = tan− Figure 10.114 Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal. Example 10.61 Finding the Magnitude and Direction of a Vector Find the magnitude and direction of the vector with initial point P(−8, 1) and terminal point Q(−2, − 5). Draw the vector. Solution First, find the position vector. We use the Pythagorean Theorem to find the magnitude. u = 〈 −2, − (−8), −5−1 〉 = 〈 6, − 6 〉 \|u\| = (6)2 + ( − 6)2 = 72 = 6 2 The direction is given as 1182 Chapter 10 Further Applications of Trigonometry = −1 ⇒ θ = tan−1(−1) tan θ = −6 6 = − 45° However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, − 45° + 360° = 315°. See Figure 10.115. Figure 10.115 Example 10.62 Showing That Two Vectors Are Equal Show that vector v with initial point at (5, −3) and terminal point at (−1, 2) is equal to vector u with initial point at (−1, −3) and terminal point at (−7, 2). Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each vector. Solution As shown in Figure 10.116, draw the vector v starting at initial (5, −3) and terminal point (−1, 2). Draw the vector u with initial point (−1, −3) and terminal point (−7, 2). Find the standard position for each. Next, find and sketch the position vector for v and u. We have v = 〈 −1 − 5, 2 − ( − 3) 〉 = 〈 −6, 5 〉 u = 〈 −7 − (−1), 2 − (−3) �
�� = 〈 −6, 5 〉 Since the position vectors are the same, v and u are the same. An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1183 \|v\| = (−1 − 5)2 + (2 − (−3))2 = (−6)2 + (5)2 = 36 + 25 = 61 \|u\| = (−7 − (−1))2 + (2 − (−3))2 = (−6)2 + (5)2 = 36 + 25 = 61 As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives tan θ = − 5 6 = − 39.8° ⇒ θ = tan−1 ⎛ ⎝− 5 6 ⎞ ⎠ However, we can see that the position vector terminates in the second quadrant, so we add 180°. Thus, the direction is − 39.8° + 180° = 140.2°. Figure 10.116 Performing Vector Addition and Scalar Multiplication Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector u = 〈 x, y 〉 as an arrow or directed line segment from the origin to the point (x, y), vectors can be situated anywhere in the plane. The sum of two vectors u and v, or vector addition, produces a third vector u+ v, the resultant vector. To find u + v, we first draw the vector u, and from the terminal end of u, we drawn the vector v. In other words, we have the initial point of v meet the terminal end of u. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in Figure 10.117. 1184 Chapter 10 Further Applications of Trigonometry Figure 10
.117 Vector subtraction is similar to vector addition. To find u − v, view it as u + (−v). Adding −v is reversing direction of v and adding it to the end of u. The new vector begins at the start of u and stops at the end point of −v. See Figure 10.118 for a visual that compares vector addition and vector subtraction using parallelograms. Figure 10.118 Example 10.63 Adding and Subtracting Vectors Given u = 〈 3, − 2 〉 and v = 〈 −1, 4 〉, find two new vectors u + v, and u − v. Solution To find the sum of two vectors, we add the components. Thus, u + v = 〈 3, − 2 〉 + 〈 −1), − 2 + 4 〉 = 〈 2, 2 〉 See Figure 10.119(a). To find the difference of two vectors, add the negative components of v to u. Thus, u + ( − v) = 〈 3, − 2 〉 + 〈 1, − 4 〉 = 〈 3 + 1, − 2 + ( − 4) 〉 = 〈 4, − 6 〉 See Figure 10.119(b). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1185 Figure 10.119 (a) Sum of two vectors (b) Difference of two vectors Multiplying By a Scalar While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector. Scalar Multiplication Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply v = 〈 a, b 〉 by k, we have Only the magnitude changes, unless k is negative, and then the vector reverses direction. kv = 〈 ka, kb 〉 Example 10.64 Performing Scal
ar Multiplication Given vector v = 〈 3, 1 〉, find 3v, 1 2 v, and −v. Solution See Figure 10.120 for a geometric interpretation. If v = 〈 3, 1 〉, then 3v = 〈 3 ⋅ 3, 3 ⋅ 1 〉 = 〈 9v = 〈 −3, −1 〉 ⋅ 3, 1 2, 1 2 〉 ⋅ 1 〉 1186 Chapter 10 Further Applications of Trigonometry Figure 10.120 Analysis Notice that the vector 3v is three times the length of v, 1 2 but in the opposite direction. v is half the length of v, and –v is the same length of v, 10.40 Find the scalar multiple 3 u given u = 〈 5, 4 〉. Example 10.65 Using Vector Addition and Scalar Multiplication to Find a New Vector Given u = 〈 3, − 2 〉 and v = 〈 −1, 4 〉, find a new vector w = 3u + 2v. Solution First, we must multiply each vector by the scalar. 3u = 3 〈 3, − 2 〉 = 〈 9, − 6 〉 2v = 2 〈 −1, 4 〉 = 〈 −2, 8 〉 w = 3u + 2v = 〈 9, − 6 〉 + 〈 −2, 8 〉 = 〈 9 − 2, − 6 + 8 〉 = 〈 7, 2 〉 Then, add the two together. So, w = 〈 7, 2 〉. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1187 Finding Component Form In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the x direction, and the vertical component is the y direction. For example, we can see in the graph in Figure 10.121 that the position vector 〈 2, 3 〉 comes from adding the vectors v1 and v2. We
have v1 with initial point (0, 0) and terminal point (2, 0). We also have v2 with initial point (0, 0) and terminal point (0, 3). v1 = 〈 2 − 0, 0 − 0 〉 = 〈 2, 0 〉 Therefore, the position vector is v2 = 〈 0 − 0, 3 − 0 〉 = 〈 0, 3 〉 v = 〈 2 + 0, 3 + 0 〉 = 〈 2, 3 〉 Using the Pythagorean Theorem, the magnitude of v1 is 2, and the magnitude of v2 is 3. To find the magnitude of v, use the formula with the position vector. \|v\| = \|v1\|2 + \|v2\|2 = 22 + 32 = 13 The magnitude of v is 13. To find the direction, we use the tangent function tan θ = y x. tan θ = v2 v1 tan θ = 3 2 θ = tan−1 ⎛ ⎝ ⎞ ⎠ = 56.3° 3 2 Figure 10.121 Thus, the magnitude of v is 13 and the direction is 56.3∘ off the horizontal. Example 10.66 Finding the Components of the Vector 1188 Chapter 10 Further Applications of Trigonometry Find the components of the vector v with initial point (3, 2) and terminal point (7, 4). Solution First find the standard position. See the illustration in Figure 10.122. v = 〈 7 − 3, 4 − 2 〉 = 〈 4, 2 〉 Figure 10.122 The horizontal component is v1 = 〈 4, 0 〉 and the vertical component is v2 = 〈 0, 2〉. Finding the Unit Vector in the Direction of v In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. We can then preserve the direction of the original vector while simplifying calculations. Unit vectors are defined in terms of components. The horizontal unit vector is written as i = 〈 1, 0 〉 and is directed along the positive horizontal axis. The vertical unit vector is written as
j = 〈 0, 1 〉 and is directed along the positive vertical axis. See Figure 10.123. Figure 10.123 The Unit Vectors If v is a nonzero vector, then v \|v\| is a unit vector in the direction of v. Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1189 Example 10.67 Finding the Unit Vector in the Direction of v Find a unit vector in the same direction as v = 〈 −5, 12〉. Solution First, we will find the magnitude. \|v\| = ( − 5)2 + (12)2 = 25 + 144 = 169 = 13 Then we divide each component by \|v\|, which gives a unit vector in the same direction as v: v \|v\| = − 5 13 i + 12 13 j or, in component form See Figure 10.124. v \|v\| = 〈 − 5 13, 12 13 〉 Figure 10.124 1190 Chapter 10 Further Applications of Trigonometry Verify that the magnitude of the unit vector equals 1. The magnitude of − 5 13 i + 12 13 j is given as 2 ⎛ ⎝− 5 13 ⎞ ⎠ + ⎛ ⎝ 12 13 ⎞ ⎠ 2 = 25 169 = 169 169 + 144 169 = 1 The vector u = 5 13 i + 12 13 j is the unit vector in the same direction as v = 〈 −5, 12 〉. Performing Operations with Vectors in Terms of i and j So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j. Vectors in the Rectangular Plane Given a vector v with initial point P = (x1, y1) and terminal point Q = (x2, y2 ), v is written as v = (x2 − x1)i + (y1 −
y2) j The position vector from (0, 0) to (a, b), where (x2 − x1) = a and (y2 − y1) = b, is written as v = ai + bj. This vector sum is called a linear combination of the vectors i and j. The magnitude of v = ai + bj is given as \|v\| = a2 + b2. See Figure 10.125. Figure 10.125 Example 10.68 Writing a Vector in Terms of i and j Given a vector v with initial point P = (2, −6) and terminal point Q = (−6, 6), write the vector in terms of i and j. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1191 Begin by writing the general form of the vector. Then replace the coordinates with the given values. v = (x2 − x1)i + (y2 − y1) j = ( − 6 − 2)i + (6 − ( − 6)) j = − 8i + 12 j Example 10.69 Writing a Vector in Terms of i and j Using Initial and Terminal Points Given initial point P1 = (−1, 3) and terminal point P2 = (2, 7), write the vector v in terms of i and j. Solution Begin by writing the general form of the vector. Then replace the coordinates with the given values. v = (x2 − x1)i + (y2 − y1) j v = (2 − ( − 1))i + (7 − 3) j = 3i + 4 j Write the vector u with initial point P = (−1, 6) and terminal point Q = (7, − 5) in terms of i and 10.41 j. Performing Operations on Vectors in Terms of i and j When vectors are written in terms of i and j, we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components. Adding and Subtracting Vectors in Rectangular Coordinates Given v = ai + bj and u = ci + dj, then v + u = (a + c)i + (b + d) j v − u = (a − c)i + (b − d) j Example 10.70 Finding the Sum of the Vect
ors Find the sum of v1 = 2i − 3 j and v2 = 4i + 5 j. Solution According to the formula, we have 1192 Chapter 10 Further Applications of Trigonometry v1 + v2 = (2 + 4)i + ( − 3 + 5) j = 6i + 2 j Calculating the Component Form of a Vector: Direction We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using i and j. For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction. Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with \|v\| replacing r. Vector Components in Terms of Magnitude and Direction Given a position vector v = 〈 x, y 〉 and a direction angle θ, cos θ = x \|v\| x = \|v\|cos θ and sin θ = y \|v\| y = \|v\|sin θ Thus, v = xi + y j = \|v\|cos θi + \|v\|sin θ j, and magnitude is expressed as \|v\| = x2 + y2. Example 10.71 Writing a Vector in Terms of Magnitude and Direction Write a vector with length 7 at an angle of 135° to the positive x-axis in terms of magnitude and direction. Solution Using the conversion formulas x = \|v\|cos θi and y = \|v\|sin θ j, we find that x = 7cos(135°)i = − 7 2 2 y = 7sin(135°) j = 7 2 2 This vector can be written as v = 7cos(135°)i + 7sin(135°) j or simplified as 10.42 A vector travels from the origin to the point (3, 5). Write the vector in terms of magnitude and direction. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1193 Finding the Dot Product of Two Vectors As we discussed earlier in the section, scalar multiplication involves multiplying a
vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses. The dot product of two vectors involves multiplying two vectors together, and the result is a scalar. Dot Product The dot product of two vectors v = 〈 a, b 〉 and u = 〈 c, d 〉 is the sum of the product of the horizontal components and the product of the vertical components. v ⋅ u = ac + bd To find the angle between the two vectors, use the formula below. u \|u\| cos θ = v \|v\| ⋅ Example 10.72 Finding the Dot Product of Two Vectors Find the dot product of v = 〈 5, 12 〉 and u = 〈 −3, 4 〉. Solution Using the formula, we have v ⋅ u = 〈 5, 12 〉 ⋅ 〈 −3, 4 〉 = 5 ⋅ ( − 3) + 12 ⋅ 4 = − 15 + 48 = 33 Example 10.73 Finding the Dot Product of Two Vectors and the Angle between Them Find the dot product of v1 = 5i + 2j and v2 = 3i + 7j. Then, find the angle between the two vectors. Solution Finding the dot product, we multiply corresponding components. v1 ⋅ v2 = 〈 5, 2 〉 ⋅ 〈 3 = 15 + 14 = 29 To find the angle between them, we use the formula cos θ = v \|v\| ⋅ u \|u\|. 1194 Chapter 10 Further Applications of Trigonometry v \|v\| ⋅ u \|u\| 〉 ⋅ 〈 3 58 + 7 58 〉 + 2 29 + 2 29 = 〈 5 29 ⋅ 3 58 + 14 = 5 29 = 15 1682 1682 ⋅ 7 58 = 29 1682 = 0.707107 cos−1(0.707107) = 45° See Figure 10.126. Figure 10.126 Example 10.74 Finding the Angle between Two
Vectors Find the angle between u = 〈 −3, 4 〉 and v = 〈 5, 12 〉. Solution Using the formula, we have This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1195 ⋅ ⋅ ⎛ ⎝ ⎞ ⎠ = v u \|v\| \|u\| = θ = cos−1 ⎛ u ⎝ \|u\| −3i + 4 j 5 ⎛ ⎝− 3 5 = − 15 65 ⎞ ⋅ 5 ⎠ + 13 + 48 65 ⋅ ⎞ v ⎠ \|v\| 5i + 12 j 13 4 5 ⋅ 12 13 ⎛ ⎝ ⎞ ⎠ = 33 65 θ = cos−1 ⎛ ⎝ = 59.5∘ ⎞ ⎠ 33 65 See Figure 10.127. Figure 10.127 Example 10.75 Finding Ground Speed and Bearing Using Vectors We now have the tools to solve the problem we introduced in the opening of the section. 1196 Chapter 10 Further Applications of Trigonometry An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See Figure 10.128. Figure 10.128 Solution The ground speed is represented by x in the diagram, and we need to find the angle α in order to calculate the adjusted bearing, which will be 140° + α. Notice in Figure 10.128, that angle BCO must be equal to angle AOC by the rule of alternating interior angles, so angle BCO is 140°. We can find x by the Law of Cosines: x2 = (16.2)2 + (200)2 − 2(16.2)(200)cos(140°) x2 = 45, 226.41 x = 45, 226.41 x = 212.7 The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines. = sin α 16.2 sin α = sin(140°) 212.7 16.2sin(140°) 212.
7 = 0.04896 sin−1(0.04896) = 2.8° Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour. Access these online resources for additional instruction and practice with vectors. • Introduction to Vectors (http://openstaxcollege.org/l/introvectors) • Vector Operations (http://openstaxcollege.org/l/vectoroperation) • The Unit Vector (http://openstaxcollege.org/l/unitvector) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1197 10.8 EXERCISES Verbal What are the characteristics of the letters that are 495. commonly used to represent vectors? 496. How is a vector more specific than a line segment? 497. What are i and j, and what do they represent? 498. What is component form? When a unit vector is expressed as 〈 a, b 〉, which 499. letter is the coefficient of the i and which the j? Algebraic Given a vector with initial point (5, 2) and terminal 500. point (−1, − 3), find an equivalent vector whose initial form point is (0, 0). Write the vector in component 〈 a, b 〉. Given a vector with initial point (−4, 2) and 501. terminal point (3, − 3), find an equivalent vector whose initial point is (0, 0). Write the vector in component form 〈 a, b 〉. Given a vector with initial point (7, − 1) and 502. terminal point (−1, − 7), find an equivalent vector whose initial point is (0, 0). Write the vector in component form 〈 a, b 〉. P1 = (8, 3), P2 = (6, 5), P3 = (11, 8), and P4 = (9, 10) Given initial point P1 = (−3, 1) and terminal point 508. P2 = (5, 2), write the vector v in terms of i and j. Given initial point P1 = (6, 0) and terminal point 509. P2 = (−1, − 3), write the vector v in
terms of i and j. For the following exercises, use the vectors u = i + 5j, v = −2i− 3j, and w = 4i − j. 510. Find u + (v − w) 511. Find 4v + 2u For the following exercises, use the given vectors to compute u + v, u − v, and 2u − 3v. 512. u = 〈 2, − 3 〉, v = 〈 1, 5 〉 513. u = 〈 −3, 4 〉, v = 〈 −2, 1 〉 Let v = −4i + 3j. Find a vector that is half the length 514. and points in the same direction as v. Let v = 5i + 2j. Find a vector that is twice the length 515. and points in the opposite direction as v. For the following exercises, find a unit vector in the same direction as the given vector. For the following exercises, determine whether the two vectors u and v are equal, where u has an initial point P1 and a terminal point P2 and v has an initial point P3 and a terminal point P4. 516. a = 3i + 4j 517. b = −2i + 5j 518. c = 10i – j P1 = (5, 1), P2 = (3, − 2), P3 = (−1, 3), and 503. P4 = (9, − 4) P1 = (2, − 3), P2 = (5, 1), P3 = (6, − 1), and 504. P4 = (9, 3) P1 = (−1, − 1), P2 = (−4, 5), P3 = (−10, 6), 505. and P4 = (−13, 12) P1 = (3, 7), P2 = (2, 1), P3 = (1, 2), and 506. P4 = (−1, − 4) 507. 519 520. u = 100i + 200j 521. u = −14i + 2j the following exercises, For direction of the vector, 0 ≤ θ < 2π. find the magnitude and 522. 〈 0, 4 〉 523. 〈 6, 5 〉 524. �
�� 2, −5 〉 1198 Chapter 10 Further Applications of Trigonometry 525. 〈 −4, −6 〉 For the following exercises, use the vectors shown to sketch 2u + v. 526. Given u = 3i − 4j and v = −2i + 3j, calculate u ⋅ v. 536. 527. Given u = −i − j and v = i + 5j, calculate u ⋅ v. Given u = 〈 −2, 4 〉 and v = 〈 −3, 1 〉, 528. calculate u ⋅ v. u = 〈 −1, 6 〉 and v = 〈 6, − 1 〉, 529. Given calculate u ⋅ v. Graphical For the following exercises, given v, draw v, 3v and 1 2 v. 530. 〈 2, −1 〉 531. 〈 −1, 4 〉 532. 〈 −3, −2 〉 537. For the following exercises, use the vectors shown to sketch u + v, u − v, and 2u. 533. 534. 535. This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, use the vectors shown to sketch u − 3v. 538. 539. Chapter 10 Further Applications of Trigonometry 1199 A 60-pound box is resting on a ramp that is inclined 12°. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) component of the force. b. Find the magnitude of the component of the force that is parallel to the ramp. A 25-pound box is resting on a ramp that is inclined 550. 8°. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) component of the force. b. Find the magnitude of the component of the force that is parallel to the ramp. Find the magnitude of the horizontal and vertical 551. components of a vector with magnitude 8 pounds pointed in a direction of 27° above the horizontal. Round to the nearest hundredth. Find the magnitude of the horizontal and vertical 552. components of the vector with magnitude 4 pounds pointed in a direction of 127° above the
horizontal. Round to the nearest hundredth. Find the magnitude of the horizontal and vertical 553. components of a vector with magnitude 5 pounds pointed in a direction of 55° above the horizontal. Round to the nearest hundredth. For the following exercises, write the vector shown in component form. 540. 541. Given initial point P1 = (2, 1) and terminal point 542. P2 = (−1, 2), write the vector v in terms of i and j, then draw the vector on the graph. Find the magnitude of the horizontal and vertical 554. components of the vector with magnitude 1 pound pointed in a direction of 8° above the horizontal. Round to the nearest hundredth. Given initial point P1 = (4, − 1) and terminal point 543. P2 = (−3, 2), write the vector v in terms of i and j. Draw the points and the vector on the graph. Given initial point P1 = (3, 3) and terminal point 544. P2 = (−3, 3), write the vector v in terms of i and j. Draw the points and the vector on the graph. Extensions For the following exercises, use the given magnitude and direction in standard position, write the vector in component form. 545. \|v\| = 6, θ = 45 ° 546. \|v\| = 8, θ = 220° 547. \|v\| = 2, θ = 300° 548. \|v\| = 5, θ = 135° 549. Real-World Applications A woman leaves home and walks 3 miles west, then 2 555. miles southwest. How far from home is she, and in what direction must she walk to head directly home? A boat leaves the marina and sails 6 miles north, then 556. 2 miles northeast. How far from the marina is the boat, and in what direction must it sail to head directly back to the marina? A man starts walking from home and walks 4 miles 557. east, 2 miles southeast, 5 miles south, 4 miles southwest, and 2 miles east. How far has he walked? If he walked straight home, how far would he have to walk? A woman starts walking from home and walks 4 558. miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk?
A man starts walking from home and walks 3 miles at 559. 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction? 1200 Chapter 10 Further Applications of Trigonometry 560. A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction? Suppose a body has a force of 10 pounds acting on it 571. to the right, 25 pounds acting on it ─135° from the horizontal, and 5 pounds acting on it directed 150° from the horizontal. What single force is the resultant force acting on the body? The condition of equilibrium is when the sum of the 572. forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45° from the horizontal. What single force is needed to produce a state of equilibrium on the body? Suppose a body has a force of 3 pounds acting on it to 573. the left, 4 pounds acting on it upward, and 2 pounds acting on it 30° from the horizontal. What single force is needed to produce a state of equilibrium on the body? Draw the vector. An airplane is heading north at an airspeed of 600 km/ 561. hr, but there is a wind blowing from the southwest at 80 km/ hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? An airplane is heading north at an airspeed of 500 km/ 562. hr, but there is a wind blowing from the northwest at 50 km/ hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground? An airplane needs to head due north, but there is a 563. wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane? An airplane needs to head due north, but there is a 564. wind blowing from
the northwest at 80 km/hr. The plane flies with an airspeed of 500 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane? 565. As part of a video game, the point (5, 7) is rotated counterclockwise about the origin through an angle of 35°. Find the new coordinates of this point. 566. As part of a video game, the point (7, 3) is rotated counterclockwise about the origin through an angle of 40°. Find the new coordinates of this point. 567. Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 10 mph relative to the car, and the car is traveling 25 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)? Two children are throwing a ball back and forth 568. straight across the back seat of a car. The ball is being thrown 8 mph relative to the car, and the car is traveling 45 mph down the road. If one child doesn't catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)? A 50-pound object rests on a ramp that is inclined 569. 19°. Find the magnitude of the components of the force parallel to and perpendicular to (normal) the ramp to the nearest tenth of a pound. Suppose a body has a force of 10 pounds acting on it 570. to the right, 25 pounds acting on it upward, and 5 pounds acting on it 45° from the horizontal. What single force is the resultant force acting on the body? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1201 CHAPTER 10 REVIEW KEY TERMS altitude a perpendicular line from one vertex of a triangle to the opposite side, or in the case of an obtuse triangle, to the line containing the opposite side, forming two right triangles ambiguous case a scenario in which more than one triangle is a valid solution for a given oblique SSA triangle Archimedes’ spiral a polar curve given by r = θ. When multiplied by a constant, the equation appears as r = aθ. As r = θ, the curve continues to widen in a spiral path
over the domain. argument the angle associated with a complex number; the angle between the line from the origin to the point and the positive real axis cardioid a member of the limaçon family of curves, named for its resemblance to a heart; its equation is given as r = a ± bcos θ and r = a ± bsin θ, where a b = 1 convex limaҫon a type of one-loop limaçon represented by r = a ± bcos θ and r = a ± bsin θ such that a b ≥ 2 De Moivre’s Theorem formula used to find the nth power or nth roots of a complex number; states that, for a positive integer n, zn is found by raising the modulus to the nth power and multiplying the angles by n dimpled limaҫon a type of one-loop limaçon represented by r = a ± bcos θ and r = a ± bsin θ such that 1 < a b < 2 dot product given two vectors, the sum of the product of the horizontal components and the product of the vertical components Generalized Pythagorean Theorem used for SAS and SSS triangles initial point the origin of a vector an extension of the Law of Cosines; relates the sides of an oblique triangle and is inner-loop limaçon a polar curve similar to the cardioid, but with an inner loop; passes through the pole twice; represented by r = a ± bcos θ and r = a ± bsin θ where a < b Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle Law of Sines states that the ratio of the measurement of one angle of a triangle to the length of its opposite side is equal to the remaining two ratios of angle measure to opposite side; any pair of proportions may be used to solve for a missing angle or side lemniscate a polar curve resembling a figure 8 and given by the equation r 2 = a2 cos 2θ and r 2 = a2 sin 2θ, a ≠ 0 magnitude the length of a vector; may represent a quantity such as speed, and is calculated using the Pythagorean Theorem modulus the absolute value of a complex number, or the distance from the origin
to the point (x, y); also called the amplitude oblique triangle any triangle that is not a right triangle one-loop limaҫon a polar curve represented by r = a ± bcos θ and r = a ± bsin θ such that a > 0, b > 0, and a b > 1; may be dimpled or convex; does not pass through the pole parameter a variable, often representing time, upon which x and y are both dependent polar axis on the polar grid, the equivalent of the positive x-axis on the rectangular grid 1202 Chapter 10 Further Applications of Trigonometry polar coordinates on the polar grid, the coordinates of a point labeled (r, θ), where θ indicates the angle of rotation from the polar axis and r represents the radius, or the distance of the point from the pole in the direction of θ polar equation an equation describing a curve on the polar grid. polar form of a complex number a complex number expressed in terms of an angle θ and its distance from the origin r; can be found by using conversion formulas x = rcos θ, y = rsin θ, and r = x2 + y2 pole the origin of the polar grid resultant a vector that results from addition or subtraction of two vectors, or from scalar multiplication rose curve a polar equation resembling a flower, given by the equations r = acos nθ and r = asin nθ; when n is even there are 2n petals, and the curve is highly symmetrical; when n is odd there are n petals. scalar a quantity associated with magnitude but not direction; a constant scalar multiplication the product of a constant and each component of a vector standard position the placement of a vector with the initial point at (0, 0) and the terminal point (a, b), represented by the change in the x-coordinates and the change in the y-coordinates of the original vector terminal point the end point of a vector, usually represented by an arrow indicating its direction unit vector a vector that begins at the origin and has magnitude of 1; the horizontal unit vector runs along the x-axis and is defined as v1 = 〈 1, 0 〉 the vertical unit vector runs along the y-axis and is defined as v2 = 〈 0, 1 〉. vector a quantity associated with both magnitude and direction, represented as a directed line segment with a starting point
(initial point) and an end point (terminal point) vector addition the sum of two vectors, found by adding corresponding components KEY EQUATIONS Law of Sines Area for oblique triangles sin α a = a sin α = sin β b = b sin β = sin γ c c sin γ Area = 1 2 = 1 2 = 1 2 bcsin α acsin β absin γ Law of Cosines a2 = b2 + c2 − 2bccos α b2 = a2 + c2 − 2accos β c2 = a2 + b2 − 2abcos γ Heron’s formula Area = s(s − a)(s − b)(s − c) (a + b + c) 2 where s = This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1203 Conversion formulas cos θ = x r → x = rcos θ y r → y = rsin θ sin θ = r 2 = x2 + y2 y tan θ = x KEY CONCEPTS 10.1 Non-right Triangles: Law of Sines • The Law of Sines can be used to solve oblique triangles, which are non-right triangles. • According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side. • There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. See Example 10.1. • The ambiguous case arises when an oblique triangle can have different outcomes. • There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. See Example 10.2 and Example 10.3. • The Law of Sines can be used to solve triangles with given criteria. See Example 10.4. • The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. See Example 10.5. • There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. See Example 10.6. 10.2 Non-right Triangles: Law of Cosines • The Law of Cosines defines the relationship among
angle measurements and lengths of sides in oblique triangles. • The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See Example 10.7 and Example 10.8. • The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See Example 10.9 and Example 10.10. • Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See Example 10.11 and See Example 10.12. 10.3 Polar Coordinates • The polar grid is represented as a series of concentric circles radiating out from the pole, or origin. • To plot a point in the form (r, θ), θ > 0, move in a counterclockwise direction from the polar axis by an angle of θ, and then extend a directed line segment from the pole the length of r in the direction of θ. If θ is negative, move in a clockwise direction, and extend a directed line segment the length of r in the direction of θ. See Example 10.13. • If r is negative, extend the directed line segment in the opposite direction of θ. See Example 10.14. • To convert from polar coordinates to rectangular coordinates, use the formulas x = rcos θ and y = rsin θ. See Example 10.15 and Example 10.16. 1204 Chapter 10 Further Applications of Trigonometry • To convert x r, sin θ = cos θ = y r, tan θ = y x, and r = x2 + y2. See Example 10.17. from rectangular coordinates to polar coordinates, use one or more of the formulas: • Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. See Example 10.18, Example 10.19, and Example 10.20. • Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation,
and then graph it in the rectangular plane. See Example 10.21, Example 10.22, and Example 10.23. 10.4 Polar Coordinates: Graphs • It is easier to graph polar equations if we can test the equations for symmetry with respect to the line θ =, the π 2 polar axis, or the pole. • There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. See Example 10.24. • Polar equations may be graphed by making a table of values for θ and r. • The maximum value of a polar equation is found by substituting the value θ that leads to the maximum value of the trigonometric expression. • The zeros of a polar equation are found by setting r = 0 and solving for θ. See Example 10.25. • Some formulas that produce the graph of a circle in polar coordinates are given by r = acos θ and r = asin θ. See Example 10.26. • The formulas that produce the graphs of a cardioid are given by r = a ± bcos θ and r = a ± bsin θ, for a > 0, b > 0, and a b = 1. See Example 10.27. • The formulas that produce the graphs of a one-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ for 1 < a b < 2. See Example 10.28. • The formulas that produce the graphs of an inner-loop limaçon are given by r = a ± bcos θ and r = a ± bsin θ for a > 0, b > 0, and a < b. See Example 10.29. • The formulas that produce the graphs of a lemniscates are given by r 2 = a2 cos 2θ and r 2 = a2 sin 2θ, where a ≠ 0. See Example 10.30. • The formulas that produce the graphs of rose curves are given by r = acos nθ and r = asin nθ, where a ≠ 0; if n is even, there are 2n petals, and if n is odd, there are n petals. See Example 10.31 and Example 10.32. • The formula that produces the graph of an
Archimedes’ spiral is given by r = θ, θ ≥ 0. See Example 10.33. 10.5 Polar Form of Complex Numbers • Complex numbers in the form a + bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Label the x-axis as the real axis and the y-axis as the imaginary axis. See Example 10.34. • The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point: \|z\| = a2 + b2. See Example 10.35 and Example 10.36. • To write complex numbers in polar form, we use the formulas x = rcos θ, y = rsin θ, and r = x2 + y2. Then, z = r(cos θ + isin θ). See Example 10.37 and Example 10.38. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1205 • To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through by r. See Example 10.39 and Example 10.40. • To find the product of two complex numbers, multiply the two moduli and add the two angles. Evaluate the trigonometric functions, and multiply using the distributive property. See Example 10.41. • To find the quotient of two complex numbers in polar form, find the quotient of the two moduli and the difference of the two angles. See Example 10.42. • To find the power of a complex number zn 10.43., raise r to the power n, and multiply θ by n. See Example • Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. See Example 10.44. 10.6 Parametric Equations • Parameterizing a curve involves translating a rectangular equation in two variables, x and y, into two equations in three variables, x, y, and t. Often, more information is obtained from a set of parametric equations. See Example 10.45, Example 10.46, and Example 10.47. • Sometimes equations are simpler to graph when written in rectangular form. By eliminating t, an equation in x and y is the result. • To eliminate t,
solve one of the equations for t, and substitute the expression into the second equation. See Example 10.48, Example 10.49, Example 10.50, and Example 10.51. • Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for t in one of the equations, and substitute the expression into the second equation. See Example 10.52. • There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation. • Find an expression for x such that the domain of the set of parametric equations remains the same as the original rectangular equation. See Example 10.53. 10.7 Parametric Equations: Graphs • When there is a third variable, a third parameter on which x and y depend, parametric equations can be used. • To graph parametric equations by plotting points, make a table with three columns labeled t, x(t), and y(t). Choose values for t in increasing order. Plot the last two columns for x and y. See Example 10.54 and Example 10.55. • When graphing a parametric curve by plotting points, note the associated t-values and show arrows on the graph indicating the orientation of the curve. See Example 10.56 and Example 10.57. • Parametric equations allow the direction or the orientation of the curve to be shown on the graph. Equations that are not functions can be graphed and used in many applications involving motion. See Example 10.58. • Projectile motion depends on two parametric equations: x = (v0 cos θ)t and y = − 16t 2 + (v0 sin θ)t + h. Initial velocity is symbolized as v0. θ represents the initial angle of the object when thrown, and h represents the height at which the object is propelled. 10.8 Vectors • The position vector has its initial point at the origin. See Example 10.59. • If the position vector is the same for two vectors, they are equal. See Example 10.60. 1206 Chapter 10 Further Applications of Trigonometry • Vectors are defined by their magnitude and direction. See Example 10.61. • If two vectors have the same magnitude and direction, they are equal. See Example 10.62. • Vector addition and subtraction result in a new vector found by adding or subtracting corresponding elements. See Example 10
.63. • Scalar multiplication is multiplying a vector by a constant. Only the magnitude changes; the direction stays the same. See Example 10.64 and Example 10.65. • Vectors are comprised of two components: the horizontal component along the positive x-axis, and the vertical component along the positive y-axis. See Example 10.66. • The unit vector in the same direction of any nonzero vector is found by dividing the vector by its magnitude. • The magnitude of a vector in the rectangular coordinate system is \|v\| = a2 + b2. See Example 10.67. • In the rectangular coordinate system, unit vectors may be represented in terms of i and j where i represents the horizontal component and j represents the vertical component. Then, v = ai + bj is a scalar multiple of v by real numbers a and b. See Example 10.68 and Example 10.69. • Adding and subtracting vectors in terms of i and j consists of adding or subtracting corresponding coefficients of i and corresponding coefficients of j. See Example 10.70. • A vector v = ai + bj is written in terms of magnitude and direction as v = \|v\|cos θi + \|v\|sin θ j. See Example 10.71. • The dot product of two vectors is the product of the i terms plus the product of the j terms. See Example 10.72. • We can use the dot product to find the angle between two vectors. Example 10.73 and Example 10.74. • Dot products are useful for many types of physics applications. See Example 10.75. CHAPTER 10 REVIEW EXERCISES Non-right Triangles: Law of Sines For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. 574. β = 50°, a = 105, b = 45 575. α = 43.1°, a = 184.2, b = 242.8 576. Solve the triangle. 577. Find the area of the triangle. 578. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 2.1 km apart, to be 25° and 49°, as shown in Figure 10.129. Find the distance of the plane from point A and the elevation of
the plane. Figure 10.129 Non-right Triangles: Law of Cosines 579. Solve the triangle, rounding to the nearest tenth, assuming α is opposite side a, β is opposite side b, and γ s opposite side c : a = 4, b = 6, c = 8. 580. Solve the triangle in Figure 10.130, rounding to the nearest tenth. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 10 Further Applications of Trigonometry 1207 591. x2 + y2 = − 2y the following exercises, convert For equation to a Cartesian equation. the given polar 592. r = 7cos θ 593. r = −2 4cos θ + sin θ For the following exercises, convert to rectangular form and graph. 594. θ = 3π 4 595. r = 5sec θ Polar Coordinates: Graphs the following exercises, For symmetry. 596. r = 4 + 4sin θ 597. r = 7 test each equation for 598. Sketch a graph of the polar equation r = 1 − 5sin θ. Label the axis intercepts. 599. Sketch a graph of the polar equation r = 5sin(7θ). 600. Sketch a graph of the polar equation r = 3 − 3cos θ Polar Form of Complex Numbers For the following exercises, find the absolute value of each complex number. 601. −2 + 6i 602. 4 − 3i Figure 10.130 581. Find the area of a triangle with sides of length 8.3, 6.6, and 9.1. 582. To find the distance between two cities, a satellite calculates the distances and angle shown in Figure 10.131 (not to scale). Find the distance between the cities. Round answers to the nearest tenth. Figure 10.131 Polar Coordinates 583. Plot the point with polar coordinates ⎛ ⎝3, ⎞ ⎠. π 6 584. Plot the point with polar coordinates ⎛ ⎝5, − 2π 3 ⎞ ⎠ 585. Convert ⎛ ⎝6, − 3π 4 ⎞ ⎠ to rectangular coordinates. 586. Convert ⎛ ⎝−2, 3π 2 ⎞ ⎠ to rectangular coordinates.
587. Convert (7, − 2) to polar coordinates. Write the complex number in polar form. 588. Convert (−9, − 4) to polar coordinates. For the following exercises, convert the given Cartesian equation to a polar equation. 589. x = − 2 590. x2 + y2 = 64 603. 5 + 9i 604. 1 2 − 3 2 i For the following exercises, convert the complex number from polar to rectangular form. 605. ⎛ z = 5cis ⎝ ⎞ ⎠ 5π 6 1208 Chapter 10 Further Applications of Trigonometry 606. z = 3cis(40°) For the following exercises, find the product z1 z2 in polar form. 607. z1 = 2cis(89°) z2 = 5cis(23°) 608. ⎛ z1 = 10cis ⎝ ⎞ ⎠ π 6 ⎛ z2 = 6cis ⎝ ⎞ ⎠ π 3 For the following exercises, find the quotient form. 609. z1 = 12cis(55°) z2 = 3cis(18°) 610. ⎛ z1 = 27cis ⎝ ⎞ ⎠ 5π 3 ⎛ z2 = 9cis ⎝ ⎞ ⎠ π 3 618. x(t) = − cos t ⎧ ⎨ y(t) = 2sin2 t ⎩ 619. Parameterize (write a parametric equation for) each using x(t) = acos t and by Cartesian equation y(t) = bsin t for x2 25 + y2 16 = 1. 620. Parameterize the line from ( − 2, 3) to (4, 7) so that the line is at ( − 2, 3) at t = 0 and (4, 7) at t = 1. Parametric Equations: Graphs z1 z2 in polar For the following exercises, make a table of values for each set of parametric equations, graph the equations, and include an orientation; then write the Cartesian equation. 621. ⎧ x(t) = 3t 2 ⎨ y(t) = 2t − 1 ⎩ 622. x(
t) = et ⎧ ⎨ y(t) = − 2e5 t ⎩ 623. x(t) = 3cos t ⎧ ⎨ y(t) = 2sin t ⎩ For the following exercises, find the powers of each complex number in polar form. 611. Find z4 when z = 2cis(70°) ⎛ 612. Find z2 when z = 5cis ⎝ ⎞ ⎠ 3π 4 624. A ball is launched with an initial velocity of 80 feet per second at an angle of 40° to the horizontal. The ball is released at a height of 4 feet above the ground. a. Find the parametric equations to model the path of the ball. b. Where is the ball after 3 seconds? c. How long is the ball in the air? For the following exercises, evaluate each root. 613. Evaluate the cube root of z when z = 64cis(210°). Vectors ⎛ 614. Evaluate the square root of z when z = 25cis ⎝ ⎞ ⎠. 3π 2 For the following exercises, plot the complex number in the complex plane. 615. 6 − 2i 616. −1 + 3i Parametric Equations For the following exercises, eliminate the parameter t to rewrite the parametric equation as a Cartesian equation. 617. x(t) = 3t − 1 ⎧ ⎨ y(t) = t ⎩ This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, determine whether the two vectors, u and v, are equal, where u has an initial point P1 and a terminal point P2, and v has an initial point P3 and a terminal point P4. 625. P4 = (9, 2) P1 = (−1, 4), P2 = (3, 1), P3 = (5, 5) and P1 = (6, 11), P2 = (−2, 8), P3 = (0, − 1) and 626. P4 = (−8, 2) the following vectors exercises, For u = 2i − j,v = 4i − 3 j, and w = − 2i + 5 j to evaluate the expression. use the
627. u − v Chapter 10 Further Applications of Trigonometry 1209 628. 2v − u + w For the following exercises, find a unit vector in the same direction as the given vector. 629. a = 8i − 6j 630. b = −3i − j the following exercises, For direction of the vector. find the magnitude and 631. 〈 6, −2 〉 632. 〈 −3, −3 〉 For the following exercises, calculate u ⋅ v. 633. u = −2i + j and v = 3i + 7j 634. u = i + 4j and v = 4i + 3j 635. Given v = 〈−3, 4 〉 draw v, 2v, and 1 2 v. 636. Given the vectors shown in Figure 10.132, sketch u + v, u − v and 3v. Figure 10.132 637. Given initial point P1 = (3, 2) and terminal point P2 = (−5, − 1), write the vector v in terms of i and j. Draw the points and the vector on the graph. CHAPTER 10 PRACTICE TEST 638. Assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve the triangle, if possible, and round each answer tenth, given β = 68°, b = 21, c = 16. to the nearest 639. Find the area of the triangle in Figure 10.133. Round each answer to the nearest tenth. Figure 10.133 1210 Chapter 10 Further Applications of Trigonometry 640. A pilot flies in a straight path for 2 hours. He then makes a course correction, heading 15° to the right of his original course, and flies 1 hour in the new direction. If he maintains a constant speed of 575 miles per hour, how far is he from his starting position? 656. Eliminate the parameter t to rewrite the following equation: parametric x(t) = t + 1 ⎧ ⎨ y(t) = 2t 2. ⎩ equations Cartesian as a 641. Convert (2, 2) the point. to polar coordinates, and then plot 657. Parameterize (write a parametric equation for) the following Cartesian equation by using x(t) = acos t and
642. Convert ⎛ ⎝2, π 3 ⎞ ⎠ to rectangular coordinates. y(t) = bsin t : x2 36 + y2 100 = 1. 643. Convert the polar equation to a Cartesian equation: x2 + y2 = 5y. 658. Graph the set of parametric equations and find the x(t) = − 2sin t ⎧ Cartesian equation: ⎨ y(t) = 5cos t ⎩. 659. A ball is launched with an initial velocity of 95 feet per second at an angle of 52° to the horizontal. The ball is released at a height of 3.5 feet above the ground. a. Find the parametric equations to model the path of the ball. b. Where is the ball after 2 seconds? c. How long is the ball in the air? For the following exercises, use the vectors u = i − 3j and v = 2i + 3j. 660. Find 2u − 3v. 661. Calculate u ⋅ v. 662. Find a unit vector in the same direction as v. 663. Given vector v has an initial point P1 = (2, 2) and terminal point P2 = (−1, 0), write the vector v in terms of i and j. On the graph, draw v, and − v. Convert 644. r = − 3csc θ. to rectangular form and graph: 645. Test the equation for symmetry: r = − 4sin⎛ ⎝2θ). 646. Graph r = 3 + 3cos θ. 647. Graph r = 3 − 5sin θ. 648. Find the absolute value of the complex number 5 − 9i. 649. Write the complex number in polar form: 4 + i. 650. Convert ⎛ rectangular form: z = 5cis ⎝ the complex number 2π 3 ⎞ ⎠. from polar to Given z1 = 8cis(36°) and z2 = 2cis(15°), evaluate each expression. 651. z1 z2 652. z1 z2 653. (z2)3 654. z1 655. Plot the complex number −5 − i in the complex plane. This content is available for free at https://cnx.org/
content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1211 11 \| SYSTEMS OF EQUATIONS AND INEQUALITIES Figure 11.1 Enigma machines like this one, once owned by Italian dictator Benito Mussolini, were used by government and military officials for enciphering and deciphering top-secret communications during World War II. (credit: Dave Addey, Flickr) Chapter Outline 11.1 Systems of Linear Equations: Two Variables 11.2 Systems of Linear Equations: Three Variables 11.3 Systems of Nonlinear Equations and Inequalities: Two Variables 11.4 Partial Fractions 11.5 Matrices and Matrix Operations 11.6 Solving Systems with Gaussian Elimination 11.7 Solving Systems with Inverses 11.8 Solving Systems with Cramer's Rule 1212 Chapter 11 Systems of Equations and Inequalities Introduction By 1943, it was obvious to the Nazi regime that defeat was imminent unless it could build a weapon with unlimited destructive power, one that had never been seen before in the history of the world. In September, Adolf Hitler ordered German scientists to begin building an atomic bomb. Rumors and whispers began to spread from across the ocean. Refugees and diplomats told of the experiments happening in Norway. However, Franklin D. Roosevelt wasn’t sold, and even doubted British Prime Minister Winston Churchill’s warning. Roosevelt wanted undeniable proof. Fortunately, he soon received the proof he wanted when a group of mathematicians cracked the “Enigma” code, proving beyond a doubt that Hitler was building an atomic bomb. The next day, Roosevelt gave the order that the United States begin work on the same. The Enigma is perhaps the most famous cryptographic device ever known. It stands as an example of the pivotal role cryptography has played in society. Now, technology has moved cryptanalysis to the digital world. Many ciphers are designed using invertible matrices as the method of message transference, as finding the inverse of a matrix is generally part of the process of decoding. In addition to knowing the matrix and its inverse, the receiver must also know the key that, when used with the matrix inverse, will allow the message to be read. In this chapter, we will investigate matrices and their inverses, and various ways to use matrices to solve systems of equations. First, however, we will study systems of equations on their own: linear and nonlinear,
and then partial fractions. We will not be breaking any secret codes here, but we will lay the foundation for future courses. 11.1 \| Systems of Linear Equations: Two Variables Learning Objectives In this section, you will: 11.1.1 Solve systems of equations by graphing. 11.1.2 Solve systems of equations by substitution. 11.1.3 Solve systems of equations by addition. 11.1.4 Identify inconsistent systems of equations containing two variables. 11.1.5 Express the solution of a system of dependent equations containing two variables. Figure 11.2 (credit: Thomas Sørenes) A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions. Introduction to Systems of Equations In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1213 infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution. In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables. 2x + y = 15 3x – y = 5 The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is
the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists. 2(4) + (7) = 15 True 3(4) − (7) = 5 True In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions. Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts. There are no points common to both lines; hence, there is no solution to the system. Types of Linear Systems There are three types of systems of linear equations in two variables, and three types of solutions. • An independent system has exactly one solution pair (x, y). The point where the two lines intersect is the only solution. • An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect. • A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations. Figure 11.3 compares graphical representations of each type of system. Figure 11.3 1214 Chapter 11 Systems of Equations and Inequalities Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution. 1. Substitute the ordered pair into each equation in the system. 2. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution. Example 11.1 Determining Whether an Ordered Pair Is a Solution to a System of Equations Determine whether the ordered pair (5, 1) is a solution to the given system of equations. x + 3y = 8 2x − 9
= y Solution Substitute the ordered pair (5, 1) into both equations. (5) + 3(1) = 8 8 = 8 True 2(5) − 9 = (1) 1=1 True The ordered pair (5, 1) satisfies both equations, so it is the solution to the system. Analysis We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines. See Figure 11.4. Figure 11.4 11.1 Determine whether the ordered pair (8, 5) is a solution to the following system. 5x−4y = 20 2x + 1 = 3y This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1215 Solving Systems of Equations by Graphing There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes. Example 11.2 Solving a System of Equations in Two Variables by Graphing Solve the following system of equations by graphing. Identify the type of system. Solution Solve the first equation for y. Solve the second equation for y. 2x + y = −8 x − y = −1 2x + y = −8 y = −2x−8 x − y = −1 y = x + 1 Graph both equations on the same set of axes as in Figure 11.5. Figure 11.5 The lines appear to intersect at the point (−3,−2). We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations. 2(−3) + (−2) = −8 −8 = −8 True (−3) − (−2) = −1 −1 = −1 True The solution to the system is the ordered pair (−3,−2), so the system is independent. 1216 Chapter 11 Systems of Equations and Inequalities 11.2 Solve the following system of equations by graphing. 2x − 5y = −25 −4x + 5y = 35 Can graphing be used if the system is inconsistent or dependent? Yes
, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system. Solving Systems of Equations by Substitution Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical. Given a system of two equations in two variables, solve using the substitution method. 1. Solve one of the two equations for one of the variables in terms of the other. 2. Substitute the expression for this variable into the second equation, then solve for the remaining variable. 3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair. 4. Check the solution in both equations. Example 11.3 Solving a System of Equations in Two Variables by Substitution Solve the following system of equations by substitution. Solution First, we will solve the first equation for y. − x + y = −5 2x − 5y = 1 −x + y = −5 y = x−5 Now we can substitute the expression x−5 for y in the second equation. 2x − 5y = 1 2x − 5(x − 5) = 1 2x − 5x + 25 = 1 − 3x = −24 x = 8 Now, we substitute x = 8 into the first equation and solve for y. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1217 −(8) + y = −5 y = 3 Our solution is (8, 3). Check the solution by substituting (8, 3) into both equations8) + (3) = − 5 2x − 5y = 1 2(
8) − 5(3) = 1 True True 11.3 Solve the following system of equations by substitution. x = y + 3 4 = 3x−2y Can the substitution method be used to solve any linear system in two variables? Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions. Solving Systems of Equations in Two Variables by the Addition Method A third method of solving systems of linear equations is the addition method. In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition. Given a system of equations, solve using the addition method. 1. Write both equations with x- and y-variables on the left side of the equal sign and constants on the right. 2. Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable. 3. Solve the resulting equation for the remaining variable. 4. Substitute that value into one of the original equations and solve for the second variable. 5. Check the solution by substituting the values into the other equation. Example 11.4 Solving a System by the Addition Method Solve the given system of equations by addition. 1218 Chapter 11 Systems of Equations and Inequalities x + 2y = −1 −x + y = 3 Solution Both equations are already set equal to a constant. Notice that the coefficient of x in the second equation, –1, is the opposite of the coefficient of x in the first equation, 1. We can add the two equations to eliminate x without needing to multiply by a constant. x + 2y = − 1 −x + y = 3 3y = 2 Now that we have eliminated x, we can solve the resulting equation for y. 3y = 2 y = 2 3 Then, we substitute this value for y into one of the original equations and solve
for x The solution to this system is ⎛ ⎝− 7 3 ⎞ ⎠., 2 3 Check the solution in the first equation. x + 2y = −1 ⎛ ⎝− 1 = −1 True Analysis We gain an important perspective on systems of equations by looking at the graphical representation. See Figure 11.6 to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1219 Figure 11.6 Example 11.5 Using the Addition Method When Multiplication of One Equation Is Required Solve the given system of equations by the addition method. 3x + 5y = −11 x − 2y = 11 Solution Adding these equations as presented will not eliminate a variable. However, we see that the first equation has 3x in it and the second equation has x. So if we multiply the second equation by −3, the x-terms will add to zero. x−2y = 11 −3(x−2y) = −3(11) Multiply both sides by −3. −3x + 6y = −33 Use the distributive property. Now, let’s add them. 3x + 5y = −11 −3x + 6y = −33 _______________ 11y = −44 y = −4 For the last step, we substitute y = −4 into one of the original equations and solve for x. 3x + 5y = − 11 3x + 5( − 4) = − 11 3x − 20 = − 11 3x = 9 x = 3 Our solution is the ordered pair (3, −4). See Figure 11.7. Check the solution in the original second equation. x − 2y = 11 (3) − 2( − 4) = 3 + 8 = 11 True 1220 Chapter 11 Systems of Equations and Inequalities Figure 11.7 11.4 Solve the system of equations by addition. 2x−7y = 2 3x + y = −20 Example 11.6 Using the Addition Method When Multiplication of Both Equations Is Required Solve the given system of equations in two variables by addition. 2x +
3y = −16 5x−10y = 30 Solution One equation has 2x and the other has 5x. The least common multiple is 10x so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate x by multiplying the first equation by −5 and the second equation by 2. Then, we add the two equations together. − 5(2x + 3y) = − 5(−16) − 10x − 15y = 80 2(5x − 10y) = 2(30) 10x − 20y = 60 −10x−15y = 80 10x−20y = 60 ________________ −35y = 140 y = −4 Substitute y = −4 into the original first equation. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1221 2x + 3(−4) = −16 2x − 12 = −16 2x = −4 x = −2 The solution is (−2, −4). Check it in the other equation. 5x−10y = 30 5(−2)−10(−4) = 30 −10 + 40 = 30 30 = 30 See Figure 11.8. Figure 11.8 Example 11.7 Using the Addition Method in Systems of Equations Containing Fractions Solve the given system of equations in two variables by addition Solution First clear each equation of fractions by multiplying both sides of the equation by the least common denominator. 6 + x 3 ⎞ ⎠ = 6(3) y ⎛ ⎝ 6 2x + y = 18 y ⎞ ⎛ ⎠ = 4(1) ⎝ 4 2x − y = 4 x 2 − 4 1222 Chapter 11 Systems of Equations and Inequalities Now multiply the second equation by −1 so that we can eliminate the x-variable. −1(2x − y) = −1(4) −2x + y = −4 Add the two equations to eliminate the x-variable and solve the resulting equation. Substitute y = 7 into the first equation. 2x + y = 18 −2x + y = −4 _____________ 2y = 14 y = 7 2x + (7) = 18 2x = 11 x = 11 2 = 7.5 The
solution is ⎛ ⎝ 11 2 ⎞ ⎠. Check it in the other equation., 7 x 2 11 2 2 11 11.5 Solve the system of equations by addition. 2x + 3y = 8 3x + 5y = 10 Identifying Inconsistent Systems of Equations Containing Two Variables Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different y -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as 12 = 0. Example 11.8 Solving an Inconsistent System of Equations Solve the following system of equations. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1223 x = 9−2y x + 2y = 13 Solution We can approach this problem in two ways. Because one equation is already solved for x, is to use substitution. the most obvious step x + 2y = 13 (9 − 2y) + 2y = 13 9 + 0y = 13 9 = 13 Clearly, this statement is a contradiction because 9 ≠ 13. Therefore, the system has no solution. The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows. x = 9−2y 2y = − We then convert the second equation expressed to slope-intercept form. x + 2y = 13 2y = − x + 13 y = − 1 2 x + 13 2 Comparing the equations, we see that they have the same slope but different y-intercepts. Therefore, the lines are parallel and do not intersect + 13 2 Analysis Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown in Figure 11.9. Figure 11.9 1224 Chapter 11 Systems of Equations and Inequalities 11.6 Solve the following system of equations in two variables. 2y−2x = 2 2y−2x = 6 Expressing the Solution of a System
of Dependent Equations Containing Two Variables Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as 0 = 0. Example 11.9 Finding a Solution to a Dependent System of Linear Equations Find a solution to the system of equations using the addition method. x + 3y = 2 3x + 9y = 6 Solution With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating x. If we multiply both sides of the first equation by −3, then we will be able to eliminate the x -variable. Now add the equations. x + 3y = 2 (−3)(x + 3y) = (−3)(2) −3x − 9y = − 6 − 3x − 9y = −6 + 3x + 9y = 6 ______________ 0 = 0 We can see that there will be an infinite number of solutions that satisfy both equations. Analysis If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form. x + 2 3 x + 3y = 2 3y = − x + 2 y = − 1 3 3x + 9y = 6 9y = −3x + See Figure 11.10. Notice the results are the same. The general solution to the system is ⎛ ⎝x, −1 3 x + 2 3 ⎞ ⎠. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1225 Figure 11.10 11.7 Solve the following system of equations in two variables. y−2x = 5 −3y + 6x = −15 Using Systems of Equations to Investigate Profits Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by
the equation R = xp, where x = quantity and p = price. The revenue function is shown in orange in Figure 11.11. The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in Figure 11.11. The x -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars. Figure 11.11 The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money. 1226 Chapter 11 Systems of Equations and Inequalities The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as P(x) = R(x) − C(x). Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses. Example 11.10 Finding the Break-Even Point and the Profit Function Using Substitution Given the cost function C(x) = 0.85x + 35,000 and the revenue function R(x) = 1.55x, find the break-even point and the profit function. Solution Write the system of equations using y to replace function notation. y = 0.85x + 35,000 y = 1.55x Substitute the expression 0.85x + 35,000 from the first equation into the second equation and solve for x. 0.85x + 35,000 = 1.55x 35,000 = 0.7x 50,000 = x Then, we substitute x = 50,000 into either the cost function or the revenue function. 1.55(50,000) = 77,500 The break-even point is (50,000, 77,500). The profit function is found using the formula P(x) = R(x) − C(x). P(x) = 1.55x − (0.85x + 35, 000) = 0.7x − 35, 000 The profit
function is P(x) = 0.7x−35,000. Analysis The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units. See Figure 11.12. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1227 Figure 11.12 We see from the graph in Figure 11.13 that the profit function has a negative value until x = 50,000, when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss. Figure 11.13 Example 11.11 Writing and Solving a System of Equations in Two Variables 1228 Chapter 11 Systems of Equations and Inequalities The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets? Solution Let c = the number of children and a = the number of adults in attendance. The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day. c + a = 2,000 The revenue from all children can be found by multiplying $25.00 by the number of children, 25c. The revenue from all adults can be found by multiplying $50.00 by the number of adults, 50a. The total revenue is $70,000. We can use this to write an equation for the revenue. We now have a system of linear equations in two variables. 25c + 50a = 70,000 c + a = 2,000 25c + 50a = 70,000 In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either c or a. We will solve for a. Substitute the expression 2,000 − c in the second equation for a and solve for c. c + a = 2,
000 a = 2,000 − c 25c + 50(2,000 − c) = 70,000 25c + 100,000 − 50c = 70,000 − 25c = −30,000 c = 1,200 Substitute c = 1,200 into the first equation to solve for a. 1,200 + a = 2,000 a = 800 We find that 1,200 children and 800 adults bought tickets to the circus that day. Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were 11.8 bought for a total of $14,200, how many children and how many adults bought meal tickets? Access these online resources for additional instruction and practice with systems of linear equations. • Solving Systems of Equations Using Substitution (http://openstaxcollege.org/l/syssubst) • Solving Systems of Equations Using Elimination (http://openstaxcollege.org/l/syselim) • Applications of Systems of Equations (http://openstaxcollege.org/l/sysapp) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1229 11.1 EXERCISES Verbal Can a system of linear equations have exactly two 1. solutions? Explain why or why not. 2. If you are performing a break-even analysis for a business and their cost and revenue equations are dependent, explain what this means for the company’s profit margins. 3. If you are solving a break-even analysis and get a negative break-even point, explain what this signifies for the company? 4. If you are solving a break-even analysis and there is no break-even point, explain what this means for the company. How should they ensure there is a break-even point? Given a system of equations, explain at 5. different methods of solving that system. least two Algebraic For the following exercises, determine whether the given ordered pair is a solution to the system of equations. 15. 16. 17. 18. 19. 20. −2x + 3y = 1.2 −3x − 6y = 1.8 x−0.2y = 1 −10x + 2y = 5 3x + 5y = 9 30x + 50y = −90 −3x
+ y = 2 12x−4y = − = 16 = 11 y = 3 6. 7. 8. 9. 5x − y = 4 x + 6y = 2 and (4, 0) −3x − 5y = 13 − x + 4y = 10 and (−6, 1) 3x + 7y = 1 2x + 4y = 0 and (2, 3) −2x + 5y = 7 2x + 9y = 7 and (−1, 1) 10. x + 8y = 43 3x−2y = −1 and (3, 5) the following exercises, For substitution. solve each system by 11. 12. 13. 14. x + 3y = 5 2x + 3y = 4 3x−2y = 18 5x + 10y = −10 4x + 2y = −10 3x + 9y = 0 2x + 4y = −3.8 9x−5y = 1.3 For the following exercises, solve each system by addition. 21. 22. 23. 24. 25. 26. 27. 28. −2x + 5y = −42 7x + 2y = 30 6x−5y = −34 2x + 6y = 4 5x − y = −2.6 −4x−6y = 1.4 7x−2y = 3 4x + 5y = 3.25 −x + 2y = −1 5x−10y = 6 7x + 6y = 2 −28x−24y = − = − 43 120 1230 29. 30. −0.2x + 0.4y = 0.6 x−2y = −3 −0.1x + 0.2y = 0.6 5x−10y = 1 For the following exercises, solve each system by any method. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 5x + 9y = 16 x + 2y = 4 6x−8y = −0.6 3x + 2y = 0.9 5x−2y = 2.25 7x−4y = 3 x − 5 12 −6x + 5 2 y = − 55 12 y = 55 2 7x−4y = 7 6 2x + 4y = 1 3 3x + 6y = 11 2x + 4y
= − 21 12 6 y = 2 y = −.2x + 1.3y = −0.1 4.2x + 4.2y = 2.1 0.1x + 0.2y = 2 0.35x−0.3y = 0 Graphical For the following exercises, graph the system of equations and state whether the system is consistent, inconsistent, or dependent and whether the system has one solution, no solution, or infinite solutions. 41. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 3x − y = 0.6 x−2y = 1.3 42. −x + 2y = 4 2x−4y = 1 43. 44. 45. x + 2y = 7 2x + 6y = 12 3x−5y = 7 x−2y = 3 3x−2y = 5 −9x + 6y = −15 Technology For the following exercises, use the intersect function on a graphing device to solve each system. Round all answers to the nearest hundredth. 46. 47. 48. 49. 50. 0.1x + 0.2y = 0.3 −0.3x + 0.5y = 1 −0.01x + 0.12y = 0.62 0.15x + 0.20y = 0.52 0.5x + 0.3y = 4 0.25x−0.9y = 0.46 0.15x + 0.27y = 0.39 −0.34x + 0.56y = 1.8 −0.71x + 0.92y = 0.13 0.83x + 0.05y = 2.1 Extensions For the following exercises, solve each system in terms of A, B, C, D, E, and F where A – F are nonzero numbers. Note that A ≠ B and AE ≠ BD. 51. 52. 53 + Ay = 1 x + By = 1 Ax + y = 0 Bx + y = 1 54. Ax + By = C x + y = 1 Chapter 11 Systems of Equations and Inequalities 1231 55. Ax + By = C Dx + Ey = F Real-World Applications For the following exercises, solve for the desired quantity. 56. A stuffed animal business has
a total cost of production C = 12x + 30 and a revenue function R = 20x. Find the break-even point. A fast-food restaurant has a cost of production 57. C(x) = 11x + 120 and a revenue function R(x) = 5x. When does the company start to turn a profit? A cell phone factory has a cost of production function 58. C(x) = 150x + 10, 000 and revenue R(x) = 200x. What is the break-even point? a 59. A musician charges C(x) = 64x + 20,000, where x is the total number of attendees at the concert. The venue charges $80 per ticket. After how many people buy tickets does the venue break even, and what is the value of the total tickets sold at that point? a cost factory A guitar production has 60. C(x) = 75x + 50,000. If the company needs to break even after 150 units sold, at what price should they sell each guitar? Round up to the nearest dollar, and write the revenue function. of For the following exercises, use a system of equations with two variables and two equations to solve. linear Find two numbers whose sum is 28 and difference is 61. 13. A number is 9 more than another number. Twice the 62. sum of the two numbers is 10. Find the two numbers. The startup cost for a restaurant is $120,000, and each 63. meal costs $10 for the restaurant to make. If each meal is then sold for $15, after how many meals does the restaurant break even? A moving company charges a flat rate of $150, and an 64. additional $5 for each box. If a taxi service would charge $20 for each box, how many boxes would you need for it to be cheaper to use the moving company, and what would be the total cost? A total of 1,595 first- and second-year college students 65. gathered at a pep rally. The number of freshmen exceeded the number of sophomores by 15. How many freshmen and sophomores were in attendance? 276 students enrolled in a freshman-level chemistry 66. class. By the end of the semester, 5 times the number of students passed as failed. Find the number of students who passed, and the number of students who failed. There were 130 faculty at a conference. If there were 67. 18 more women than men attending, how
many of each gender attended the conference? A jeep and BMW enter a highway running east-west at 68. the same exit heading in opposite directions. The jeep entered the highway 30 minutes before the BMW did, and traveled 7 mph slower than the BMW. After 2 hours from the time the BMW entered the highway, the cars were 306.5 miles apart. Find the speed of each car, assuming they were driven on cruise control. If a scientist mixed 10% saline solution with 60% 69. saline solution to get 25 gallons of 40% saline solution, how many gallons of 10% and 60% solutions were mixed? An investor earned triple the profits of what she earned 70. last year. If she made $500,000.48 total for both years, how much did she earn in profits each year? An investor who dabbles in real estate invested 1.1 71. million dollars into two land investments. On the first investment, Swan Peak, her return was a 110% increase on the money she invested. On the second investment, Riverside Community, she earned 50% over what she invested. If she earned $1 million in profits, how much did she invest in each of the land deals? If an investor invests a total of $25,000 into two bonds, 72. one that pays 3% simple interest, and the other that pays 27 % interest, and the investor earns $737.50 annual 8 interest, how much was invested in each account? If an investor invests $23,000 into two bonds, one that 73. pays 4% in simple interest, and the other paying 2% simple interest, and the investor earns $710.00 annual interest, how much was invested in each account? CDs cost $5.96 more than DVDs at All Bets Are Off 74. Electronics. How much would 6 CDs and 2 DVDs cost if 5 CDs and 2 DVDs cost $127.73? A store clerk sold 60 pairs of sneakers. The high-tops 75. sold for $98.99 and the low-tops sold for $129.99. If the receipts for the two types of sales totaled $6,404.40, how many of each type of sneaker were sold? A concert manager counted 350 ticket receipts the day 76. after a concert. The price for a student ticket was $12.50, and the price for an adult ticket was $16.00. The register confirms that $5,075 was taken in. How many student tickets and
adult tickets were sold? Admission into an amusement park for 4 children and 2 77. adults is $116.90. For 6 children and 3 adults, the admission is $175.35. Assuming a different price for children and 1232 Chapter 11 Systems of Equations and Inequalities adults, what is the price of the child’s ticket and the price of the adult ticket? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1233 11.2 \| Systems of Linear Equations: Three Variables Learning Objectives In this section, you will: 11.2.1 Solve systems of three equations in three variables. 11.2.2 Identify inconsistent systems of equations containing three variables. 11.2.3 Express the solution of a system of dependent equations containing three variables. Figure 11.14 (credit: “Elembis,” Wikimedia Commons) John received an inheritance of $12,000 that he divided into three parts and invested in three ways: in a money-market fund paying 3% annual interest; in municipal bonds paying 4% annual interest; and in mutual funds paying 7% annual interest. John invested $4,000 more in municipal funds than in municipal bonds. He earned $670 in interest the first year. How much did John invest in each type of fund? Understanding the correct approach to setting up problems such as this one makes finding a solution a matter of following a pattern. We will solve this and similar problems involving three equations and three variables in this section. Doing so uses similar techniques as those used to solve systems of two equations in two variables. However, finding solutions to systems of three equations requires a bit more organization and a touch of visual gymnastics. Solving Systems of Three Equations in Three Variables In order to solve systems of equations in three variables, known as three-by-three systems, the primary tool we will be using is called Gaussian elimination, named after the prolific German mathematician Karl Friedrich Gauss. While there is no definitive order in which operations are to be performed, there are specific guidelines as to what type of moves can be made. We may number the equations to keep track of the steps we apply. The goal is to eliminate one variable at a time to achieve upper triangular form, the ideal form for a three-by-three system because it allows for straightforward back-substitution to find a solution (x
, y, z), which we call an ordered triple. A system in upper triangular form looks like the following: Ax + By + Cz = D Ey + Fz = G Hz = K The third equation can be solved for z, and then we back-substitute to find y and x. To write the system in upper triangular form, we can perform the following operations: 1. Interchange the order of any two equations. 2. Multiply both sides of an equation by a nonzero constant. 3. Add a nonzero multiple of one equation to another equation. 1234 Chapter 11 Systems of Equations and Inequalities The solution set to a three-by-three system is an ordered triple ⎧ ⎬. Graphically, the ordered triple defines the point that is the intersection of three planes in space. You can visualize such an intersection by imagining any corner in a rectangular room. A corner is defined by three planes: two adjoining walls and the floor (or ceiling). Any point where two walls and the floor meet represents the intersection of three planes. ⎨(x, y, z)⎫ ⎩ ⎭ Number of Possible Solutions Figure 11.15 and Figure 11.16 illustrate possible solution scenarios for three-by-three systems. • Systems that have a single solution are those which, after elimination, result in a solution set consisting of an ⎬. Graphically, the ordered triple defines a point that is the intersection of three planes ordered triple ⎧ ⎨(x, y, z)⎫ ⎭ ⎩ in space. • Systems that have an infinite number of solutions are those which, after elimination, result in an expression that is always true, such as 0 = 0. Graphically, an infinite number of solutions represents a line or coincident plane that serves as the intersection of three planes in space. • Systems that have no solution are those that, after elimination, result in a statement that is a contradiction, such as 3 = 0. Graphically, a system with no solution is represented by three planes with no point in common. Figure 11.15 (a)Three planes intersect at a single point, representing a three-by-three system with a single solution. (b) Three planes intersect in a line, representing a three-by-three system with infinite solutions. Figure 11.16 All three figures represent three-by-three systems with no solution. (a) The
three planes intersect with each other, but not at a common point. (b) Two of the planes are parallel and intersect with the third plane, but not with each other. (c) All three planes are parallel, so there is no point of intersection. Example 11.12 Determining Whether an Ordered Triple Is a Solution to a System Determine whether the ordered triple (3, −2, 1) is a solution to the system. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1235 x + y + z = 2 6x − 4y + 5z = 31 5x + 2y + 2z = 13 Solution We will check each equation by substituting in the values of the ordered triple for x, y, and z. x + y + z = 2 (3) + (−2) + (1) = 2 True 6x−4y + 5z = 31 6(3)−4(−2) + 5(1) = 31 18 + 8 + 5 = 31 True 5x + 2y + 2z = 13 5(3) + 2(−2) + 2(1) = 13 15−4 + 2 = 13 True The ordered triple (3, −2, 1) is indeed a solution to the system. Given a linear system of three equations, solve for three unknowns. 1. Pick any pair of equations and solve for one variable. 2. Pick another pair of equations and solve for the same variable. 3. You have created a system of two equations in two unknowns. Solve the resulting two-by-two system. 4. Back-substitute known variables into any one of the original equations and solve for the missing variable. Example 11.13 Solving a System of Three Equations in Three Variables by Elimination Find a solution to the following system: x−2y + 3z = 9 (1) − x + 3y − z = −6 (2) 2x−5y + 5z = 17 (3) Solution There will always be several choices as to where to begin, but the most obvious first step here is to eliminate x by adding equations (1) and (2). x − 2y + 3z = 9 (1) − x + 3y − z = −6 (2) y + 2z = 3 (3)
The second step is multiplying equation (1) by −2 and adding the result to equation (3). These two steps will eliminate the variable x. −2x + 4y − 6z = −18 (1) multiplied by − 2 2x − 5y + 5z = 17 (3) ____________________________________ − y − z = −1 (5) In equations (4) and (5), we have created a new two-by-two system. We can solve for z by adding the two equations. 1236 Chapter 11 Systems of Equations and Inequalities y + 2z = 3 (4) −y − z = − 1 (5) z = 2 (6) Choosing one equation from each new system, we obtain the upper triangular form: x−2y + 3z = 9 (1) y + 2z = 3 (4) z = 2 (6) Next, we back-substitute z = 2 into equation (4) and solve for y. y + 2(21 Finally, we can back-substitute z = 2 and y = −1 into equation (1). This will yield the solution for x. x−2(−1) + 3(2 The solution is the ordered triple (1, −1, 2). See Figure 11.17. Figure 11.17 Example 11.14 Solving a Real-World Problem Using a System of Three Equations in Three Variables In the problem posed at the beginning of the section, John invested his inheritance of $12,000 in three different funds: part in a money-market fund paying 3% interest annually; part in municipal bonds paying 4% annually; and the rest in mutual funds paying 7% annually. John invested $4,000 more in mutual funds than he invested in municipal bonds. The total interest earned in one year was $670. How much did he invest in each type of fund? Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1237 To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts: x = amount invested in money-market fund y = amount invested in municipal bonds z = amount invested in mutual funds The first equation indicates that the sum of the three principal amounts is $12,000. x + y + z
= 12,000 We form the second equation according to the information that John invested $4,000 more in mutual funds than he invested in municipal bonds. z = y + 4,000 The third equation shows that the total amount of interest earned from each fund equals $670. Then, we write the three equations as a system. 0.03x + 0.04y + 0.07z = 670 x + y + z = 12,000 − y + z = 4,000 0.03x + 0.04y + 0.07z = 670 To make the calculations simpler, we can multiply the third equation by 100. Thus, x + y + z = 12,000 (1) − y + z = 4,000 (2) 3x + 4y + 7z = 67,000 (3) Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up. x + y + z = 12,000 3x + 4y + 7z = 67,000 − y + z = 4,000 Step 2. Multiply equation (1) by −3 and add to equation (2). Write the result as row 2. x + y + z = 12,000 y + 4z = 31,000 − y + z = 4,000 Step 3. Add equation (2) to equation (3) and write the result as equation (3). x + y + z = 12,000 y + 4z = 31,000 5z = 35,000 Step 4. Solve for z in equation (3). Back-substitute that value in equation (2) and solve for y. Then, backsubstitute the values for z and y into equation (1) and solve for x. 1238 Chapter 11 Systems of Equations and Inequalities 5z = 35,000 z = 7,000 y + 4(7,000) = 31,000 y = 3,000 x + 3,000 + 7,000 = 12,000 x = 2,000 John invested $2,000 in a money-market fund, $3,000 in municipal bonds, and $7,000 in mutual funds. 11.9 Solve the system of equations in three variables. 2x + y−2z = −1 3x−3y − z = 5 x−2y + 3z = 6 Identifying Inconsistent Systems of Equ
ations Containing Three Variables Just as with systems of equations in two variables, we may come across an inconsistent system of equations in three variables, which means that it does not have a solution that satisfies all three equations. The equations could represent three parallel planes, two parallel planes and one intersecting plane, or three planes that intersect the other two but not at the same location. The process of elimination will result in a false statement, such as 3 = 7 or some other contradiction. Example 11.15 Solving an Inconsistent System of Three Equations in Three Variables Solve the following system. x−3y + z = 4 (1) − x + 2y−5z = 3 (2) 5x−13y + 13z = 8 (3) Solution Looking at the coefficients of x, we can see that we can eliminate x by adding equation (1) to equation (2). x−3y + z = 4 (1) −x + 2y−5z = 3 (2) − y−4z = 7 (4) Next, we multiply equation (1) by −5 and add it to equation (3). −5x + 15y − 5z = −20 5x − 13y + 13z = 8 ______________________________________ 2y + 8z = −12 (1) multiplied by −5 (3) (5) Then, we multiply equation (4) by 2 and add it to equation (5). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 11 Systems of Equations and Inequalities 1239 −2y − 8z = 14 (4) multiplied by 2 2y + 8z = − 12 (5) _______________________________________ 0 = 2 The final equation 0 = 2 is a contradiction, so we conclude that the system of equations in inconsistent and, therefore, has no solution. Analysis In this system, each plane intersects the other two, but not at the same location. Therefore, the system is inconsistent. 11.10 Solve the system of three equations in three variables. x + y + z = 2 y−3z = 1 2x + y + 5z = 0 Expressing the Solution of a System of Dependent Equations Containing Three Variables We know from working with systems of equations in two variables that a dependent system of equations has an infinite number of solutions. The same is true for dependent systems of equations in

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