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°, we can then find the third angle of the triangle. The complete set of angles and sides is γ = 180° − 30° − 56.3° ≈ 93.7° α ≈ 56.3° β = 30° γ ≈ 93.7° a = 10 b ≈ 6.013 c = 12 Try It #1 Find the missing side and angles of the given triangle: α = 30°, b = 12, c = 24. SECTION 8.2 non-right triangles: law oF cosines 661 E... |
6,000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5,050 feet from the first tower and 2,420 feet from the second tower. Determine the position of the cell phone north and east of the first towe... |
,998 feet north of the first tower, and 1,998 feet from the highway. Example 4 Calculating Distance Traveled Using a SAS Triangle Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The dia... |
= √ Area = √ —— s(s − a)(s − b)(s − c) 16(16 − 10)(16 − 15)(16 − 7) —— The area is approximately 29.4 square units. Area ≈ 29.4 Try It #3 Use Heron’s formula to find the area of a triangle with sides of lengths a = 29.7 ft, b = 42.3 ft, and c = 38.4 ft. Example 6 Applying Heron’s Formula to a Real-World Problem A Chic... |
what s represents in Heron’s formula. 4. Explain the relationship between the Pythagorean Theorem and the Law of Cosines. 5. When must you use the Law of Cosines instead of the Pythagorean Theorem? AlGeBRAIC For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. If po... |
Round to the nearest hundredth. 27. Find the area of a triangle with sides of length 18 in, 28. Find the area of a triangle with sides of length 20 cm, 21 in, and 32 in. Round to the nearest tenth. 26 cm, and 37 cm. Round to the nearest tenth. 1 1 1 _ _ _ m, c = m, b = 29. a = m 4 3 2 31. a = 1.6 yd, b = 2.6 yd, c = 4... |
inscribed in a circle with a radius of 8 inches. (See Figure 12.) Find the perimeter of the octagon. 56. A regular pentagon is inscribed in a circle of radius 12 cm. (See Figure 13.) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter. Figure 12 Figure 13 For the following exercises, suppose... |
2 hours. 4 mph 18 mph Figure 17 68. A triangular swimming pool measures 40 feet on one side and 65 feet on another side. These sides form an angle that measures 50°. How long is the third side (to the nearest tenth)? 69. A pilot flies in a straight path for 1 hour 30 min. She then makes a course correction, heading 10... |
110 feet on one side and 250 feet on another; the included angle measures 85°. Round to the nearest whole square foot. 670 CHAPTER 8 Further applications oF trigonometry leARnInG OBjeCTIVeS In this section, you will: • Plot points using polar coordinates. • Convert from polar coordinates to rectangular coordinates. • ... |
4 –1 –2 –3 –4 Polar Grid Figure 2 SECTION 8.3 polar coordinates 671 Example 1 Plotting a Point on the Polar Grid π _ on the polar grid. Plot the point 3, 2 π _ Solution The angle is found by sweeping in a counterclockwise direction 90° from the polar axis. The point is located 2 π _ direction, as shown in Figure 3... |
= rcos θ y _ sin θ = r → y = rsin θ Dropping a perpendicular from the point in the plane to the x-axis forms a right triangle, as illustrated in Figure 5. An easy way to remember the equations above is to think of cos θ as the adjacent side over the hypotenuse and sin θ as the opposite side over the hypotenuse. y (x, ... |
5 –4 –3 –2 Figure 7 Try It #3 Write the polar coordinates −1, 2π _ as rectangular coordinates. 3 Converting from Rectangular Coordinates to Polar Coordinates To convert rectangular coordinates to polar coordinates, we will use two other familiar relationships. With this conversion, however, we need to be aware tha... |
of 3 √ 4 5π ___ 2, 4 5π _ indicates a move 4 is 2. However, the angle 5π _ 4 2, − 2, — — — — located in the third quadrant and, as r is negative, we extend the directed line segment in the opposite direction, into the π 7π _ _ first quadrant. This is the same point as 3 √, from. 4 4 2, is the same. The radius,... |
2 θ = 1. Thus, x 2 + y 2 = 9, r = 3, and r = −3 should generate the same graph. See Figure 10. r = ± 3 Use the square root property. –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (a) 21 3 4 5 x 1 2 3 4 (b) To graph a circle in rectangular form, we must first solve for y. Figure 10 (a) Cartesian form x 2 + y 2 = 9 (b) Pola... |
rite the Cartesian equation y = 3x + 2 as a polar equation. Solution We will use the relationships x = rcos θ and y = rsin θ. y = 3x + 2 rsin θ = 3rcos θ + 2 rsin θ − 3rcos θ = 2 r(sin θ − 3cos θ) = 2 r = 2 __ sin θ − 3cos θ Isolate r. Solve for r. Try It #4 Rewrite the Cartesian equation y 2 = 3 − x 2 in polar form. I... |
use substitution. In order to replace r with x and y, we must use the expression x 2 + y 2 = r 2. r = 3 _ 1 − 2cos θ r(1 − 2cos θ2x = 3 x _ Use cos θ = r to eliminate θ. r = 3 + 2x Isolate r. r 2 = (3 + 2x) 2 Square both sides. x 2 + y 2 = (3 + 2x) 2 Use x 2 + y 2 = r 2. The Cartesian equation is x 2 + y 2 = (3 + 2x) ... |
) − y 2 = −9 + 12 Organize terms to complete the square for x. 3(x 2 + 4x + 4) − y 2 = −9 + 12 3(x + 2)2 − y 2 = 3 (x + 2)2 − y 2 _ = 1 3 Try It #5 Rewrite the polar equation r = 2sin θ in Cartesian form. Example 11 Rewriting a Polar Equation in Cartesian Form Rewrite the polar equation r = sin(2θ) in Cartesian form. ... |
θ for the point. 6. 7, 7π _ 6 7. (5, π) π _ 8. 6, − 4 π _ 9. −3, 6 10. 4, 7π _ 4 For the following exercises, convert the given Cartesian coordinates to polar coordinates with r > 0, 0 ≤ θ <2π. Remember to consider the quadrant in which the given point is located. 11. (4, 2) 12. (−4, 6) 13. (3, −5) 14.... |
π _ 45. −2, 3 −5π _ 4 50. 4, π _ 46. −1, − 2 51. 3, 5π _ 6 47. 3.5, 7π _ 4 52. −1.5, 7π _ 6 π _ 48. −4, 3 π _ 53. −2, 4 π _ 49. 5, 2 54. 1, 3π _ 2 For the following exercises, convert the equation from rectangular to polar form and graph on the polar axis. 55. 5x − y = 6 59. x =... |
sc θ; a > 0. 78. What polar equations will give an oblique line? For the following exercises, graph the polar inequality. 79 81 83. 0 ≤ θ ≤ 3 75. Describe the graph of r = asec θ; a < 0. 77. Describe the graph of r = acsc θ; a < 0. π _ 803 823 < r < 2 < θ ≤ 84. − 3 6 SECTION 8.4 polar coordinates: graphs 681 leARnInG O... |
of symmetry to polar equations. Further, we will use symmetry (in addition to plotting key points, zeros, and maximums of r) to determine the graph of a polar equation. π _ In the first test, we consider symmetry with respect to the line θ = (y-axis). We replace (r, θ) with (−r, −θ) to 2 determine if the new equation ... |
exists by plotting reflecting points across the apparent axis of symmetry or the pole. Testing for symmetry is a technique that simplifies the graphing of polar equations, but its application is not perfect. symmetry tests A polar equation describes a curve on the polar grid. The graph of a polar equation can be evalu... |
Replacing r with −r changes the equation and fails the test. The graph may or may not be symmetric with respect to the pole. −r = 2sin θ r = −2sin θ ≠ 2sin θ Failed Table 1 SECTION 8.4 polar coordinates: graphs 683 Analysis Using a graphing calculator, we can see that the equation r = 2sin θ is a circle centered at (0... |
= 2 would be polar axis intercepts, which may be helpful in drawing the graph and identifying the curve of a polar equation. Example 2 Finding Zeros and Maximum Values for a Polar Equation Using the equation in Example 1, find the zeros and maximum | r | and, if necessary, the polar axis intercepts of r = 2sin θ. Solu... |
esian coordinates, test the given equation for symmetry and find the zeros and maximum values of | r | : r = 3cos θ. Investigating Circles Now we have seen the equation of a circle in the polar coordinate system. In the last two examples, the same equation was used to illustrate the properties of symmetry and demonstra... |
4cos θ r = 4cos(0) r = 4(1) = 4 The maximum value of the equation is 4. A key point to plot is (4, 0). As r = 4cos θ is symmetric with respect to the polar axis, we only need to calculate r-values for θ over the interval [0, π]. Points in the upper quadrant can then be reflected to the lower quadrant. Make a table of ... |
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... |
b r = a + bcosθ (a) r = a − bcosθ (b) r = a + bsinθ (c) r = a − bsinθ (d) Figure 9 Dimpled limaçons How To… 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 symmetr... |
.4 π _ 2 1 2π _ 3 1.4 5π _ 6 2.5 π 4 7π _ 6 5.5 4π _ 3 6.6 3π _ 2 7 5π _ 3 6.6 11π _ 6 5.5 2π 4 The graph is shown in Figure 10. Table 5 1 ) –4 Figure 10 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. ... |
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 6. θ r 0 7 π __ 6 6.3 π __ 3 4.5 π __ 2 2 2π ___ 3 5π ___ 6 −0.5 −2.3 π −3 7... |
cos u 0 = cos u π _ cos− 2θ = 2 π _ θ = 4 Substitute 2θ back in for u. π _ is a zero of the equation. So, the point 0, 4 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, r2 = 4cos(0) r2 = 4(1 We have a maximum at (2, 0). Since this graph is symme... |
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 π _ symmetric with respect to the polar axis, but also with respect to the line θ = and the pole. 2 Now we will find the zeros. First make the substitution u = 4θ. 0 ... |
of the equation shows symmetry with respect to the line θ =. Next, find the zeros and maximum. 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, π _ r = 2sin 5 ⋅ 2 r = 2(1) = 2 Thu... |
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. θ π _ 4 0.785 π _ 2 1.57 π 2π 3π _ 2 4.71 7π _ 4 5.50 r 3.14 Table 10 Notice that the... |
n petals (b) r = acos nθ r = asin nθ n odd, n petals (c) r = θ θ ≥ 0 (d) Figure 21 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://openstaxcolleg... |
θ 25. r = 5 + 4cos θ 26. r = 10 + 9cos θ 27. r = 1 + 3sin θ 28. r = 2 + 5sin θ 29. r = 5 + 7sin θ 30. r = 2 + 4cos θ 31. r = 5 + 6cos θ 32. r 2 = 36cos(2θ) 33. r 2 = 10cos(2θ) 34. r 2 = 4sin(2θ) 35. r 2 = 10sin(2θ) 36. r = 3sin(2θ) 40. r = 4sin(5θ) TeCHnOlOGY 37. r = 3cos(2θ) 38. r = 5sin(3θ) 39. r = 4sin(4θ) 41. r = ... |
0, 8π], [0, 12π], and [0, 16π]. Describe the effect of increasing the width of the domain. 60. On a graphing utility, graph and sketch 5 _ θ r = sin θ + sin 2 3 on [0, 4π]. 61. On a graphing utility, graph each polar equation. 62. On a graphing utility, graph each polar equation. Explain the similarities and di... |
form. • Find roots of complex numbers in polar form. 8.5 POlAR FORM OF COMPlex nUMBeRS “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 ... |
Imaginary (2 + 4i) |z| = 20 21 3 4 5 Real –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 2 absolute value of a complex number Given z = x + yi, a complex number, the absolute value of z is defined as ∣ z ∣ = √ It is the distance from the origin to the point (x, y). — x2 + y2 Notice that the absolute value of a real num... |
applications oF trigonometry 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 ha... |
of r √ — — x2+ y2 (−4)2 + (42) 32 2 — — Find the angle θ using the formula: Thus, the solution is 4 √ — 2 cis 3π _ . 4 — cos θ = x _ cos θ = r −4 ______ 2 4 √ cos θ = − 1 _____ 2 √ 3π θ = cos−1 − 1 _____ _ = 4 2 √ — — Try It #5 _ 3 + i in polar form. Write z = √ Converting a Complex number from Polar to Rectang... |
13 13 = 12 + 5i The rectangular form of the given number in complex form is 12 + 5i. Try It #6 Convert the complex number to rectangular form: z = 4 cos 11π _ 6 + isin 11π _ 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 oper... |
in θ1) and z2 = r2(cos θ2 + isin θ2), then the quotient of these numbers is z1 __ z2 z1 __ z2 r1 __ r2 [cos(θ1 −θ2) + isin(θ1 − θ2)], z2 ≠ 0 r1 __ r2 cis(θ1 − θ2), z2 ≠ 0 = = Notice that the moduli are divided, and the angles are subtracted. How To… Given two complex numbers in polar form, find the quotient.. 1. Divide... |
(nθ) + isin(nθ)] zn = r n cis(nθ) where n is a positive integer. 7 04 CHAPTER 8 Further applications oF trigonometry Example 10 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 fo... |
.5 polar Form oF complex numBers 705 Example 11 Finding the nth Root of a Complex Number Evaluate the cube roots of z = 8 cos Solution We have 2π + isin ___ 3 2π . ___ 3 1 1 __ __ = 8 z 3 3 cos 2π ___ 3 _ 3 + + isin 2kπ ____ 3 2π ___ 3 _ 3 + 2kπ ____ 3 1 __ = 2 cos z 3 2π ___ + 9 2kπ ____ 3 ... |
π ___ 9 + 6π ___ 9 = 8π ___ 9 Try It #8 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) •... |
8 — 15 cis π _ 12 For the following exercises, find z1 _ in polar form. z2 29. z1 = 21cis(135°); z2 = 3cis(65°) 31. z1 = 15cis(120°); z2 = 3cis(40°) 33. z1 = 5 √ — 2 cis(π); z2 = √ — 2 cis 2π _ 3 24. z1 = √ — 2 cis(205°); z2 = 2 √ — 2 cis(118°) π π _ _ ; z2 = 5cis 26. z1 = 3cis 4 6 π π _ _ ; z2 = 2ci... |
. Evaluate the square root of z when z = 16cis(100°). 44. Evaluate the square root of z when z = 32cis(π). GRAPHICAl For the following exercises, plot the complex number in the complex plane. 46. 2 + 4i 47. −3 − 3i 49. −1 − 5i 52. −4 55. 1 − 4i TeCHnOlOGY 50. 3 + 2i 53. 6 − 2i 48. 5 − 4i 51. 2i 54. −2 + i For the follo... |
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. ... |
for the graph of a circle is not a function. r2 − x2, or two equations: y1 = √ — — — SECTION 8.6 parametric eQuations 709 y Vertical line test on circle r2 = x2 + y2 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 Figure 2 However, if we were to graph each equation on its own, each one would pass the vertical line ... |
points are generated as t increases. –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (a) 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (b) 21 3 4 5 x Figure 3 (a) Parametric y(t ) = t 2 − 1 (b) Rectangular y = x 2 − 1 7 10 CHAPTER 8 Further applications oF trigonometry Analysis The arrows indicate the direction in wh... |
Figure 4. 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... |
5 x = 2(0) − 5 = −5 x = 2(1) − 5 = −3 x = 2(2) − 5 = −1 x = 2(3) − 5 = 1 x = 2(4) − 5 = 3 Table 3 y(t) = −t + 3 y = −(0) + 3 = 3 y = −(1) + 3 = 2 y = −(2) + 3 = 1 y = −(3) + 3 = 0 y = −(4) + 3 = − 1 From this table, we can create three graphs, as shown in Figure 5. –6 –5 –4 –3 –2 x 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 – 6 (a... |
lynomial, 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 4 Eliminating the Parameter in Polynomials Giv... |
shown in Figure 7(b) and has only one restriction on the domain, x ≠ 0. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 (a) x(t) = e−t y(t) = 3et 21 3 4 5 6 x –6 –5 –4 –3 –2 Figure 7 y = 3 x 21 1 –1 –2 –3 –4 –5 –6 (b) Example 6 Eliminating the Parameter in Logarithmic Equations — Eliminate the parameter and write as... |
t ≤ 2π and sketch the graph. Solution Solving for cos t and sin t, we have x(t) = 4cos t y(t) = 3sin t x = 4cos t x _ = cos t 4 y = 3sin t y _ = sin t 3 Next, use the Pythagorean identity and make the substitutions. 2 2 x _ 4 cos2 t + sin2 t = 1 y + _ 3 y2 x2 _ _ 9 16 = 1 = 1 + The graph for the equation is sh... |
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... |
x – –56 –4 –3 –2 Figure 9 Rectangular y = (x + 3)2 + 1 21 1 –1 –2 –3 –4 (b) 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:... |
2 cos2 t y(t) = −sin t x(t) = 2t − 1 y(t) = t3 − 2 9. 13. 17. 21. 25. { { { { { 10. 14. 18. 22. { { { { x(t) = 3t − 1 y(t) = 2t 2 x(t) = log (2t) y(t) = √ — t − 1 x(t) = t 5 y(t) = t10 x(t) = cos t + 4 y(t) = 2 sin2 t For the following exercises, rewrite the parametric equation as a Cartesian equation by building an x... |
that the line is at (−1, 0) at t = 0, and at (3, −2) at t = 1. 41. Parameterize the line from (4, 1) to (6, −2) so that the line is at (4, 1) at t = 0, and at (6, −2) at t = 1. 7 18 CHAPTER 8 Further applications oF trigonometry TeCHnOlOGY For the following exercises, use the table feature in the graphing calculator t... |
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 1 Parametric equations can model the path of a proje... |
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. Try It #1 Sketch the graph of the parametric equations x = √ — t, y = 2t + 3, 0 ≤ t ≤ 3. Example 2 Sketching the Graph of Trigonometric Parametric Equations Construct... |
4sin 5π = −2 √ ___ 3 — 3 x = 2cos 11π ____ 6 = √ — 3 y = 4sin 11π ____ 6 = −2 x = 2cos(2π) = 2 y = 4sin(2π) = 0 Table 2 t 0 π __ 6 π __ 3 π __ 2 2π ___ 3 5π ___ 6 π 7π ___ 6 4π ___ 3 3π ___ 2 5π ___ 3 11π ____ 6 2π Figure 3 shows the graph. (– 3, 2) t = 5π 6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (0, 4)... |
5cos(4) ≈ −3.3 x = 5cos(5) ≈ 1.4 x = 5cos(−1) ≈ 2.7 x = 5cos(−2) ≈ −2.1 x = 5cos(−3) ≈ −4.95 x = 5cos(−4) ≈ −3.3 x = 5cos(−5) ≈ 1.4 Table 3 y = 2sin t y = 2sin(0) = 0 y = 2sin(1) ≈ 1.7 y = 2sin(2) ≈ 1.8 y = 2sin(3) ≈ 0.28 y = 2sin(4) ≈ −1.5 y = 2sin(5) ≈ −1.9 y = 2sin(−1) ≈ −1.7 y = 2sin(−2) ≈ −1.8 y = 2sin(−3) ≈ −0.2... |
in a dashed style colored red. Clearly, both forms produce the same graph. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 5 Example 4 Graphing Parametric Equations and Rectangular Equations on the Coordinate System Graph the parametric equations x = t + 1 and y = √ coordinate system. — t, t ≥ 0, and the r... |
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. How To… 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 f... |
above the ground. SECTION 8.7 parametric eQuations: graphs 725 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, y = −16t 2 +(140sin(45°))t + 3 y = 0 t = 6.2173 Set y (t) = 0 and solve the quadratic. When t = 6.2173 seconds, the ball has hit the ground. ... |
? GRAPHICAl For the following exercises, graph each set of parametric equations by making a table of values. Include the orientation on the graph. x(t) = t y(t) = t 2 − 1 x(t) = t − 1 y(t) = t 2 7. 6. x y { { −3 −2 −1 0 1 2 t x y t −3 −2 −1 0 1 2 3 x(t) = 2 + t y(t) = 3 − 2t 8(t) = −2 − 2t y(t) = 3 + t 9(t) = t3 y(t) =... |
27. { For the following exercises, graph the equation and include the orientation. 28. x = t2, y = 3t, 0 ≤ t ≤ 5 29. x = 2t, y = t2, −5 ≤ t ≤ 5 _ 31. x(t) = −t, y(t) = √ t, t ≥ 5 25 − t2, 0 < t ≤ 5 π __ 32. x(t) = −2cos t, y = 6sin t 0 ≤ t ≤ π 33. x(t) = −sec t, y = tan t, − 2 30. x = t, y = √ π __ < t < 2 — For the f... |
, 0), major axis of length 10, minor axis of length 6, and a counterclockwise orientation. For the following exercises, use a graphing utility to graph on the window [−3, 3] by [−3, 3] on the domain [0, 2π) for the following values of a and b, and include the orientation. x(t) = sin(at) y(t) = sin(bt) { 47. a = 1, b = ... |
along one of the axes. What controls which axis the graph creeps along? 61. Explain the effect on the graph of the parametric 62. Explain the effect on the graph of the parametric equation when we switched sin t and cos t. equation when we changed the domain. exTenSIOnS 63. An object is thrown in the air with vertical... |
cos t − 2sin(6t) on the domain [0, 2π]. 73. A rose: { x(t) = 5sin(2t) sin t y(t) = 5sin(2t) cos t on the domain [0, 2π]. SECTION 8.8 vectors 729 leARnInG OBjeCTIVeS In this section, you will: • View vectors geometrically. • Find magnitude and direction. • Perform vector addition and scalar multiplication. • Find the co... |
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 ... |
see that the position vector is 〈4, 1〉. Figure 3 Example 2 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 ... |
6, − 6〉 We use the Pythagorean Theorem to find the magnitude. | u | = √ — — (6) 2 + (−6) 2 72 2 — = √ = 6 √ 7 32 CHAPTER 8 Further applications oF trigonometry The direction is given as tan θ = −6 ___ = −1 ⇒ θ = tan 6 = −45° −1 (−1) However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positi... |
— — — — As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives 5 5 _ _ ⇒ θ = tan−1 − tan θ = − 6 6 = −39.8° 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.... |
To find the sum of two vectors, we add the components. Thus, u + v = 〈3, − 2〉 + 〈−1, 4〉 = 〈3 + ( − 1), − 2 + 4〉 = 〈2, 2〉 See Figure 10(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〉 7 34 CHAPTER 8 Further applicatio... |
�� 3v v –v v1 2 Figure 11 1 _ v is half the length of v, and −v is the same length of v, Analysis Notice that the vector 3v is three times the length of v, 2 but in the opposite direction. SECTION 8.8 vectors 735 Try It #2 Find the scalar multiple 3u given u = 〈5, 4〉. Example 7 Using Vector Addition and Scalar Multipli... |
v2 is 3. To find the magnitude of v, use the formula with the position vector. __________ | v | = √ = √ ∣ v1 ∣ 2 + ∣ v2 ∣ 2 22 + 32 — The magnitude of v is √ — y _ 13. To find the direction, we use the tangent function tan θ = x. = √ — 13 tan θ = v2 _ v1 3 _ tan θ = 2 3 _ = 56.3° θ = tan−1 2 7 36 CHAPTER 8 Further... |
8.8 vectors 737 the unit vectors If v is a nonzero vector, then is a unit vector in the direction of v. Any vector divided by its magnitude is a v _ ∣ v ∣ 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. Example 9 Finding the Un... |
= (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 16. v = ai + bj bj ai Figure 16 Example... |
and v2 = 4i + 5j. Solution According to the formula, we have 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... |
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. 7 40 CHAPTER 8 Further applications oF trigonometry dot product The dot product of two vectors v = 〈a, b〉 and u = 〈c, d〉 is the sum of the product of the horizon... |
Find the angle between u = 〈−3, 4〉 and v = 〈5, 12〉. Solution Using the formula, we have ⋅ θ = cos−3i + 4j _ ⋅ 5 v _ | v | 5i + 12j _ 13 13 4 _ + ⋅ 5 12 _ 13 = − 15 _ 65 + 48 _ 65 = 33 _ 65 θ = cos−1 33 _ 65 See Figure 18. = 59.5° y 12 11 10 9 8 7 6 5 4 3 2 1 59.5° –6 –5 –4 –3 –2 –1–1 21 3 4 5 6 x Figure 18... |
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) SECTION 8.8 section ex... |
P2 = (−1, − 3), write the vector v in terms of i and j. 10. P1 = (2, −3), P2 = (5, 1), P3 = (6, − 1), and P4 = (9, 3) 12. P1 = (3, 7), P2 = (2, 1), P3 = (1, 2), and P4 = (−1, − 4) 14. Given initial point P1 = (−3, 1) and terminal point P2 = (5, 2), write the vector v in terms of i and j. For the following exercises, u... |
. 34. Given u = 〈−2, 4〉 and v = 〈−3, 1〉, calculate u ⋅ v. 35. Given u = 〈−1, 6〉 and v = 〈6, − 1〉, calculate u ⋅ v. 7 44 CHAPTER 8 Further applications oF trigonometry GRAPHICAl 1 __ v. For the following exercises, given v, draw v, 3v and 2 36. 〈2, −1〉 37. 〈−1, 4〉 38. 〈−3, −2〉 For the following exercises, use the vector... |
of the normal (perpendicular) 8°. Rounding to the nearest tenth, a. Find the magnitude of the normal (perpendicular) component of the force. component of the force. b. Find the magnitude of the component of the force b. Find the magnitude of the component of the force that is parallel to the ramp. that is parallel to ... |
south of west. If she walked straight home, how far would she have to walk, and in what direction? 7 46 CHAPTER 8 Further applications oF trigonometry 67. An airplane is heading north at an airspeed of 68. An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km... |
Find the magnitude of the components of the force parallel to and perpendicular to (normal) the ramp to the nearest tenth of a pound. 76. Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it upward, and 5 pounds acting on it 45° from the horizontal. What single force is the resulta... |
Theorem formula used to find the nth power or nth roots of a complex number; states that, for a positive integer n, z n is found by raising the modulus to the nth power and multiplying the angles by n a _ < 2 dimpled limaçon a type of one-loop limaçon represented by r = a ± bcos θ and r = a ± bsin θ such that 1 < b do... |
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 angl... |
β c2 = a2 + b2 − 2abcos γ Heron’s formula Area = √ —— s(s − a)(s − b)(s − c) where s = (a + b + c) _ 2 Conversion formulas Key Concepts 8.1 Non-right Triangles: Law of Sines x _ r → x = rcos θ cos θ = y _ r → y = rsin θ sin θ = r2 = x2 + y2 y _ tan θ = x • The Law of Sines can be used to solve oblique triangles, which... |
on’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See Example 5 and See Example 6. 8.3 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),... |
circle in polar coordinates are given by r = acos θ and r = asin θ. See Example 3. • The formulas that produce the graphs of a cardioid are given by r = a ± bcos θ and r = a ± bsin θ, for a > 0, a _ = 1. See Example 4. b > 0, and b 7 50 CHAPTER 8 Further applications oF trigonometry • The formulas that produce the gra... |
rectangular form, first evaluate the trigonometric functions. Then, multiply through by r. See Example 6 and Example 7. • 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 8. • To f... |
curve. See Example 3 and Example 4. • 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 5. • Projectile motion depends on two parametric equations: x = (v0 cos θ)... |
the dot product to find the angle between two vectors. Example 15 and Example 16. • Dot products are useful for many types of physics applications. See Example 17. 7 52 CHAPTER 8 Further applications oF trigonometry CHAPTeR 8 ReVIeW exeRCISeS nOn-RIGHT TRIAnGleS: lAW OF SIneS For the following exercises, assume α is o... |
13. Convert −2, ___ 2 15. Convert (−9, −4) to polar coordinates. For the following exercises, convert the given Cartesian equation to a polar equation. 16. x = − 2 17. x2 + y2 = 64 18. x2 + y2 = − 2y For the following exercises, convert the given polar equation to a Cartesian equation. 19. r = 7cos θ 20. r = −2 ____... |
4 when z = 2cis(70°) For the following exercises, evaluate each root. 40. Evaluate the cube root of z when z = 64cis(210°). 3π 39. Find z2 when z = 5cis ___ 4 3π . 41. Evaluate the square root of z when z = 25cis ___ 2 For the following exercises, plot the complex number in the complex plane. 42. 6 − 2i 43. −1 +... |
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