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�nitions of Re(z) and Im(z), we can still give r and θ special names in relation to z. Deﬁnition 11.2. The Modulus and Argument of Complex Numbers: Let z = a + bi be a complex number with a = Re(z) and b = Im(z). Let (r, θ) be a polar representation of the point with rectangular coordinates (a, b) where r ≥ 0. The modu...
If z = 0 then the point in question is not the origin, so all of these angles θ are coterminal. Since coterminal angles are exactly 2π radians apart, we are guaranteed that only one of them lies in the interval (−π, π], and this angle is what we call the principal argument of z, Arg(z). In fact, the set arg(z) of all ...
� 6 + 2πk 3 6 + 2πk \| k is an integer. Of these values, only θ = − π for integers k. Hence, arg(z) = − π satisﬁes the requirement that −π < θ ≤ π, hence Arg(z) = − π 6. 3, so. The complex number z = −2 + 4i has Re(z) = −2, Im(z) = 4, and is associated with the point P (−2, 4). Our next task is to ﬁnd a polar representa...
0, 3) lies 3 units away from the origin on the positive y-axis. Hence, r = \|z\| = 3 and θ = π 2 + 2πk for integers k. We get arg(z) = π 2 + 2πk \| k is an integer and Arg(z) = π 2. 4. As in the previous problem, we write z = −117 = −117 + 0i so Re(z) = −117 and Im(z) = 0. The number z = −117 corresponds to the point (−11...
� Section 11.4, we know r2 = a2 + b2 so that r = ± a2 + b2. Since we require r ≥ 0, then it must be a2 + b2. Using the distance formula, we ﬁnd the distance a2 + b2, which means \|z\| = that r = √ √ 994 Applications of Trigonometry √ a2 + b2, establishing the ﬁrst property.5 For the second property, note from (0, 0) to (...
z\|k+1. As is customary with induction proofs, is to show that P (k + 1) is true, namely we ﬁrst try to reduce the problem in such a way as to use the Induction Hypothesis. = \|z\| = \|z\|1. Next, assume P (k) is true. That is, assume zk+1 zk zk+1 Properties of Exponents \|z\| Product Rule zkz = zk = = \|z\|k\|z\| = \|z\|k+1 Proper...
then tan(θ) = Im(z) Re(z). 2 + 2πk \| k is an integer. 2 + 2πk \| k is an integer. If Re(z) = 0 and Im(z) > 0, then arg(z) = π If Re(z) = 0 and Im(z) < 0, then arg(z) = − π If Re(z) = Im(z) = 0, then z = 0 and arg(z) = (−∞, ∞). To prove Theorem 11.15, suppose z = a + bi for real numbers a and b. By deﬁnition, a = Re(z) ...
cos(θ) + i sin(θ)’ is abbreviated cis(θ) so we can write z = rcis(θ). Since r = \|z\| and θ ∈ arg(z), we get Deﬁnition 11.3. A Polar Form of a Complex Number: Suppose z is a complex number and θ ∈ arg(z). The expression: is called a polar form for z. \|z\|cis(θ) = \|z\| [cos(θ) + i sin(θ)] 996 Applications of Trigonometry Si...
Writing 3 = 3 + 0i, we get Re(z) = 3 and (d) Lastly, we have z = cis π 2 Im(z) = 0, which makes sense seeing as 3 is a real number. = cos π 2 = i. Since i = 0 + 1i, we get Re(z) = 0 and Im(z) = 1. Since i is called the ‘imaginary unit,’ these answers make perfect sense. + i sin π 2 2. To write a polar form of a comple...
\|w\|cis(β). Then Product Rule: zw = \|z\|\|w\|cis(α + β) Power Rule (DeMoivre’s Theorem) : zn = \|z\|ncis(nθ) for every natural number n Quotient Rule: z w = \|z\| \|w\| cis(α − β), provided \|w\| = 0 The proof of Theorem 11.16 requires a healthy mix of deﬁnition, arithmetic and identities. We ﬁrst start with the product rule. zw ...
= \|z\|kcis(kθ) for some k ≥ 1. Our goal is to show that P (k + 1) is true, or that zk+1 = \|z\|k+1cis((k + 1)θ). We have zk+1 = zkz = \|z\|kcis(kθ) (\|z\|cis(θ)) = \|z\|k\|z\| cis(kθ + θ) = \|z\|k+1cis((k + 1)θ) Properties of Exponents Induction Hypothesis Product Rule 9Compare this proof with the proof of the Power Rule in Theore...
(α) sin(β)] Rearrange and Factor = cos(α − β) + i sin(α − β) = cis(α − β) Diﬀerence Identities Deﬁnition of ‘cis’ If we call the denominator D then we get D = [cos(β) + i sin(β)] [cos(β) − i sin(β)] = cos2(β) − i cos(β) sin(β) + i cos(β) sin(β) − i2 sin2(β) Expand = cos2(β) − i2 sin2(β) Simplify = cos2(β) + sin2(β) Aga...
π √ 3 After simplifying, we get zw = −4 3 + 4i. = 3. For w = −1 + i √ 3 −1 = − we have \|w\| = w lies in Quadrant II, θ = 2π 1. We get zw = 4cis π 6. We can now proceed. + i sin 5π 6. = 8 cos 5π 6 = 8cis 5π 6 Re(z) = 2 (−1)2 + ( 6 + 2π 3. Since √ √ √ √ 6 2 3 3 2. We use DeMoivre’s Theorem which yields w5 = 2cis 2π 3 + i ...
11.16 to ﬁnd and simplify z w using their polar forms as opposed to starting with 2, rationalizing the denominator, and so forth. 3 + 2i)(−1 + i √ √ √ √ 3+2i √ 3 −1+i There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem 11....
from Example 11.7.3, we arrive at z w by ﬁrst halving the distance from 0 to z, then rotating clockwise 2π w = \|z\| 3 radians. Imaginary Axis Imaginary Axis 3i 2i i z = 4cis π 6 1 \|w\| z = 2cis π 6 0 1 2 3 Real Axis i −i −2i 1 \|w\| z = 2cis π 6 0 1 2 3 Real Axis zw = 2cis π 6 2π 3 Dividing z by \|w\| = 2. Rotating clockwis...
, counting multiplicity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express z = 8 in polar form. Since z = 8 lies 8 units away on the pos...
and w2 = 2cis 4π √ and 2, 4π 3 3 and w2 = −1 − i √ 3 Theorem 11.17. The nth roots of a Complex Number: Let z = 0 be a complex number with polar form z = rcis(θ). For each natural number n, z has n distinct nth roots, which we denote by w0, w1,..., wn − 1, and they are given by the formula √ wk = n rcis θ n + 2π n k Th...
an integer multiple of 2π, (k − j) must be a multiple of n. But because of the restrictions on k and j. (Think this through.) Hence, (k − j) is a positive number less than n, so it cannot be a multiple of n. As a result, wk and wj are diﬀerent complex numbers, and we are done. By Theorem 3.14, we know there at most n ...
π four fourth roots of z to be w0 = 4 √ √ 4. and w3 = 4, w2 = 4 16cis π 2cis 3π 4 (3) = 2cis 7π 4 (2) = 2cis 5π 16cis −i 2, w2 = − Converting these to rectangular form gives w0 = 2 √ and w3 = 4 + 2π √ 2+i 2, w1 = − 4 + 2π 2 − i 2+i √ 2. 11.7 Polar Form of Complex Numbers 1003 √ √ 3. For z = 2cis π 12 2, we have z = 2ci...
th root of the modulus and divide the argument by n. This gives the ﬁrst root w0. Each succeessive root is found by adding 2π n to the argument, which amounts to rotating w0 by 2π n radians. This results in n roots, spaced equally around the complex plane. As an example of this, we plot our answers to number 2 in Examp...
z = 6cis(0) 22. z = 2cis π 6 3π 4 3π 2 √ 26. z = 6cis √ 30. z = 13cis √ 23. z = 7 2cis π 4 24. z = 3cis 27. z = 9cis (π) 28. z = 3cis π 2 4π 3 31. z = 1 2 cis 32. z = 12cis − π 3 7π 4 7π 8 34. z = 2cis 10cis arctan √ 3 arctan − √ 1 3 2 5 12 4 3 √ 36. z = 37. z = 15cis (arctan (−2)) 38. z = 39. z = 50cis π − arctan 7 2...
= 4i 66. the two square roots of z = −25i 67. the two square roots of z = 1 + i √ 3 68. the two square roots of 5 2 − 5 √ 3 2 i 69. the three cube roots of z = 64 70. the three cube roots of z = −125 71. the three cube roots of z = i 72. the three cube roots of z = −8i 73. the four fourth roots of z = 16 74. the four ...
by showing that if w = 0 than 1 w = 1 \|w\|. 81. Recall from Section 3.4 that given a complex number z = a+bi its complex conjugate, denoted z, is given by z = a − bi. (a) Prove that \|z\| = \|z\|. √ (b) Prove that \|z\| = (c) Show that Re(z) = zz z + z 2 and Im(z) = z − z 2i (d) Show that if θ ∈ arg(z) then −θ ∈ arg (z). Int...
z = reit where θ = t radians. (e) Show that eiπ + 1 = 0. (This famous equation relates the ﬁve most important constants in all of Mathematics with the three most fundamental operations in Mathematics.) (f) Show that cos(t) = eit + e−it 2 and that sin(t) = eit − e−it 2i for all real numbers t. 11.7 Polar Form of Comple...
7. z = − 2 − 1 arg(z) = 7π 8. z = −3 − 3i = 3 arg(z) = 5π √ 3 2, 6, Re(z) = − 2 i = cis 7π Im(z) = − 1 2, 6 + 2πk \| k is an integer and Arg(z) = − 5π 6., Re(z) = −3, Im(z) = −3, 4 + 2πk \| k is an integer and Arg(z) = − 3π 4. 2cis 5π 4 √ \|z\| = 1 √ \|z\| = 3 2 9. z = −5i = 5cis 3π 2 arg(z) = 3π √, Re(z) = 0, Im(z) = −5, \|...
, \|z\| = arg(z) = arctan + 2πk \| k is an integer and Arg(z) = arctan 3cis √ 2 2 √ 3 √ 2 2. Applications of Trigonometry 15. z = −7 + 24i = 25cis π − arctan 24 7, Re(z) = −7, Im(z) = 24, \|z\| = 25 arg(z) = π − arctan 24 7 + 2πk \| k is an integer and Arg(z) = π − arctan 24 7. 16. z = −2 + 6i = 2 √ 10cis (π − arctan (3)), R...
z) = {arctan (−3) + 2πk \| k is an integer} and Arg(z) = arctan (−3) = − arctan(3). 21. z = 6cis(0) = 6 √ 23. z = 7 25. z = 4cis 2π 3 2cis π 4 = −2 + 2i = 7 + 7i √ 3 27. z = 9cis (π) = −9 √ 3 + i = 22. z = 2cis π 6 24. z = 3cis π 2 6cis 3π 4 26. z = √ = 3i √ = − √ 3 + i 3 28. z = 3cis 4π 3 √ 30. z = = − 3 2 − 3i √ √ 3 2...
2cis 11π 46. z5w2 = 8748cis − π 3 12 42. z w = 1 2 cis − 11π 12 45. w3 = 216cis − 3π 4 48. z2 w = 3 2 cis − π 12 49. w z2 = 2 3 cis π 12 51. w2 z3 = 4 √ 3 cis(π) 54. (− 3 − i)3 = −8i 52. w z 6 = 64cis − π 2 55. (−3 + 3i)4 = −324 √ 56. ( 3 + i)4 = −8 + 8i √ 3 57. 5 2 + 5 2 i3 = − 125 4 + 125 58. 59. 3 2 − 3 2 i3 = − 27...
= 125cis (π) we have w0 = 5cis w1 = 5cis (π) = −5 w2 = 5cis 5π 71. Since z = i = cis π 2 we have w0 = cis w1 = cis 5π w2 = cis 3π 2 = −i 72. Since z = −8i = 8cis 3π 2 we have w0 = 2cis π 2 = 2i w1 = 2cis 7π 6 √ = − 3 − i w2 = cis 11π 6 = √ 3 − i 73. Since z = 16 = 16cis (0) we have w0 = 2cis (0) = 2 w2 = 2cis (π) = −2...
12 = 2cis π 12 2+ 2 2− 2 6− 6 4 + i − √ 2 √ w1 = 3 2cis 3π 4 √ w2 = 3 2cis 17π 12 √ − √ 6 √ = 3 2 2− 4 √ √ 3 2− 2 + i √ 3 √ 2+ 2 11.7 Polar Form of Complex Numbers 1011 5 78. w0 = cis(0) = 1 w1 = cis 2π w2 = cis 4π w3 = cis 6π w4 = cis 8π ≈ 0.309 + 0.951i ≈ −0.809 + 0.588i ≈ −0.809 − 0.588i ≈ 0.309 − 0.951i 5 5 5 79. ...
point P is called the initial point or tail of v and the point Q is called the terminal point or head of v. Since we can reconstruct −−→ v completely from P and Q, we write v = P Q, where the order of points P (initial point) and Q (terminal point) is important. (Think about this before moving on.) Q (4, 6) P (1, 2) −...
v is represented by a directed line segment with initial point P (x0, y0) and terminal point Q (x1, y1). The component form of v is given by −−→ P Q = x1 − x0, y1 − y0 v = Using the language of components, we have that two vectors are equal if and only if their corresponding components are equal. That is, v1, v2 = v 1...
. From the vector diagram, we get a triangle, the lengths of whose sides are the magnitude of v, which is 175, the magnitude of w, which is 35, and the magnitude of v + w, which we’ll call c. From the given bearing information, we go through the usual geometry to determine that the angle between the sides of length 35 ...
such as Linear Algebra, vectors are deﬁned as 1 × n or n × 1 matrices, depending on the situation. 11.8 Vectors 1015 v + w = 3, 4 + 1, −2 = 3 + 1, 4 + (−2) = 4, 2 To visualize this sum, we draw v with its initial point at (0, 0) (for convenience) so that its terminal point is (3, 4). Next, we graph w with its initial ...
the deﬁnition of vector addition.9 For the commutative property, we note that if v = v1, v2 and w = w1, w2 then v + w = v1, v2 + w1, w2 = v1 + w1, v2 + w2 = w1 + v1, w2 + v2 = w + v Geometrically, we can ‘see’ the commutative property by realizing that the sums v + w and w + v are the same directed diagonal determined...
− w = v + (− w) = v1, v2 + −w1, −w2 = v1 + (−w1), v2 + (−w2) = v1 − w1, v2 − w2 In other words, like vector addition, vector subtraction works component-wise. To interpret the vector v − w geometrically, we note w + (v − w) = w + (v + (− w)) Deﬁnition of Vector Subtraction = w + ((− w) + v) Commutativity of Vector Add...
vector v and scalars k and r, (kr)v = k(rv). Identity Property: For all vectors v, 1v = v. Additive Inverse Property: For all vectors v, −v = (−1)v. Distributive Property of Scalar Multiplication over Scalar Addition: For every vector v and scalars k and r, (k + r)v = kv + rv Distributive Property of Scalar Multiplica...
5v + [(−1)(2)] (v + 1, −2) = 0 5v + (−2) (v + 1, −2) = 0 5v + [(−2)v + (−2) 1, −2] = 0 5v + [(−2)v + (−2)(1), (−2)(−2)] = 0 [5v + (−2)v] + −2, 4 = 0 (5 + (−2))v + −2, 4 = 0 3v + −2, 4 = 0 (3v + −2, 4) + (− −2, 4) = 0 + (− −2, 4) 3v + [−2, 4 + (− −2, 4)] = 0 + (−1) −2, 4 3v + 0 = 0 + (−1)(−2), (−1)(4) 3v = 2, −4 1 3 1 ...
8. Suppose v is a vector with component form v = v1, v2. Let (r, θ) be a polar representation of the point with rectangular coordinates (v1, v2) with r ≥ 0. The magnitude of v, denoted v, is given by v = r = 1 + v2 v2 2 If v = 0, the (vector) direction of v, denoted ˆv is given by ˆv = cos(θ), sin(θ) Taken together, we...
= v2 = 0. Hence, v = 0 if and only if v = 0, 0 = 0, as required. The second property is a result of the deﬁnition of magnitude and scalar multiplication along with a propery of radicals. If v = v1, v2 and k is a scalar then 10If this all looks familiar, it should. The interested reader is invited to compare Deﬁnition ...
v (b) v − 2 w (c) v − 2 w (d) ˆw Solution. 1. We are told that v = 5 and are given information about its direction, so we can use the formula v = vˆv to get the component form of v. To determine ˆv, we appeal to Deﬁnition 11.8. We are told that v lies in Quadrant II and makes a 60◦ angle with the negative x-axis, so th...
π 3 √ √ 3 3. (a) Since we are given the component form of v, we’ll use the formula ˆv = 5 3, 4, we have v = 25 = 5. Hence, ˆv = 1 32 + 42 = √ √ 5 1 v. v. For (b) We know from our work above that v = 5, so to ﬁnd v−2 w, we need only ﬁnd w. 12 + (−2)2 = Since w = 1, −2, we get w = 5. Hence. (c) In the expression v −2 w, ...
. If we regard the airport as being 11.8 Vectors 1023 at the origin, the positive y-axis acting as due north and the positive x-axis acting as due east, we see that the vectors v and w are in standard position and their directions correspond to the angles 50◦ and −30◦, respectively. Hence, the component form of v = 175...
be approximately N51◦E. y (N) y (N) v v v + w 40◦ 50◦ −30◦ w 60◦ x (E) θ w x (E) In part 3d of Example 11.8.4, we saw that ˆw = 1. Vectors with length 1 have a special name and are important in our further study of vectors. Deﬁnition 11.9. Unit Vectors: Let v be a vector. If v = 1, we say that v is a unit vector. 13Ke...
while ˆ represents the positive y-direction. We have the following ‘decomposition’ theorem.17 Theorem 11.21. Principal Vector Decomposition Theorem: Let v be a vector with component form v = v1, v2. Then v = v1ˆı + v2ˆ. The proof of Theorem 11.21 is straightforward. Since ˆı = 1, 0 and ˆ = 0, 1, we have from the de...
30◦ 60◦ w 50 lbs. We have three forces acting on the speaker: the weight of the speaker, which we’ll call w, pulling the speaker directly downward, and the forces on the support rods, which we’ll call T1 and T2 (for ‘tensions’) acting upward at angles 60◦ and 30◦, respectively. We are looking for the tensions on the s...
2 3 √ + T1 2 √ T2 2 3 − = 0 T2 2 − 50 = 0 From (E1), we get T1 = T2 which yields 2 T2 − 50 = 0. Hence, T2 = 25 pounds and T1 = T2 3. Substituting that into (E2) gives ( T2 √ 2 3 = 25 + T2 3 √ 3 pounds. 2 − 50 = 0 √ √ √ 3) 11.8 Vectors 11.8.1 Exercises 1027 In Exercises 1 - 10, use the given pair of vectors v and w to ...
v lies in Quadrant II and makes a 30◦ angle with the negative x-axis √ 16. v = 2 3; when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the positive y-axis 17. v = 7 2 ; when drawn in standard position v lies along the negative x-axis √ 18. v = 5 6; when drawn in standard position v lies i...
in standard position v makes a 304.5◦ angle with the positive x-axis In Exercises 32 - 52, for the given vector v, ﬁnd the magnitude v and an angle θ with 0 ≤ θ < 360◦ so that v = v cos(θ), sin(θ) (See Deﬁnition 11.8.) Round approximations to two decimal places. 32. v = 1, √ 35. v = − √ 2, 3 √ 2 33. v = 5, 5 36 38. v ...
100 miles away at a bearing of S20◦E from port, in three hours, what speed and heading should she set to take into account the ocean current? Round the speed to the nearest mile per hour and express the heading as a bearing, rounded to the nearest tenth of a degree. HINT: If v denotes the velocity of the HMS Sasquatch...
three decimal places. 61. Two drunken college students have ﬁlled an empty beer keg with rocks and tied ropes to it in order to drag it down the street in the middle of the night. The stronger of the two students pulls with a force of 100 pounds at a heading of N77◦E and the other pulls at a heading of S68◦E. What for...
y = 2x − 4. (Hint: Show that v0 + ts = t, 2t − 4 for any real number t.) Now consider the non-vertical line y = mx + b. Repeat the previous analysis with v0 = 0, b and let s = 1, m. Thus any non-vertical line can be thought of as a collection of terminal points of the vector sum of 0, b (the position vector of the y-i...
, 0, vector v + w = 6, scalar wˆv = −2 √ 3, 2, vector w − 2v = −2, −1, vector v + w = 2, scalar wˆv = 3 5 5, 4, vector √ √ − 3 2 2, 3 2 2 7. v + w = 0, 0, vector w − 2v =, vector v + w = 0, scalar v w − wv = − √ √ 2, 2, vector v + w = 2, scalar wˆv = √ 2 2, − √ 2 2, vector 1032 Applications of Trigonometry 8, vector w ...
62, 1097.77 29. v ≈ −386.73, −230.08 32. v = 2, θ = 60◦ 30. v ≈ −52.13, −160.44 √ 33. v = 5 2, θ = 45◦ 31. v ≈ 14.73, −21.43 34. v = 4, θ = 150◦ 35. v = 2, θ = 135◦ 36. v = 1, θ = 225◦ 38. v = 6, θ = 0◦ 39. v = 2.5, θ = 180◦ 41. v = 10, θ = 270◦ 42. v = 5, θ ≈ 53.13◦ 44. v = 5, θ ≈ 143.13◦ 45. v = 25, θ ≈ 106.26◦ 37. v...
weight that can be held by the cables in that conﬁguration is about 133 pounds. 60. The tension on the left hand cable is 285.317 lbs. and on the right hand cable is 92.705 lbs. 61. The weaker student should pull about 60 pounds. The net force on the keg is about 153 pounds. 62. The resultant force is only about 296 p...
= v1, v2 and w = w1, w2. Then v · w = v1, v2 · w1, w2 = v1w1 + v2w2 Deﬁnition of Dot Product = w1v1 + w2v2 = w1, w2 · v1, v2 Deﬁnition of Dot Product = w · v Commutativity of Real Number Multiplication The distributive property is proved similarly and is left as an exercise. For the scalar property, assume that v = v1...
]) = (v + [− w]) · v + (v + [− w]) · [− w] = v · (v + [− w]) + [− w] · (v + [− w]) = v · v + v · [− w] + [− w] · v + [− w] · [− w] = v · v + v · [(−1) w] + [(−1) w] · v + [(−1) w] · [(−1) w] = v · v + (−1)(v · w) + (−1)( w · v) + [(−1)(−1)]( w · w) = v · v + (−1)(v · w) + (−1)(v · w(v · w) + w · w = v2 − 2(v · w) + w2 ...
and w are nonzero 1036 Applications of Trigonometry We prove Theorem 11.23 in cases. If θ = 0, then v and w have the same direction. It follows1 that there is a real number k > 0 so that w = kv. Hence, v · w = v · (kv) = k(v · v) = kv2 = kvv. Since k > 0, k = \|k\|, so kv = \|k\|v = kv by Theorem 11.20. Hence, kvv = v(kv)...
, so are v and w. Hence, we may divide both sides of v · w = v w cos(θ) by v w to get cos(θ) = v· w v w. Since 0 ≤ θ ≤ π by deﬁnition, the values of θ exactly match the. Using Theorem 11.22, we can rewrite v· w range of the arccosine function. Hence, θ = arccos v w = ˆv · ˆw, giving us the alternative formula θ = arcco...
−4 · 2, 1 = 6 − 4 = 2. Also v = √ √ w = 22 + 12 = 5, so θ = arccos = arccos isn’t the cosine of one 32 + (−4)2 = √ 25 = 5 and 2 √ 5 5 of the common angles, we leave our answer as θ = arccos 2 √ 5 25 √ 5 25. Since 2 2 √ 5 25. The vectors v = 2, 2, and w = 5, −5 in Example 11.9.2 are called orthogonal and we write v ⊥ w...
and show v1 ⊥ v2 if and only if m1m2 = −1. To that end, we substitute x = 0 and x = 1 into y = m1x + b1 to ﬁnd two points which lie on L1, namely P (0, b1) 2Note that there is no ‘zero product property’ for the dot product since neither v nor w is 0, yet v · w = 0. 3See Exercise 2.1.1 in Section 2.1. 1038 Applications...
pˆp. Since we want ˆp to have the same direction as w, we have ˆp = ˆw. To determine p, we make use of Theorem 10.4 as applied to the right triangle ORT. We ﬁnd cos(θ) = p v, or p = v cos(θ). To get things in terms of just v and w, we use Theorem 11.23 to get p = v cos(θ) = v w cos(θ) w. Using Theorem 11.22, we rewrit...
��nition 11.12. Let v and w be nonzero vectors. The orthogonal projection of v onto w, denoted proj w(v) is given by proj w(v) = (v · ˆw) ˆw. Deﬁnition 11.12 gives us a good idea what the dot product does. The scalar v · ˆw is a measure of how much of the vector v is in the direction of the vector w and is thus called ...
what remains to check is the orthogonality condition. Consider the vector q whose initial point is the terminal point of p and whose terminal point is the terminal point of v3 −2 −1 1 From the deﬁnition of vector arithmetic, p + q = v, so that q = v − p. In the case of Example 11.9.4, v = 1, 8 and p = −3, 6, so q = 1,...
p − p ) = 0. Now there are scalars k and k so that p = k w and p = k w. This means w · (p − p ) = w · (k w − k w) = w · ([k − k ] w) = (k − k )( w · w) = (k − k ) w2. Since w = 0, w2 = 0, which means the only way w · (p − p ) = (k − k ) w2 = 0 is for k − k = 0, or k = k. This means p = k w = k w = p. With, it must be t...
F and −−→ P Q = F −−→ P Q cos(θ), W = F · −−→ P Q. Example 11.9.5. Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a 30◦ angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds a...
, −9 2. v = −6, −5 and w = 10, −12 3. v = 1, √ 3 and w = 1, − √ 3 5. v = −2, 1 and w = 3, 6 4. v = 3, 4 and w = −6, −8 6. v = −3 √ 3, 3 and w = − 3, −1 √ 7. v = 1, 17 and w = −1, 0 8. v = 3, 4 and w = 5, 12 9. v = −4, −2 and w = 1, −5 10. v = −5, 6 and w = 4, −7 11. v = −8, 3 and w = 2, 6 12. v = 34, −91 and w = 0, 1 1...
with a force of 100 pounds on a rope which makes a 13◦ angle with the direction of motion. (In this case, the keg was being pulled due east and the student’s heading was N77◦E.) Find the work done by this student if the keg is dragged 42 feet. 1044 Applications of Trigonometry 25. Find the work done pushing a 200 poun...
an added bonus, we can now show that the Triangle Inequality \|z + w\| ≤ \|z\| + \|w\| holds for all complex numbers z and w as well. Identify the complex number z = a + bi with the vector u = a, b and identify the complex number w = c + di with the vector v = c, d and just follow your nose! 6It is also known by other names...
− 248 q = − 77 65, − 44 65 65, 434 65 1046 Applications of Trigonometry 11. v = −8, 3 and w = 2, 6 12. v = 34, −91 and w = 0, 1 v · w = 2 θ ≈ 87.88◦ proj w(v) = 1 10, 3 q = − 81 10, 27 10 10 v · w = −91 θ ≈ 159.51◦ proj w(v) = 0, −91 q = 34, 0 13. v = 3ˆı − ˆ and w = 4ˆ 14. v = −24ˆı + 7ˆ and w = 2ˆı v · w = −4 θ ≈...
j w(v) = 1− 4 q = √ 3, 1− 4 3+1 4 √ 3 √ 3+1 4, − 21. (1500 pounds)(300 feet) cos (0◦) = 450, 000 foot-pounds 22. (10 pounds)(3 feet) cos (0◦) = 30 foot-pounds 11.9 The Dot Product and Projection 1047 23. (13 pounds)(25 feet) cos (15◦) ≈ 313.92 foot-pounds 24. (100 pounds)(42 feet) cos (13◦) ≈ 4092.35 foot-pounds 25. (2...
with an orientation and the arrows on C indicate motion in the direction of increasing values of t. In this case, our bug starts at the point O, travels upwards to the left, then loops back around to cross its path2 at the point Q and ﬁnally heads oﬀ into the ﬁrst quadrant. It is important to note that the curve itsel...
= 2t − 1 for t ≥ −2. Solution. We follow the same procedure here as we have time and time again when asked to graph anything new – choose friendly values of t, plot the corresponding points and connect the results in a pleasing fashion. Since we are told t ≥ −2, we start there and as we plot successive points, we draw...
−5 describes the same curve as the given parametric equations. The one piece of information we can never recover after eliminating the parameter is the orientation of the curve. y+1 2 2 Eliminating the parameter and obtaining an equation in terms of x and y, whenever possible, can be a great help in graphing curves de...
, since that corresponds to a relative minimum4 on the graph of y = 2t2. Plugging in t = −1 gives the point (−1, 2), t = 0 gives (0, 0) and t = 1 gives (1, 2). More generally, we see that x = t3 is increasing over the entire interval [−1, 1] whereas y = 2t2 is decreasing over the interval [−1, 0] and then increasing ov...
of t to get a sense of the orientation of the curve. Since t lies in the exponent here, ‘friendly’ values of t involve natural logarithms. Starting with t = ln(1) = 0 we get5 (2, 1), for t = ln(2) we get 1, 1. Since t is ranging over the unbounded interval [0, ∞), 4 we take the time to analyze the end behavior of both...
�). We ﬁnd that the range of x is (0, 1] while the range of 2, 2, t = π y is [1, ∞). Plotting a few friendly points, we see that t = π 2, 2. Since t = 0 and t = π aren’t included in the 6 returns us to 1 gives (1, 1) and t = 5π domain for t, (because y = csc(t) is undeﬁned at these t-values), we analyze the behavior of...
hugging the vertical asymptote x = 0 of the graph of y = 1 x. We see that the parametrization given above traces out the portion of y = 1 x for 0 < x ≤ 1 twice as t runs through the interval (0, π). = sin(t), 0 < t < π y = csc(t), 0 < t < π {x = sin(t), y = csc(t), 0 < t < π 4. Proceeding as above, we set about graphi...
1 9 + y2 4 = 1. From Section 3 7.4, we know that the graph of this equation is an ellipse centered at (1, 0) with vertices at (−2, 0) and (4, 0) with a minor axis of length 4. Our parametric equations here are tracing out three-quarters of this ellipse, in a counter-clockwise direction. = 1, or (x−1)1 π 2 π t 3π 2 −2 x...
answers. 1. y = x2 from x = −3 to x = 2 2. y = f −1(x) where f (x) = x5 + 2x + 1 3. The line segment which starts at (2, −3) and ends at (1, 5) 4. The circle x2 + 2x + y2 − 4y = 4 4 + y2 5. The left half of the ellipse x2 9 = 1 Solution. 1. Since y = x2 is written in the form y = f (x), we let x = t and y = f (t) = t2...
check.7 7Provided you followed the inverse function theory, of course. 1054 Applications of Trigonometry 3. To parametrize line segment which starts at (2, −3) and ends at (1, 5), we make use of the formulas x = x0 +(x1 −x0)t and y = y0 +(y1 −y0)t for 0 ≤ t ≤ 1. While these equations at ﬁrst glance are quite a handful...
the squares, we get (x + 1)2 + (y − 2)2 = 9, or (x+1)2 9 = 1. Once again, the formulas x = h + a cos(t) and y = k + b sin(t) can be a challenge to memorize, but they come from the Pythagorean Identity cos2(t) + sin2(t) = 1. In the equation (x+1)2 3. Rearranging these last two equations, we get x = −1 + 3 cos(t) and y ...
through the usual analysis as demonstrated in Example 11.10.2 to show that the entire circle is covered as t ranges through the interval [0, 2π). 8Compare and contrast this with Exercise 65 in Section 11.8. 11.10 Parametric Equations 1055 5. In the equation x2 4 + y2 9 = 1, we can either use the formulas above or thin...
start of the parameter t ahead by c units. We demonstrate these techniques in the following example. Example 11.10.4. Find a parametrization for the following curves. 1. The curve which starts at (2, 4) and follows the parabola y = x2 to end at (−1, 1). Shift the parameter so that the path starts at t = 0. 2. The two ...
second part so it starts at t = 1. Our current description of the second part starts at t = 0, so we introduce a ‘time delay’ of 1 unit to the second set of parametric equations. Replacing t with (t − 1) in the second set of parametric equations gives {x = 3 + 2(t − 1), y = 4 − 4(t − 1) for 0 ≤ t − 1 ≤ 1. Simplifying ...
traces out the Unit Circle clockwise starting at the point (0, −1). We now shift the parameter by introducing a ‘time delay’ of 3π 2 units by replacing every occurrence of t with t − 3π 2. This 2 simpliﬁes10 to {x = − sin(t), y = − cos(t) for 0 ≤ t < 2π, as required.. We get x = cos t − 3π 2, y = − sin t − 3π 2 2 ≤ t ...
circle is always r units above the x-axis. Putting these two facts together, we have that at time t, the center of the circle is at the point (vt, r). From Section 10.1.1, we know v = rθ t, or vt = rθ. Hence, the center of the circle, in terms of the parameter θ, is (rθ, r). As a result, we need to modify the equation...
language of Section 1.7, we are stretching the graph by a factor of r in both the x- and y-directions. Hence, we multiply both the x- and y-coordinates of points on the graph by r. r in the equation for the Unit Circle x2 + y2 = 1, we obtain x r and y with y 2 + y r r 12Does this seem familiar? See Example 11.1.1 in S...
0 ≤ t ≤ 2π x = 2 cos(t) y = sec(t) for 0 ≤ t < π 2 x = sec(t) y = tan(t) for − π 2 < t < π 2 x = tan(t) y = 2 sec(t) for − π 2 < t < π 2 x = cos(t) y = t for 0 ≤ t ≤ π 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. x = 4t − 1 y = 3 − 4t for + 2t − t2 18 − t2 x = 1 9 y = 1 3 t for 0 ≤ t ≤ 3 for t ≥ −3 x = t3 y = t for − ∞ < t < ∞...
4 − x2 from (−2, 0) to (2, 0). 28. the curve y = 4 − x2 from (−2, 0) to (2, 0) (Shift the parameter so t = 0 corresponds to (−2, 0).) 29. the curve x = y2 − 9 from (−5, −2) to (0, 3). 30. the curve x = y2 − 9 from (0, 3) to (−5, −2). (Shift the parameter so t = 0 corresponds to (0, 3).) 31. the circle x2 + y2 = 25, or...
. 42. Use your results from Exercises 3 and 4 in Section 11.1 to ﬁnd the parametric equations which model a passenger’s position as they ride the London Eye. 11.10 Parametric Equations 1061 Suppose an object, called a projectile, is launched into the air. Ignoring everything except the force gravity, the path of the pr...
. If the weight is released 5 feet above the ground at an angle of 85◦ with respect to the horizontal and the sheaf reaches a maximum height of 31.5 feet, use your results from part 44 to determine how fast the sheaf was launched into the air. (Once again, use g = 32 ft. s2.) 46. Suppose θ = π formula given for y(t) ab...
or this entry on the hyperbolic functions) and spend some time reliving the thrills of trigonometry with these ‘hyperbolic’ functions. 52. If these functions look familiar, they should. Enjoy some nostalgia and revisit Exercise 35 in Section 6.5, Exercise 47 in Section 6.3 and the answer to Exercise 38 in Section 6.4....
. x = 2 tan(t) y = cot(t) y for 15. x = sec(t) y = tan(t) for − π 2 < t < π 2 y 16. x = sec(t) y = tan(t) < t < 3π 2 π 2 for y 4 3 2 1 −1 −2 −3 −4 −3 −2 −1 −1 −2 −3 −4 1066 Applications of Trigonometry 17. x = tan(t) y = 2 sec(t) π 2 for − y < t < π 2 18. x = tan(t) y = 2 sec(t) < t < 3π 2 π 2 for y 4 3 2 1 −2 −1 1 2 x...
x = 3 − 5t y = −5 + 7t for 0 ≤ t ≤ 1 for − 2 ≤ t ≤ 2 for − 2 ≤ t ≤ 3 for 0 ≤ t < 2π x = t y = 4 − t2 x = t2 − 9 y = t x = 5 cos(t) y = 5 sin(t) x = 3 cos(t) x = 3 + √ y = −1 + x = 2 cos(t) 25. 27. 29. 31. 33. 35. 37. 38. y = 3 + 3 sin(t) for 0 ≤ t < 2π 117 cos(t) √ 117 sin(t) for 0 ≤ t < 2π 36. for 0 ≤ t < 2π y = 3 si...
◦)t + 6 for t ≥ 0. To ﬁnd when the hammer hits the ground, we solve y(t) = 0 and get t ≈ −0.23 or 1.61. Since t ≥ 0, the hammer hits the ground after approximately t = 1.61 seconds after it was launched into the air. To ﬁnd how far away the hammer hits the ground, we ﬁnd x(1.61) ≈ 39.48 feet from where it was thrown in...
1070 trigonometry friendly deﬁnition of, 828 graph of, 827 properties of, 828 arccosine deﬁnition of, 820 graph of, 819 properties of, 820 arccotangent deﬁnition of, 824 graph of, 824 properties of, 824 arcsecant calculus friendly deﬁnition of, 831 graph of, 830 properties of, 831 trigonometry friendly deﬁnition of, 8...
498 from slicing a cone, 495 radius of, 498 standard equation, 498 standard equation, alternate, 519 circular function, 744 cis(θ), 995 coeﬃcient of determination, 226 cofactor, 616 Cofunction Identities, 773 common base, 420 common logarithm, 422 commutative property function composition does not have, 366 matrix add...
s Rule, 619 curve orientated, 1048 cycloid, 1056 decibel, 431 decimal degrees, 695 decreasing function formal deﬁnition of, 101 intuitive deﬁnition of, 100 degree measure, 694 degree of a polynomial, 236 DeMoivre’s Theorem, 997 dependent system, 554 dependent variable, 55 depreciation, 420 Descartes’ Rule of Signs, 273...
54 linear of two variables, 549 even function, 95 Even/Odd Identities, 770 exponential function algebraic properties of, 437 change of base formula, 442 common base, 420 deﬁnition of, 418 graphical properties of, 419 inverse properties of, 437 natural base, 420 one-to-one properties of, 437 solving equations with, 448 ...

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