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 x1 − 4x3 + x4 − x6 = 6 x2 + 2x3 = 1 x4 + 3x5 − x6 = 8 x5 + 9x6 = 10 Our goal henceforth will be to transform a given system of linear equations into triangular form using the moves in Theorem 8.1. Example 8.1.2. Use Theorem 8.1 to put the following systems into triangular form and then solve the system if possible. C...
E2 get y = 1 2 t + 1250 3 . This system is consistent, dependent and its solution set is {− 3 3 , t \| − ∞ < t < ∞}. While this answer checks algebraically, we have neglected to take into account that x, y and w, being amounts of acid and water, need to be nonnegative. That is, x ≥ 0, y ≥ 0 and w ≥ 0. The 3 ≥ 0 or t ≥ −...
t in E2, since y is not present, we record its coeﬃcient as 0. The matrix associated with this system is (E1) → (E2) → (E3) →   x 2 10 4 −9 y c z 3 −1 1 0 −1 2 2 5   568 Systems of Equations and Matrices This matrix is called an augmented matrix because the column containing the constants is appended to the matrix ...
in our next example. Example 8.2.3. Find the quadratic function passing through the points (−1, 3), (2, 4), (5, −2). Solution. According to Deﬁnition 2.5, a quadratic function has the form f (x) = ax2 +bx+c where a = 0. Our goal is to ﬁnd a, b and c so that the three given points are on the graph of f . If (−1, 3) is ...
ty is the deﬁnition of A + B, the third equality holds by the commutative law of real number addition, and the fourth equality is the deﬁnition of B + A. In other words, matrix addition is commutative because real number addition is. A similar argument shows the associative property of matrix addition also holds, inher...
A and let Cj denote the jth column of B. The product of A and B, denoted AB, is the matrix deﬁned by that is AB = [Ri · Cj]m×r      AB = R1 · C1 R1 · C2 R2 · C1 R2 · C2 . . . R1 · Cr . . . R2 · Cr ... ... ... Rm · C1 Rm · C2 . . . Rm · Cr      There are a number of subtleties in Deﬁnition 8.10 which warrant c...
n the country of Sasquatchia, all 150 residents get one of the two local newspapers. Market research has shown that in any given week, 90% of those who subscribe to the Pedimaxus Tribune want to keep getting it, but 10% want to switch to the Sasquatchia Picayune. Of those who receive the Picayune, 80% want to continue ...
1 0 17 0 1 − 3 17 which gives x1 = 4 17 . To solve the second system, we use the exact same row operations, in the same order, to put its augmented matrix into reduced row echelon form (Think about why that works.) and we obtain 17 and x3 = − 3 2 −3 0 4 1 3 Gauss Jordan Elimination −−−−−−−−−−−−−−−−→ 1 0 0 1 3 17 2 17 w...
Ohm’s Law and Kirchhoﬀ’s Voltage Law, we can relate the voltage supplied to the circuit by the two batteries to the voltage drops across the six resistors in order to ﬁnd the four ‘mesh’ currents: i1, i2, i3 and i4, measured in milliamps, mA. If we think of electrons ﬂowing through the circuit, we can think of the volt...
your work to Exercise 31 in Section 5.2. 612 Systems of Equations and Matrices 8.4.2 Answers 1. A−1 = −2 3 1 2 − 1 2 3. C is not invertible 5. E−1 =   −1 8 4 1 −3 −1 1 −6 −3   2. B−1 = 3 7 5 12 4. D−1 = 9 2 − 1 2 8 −1 6. F −    16 3 0 2 − 35 −90 − 1 2 0 1 5 0 −7 −36      0 7 2 0 1 7. G is not invertible 8. ...
te matrices can vary from the determinant of A by at most a nonzero multiple. This means that if det(A) = 0, then the determinant of A’s reduced row echelon form must also be nonzero, which, according to Deﬁnition 8.4 means that all the main diagonal entries on A’s reduced row echelon form must be 1. That is, A’s reduc...
d of card does he own? 20. How much of a 5 gallon 40% salt solution should be replaced with pure water to obtain 5 gallons of a 15% solution? 21. How much of a 10 liter 30% acid solution must be replaced with pure acid to obtain 10 liters of a 50% solution? 22. Daniel’s Exotic Animal Rescue houses snakes, tarantulas an...
this opening example, resolving a rational function into partial fractions takes two steps: ﬁrst, we need to determine the form of the decomposition, and then we need to determine the unknown coeﬃcients which appear in said form. Theorem 3.16 guarantees that any polynomial with real coeﬃcients can be factored over the ...
7. 2x x2 − 1 11x2 − 5x − 10 5x3 − 5x2 −x2 + 15 4x4 + 40x2 + 36 5x4 − 34x3 + 70x2 − 33x − 19 (x − 3)2 −7x2 − 76x − 208 x3 + 18x2 + 108x + 216 4x3 − 9x2 + 12x + 12 x4 − 4x3 + 8x2 − 16x + 16 8. 10. 12. 14. 16. 18. −7x + 43 3x2 + 19x − 14 −2x2 + 20x − 68 x3 + 4x2 + 4x + 16 −21x2 + x − 16 3x3 + 4x2 − 3x + 2 x6 + 5x5 + 16x4 ...
e the previous problem, there seems to be no avoiding substitution and a bit of algebraic unpleasantness. Solving y + 4e2x = 1 for y, we get y = 1 − 4e2x which, when substituted into the second equation, yields 1 − 4e2x2 + 2ex = 1. After expanding and gathering like terms, we get 16e4x − 8e2x + 2ex = 0. Factoring gives...
rices 8.7.1 Exercises In Exercises 1 - 6, solve the given system of nonlinear equations. Sketch the graph of both equations on the same set of axes to verify the solution set. x2 − y = 4 x2 + y2 = 4 x2 + y2 = 16 9x2 − 16y2 = 144 1. 4. 2. 5. x2 + y2 = 4 x2 − y = 5 x2 + y2 = 16 16 x2 = 1 9 y2 − 1 1 x2 + y2 = 16 16x2 + 4y...
in the same way we graph functions. For example, if we wish to graph the sequence {bk}∞ k=0 from Example 9.1.1, we graph the equation y = b(k) for the values k ≥ 0. That is, we plot the points (k, b(k)) for the values of k in the domain, k = 0, 1, 2, . . .. The resulting collection of points is the graph of the sequen...
rom one term to the next by subtracting 4 which is the same as adding −4. This means we have an arithmetic sequence on our hands. Using Equation 9.1 with a = 5 and d = −4, we get the nth denominator by the formula dn = 5 + (n − 1)(−4) = 9 − 4n for n ≥ 1. Our ﬁnal answer is an = 2 9−4n , n ≥ 1. 3. The sequence as given ...
009 + . . . 0 zeros + − . . . + 1 117 1 2 + − Solution. 1. (a) We substitute k = 1 into the formula 13 100k and add successive terms until we reach k = 4. 4 k=1 13 100k = 13 1002 + 13 1001 + 13 1003 + = 0.13 + 0.0013 + 0.000013 + 0.00000013 = 0.13131313 13 1004 (b) Proceeding as in (a), we replace every occurrence of n...
5.75 Our ﬁnal answer is $50,225.75. 2. To ﬁnd how long it will take for the annuity to grow to $100,000, we set A = 100000 and solve for t. We isolate the exponential and take natural logs of both sides of the equation. 100000 = 50 (1 + 0.005)12t − 1 0.005 10 = (1.005)12t − 1 (1.005)12t = 11 ln (1.005)12t = ln(11) 12t ...
on. 1. The sum formula for arithmetic sequences: n (a + (j − 1)d) = j=1 2. For a complex number z, (z)n = zn for n ≥ 1. n 2 (2a + (n − 1)d). 3. 3n > 100n for n > 5. 4. Let A be an n × n matrix and let A be the matrix obtained by replacing a row R of A with cR for some real number c. Use the deﬁnition of determinant to ...
he ﬁrst row and ﬁrst column of A. Since A is lower triangular, so is A11 and, as such, the induction hypothesis applies to A11. In other words, det (A11) is the product of the entries along A11’s main diagonal. Now, the entries on the main diagonal of A11 are the entries a22, a33, . . . , a(k+1)(k+1) from the main diag...
b = a + b Now we assume that P (k) is true. That is, we assume that we can expand (a + b)k using the formula given in Theorem 9.4 and attempt to show that P (k + 1) is true. 4and a fair amount of tenacity and attention to detail. 686 Sequences and the Binomial Theorem (a + b)k+1 = (a + b)(a + b)k k = (a + b) k j ak−jb...
om 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 addition, ...
modulus of a complex number deﬁnition of, 991 properties of, 993 multiplicity eﬀect on the graph of a polynomial, 245, 249 of a zero, 244 natural base, 420 natural logarithm, 422 natural number deﬁnition of, 2 set of, 2 negative angle, 698 Newton’s Law of Cooling, 421, 474 Newton’s Law of Universal Gravitation, 351 obl...
ace the endpoint, we proceed as we did in Example 10.1.1, successively halving the angle measure until we ﬁnd 5π 8 ≈ 1.96 which tells us our arc extends just a bit beyond the quarter mark into Quadrant III. 706 Foundations of Trigonometry 4. Since 117 is positive, the arc corresponding to t = 117 begins at (1, 0) and p...
d in radians, and subtended arc of length s. Prove that the area of the shaded sector is A = 1 (Hint: Use the proportion s circumference of the circle .) A area of the circle = 2 r2θ. r s θ r 20Source: Cedar Point’s webpage. 10.1 Angles and their Measure 711 In Exercises 57 - 62, use the result of Exercise 56 to comput...
t on the terminal side of θ which lies on the Unit Circle. Note that the terminal side of θ lies π 6 radians short of one half revolution. In Example 10.2.1, we determined that cos π 2 . This means 6 2 and sin π = 1 = √ 3 6 4This is unfortunate from a ‘function notation’ perspective. See Section 10.6. 5See Sections 1.1...
forth in the previous examples, we ﬁnd θ = π and θ = 3π θ = π 2 isn’t a reference angle we can nonetheless use it to ﬁnd our 2 + 2πk 2 + 2πk for integers, k. While this solution is correct, it can be shortened to 2 + πk for integers k. (Can you see why this works from the diagram?) One of the key items to take from Exa...
osine and Sine 735 Theorem 10.5. Domain and Range of the Cosine and Sine Functions: • The function f (t) = cos(t) • The function g(t) = sin(t) – has domain (−∞, ∞) – has range [−1, 1] – has domain (−∞, ∞) – has range [−1, 1] Suppose, as in the Exercises, we are asked to solve an equation such as sin(t) = − 1 2 . As we ...
7 in feet and t is measured in seconds radians 127 t. Here x and y are measured 744 Foundations of Trigonometry 10.3 The Six Circular Functions and Fundamental Identities In section 10.2, we deﬁned cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the...
worth memorizing and they, along with some of their other common forms, are summarized in the following theorem. cos2(θ) = 1 10.3 The Six Circular Functions and Fundamental Identities 749 Theorem 10.8. The Pythagorean Identities: 1. cos2(θ) + sin2(θ) = 1. Common Alternate Forms: 1 − sin2(θ) = cos2(θ) 1 − cos2(θ) = sin...
he Clocktower 2. Sketching the problem situation below, we ﬁnd ourselves with two unknowns: the height h of the tree and the distance x from the base of the tree to the ﬁrst observation point. 7Named in honor of Raymond Q. Armington, Lakeland’s Clocktower has been a part of campus since 1972. 10.3 The Six Circular Func...
cent to θ? 71. If θ = 15◦ and the hypotenuse has length 10, how long is the side opposite θ? 72. If θ = 87◦ and the side adjacent to θ has length 2, how long is the side opposite θ? 73. If θ = 38.2◦ and the side opposite θ has lengh 14, how long is the hypoteneuse? 74. If θ = 2.05◦ and the hypotenuse has length 3.98, h...
n θ = 50. cot(θ) = 0 when θ = π 6 π 2 51. tan(θ) = −1 when θ = + 2πk or θ = 5π 6 + 2πk for any integer k. + πk for any integer k 3π 4 + πk for any integer k 52. sec(θ) = 0 never happens 53. csc(θ) = − 1 2 never happens 54. sec(θ) = −1 when θ = π + 2πk = (2k + 1)π for any integer k 55. tan(θ) = − √ 3 when θ = 2π 3 + πk ...
m 10.15. Sum and Diﬀerence Identities for Sine: For all angles α and β, sin(α + β) = sin(α) cos(β) + cos(α) sin(β) sin(α − β) = sin(α) cos(β) − cos(α) sin(β) 774 Example 10.4.2. 1. Find the exact value of sin 19π 12 Foundations of Trigonometry 2. If α is a Quadrant II angle with sin(α) = 5 13 , and β is a Quadrant III ...
t into the identity we are asked to prove. The identity we are asked to start with is sin(2θ) = 2 tan(θ) 1+tan2(θ) . If we are to use this to derive an identity for tan θ , it seems reasonable to proceed by replacing each occurrence of θ with θ 2 2 2 tan θ 2 1 + tan2 θ 2 2 tan θ 2 1 + tan2 θ 2 sin 2 θ 2 sin(θ) = = We n...
sin(α + β) = (d) cos(α − β) = (f) tan(α − β) = √ 2 √ 28 − 30 2 −4 + 7 30 √ √ 28 + 4 − 7 2 2 = − 24. (a) sin(α + β) = 16 65 (b) cos(α − β) = 33 65 (c) tan(α − β) = 63 + 100 √ 2 41 56 33 788 Foundations of Trigonometry 25. (a) csc(α − β) = − 5 4 (b) sec(α + β) = 125 117 (c) cot(α + β) = 117 44 39. cos(75◦) = 41. cos(67.5...
(5, 4) y 4 3 2 1 −1 −2 1 2 3 4 5 x One cycle is graphed on [1, 5] so the period is the length of that interval which is 4. One cycle of y = f (x). 2. Proceeding as above, we set the argument of the sine, π − 2x, equal to each of our quarter marks and solve for x. 794 Foundations of Trigonometry a π − 2x = a x π π − 2x...
his criteria is . We can easily φ = π check our answer using the sum formula for cosine 3 . Hence, one way to write f (x) as a sinusoid is f (x) = 2 cos 2x + π 3, we get A2 = 12+( √ 2 and sin(φ) = √ √ √ 3 f (x) = 2 cos 2x + π 3 = 2 cos(2x) cos π 3 cos(2x) 1 − sin(2x) 2 √ 3 sin(2x) = 2 = cos(2x) − − sin(2x) sin π 3 √ 3 ...
. y 1 −1 x 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π cot(x) undeﬁned (x, cot(x)) 1 0 −1 undeﬁned 1 0 −1 undeﬁned 1 3π 4 , 1 5π 3π 2 , 0 4 , −1 7π π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π x The graph of y = cot(x) over [0, 2π]. From these data, it clearly appears as if the period of cot(x) is π, and we leave it to the reader to prove ...
f (x) = √ 2 sin(x) + √ 2 cos(x) + 1 = 2 sin x + π 4 + 1 = 2 cos x + 7π 4 + 1 26. f (x) = 3 √ 3 sin(3x) − 3 cos(3x) = 6 sin 3x + 11π 6 = 6 cos 3x + 4π 3 27. f (x) = − sin(x) + cos(x) − 2 = √ 2 sin x + 3π 4 − 2 = √ x + 2 cos π 4 − 2 28. f (x) = − 1 2 sin(2x) − √ 3 2 cos(2x) = sin 2x + 4π 3 = cos 2x + 5π 6 29. f (x) = 2 ...
xt pair of functions we wish to discuss are the inverses of tangent and cotangent, which are named arctangent and arccotangent, respectively. First, we restrict f (x) = tan(x) to its fundamental cycle on − π to obtain f −1(x) = arctan(x). Among other things, note that the 2 , π 2 vertical asymptotes x = − π 2 and x = π...
tions on t. To relate sec(t) to tan(t), we use the identity 1 + tan2(t) = sec2(t). This is valid for all values of t under consideration, √ and when we substitute sec(t) = x, we get 1 + tan2(t) = x2. Hence, tan(t) = ± x2 − 1. 2 , π then then tan(t) ≥ 0; if, on the the other hand, t belongs to π If t belongs to 0, π 2 t...
us problem by letting 2 . Then t is a real number with csc(t) = − 3 t = arccsc − 3 2 . Since csc(t) < 0, we have 2 that π < θ ≤ 3π 2 , so t corresponds to a Quadrant III angle, θ. As above, we let α be 2 , which means α = arccsc 3 the reference angle for θ. Then 0 < α < π 2 radians. Since the argument of arccosecant is...
(see Section 10.6.3). With practice, the inverse trigonometric functions will become as familiar to you as the square root function. Speaking of practice . . . 10.6 The Inverse Trigonometric Functions 841 10.6.5 Exercises In Exercises 1 - 40, ﬁnd the exact value. 1. arcsin (−1) 2. arcsin − √ 3 2 5. arcsin (0) 6. arcsi...
(c) Use BAD to show that β = arctan(2) (d) Use BCD to show that γ = arctan(3) (e) Use the fact that O, B and C all lie on the x-axis to conclude that α + β + γ = π. Thus arctan(1) + arctan(2) + arctan(3) = π. 10.6 The Inverse Trigonometric Functions 849 10.6.6 Answers 1. arcsin (−1) = − π 2 2. arcsin − √ 3 2 = − π 3 3...
e cot(u) = c for c = 0, convert to tangent and solve as above. If c = 0, the solution to cot(u) = 0 is u = π 2 + πk for integers k. 2 and ﬁnd the solution x = π 2 , we know the solutions take the form u = π 6 + 2πk for integers k. How do we solve something like sin(3x) = 1 Using the above guidelines, we can comfortably...
in the interval [0, 2π). Verify your solutions on [0, 2π) graphically. 1. 3 sin3(x) = sin2(x) 3. cos(2x) = 3 cos(x) − 2 5. cos(3x) = cos(5x) 7. sin(x) cos x 2 + cos(x) sin x 2 = 1 Solution. 2. sec2(x) = tan(x) + 3 4. cos(3x) = 2 − cos(x) 6. sin(2x) = √ 3 cos(x) 8. cos(x) − √ 3 sin(x) = 2 1. We resist the temptation to...
s(x) as sin(2x) − cos(x) > 0 and let f (x) = sin(2x) − cos(x). Our original inequality is thus equivalent to f (x) > 0. The domain of f is all real numbers, so we can advance to ﬁnding the zeros of f . Setting f (x) = 0 yields sin(2x) − cos(x) = 0, which, by way of the double angle identity for sine, becomes 2 sin(x) c...
= u To check using our calculator, we need to graph y = 3 arcsec(2x − 1) + π. To do so, we make from Theorems 10.28 and 10.29.16 We see the graph use of the identity arcsec(u) = arccos 1 u of y = 3 arccos + π and the horizontal line y = 2π intersect at 3 1 2 = 1.5. 2x−1 16Since we are checking for solutions where arcse...
9. x = − 1 2 61. x = 2 3 63. x = 2 √ 2 65. x = ± √ 3 2 67. x = −1, 0 , 6π 7 , 4π 7 −1 + 2 8π 7 √ 5 , 10π 7 , 12π 7 , π 5 , 3π 5 , π, , 7π 5 ≈ 0.6662, π − arcsin 9π 5 −1 + 2 √ 5 ≈ 2.4754 60. x = −1 62. x = − √ 3 2 64. x = 6 66. x = 1 2 √ 68. x = − 3 69. [π, 2π] 71. 73. 0, π 3 ∪ π 4 , 3π 4 , 2π 3 5π 4 ∪ , 7π 4 4π 3 ∪ 5π ...
applications of sinusoids in Science and Engineering is the study of harmonic motion. The equations for harmonic motion can be used to describe a wide range of phenomena, from the motion of an object on a spring, to the response of an electronic circuit. In this subsection, we restrict our attention to modeling a simp...
+ 3) √ 2 cos(2t) + (t + 3) √ x(t) using a graphing utility. 2 sin(2t) in the form x(t) = A(t) sin(ωt + φ). Graph 3. Find the period of x(t) = 5 sin(6t) − 5 sin (8t). Graph x(t) using a graphing utility. Solution. √ √ √ 1. We start rewriting x(t) = 5e−t/5 cos(t) + 5e−t/5 3 sin(t) by factoring out 5e−t/5 from both terms ...
4. x(t) = 67.5 cos π 15 t − π 2 = 67.5 sin π 15 t 5. h(t) = 28 sin 2π + 30 2 3 t − π ft. and m = 5 16 slugs 6. (a) k = 5 lbs. (b) x(t) = sin 4t + π 2 . The object ﬁrst passes through the equilibrium point when t = π 8 ≈ 0.39 seconds after the motion starts. At this time, the object is heading upwards. (c) x(t) = √ 2 2 ...
with α = 30◦, or sin(γ) = 4 sin(30◦) case, we have sin(γ) 4 = sin(30◦) = 2 3 3 900 Applications of Trigonometry 3 3 : γ = arcsin 2 we must have that 0◦ < γ < 150◦. There are two angles γ that fall in this range and have radians ≈ 138.19◦. At radians ≈ 41.81◦ and γ = π − arcsin 2 sin(γ) = 2 3 this point, we pause to se...
, a = 33.33 4. α = 95◦, β = 62◦, a = 33.33 5. α = 117◦, a = 35, b = 42 6. α = 117◦, a = 45, b = 42 7. α = 68.7◦, a = 88, b = 92 8. α = 42◦, a = 17, b = 23.5 9. α = 68.7◦, a = 70, b = 90 10. α = 30◦, a = 7, b = 14 11. α = 42◦, a = 39, b = 23.5 12. γ = 53◦, α = 53◦, c = 28.01 13. α = 6◦, a = 57, b = 100 14. γ = 74.6◦, c ...
lent to 4.004◦ 22. arctan 7 100 23. About 17% 24. About 53 feet 11.2 The Law of Sines 909 25. (a) θ = 180◦ (b) θ = 353◦ (c) θ = 84.5◦ (d) θ = 270◦ (e) θ = 121.25◦ (f) θ = 197◦1848 (g) θ = 45◦ (h) θ = 225◦ 26. The Colonel is about 3193 feet from the campﬁre. Sarge is about 2525 feet to the campﬁre. 27. The distance from...
simplify computations, we start by manipulating the expression for A2. A2 = 2 ab sin(γ) 1 2 a2b2 sin2(γ) = = 1 4 a2b2 4 1 − cos2(γ) since sin2(γ) = 1 − cos2(γ). The Law of Cosines tells us cos(γ) = a2+b2−c2 2ab , so substituting this into our equation for A2 gives A2 = = = = = = = = 1 − 1 − 2 1 − cos2(γ) a2 + b2 − c2 2...
with anything in Mathematics, the more ways you have to look at something, the better. The authors encourage the reader to take time to think about both approaches to plotting points given in polar coordinates. 920 Applications of Trigonometry Essentially we are locating a point on the terminal side of 5π 6 which is 4 ...
r2 = (−3)2 + (−3)2 = 18 2. We ﬁnd 4 . Since Q lies in Quadrant III, 4 , which satisﬁes the requirement that 0 ≤ θ < 2π. Our ﬁnal answer is . To check, we ﬁnd x = r cos(θ) = (3 = −3 so r = ± tan(θ) = −3 we choose θ = 5π (r, θ) = 3 −3 = 1, which means θ has a reference angle of π 2. Since we are asked for r ≥ 0, we choo...
reason to pause. Are there points with coordinates (r, θ) which satisfy r2 = r2 + r cos(θ)2 but do not satisfy r = r2 + r cos(θ)? Suppose (r, θ) satisﬁes r2 = r2 + r cos(θ)2 . Then r = ± (r)2 + r cos(θ). If we have that r = (r)2+r cos(θ), we are done. What if r = − (r)2 + r cos(θ) = −(r)2−r cos(θ)? We claim that the co...
ndependent variable, r as the dependent variable, evaluate r = f (θ) at some ‘friendly’ values of θ and plot the resulting points.2 We generate the table below. θ 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π r = 6 cos(θ) 6 √ 3 2 3 √ √ (r, θ) (6, 0) 2, π 4 0, π 2 2, 3π 4 (−6, π) √ 2, 5π 4 0, 3π 2 √ 2, 7π 4 (6, 2π) 0 2 −3 √ −3 −6 ...
ﬁned. 950 Applications of Trigonometry examples, the lines θ = π serve as guides for us to draw the curve as is passes through the origin. 4 , which are the zeros of the functions r = ±4 4 and θ = 3π cos(2θ), θ = 3π 4 π 4 π 2 3π 4 4 3 π θ r = 4 cos(2θ) and r = −4cos(2θ) As we plot points corresponding to values of θ ou...
r, θ) which satisﬁes both r = 3 sin θ . Equating these two expressions for r gives 2 . While normally we discourage dividing by a variable = 3 cos θ the equation 3 sin θ 2 2 expression (in case it could be 0), we note here that if 3 cos θ = 0, then for our equation 2 = 0 as well. Since no angles have both cosine and si...
r understanding of the polar coordinate system, we would be derelict in our duties if we totally ignored the graphing calculator. Indeed, there are some important polar curves which are simply too difﬁcult to graph by hand and that makes the calculator an important tool for your further studies in Mathematics, Science ...
ion, we revisit our friends the Conic Sections which we began studying in Chapter 7. Our ﬁrst task is to formalize the notion of rotating axes so this subsection is actually a follow-up to Example 8.3.3 in Section 8.3. In that example, we saw that the graph of y = 2 x is actually a hyperbola. More speciﬁcally, it is th...
= 7 24 , we have tan(2θ) = 24 7 , which 7 . Using the double angle identity for tangent, we have gives 24 tan2(θ) + 14 tan(θ) − 24 = 0. Factoring, we get 2(3 tan(θ) + 4)(4 tan(θ) − 3) = 0 which gives tan(θ) = − 4 4 . While either of these values of tan(θ) satisﬁes the equation . To 4 , since this produces an acute ang...
reduces to 7.5 gives the eccentricity is e in this case as well. If e = 1, the equation r = . This is a parabola with vertex r = − d 2 , the focus is (0, 0), the focal diameter is 2d and the directrix is x = −d, as required. Hence, we have shown that in all cases, our ‘new’ understanding of ‘conic section’, ‘focus’, ‘e...
, 0) minor axis length 2 √ 3 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 12. r = 2 1+sin(θ) is a parabola directrix y = 2 , vertex (0, 1) focus (0, 0), focal diameter 4 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 990 Applications of Trigonometry 13. r = 4 1+3 cos(θ) is a hyperbola directrix x = 4 center 3 conjugate axis l...
means \|zn\| = \|z\|n is true for all natural numbers n. Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume w = 0 (so \|w\| = 0) and we get (z) 1 w Product Rule. = = \|z\| z w 1 w 5Since the absolute value \|x\| of a real number x can be viewed as the distance from x to 0...
Indeed, a lot of work was needed to convert the numbers z and w in Example 11.7.3 into polar form, compute their product, and convert back to rectangular form – certainly more work than is required to multiply out zw = (2 3) the old-fashioned way. However, Theorem 11.16 pays huge dividends when computing powers of comp...
is w1 2i i w0 −2 −1 0 1 2 Real Axis w2 −i −2i w3 The four fourth roots of z = −16 equally spaced 2π 4 = π 2 around the plane. We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calc...
have √ 66. Since z = −25i = 25cis 3π 2 we have w0 = 5cis 3π w1 = 2cis 5π 4 √ = − √ 2 − i 2 w1 = 5cis 7π 67. Since z = 1 + i √ 3 = 2cis π 3 we have w0 = √ 2cis w1 = √ 2cis 7π 68. Since z = 5 2 − 5 3 2 i = 5cis 5π 3 we have √ w0 = √ 5cis 5π 6 = − √ 15 2 + √ 5 2 i w1 = √ 5cis 11π 6 = √ 15 2 − √ 5 2 i 69. Since z = 64 = 6...
ectors u, v and w, (u + v) + w = u + (v + w). Identity Property: The vector 0 acts as the additive identity for vector addition. That is, for all vectors v. Inverse Property: Every vector v has a unique additive inverse, denoted −v. That is, for every vector v, there is a vector −v so that v + (−v) = (−v) + v = 0. 1016...
). (See the diagram below.) Thus ˆv = cos (120◦) , sin (120◦) = , so v = vˆv = √ − . 11Of course, to go from v = vˆv to ˆv = 1 v v, we are essentially ‘dividing both sides’ of the equation by the scalar v. The authors encourage the reader, however, to work out the details carefully to gain an appreciation of the proper...
angle with the positive x-axis 13. v = 2 3 ; when drawn in standard position v lies in Quadrant I and makes a 60◦ angle with the positive y-axis 14. v = 12; when drawn in standard position v lies along the positive y-axis 15. v = 4; when drawn in standard position v lies in Quadrant II and makes a 30◦ angle with the ne...
62.59 28. v ≈ 5164.62, 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 = 2...
d note that since v1 is and Q(1, m1 + b1). We let v1 = determined by two points on L1, it may be viewed as lying on L1. Hence it has the same direction as L1. Similarly, we get the vector v2 = 1, m2 which has the same direction as the line L2. Hence, L1 and L2 are perpendicular if and only if v1 ⊥ v2. According to Theo...
imal places. 24. In Exercise 61 in Section 11.8, two drunken college students have ﬁlled an empty beer keg with rocks which they drag down the street by pulling on two attached ropes. The stronger of the two students pulls with a force of 100 pounds on a rope which makes a 13◦ angle with the direction of motion. (In th...
hand, the parametrization In this case, it is often best involves the trigonometric functions, the strategy changes slightly. to solve for the trigonometric functions and relate them using an identity. We demonstrate these techniques in the following example. Example 11.10.2. Sketch the curves described by the followi...
s. Our ﬁnal answer to this problem is x = t5 + 2t + 1, y = t for −∞ < t < ∞. As in the previous problem, our solution is trivial to 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 ...
-coordinate up r units12 (by adding r to the expression for y). We get {x = −r sin(θ) + rθ, y = −r cos(θ) + r , which can be written as {x = r(θ − sin(θ)), y = r(1 − cos(θ)) . Since the motion starts at θ = 0 and proceeds indeﬁnitely, we set θ ≥ 0. We end the section with a demonstration of the graphing calculator. Exa...
≤ π 3 2 1 −3 −2 −1 1 2 3 x 11. x = −1 + 3 cos(t) y = 4 sin(t) for 0 ≤ t ≤ 2π 12. x = 3 cos(t) y = 2 sin(t) + 1 for π 2 ≤ t ≤ 2π y 4 3 2 1 1 2 x −4 −3 −2 −1 −1 −2 −3 −4 y 3 2 1 −3 −1 −1 1 3 x 11.10 Parametric Equations 1065 13. x = 2 cos(t) y = sec(t) y for 0 ≤ t < π 2 14. x = 2 tan(t) y = cot(t) y for 15. x = sec(t) y...
tion of, 360 properties of, 367 constant, 100, 156 continuous, 241 cost, 82 decreasing, 100 deﬁnition as a relation, 43 dependent variable of, 55 diﬀerence, 76 diﬀerence quotient, 79 domain, 45 even, 95 exponential, 418 Fundamental Graphing Principle, 93 identity, 168 increasing, 100 independent variable of, 55 inverse...
row echelon form, 569 row operations for a matrix, 568 scalar multiplication matrix associative property of, 581 deﬁnition of, 580 distributive properties of, 581 properties of, 581 vector associative property of, 1018 deﬁnition of, 1017 distributive properties of, 1018 properties of, 1018 scalar projection, 1039 seca...
1 2 3 4 x −1 −2 −3 −4 The horizontal number line is usually called the x-axis while the vertical number line is usually called the y-axis.6 As with the usual number line, we imagine these axes extending oﬀ indeﬁnitely in both directions.7 Having two number lines allows us to locate the positions of points oﬀ of the num...
M and Q. This suﬃces to show that Equation 1.2 gives the coordinates of the midpoint. Example 1.1.6. Find the midpoint of the line segment connecting P (−2, 3) and Q(1, −3). Solution. M = = = x0 + x1 , 2 (−2) + 1 2 y0 + y1 2 , 3 + (−3 The midpoint is − 1 2 , 0. We close with a more abstract application of the Midpoint ...
3 4 x −4 −3 −2 −1 1 2 3 4 x The graph of A The graph of HLS1 3. HLS2 is hauntingly similar to HLS1. In fact, the only diﬀerence between the two is that instead of ‘−2 ≤ x ≤ 4’ we have ‘−2 ≤ x < 4’. This means that we still get a horizontal line segment which includes (−2, 3) and extends to (4, 3), but we do not include...
t for symmetry. Plot additional points as needed to complete the graph. Solution. To look for x-intercepts, we set y = 0 and solve (x − 2)2 + y2 = 1 (x − 2)2 + 02 = 1 (x − 2)2 = 1 √ (x − 2)2 = 1 x − 2 = ±1 extract square roots x = 2 ± 1 x = 3, 1 We get two answers for x which correspond to two x-intercepts: (1, 0) and ...
s on the graph but (4, −4) is not) The graph is symmetric about the origin 1.2 Relations 45. y = √ x − 2 x-intercept: (2, 0) The graph has no y-intercepts √ 46-intercept: (−3, 0) y-intercept: (0, 2) 39 y (x, y) 0 (2, 0) 1 (3, 1) (6, 2) 2 3 (11, 3) x 2 3 6 11 10 11 x The graph is not symmetric about the x-axis (e.g. (3,...
umbers less than or equal to 4, or, in interval notation, (−∞, 4]. 1.3 Introduction to Functions 47 All functions are relations, but not all relations are functions. Thus the equations which described the relations in Section1.2 may or may not describe y as a function of x. The algebraic representation of functions is ...
12. We can check our formula by replacing x with 5 to get g(5) = 3(5) + 12 = 15 + 12 = 27 . Most of the functions we will encounter in College Algebra will be described using formulas like the ones we developed for f (x) and g(x) above. Evaluating formulas using this function notation is a key skill for success in this...
ity of the situation limits what g can be, and 7See Sections 2.5, 11.1, and 6.5, respectively. 8In Carl’s humble opinion, of course . . . 9You could get close... within a certain speciﬁed margin of error, perhaps. 10or, ‘explicit domain’ 11Maybe this means returning a pound of grapes? 1.4 Function Notation 61 these lim...
nd pens. Find and interpret C(0), C(2) and C(5). 70. Using data from the Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by F (t) = −0.0076t2 + 0.45t + 16, 0 ≤ t ≤ 28, where t is the number of years since 1980. Use your calculator to ﬁnd F ...
pens, it costs $2000. 70. F (0) = 16.00, so in 1980 (0 years after 1980), the average fuel economy of passenger cars in the US was 16.00 miles per gallon. F (14) = 20.81, so in 1994 (14 years after 1980), the average fuel economy of passenger cars in the US was 20.81 miles per gallon. F (28) = 22.64, so in 2008 (28 yea...
s convention later in the text,5 we will hold with tradition at this point and consider the price p as a function of the number of items sold, x. That is, we regard x as the independent variable and p as the dependent variable and speak of the price-demand function, p(x). Hence, p(x) returns the price charged per item ...
9. For f (x) = x2 and g(x) = 1 x2 (f + g)(2) = 17 4 (f g) 1 2 = 1 (f − g)(−1) = 0 f g (0) is undeﬁned. (g − f )(1) = 0 g f (−2) = 1 16 10. For f (x) = x2 + 1 and g(x) = 1 x2+1 (f + g)(2) = 26 5 (f g) 1 2 = 1 (f − g)(−1) = 3 2 f g (0) = 1 (g − f )(1) = − 3 2 g f (−2) = 1 25 11. For f (x) = 2x + 1 and g(x) = x − 2 (f + g...
lutions to the equation f (x) = 0. In other words, x is a zero of f if and only if (x, 0) is an x-intercept of the graph of y = f (x). Of the three symmetries discussed in Section 1.2.1, only two are of signiﬁcance to functions: symmetry about the y-axis and symmetry about the origin.2 Recall that we can test whether t...
1.10 apply? Now let’s turn our attention to a few of the points on the graph. Clearly the point (−2, 4.5) does not have the largest y value of all of the points on the graph of f − indeed that honor goes to (6, 5.5) − but (−2, 4.5) should get some sort of consolation prize for being ‘the top of the hill’ between x = −4...

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