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156 Chapter 5 • The Integral §5.5 Example 5.34 Evaluate Z1 0 dx px . y x 0 1 y = 1 px Solution: Since x = 0 is a vertical asymptote for y = 1 px, Z1 0 dx px = lim c→0+ Z1 c dx px = lim c→0+ à 2 p x ¯¯¯¯ 1 c ! = lim c→0+ (2 −2pc) = 2 −0 = 2 . This means that the area under the curve y = 1/px over the interval (0,1]—as s... | ElementaryCalculus_Page_166_Chunk3701 |
Improper Integrals • Section 5.5 157 The following test for convergence or divergence is sometimes helpful: Comparison Test for Improper Integrals: (a) If | f (x)| ≤g(x) for all x in [a,∞), and if R∞ a g(x)dx is convergent, then R∞ a f (x)dx is convergent. (b) If f (x) ≥g(x) ≥0 for all x in [a,∞), and if R∞ a g(x)dx is... | ElementaryCalculus_Page_167_Chunk3702 |
158 Chapter 5 • The Integral §5.5 Exercises A For Exercises 1-15, evaluate the given improper integral. 1. Z∞ 1 dx x3 2. Z1 0 dx 3px 3. Z∞ 0 e−x dx 4. Z∞ 0 e−2x dx 5. Z1 −1 dx x 6. Z∞ 0 xe−x2 dx 7. Z0 −∞ 2x dx 8. Zπ/2 0 tanx dx 9. Z1 0 ln x x dx 10. Z1 −1 dx p 1−x2 11. Z3 0 ⌈x⌉dx 12. Z∞ −∞ dx x2 + 4 13. Z1 0 dx (x−1)3 ... | ElementaryCalculus_Page_168_Chunk3703 |
CHAPTER 6 Methods of Integration 6.1 Integration by Parts In physics and engineering the Gamma function1 Γ(t), defined by Γ(t) = Z∞ 0 xt−1 e−x dx for all t > 0, has found many uses. Evaluating Γ(2) entails integrating the function f (x) = x e−x. No formula or substitution you have learned so far would be of help. Differ... | ElementaryCalculus_Page_169_Chunk3704 |
160 Chapter 6 • Methods of Integration §6.1 Integration by parts is just the Product Rule for derivatives in integral form, typically used when the integral R v du would be simpler than the original integral R u dv. Example 6.1 Use integration by parts to evaluate Z x e−x dx . Use the answer to evaluate Γ(2). Solution:... | ElementaryCalculus_Page_170_Chunk3705 |
Integration by Parts • Section 6.1 161 Example 6.2 showed the importance of an appropriate choice for u and dv. There are some rough guidelines for that choice—as in Example 6.1—but no rules that are guaranteed to always work. It might not be clear when you should even attempt integration by parts. Example 6.3 Evaluate... | ElementaryCalculus_Page_171_Chunk3706 |
162 Chapter 6 • Methods of Integration §6.1 Example 6.5 Evaluate Z x2 e−x dx . Solution: This integral appears similar to the one in Example 6.1, so choose dv = e−x dx and u = x2. Then du = 2xdx and v = R e−x dv = −e−x. Now integrate by parts: Z udv = uv − Z vdu Z x2 e−x dx = x2 · (−e−x) − Z −e−x ·2x dx = −x2 e−x + 2 Z... | ElementaryCalculus_Page_172_Chunk3707 |
Integration by Parts • Section 6.1 163 The above procedure is called the tabular method for integration by parts, since it can be shown in a table (the arrows indicate multiplication): u dv u1 dv1 u2 v1 (+) + u1 v1 u3 v2 (–) −u2 v2 u3 v3 (+) + u3 v3 u4 v4 (–) −u4 v4 ... ... ... The idea is to differentiate down the u c... | ElementaryCalculus_Page_173_Chunk3708 |
164 Chapter 6 • Methods of Integration §6.1 Integration by parts can sometimes result in the original integral reappearing, allowing it to be combined with the original integral. Example 6.7 Evaluate Z sec3 x dx . Solution: Let u = sec x and dv = sec2 x dx, so that du = sec x tan x dx and v = R dv = R sec2 xdx = tan x.... | ElementaryCalculus_Page_174_Chunk3709 |
Integration by Parts • Section 6.1 165 In Example 6.1 integration by parts was used in evaluating an improper integral. In general, in definite or improper integrals where a and b are real numbers or ±∞, Zb a u dv = uv ¯¯¯¯ b a − Zb a v du . Example 6.9 Evaluate Z1 0 x3p 1−x2 dx . Solution: Since x3p 1−x2 = x2 · x p 1−x... | ElementaryCalculus_Page_175_Chunk3710 |
166 Chapter 6 • Methods of Integration §6.1 28. For the Gamma function Γ(t) show the following: (a) Γ(t+1) = tΓ(t) for all t > 0. (Hint: Use integration by parts.) (b) Γ(n) = (n−1)! for all positive integers n. (Hint: Use part (a) and induction.) Note that by part (b) the Gamma function can be thought of as an extensio... | ElementaryCalculus_Page_176_Chunk3711 |
Trigonometric Integrals • Section 6.2 167 6.2 Trigonometric Integrals In engineering applications you sometimes encounter integrals of the form Z cos (αt+φ1) cos (βt+φ2) dt where αt + φ1 and βt + φ2 are different angles (e.g. when the voltage and current are out of phase in an AC circuit). In general, integrals involvi... | ElementaryCalculus_Page_177_Chunk3712 |
168 Chapter 6 • Methods of Integration §6.2 On occasion you might need to integrate trigonometric functions raised to powers higher than two. For the sine function raised to odd powers of the form 2n +1 (for n ≥1), the trick is to replace sin2 x by 1−cos2 x, so that Z sin2n+1 x dx = Z (sin2 x)n sin x dx = Z (1−cos2 x)n... | ElementaryCalculus_Page_178_Chunk3713 |
Trigonometric Integrals • Section 6.2 169 Integrals of the form R sinm x cosn x dx, where either m or n is odd, can be evaluated using the above trick for the function having the odd power. Example 6.12 Evaluate Z sin2 x cos3 x dx . Solution: Replace cos2 x by 1−sin2 x, then let u = sin x so that du = cos x dx: Z sin2 ... | ElementaryCalculus_Page_179_Chunk3714 |
170 Chapter 6 • Methods of Integration §6.2 Similar methods can be used for integrals of the form R secm x tann x dx when either m is even or n is odd. For an even power m = 2k +2, use sec2 x = 1+tan2 x for all but two of the m powers of sec x, then use the substitution u = tan x, so that du = sec2 x dx. This results i... | ElementaryCalculus_Page_180_Chunk3715 |
Trigonometric Integrals • Section 6.2 171 For some trigonometric integrals try putting everything in terms of sines and cosines. Example 6.15 Evaluate Z cot4 x csc5 x dx . Solution: Put cot x and csc x in terms of sin x and cos x: Z cot4 x csc5 x dx = Z cos4 x sin5 x sin4 x dx = Z cos4 x sin x dx (now let u = cos x, du... | ElementaryCalculus_Page_181_Chunk3716 |
172 Chapter 6 • Methods of Integration §6.3 6.3 Trigonometric Substitutions One of the fundamental formulas in geometry is for the area A of a circle of radius r: A = πr2. The calculus-based proof of that formula uses a definite integral evaluated by means of a trigonometric substitution, as will now be demonstrated. Us... | ElementaryCalculus_Page_182_Chunk3717 |
Trigonometric Substitutions • Section 6.3 173 For an indefinite integral of the general form Rp a2 −u2 du, the same calculation as above with the substitutions u = acos θ and du = −asin θ dθ yields Zp a2 −u2 du = Zp a2 −a2 cos2 θ (−a sin θ) dθ = −a2 Z sin2 θ dθ = −a2 Z 1−cos 2θ 2 dθ = −a2 2 θ + a2 sin 2θ 4 + C , which i... | ElementaryCalculus_Page_183_Chunk3718 |
174 Chapter 6 • Methods of Integration §6.3 Example 6.16 Evaluate Zp 9−4x2 dx . Solution: The integrand is of the form p a2 −u2 with a = 3 and u = 2x, so that du = 2dx. Then dx = 1 2 du and so: Zp 9−4x2 dx = 1 2 Zp a2 −u2 du = 1 2 µ a2 2 sin−1 ³u a ´ + 1 2 u p a2 −u2 ¶ + C = 9 4 sin−1 µ2x 3 ¶ + 1 2 x p 9−4x2 + C In gen... | ElementaryCalculus_Page_184_Chunk3719 |
Trigonometric Substitutions • Section 6.3 175 Example 6.17 Evaluate Z dx (1+ x2)2 . Solution: Notice that this integral cannot be evaluated by using the Power Formula with the substi- tution u = 1 + x2 (why?). Integration by parts does not look promising, either. So try a trigonometric substitution. The integrand conta... | ElementaryCalculus_Page_185_Chunk3720 |
176 Chapter 6 • Methods of Integration §6.3 Example 6.18 Evaluate Z dx (4x2 +8x−5)3/2 . Solution: This integral cannot be evaluated by using the Power Formula, so try a trigonometric substi- tution. Complete the square on the expression 4x2 +8x−5: 4x2 + 8x −5 = 4(x2 +2x) −5 = 4(x2 +2x+1) −5 −4 = 4(x+1)2 −9 This express... | ElementaryCalculus_Page_186_Chunk3721 |
Trigonometric Substitutions • Section 6.3 177 Exercises A For Exercises 1-16, evaluate the given integral. 1. Zp 9+4x2 dx 2. Zp 2−3x2 dx 3. Zp 4x2 −9 dx 4. Zp x2 +2x+10 dx 5. Z p 1−x2 x2 dx 6. Z x2 dx p x2 −9 7. Z dx x p 1+ x2 8. Z dx x2p a2 + x2 (a > 0) 9. Z x3 dx p x2 +4 10. Z dx (4x2 −9)3/2 11. Z dx (9+4x2)2 12. Z x... | ElementaryCalculus_Page_187_Chunk3722 |
178 Chapter 6 • Methods of Integration §6.4 6.4 Partial Fractions In the last two sections some trigonometric integrals were simplified by using various trigono- metric identities. For integrals of rational functions—quotients of polynomials—some alge- braic identities (e.g. x2 −a2 = (x −a)(x + a)) will be useful in the... | ElementaryCalculus_Page_188_Chunk3723 |
Partial Fractions • Section 6.4 179 The partial fraction method can be discussed in general, and its assumptions proved4, but only the simplest cases—linear and quadratic factors— will be considered here. In all cases it will be assumed that the degree of the polynomial in the numerator of the rational function is less... | ElementaryCalculus_Page_189_Chunk3724 |
180 Chapter 6 • Methods of Integration §6.4 If one linear factor in the denominator is repeated more than once, and all other factors are distinct, then use the following decomposition: Case 2 - One repeated linear factor + distinct linear factors: A rational function p(x) q(x) such that degree(p(x)) < degree(q(x)), wh... | ElementaryCalculus_Page_190_Chunk3725 |
Partial Fractions • Section 6.4 181 Example 6.21 Evaluate Z dx x2(x+1)2 . Solution: Expanding Case 2 to two repeated factors, 1 x2(x+1)2 = A x + B x2 + C x+1 + D (x+1)2 = Ax(x+1)2 + B(x+1)2 + Cx2(x+1) + Dx2 x2(x+1)2 = (A +C)x3 + (2A +B +C + D)x2 + (A +2B)x + B x2(x+1) so that constant term : B = 1 coefficient of x : A +... | ElementaryCalculus_Page_191_Chunk3726 |
182 Chapter 6 • Methods of Integration §6.4 Example 6.22 Evaluate Z dx (x2 +1)(x2 +4) . Solution: Neither x2 +1 nor x2 +4 has real roots, so by Case 3, 1 (x2 +1)(x2 +4) = Ax+B x2 +1 + Cx+ D x2 +4 = (Ax+B)(x2 +4) + (Cx+ D)(x2 +1) (x2 +1)(x2 +4) = (A +C)x3 + (B + D)x2 + (4A +C)x + (4B + D) (x2 +1)(x2 +4) so that coefficie... | ElementaryCalculus_Page_192_Chunk3727 |
Partial Fractions • Section 6.4 183 Example 6.23 Evaluate Z dx (x2 +1)2 (x2 +4) . Solution: Neither x2 +1 nor x2 +4 has real roots, and x2 +1 is repeated, so by Case 4, 1 (x2 +1)2 (x2 +4) = Ax+B x2 +1 + Cx+ D (x2 +1)2 + Ex+ F x2 +4 = (Ax+B)(x2 +1)(x2 +4) + (Cx+ D)(x2 +4) + (Ex+ F)(x2 +1)2 (x2 +1)(x2 +4) with the right ... | ElementaryCalculus_Page_193_Chunk3728 |
184 Chapter 6 • Methods of Integration §6.4 Exercises A For Exercises 1-12, evaluate the given integral. 1. Z dx x2 −x 2. Z x+1 x2 −x dx 3. Z dx 2x2 +3x−2 4. Z dx x2 + x−6 5. Z dx x4 −x2 6. Z x (x−2)3 dx 7. Z x−1 x2(x+1) dx 8. Z x2 (x−1)2 dx 9. Z x−2 x2(x−1)2 dx 10. Z dx x4 + x2 11. Z dx x4 +5x2 +4 12. Z (x−1)2 (x2 +1)... | ElementaryCalculus_Page_194_Chunk3729 |
Miscellaneous Integration Methods • Section 6.5 185 6.5 Miscellaneous Integration Methods The integration methods presented so far are considered “standard,” meaning every calculus student should know them. This section will discuss a few additional methods, some more common than others. One such method is the Leibniz ... | ElementaryCalculus_Page_195_Chunk3730 |
186 Chapter 6 • Methods of Integration §6.5 What was actually done in the above example? A known integral, Z eαx dx = 1 α eαx + C , was differentiated with respect to α via the Leibniz rule to produce a new integral, Z x eαx dx = 1 α x eαx − 1 α2 eαx + C , with the constant α treated temporarily—only during the differe... | ElementaryCalculus_Page_196_Chunk3731 |
Miscellaneous Integration Methods • Section 6.5 187 Example 6.25 Show that Z∞ 0 e−x2 dx = 1 2 p π . Solution: Let I = R∞ 0 e−x2 dx. The integral is convergent, since by Exercise 11 in Section 4.4, for all x ex2 ≥1 + x2 ⇒ 0 ≤e−x2 ≤ 1 1+ x2 implies I is convergent by the Comparison Test, since R∞ 0 1 1+x2 dx is convergen... | ElementaryCalculus_Page_197_Chunk3732 |
188 Chapter 6 • Methods of Integration §6.5 One immediate consequence of Example 6.25 is that Z∞ −∞ e−x2 dx = p π since e−x2 is an even function. The following example shows another consequence, as well as how useful substitutions can be in writing integrals in a different form. Example 6.26 Show that the Gamma functio... | ElementaryCalculus_Page_198_Chunk3733 |
Miscellaneous Integration Methods • Section 6.5 189 Another application of substitutions in integrals is in the evaluation of fractional deriva- tives. Recall from Section 1.6 that the zero-th derivative of a function is just the function itself, and that derivatives of order n are well-defined for integer values n ≥1. ... | ElementaryCalculus_Page_199_Chunk3734 |
190 Chapter 6 • Methods of Integration §6.5 Recall from Section 6.3 that the trigonometric substitution x = r cos θ—or its sister substi- tution x = r sin θ—was motivated by trying to find the area of a circle of radius r. To simplify matters, let r = 1 so that points on the unit circle can be identified with the angle θ... | ElementaryCalculus_Page_200_Chunk3735 |
Miscellaneous Integration Methods • Section 6.5 191 Figure 6.5.2 shows only positive slopes—reflect the picture about the x-axis for negative slopes. The figure shows that sin 1 2θ = t p 1+ t2 and cos 1 2θ = 1 p 1+ t2 so that by the double-angle identities for sine and cosine, sin θ = 2 sin 1 2θ cos 1 2θ = 2 t p 1+ t2 1 ... | ElementaryCalculus_Page_201_Chunk3736 |
192 Chapter 6 • Methods of Integration §6.5 Example 6.30 Evaluate Z dθ 3 sin θ + 4 cos θ . Solution: Using t = tan 1 2θ, the integral becomes Z dθ 3 sin θ + 4 cos θ = Z 2dt 1+t2 3 2t 1+t2 + 4 1−t2 1+t2 = Z −1 2t2 −3t−2 dt = Z −1 (2t+1)(t−2) dt = Zµ A 2t+1 + B t−2 ¶ dt where coefficient of t : A + 2B = 0 ⇒ A = −2B consta... | ElementaryCalculus_Page_202_Chunk3737 |
Miscellaneous Integration Methods • Section 6.5 193 Exercises A For Exercises 1-12, evaluate the given integral. 1. Z 1 −2 cos θ sin θ dθ 2. Z dθ 3 −5 sin θ 3. Z dθ 2 −sin θ 4. Z dθ 4 + sin θ 5. Z sin θ 2 −sin θ dθ 6. Z dθ 5 −3 cos θ 7. Z dθ 1 + sin θ −cos θ 8. Z dθ 1 −sin θ + cos θ 9. Z cot θ 1 + sin θ dθ 10. Z 1 −cos... | ElementaryCalculus_Page_203_Chunk3738 |
194 Chapter 6 • Methods of Integration §6.5 19. Show that Z∞ 0 ln x 1+ x2 dx = 0. 20. Show that Z∞ 0 xa ax dx = Γ(a+1) (ln a)a+1 for a > 1. 21. Use the result from Example 6.28 to show that d1/2 dx1/2 µ d1/2 dx1/2 (x) ¶ = 1 = d dx (x) . 22. Calculate d1/2 dx1/2 (c) for all constants c. 23. Calculate d1/3 dx1/3 (x) . 24... | ElementaryCalculus_Page_204_Chunk3739 |
Numerical Integration Methods • Section 6.6 195 6.6 Numerical Integration Methods Section 5.2 showed how to obtain exact values for definite integrals of some simple functions (low-degree polynomials) by using areas of rectangles. For functions with no closed-form an- tiderivative, the rectangle method typically produce... | ElementaryCalculus_Page_205_Chunk3740 |
196 Chapter 6 • Methods of Integration §6.6 The syntax in the above commands can be explained with some examples. The following command creates 4 equally spaced points in the interval [1,7] (including x = 1 and x = 7), thus dividing [1,7] into 3 subintervals each of length (7−1)/3 = 2: octave> linspace(1,7,4) ans = 1 3... | ElementaryCalculus_Page_206_Chunk3741 |
Numerical Integration Methods • Section 6.6 197 Before the advent of modern computing, the rectangle method was considered inefficient, and so alternative methods were created. Two such methods are the trapezoid rule and Simpson’s rule. The idea behind both methods is to take advantage of a nonlinear function’s changing... | ElementaryCalculus_Page_207_Chunk3742 |
198 Chapter 6 • Methods of Integration §6.6 Example 6.32 Approximate the value of Zpπ 0 sin(x2) dx by using the trapezoid rule and Simpson’s rule with n = 105 subintervals. Solution: Since x0 = 0 and xn = pπ, then y0 = sin(x2 0) = sin 0 = 0 and yn = sin(x2 n) = sin π = 0. Thus, y0 and yn contribute nothing to the summa... | ElementaryCalculus_Page_208_Chunk3743 |
Numerical Integration Methods • Section 6.6 199 Notice that the approximations in the rectangle method, the trapezoid rule and Simpson’s rule can all be written as linear combinations of function values f (ai) multiplied by “weights” wi: Zb a f (x) dx ≈ n X i=0 wi f (ai) For example, the weights in Simpson’s rule are w... | ElementaryCalculus_Page_209_Chunk3744 |
200 Chapter 6 • Methods of Integration §6.6 Example 6.33 Approximate the value of Z2 0 dx 1+ x3 by using Gaussian quadrature with n = 4 points. Solution: For a = 0 and b = 2, use the substitution u = 1 b−a(2x−a−b) = x−1, so that x = u+1 and dx = du. Thus, g(u) = f (u+1) = 1 1+(u+1)3 . Using n = 4 in Table 6.1, the poin... | ElementaryCalculus_Page_210_Chunk3745 |
Numerical Integration Methods • Section 6.6 201 Example 6.34 Approximate the value of Z∞ 0 x5 e−x dx by using Gaussian quadrature with n = 3 points in Table 6.2. Solution: For n = 3, Table 6.2 gives a1 = 0.415775, a2 = 2.294280, a3 = 6.289945, and w1 = 0.711093, w2 = 0.278518, w3 = 0.010389. Then for f (x) = x5, Z∞ 0 x... | ElementaryCalculus_Page_211_Chunk3746 |
CHAPTER 7 Analytic Geometry and Plane Curves 7.1 Ellipses If you were to ask a random person “What is a circle?” a typical response would be to kick the can down the road: “Something that’s round.” There is a simple definition: A circle is the set of all points in a plane that are a fixed distance from a fixed point in th... | ElementaryCalculus_Page_212_Chunk3747 |
Ellipses • Section 7.1 203 The circle’s definition makes it easy to imagine its shape, especially for anyone who has drawn a circle with a compass. The definition of the ellipse, on the other hand, might not im- mediately suggest an “oval” shape. Its shape becomes apparent when physically constructing an ellipse by hand,... | ElementaryCalculus_Page_213_Chunk3748 |
204 Chapter 7 • Analytic Geometry and Plane Curves §7.1 Ellipses appear in nature (e.g. the orbits of planets around the Sun) and in many applica- tions. The ancient Greeks were able to derive many properties of the ellipse from its purely geometric definition.1 Nowadays those properties are typically derived using meth... | ElementaryCalculus_Page_214_Chunk3749 |
Ellipses • Section 7.1 205 x y −c c −a a 0 b −b x2 a2 + y2 b2 = 1 Figure 7.1.4 The graph of the resulting ellipse x2 a2 + y2 b2 = 1 with a > b > 0 and foci at (±c,0) is shown in Figure 7.1.4. Since the the x-axis is principal axis then the vertexes are found by setting y = 0: x = ±a. The vertexes are thus (±a,0), so th... | ElementaryCalculus_Page_215_Chunk3750 |
206 Chapter 7 • Analytic Geometry and Plane Curves §7.1 Example 7.1 Find the area inside the ellipse x2 a2 + y2 b2 = 1. Solution: By symmetry the area will be four times the area in the first quadrant. Solving for y in the equation of the ellipse gives y2 = b2 −b2x2 a2 ⇒ y = b s 1−x2 a2 = b a p a2 −x2 for the upper hemi... | ElementaryCalculus_Page_216_Chunk3751 |
Ellipses • Section 7.1 207 Notice that Fermat’s Principle is equivalent to saying that the angles α1 and α2 that the light’s path makes with the normal line through the point of reflection are equal, since each angle would equal 90◦−θ, as in Figure 7.1.8(a): tangent line normal line A B P θ θ α1 α2 (a) α1 = α2 x y F1 −c... | ElementaryCalculus_Page_217_Chunk3752 |
208 Chapter 7 • Analytic Geometry and Plane Curves §7.1 By the distance formula, the distance F1P from F1 = (−ea,0) to P = (x0, y0) is given by (F1P)2 = (x0 + ea)2 + y2 0 = x2 0 + 2eax0 + e2a2 + (1−e2)(a2 −x2 0) (by formula (7.1)) (F1P)2 = a2 + 2eax0 + e2x2 0 = (a+ ex0)2 F1P = a+ ex0 . Similarly, the distance F2P from ... | ElementaryCalculus_Page_218_Chunk3753 |
Ellipses • Section 7.1 209 Exercises A 1. Construct an ellipse using the procedure shown in Figure 7.1.2. Place the two pins 7in apart and use a 10in piece of string. For Exercises 2-6, sketch the graph of the given ellipse, indicate the major and minor axes and exact locations of the foci and vertexes, and find the ecc... | ElementaryCalculus_Page_219_Chunk3754 |
210 Chapter 7 • Analytic Geometry and Plane Curves §7.2 7.2 Parabolas As with ellipses, you have seen parabolas (e.g. y = x2) and some of their applications (e.g. projectile trajectories), but perhaps without knowing their purely geometric definition. The alternative definition of an ellipse described in Exercise 13 in S... | ElementaryCalculus_Page_220_Chunk3755 |
Parabolas • Section 7.2 211 x y (0, p) y = −p d1 d2 (x, y) (x,−p) 0 −p To derive the equation of a parabola in the xy-plane, start with the simple case of the focus on the y-axis at (0, p), with p > 0, and the line y = −p as the directrix, as in the figure on the right. The vertex is then at the origin (0,0). Pick a poi... | ElementaryCalculus_Page_221_Chunk3756 |
212 Chapter 7 • Analytic Geometry and Plane Curves §7.2 The slope of the parabola 4py = x2 is dy dx = 2x 4p = x 2p, so that the equation of the tangent line to the parabola at a point (x0, y0) is: y −y0 = x0 2p (x−x0) 2p(y−y0) = x0x −x2 0 2py −2py0 = x0x −4py0 2p(y+ y0) = x0x (7.4) Likewise, switching the roles of x an... | ElementaryCalculus_Page_222_Chunk3757 |
Parabolas • Section 7.2 213 Example 7.2 Suppose that an object is launched from the ground with an initial velocity v0 and at varying angles with the ground. Show that the family of all the possible trajectories—which are parabolic—form a region whose boundary (called the envelope of the trajectories) is itself a parab... | ElementaryCalculus_Page_223_Chunk3758 |
214 Chapter 7 • Analytic Geometry and Plane Curves §7.2 Now let P be a point in the first quadrant of the xy-plane below the common directrix y = v2 0 2g, denoted by D. Then P can be either inside, outside, or on the envelope, as in Figure 7.2.5: D y x P O v2 0 2g v2 0/g (a) Inside the envelope D y x P O v2 0 2g v2 0/g ... | ElementaryCalculus_Page_224_Chunk3759 |
Parabolas • Section 7.2 215 In Example 7.2 all the trajectories were in the xy-plane only. Removing that restriction, so that trajectories in all vertical planes through the y-axis are possible, would result in a solid paraboloid consisting of all possible trajectories from the origin. Parabolas also appear in suspensi... | ElementaryCalculus_Page_225_Chunk3760 |
216 Chapter 7 • Analytic Geometry and Plane Curves §7.3 7.3 Hyperbolas In the previous two sections you have seen curves with eccentricity e = 0 (circles), 0 < e < 1 (ellipses) and e = 1 (parabolas). The remaining case is e > 1: the hyperbola, whose definition is similar to the second definition of the ellipse. A hyperbo... | ElementaryCalculus_Page_226_Chunk3761 |
Hyperbolas • Section 7.3 217 D F P G Figure 7.3.3 Hyperbola: PF PG = e > 1 Figure 7.3.3 illustrates the definition of a hyperbola, con- sisting of points P whose distance PF from the focus F ex- ceeds the distance PG to the directrix D in a way so that the ratio PF PG is always the same constant e > 1 (the eccen- tricit... | ElementaryCalculus_Page_227_Chunk3762 |
218 Chapter 7 • Analytic Geometry and Plane Curves §7.3 In Figure 7.3.5 the vertexes are (±a,0), the x-axis is the transverse axis, the center is the origin (0,0), and the conjugate axis is the y-axis. Note that the existence of two foci and directrices—when the definition of the hyperbola mentioned only a focus and a d... | ElementaryCalculus_Page_228_Chunk3763 |
Hyperbolas • Section 7.3 219 There is another way to define a hyperbola, in terms of two foci: A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances from two fixed points (the foci) is a positive constant. Figure 7.3.7 illustrates the above definition with foci F1 ... | ElementaryCalculus_Page_229_Chunk3764 |
220 Chapter 7 • Analytic Geometry and Plane Curves §7.3 That slope will be used in proving the reflection property for the hyperbola: Light shone from one focus will reflect off the hyperbola in the opposite direction from the other focus. Figure 7.3.9 shows the light’s path from focus F2 as it reflects at the point P alo... | ElementaryCalculus_Page_230_Chunk3765 |
Hyperbolas • Section 7.3 221 Thus, since tan α2 = −tan(180◦−α2) = −y0 x−c, tan θ1 = −tan θ + tan α2 1 −tan θ tan α2 = − b2x0 a2 y0 + −y0 x0 −c 1 −b2x0 a2 y0 · −y0 x0 −c = − b2x2 0 −a2 y2 0 −b2x0c ✭✭✭✭✭ a2 y0 (x0 −c) (a2 + b2) x0 y0 −a2y0c ✭✭✭✭✭ a2 y0 (x0 −c) = −a2b2 −b2x0c c2x0y0 −a2y0c = b2✘✘✘✘✘ (a2 −x0c) cy0✘✘✘✘✘ (a2... | ElementaryCalculus_Page_231_Chunk3766 |
222 Chapter 7 • Analytic Geometry and Plane Curves §7.3 D F Q P G Pc P0 A α β β C Figure 7.3.13 For the case of a right circular double cone (i.e. the base of each nappe is a circle in a plane perpendicular to the axis of the cone7), let β be the complement of the angle that the cone makes with its axis, as in Figure 7... | ElementaryCalculus_Page_232_Chunk3767 |
Hyperbolas • Section 7.3 223 Exercises A 1. Construct a hyperbola using the procedure shown in Figure 7.3.8. Place the two focus pins 7in apart and use a 9in piece of string attached to a 12in ruler. For Exercises 2-6, sketch the graph of the given hyperbola, indicate the exact locations of the foci and vertexes, indic... | ElementaryCalculus_Page_233_Chunk3768 |
224 Chapter 7 • Analytic Geometry and Plane Curves §7.4 7.4 Translations and Rotations For convenience the ellipses, parabolas and hyperbolas in the previous sections were centered at the origin and had their foci on one of the coordinate axes. In general the center and foci of those curves can be moved anywhere by mea... | ElementaryCalculus_Page_234_Chunk3769 |
Translations and Rotations • Section 7.4 225 Parabola: For p ̸= 0, an equation of the form (x−h)2 = 4p(y−k) describes a parabola with vertex (h,k) and focus (h,k+ p). The directrix is the line y = k−p and the axis is the line x = h. Likewise, an equation of the form (y−k)2 = 4p(x−h) describes a parabola with vertex (h,... | ElementaryCalculus_Page_235_Chunk3770 |
226 Chapter 7 • Analytic Geometry and Plane Curves §7.4 Example 7.5 For a ̸= 0 and constants b and c, find the vertex, focus and directrix of the parabola y = ax2 + bx+ c. Solution: The idea here is to write y = ax2 + bx+ c in the form (x−h)2 = 4p(y−k) for some h, k, and p, by completing the square: ax2 + bx + c = y a µ... | ElementaryCalculus_Page_236_Chunk3771 |
Translations and Rotations • Section 7.4 227 Example 7.6 Find the equation of the ellipse x2 4 + y2 = 1 when rotated 45◦counterclockwise about the origin. Simplify the equation. x y x′ y′ 45◦ 0 2 −2 2 −2 Figure 7.4.4 Solution: For θ = 45◦the substitutions are: x 7→x cos θ + y sin θ = x cos 45◦+ y sin 45◦= x+ y p 2 y 7→... | ElementaryCalculus_Page_237_Chunk3772 |
228 Chapter 7 • Analytic Geometry and Plane Curves §7.4 The following result can be used for determining the type of conic section described by a second-degree equation:10 The graph of Ax2+Bxy+Cy2+Dx+Ey+F = 0 (with A, B, C not all zero) describes a curve whose type is based on the sign of B2 −4AC: (a) B2 −4AC < 0: an e... | ElementaryCalculus_Page_238_Chunk3773 |
Translations and Rotations • Section 7.4 229 x y x′ y′ 64.3◦ 0 3 −3 3 −3 Substitute those expressions into 5x2 +4xy+8y2 −36 = 0: 5 µ x′ −2y′ p 5 ¶2 + 4 µ x′ −2y′ p 5 ¶ µ2x′ + y′ p 5 ¶ + 8 µ2x′ + y′ p 5 ¶2 −36 = 0 45x′2 + 20y′2 = 180 x′2 4 + y′2 9 = 1 So the curve’s equation in the x′y′-plane is x′2 4 + y′2 9 = 1. This ... | ElementaryCalculus_Page_239_Chunk3774 |
230 Chapter 7 • Analytic Geometry and Plane Curves §7.5 7.5 Hyperbolic Functions In some textbooks you might see the sine and cosine functions called circular functions, since any point on the unit circle x2 + y2 = 1 can be defined in terms of those functions (see Figure 7.5.1). Those definitions motivate a similar idea ... | ElementaryCalculus_Page_240_Chunk3775 |
Hyperbolic Functions • Section 7.5 231 The y-coordinate of P can then be found: y = p x2 −1 = sµ ea + e−a 2 ¶2 −1 = s e2a +2+ e−2a 4 −4 4 = s e2a −2+ e−2a 4 = sµ ea −e−a 2 ¶2 , so since a ≥0 y = ea −e−a 2 = sinh a where sinh a is the hyperbolic sine of a. All six hyperbolic functions can now be defined in general, analo... | ElementaryCalculus_Page_241_Chunk3776 |
232 Chapter 7 • Analytic Geometry and Plane Curves §7.5 The hyperbolic functions satisfy the following identities: cosh2 x −sinh2 x = 1 tanh2 x + sech2 x = 1 coth2 x −csch2 x = 1 sinh(−x) = −sinh x cosh(−x) = cosh x tanh(−x) = −tanh x sech(−x) = sech x csch(−x) = −csch x coth(−x) = −coth x sinh(u ± v) = sinh u cosh v ±... | ElementaryCalculus_Page_242_Chunk3777 |
Hyperbolic Functions • Section 7.5 233 The derivatives of the hyperbolic functions and their integral equivalents are: d dx (sinh x) = cosh x d dx (csch x) = −csch x coth x d dx (cosh x) = sinh x d dx (sech x) = −sech x tanh x d dx (tanh x) = sech2 x d dx (coth x) = −csch2 x Z cosh x dx = sinh x + C Z csch x coth x dx ... | ElementaryCalculus_Page_243_Chunk3778 |
234 Chapter 7 • Analytic Geometry and Plane Curves §7.5 Example 7.11 In the classical theory of paramagnetism, the total number n of molecules in a gas subject to a magnetic field of strength H is n = 2πN Zπ 0 e µH kT cos θ sin θ dθ , where N is the number of molecules per unit solid angle having zero potential energy, ... | ElementaryCalculus_Page_244_Chunk3779 |
Hyperbolic Functions • Section 7.5 235 The inverse hyperbolic functions can be expressed in terms of the natural logarithm: sinh−1 x = ln(x+ p x2 +1) cosh−1 x = ln(x+ p x2 −1) for x ≥1 tanh−1 x = 1 2 ln 1+ x 1−x for |x| < 1 coth−1 x = 1 2 ln x+1 x−1 for |x| > 1 sech−1 x = ln 1+ p 1−x2 x for 0 < x ≤1 csch−1 x = ln à 1 x... | ElementaryCalculus_Page_245_Chunk3780 |
236 Chapter 7 • Analytic Geometry and Plane Curves §7.5 d dx (sinh−1 x) = 1 p x2 +1 d dx (cosh−1 x) = 1 p x2 −1 for x ≥1 d dx (tanh−1 x) = 1 1−x2 for |x| < 1 d dx (coth−1 x) = 1 1−x2 for |x| > 1 d dx (sech−1 x) = −1 x p 1−x2 for 0 < x ≤1 d dx (csch−1 x) = −1 |x| p 1+ x2 for x ̸= 0 For example, here is one way to find th... | ElementaryCalculus_Page_246_Chunk3781 |
Hyperbolic Functions • Section 7.5 237 23. Use Exercise 22 and the addition formula for sinh x to show that d dx (sinh x) = cosh x. 24. Sketch the graph of f (x) = e−2x sinh x. Find all local maxima and minima, inflection points, and vertical or horizontal asymptotes. 25. Denoting the speed of light by c, the Lorentz tr... | ElementaryCalculus_Page_247_Chunk3782 |
238 Chapter 7 • Analytic Geometry and Plane Curves §7.5 31. Continuing Example 7.11, the total magnetic moment M is defined as M = 2πNµ Zπ 0 e µH kT cos θ sin θ cos θ dθ . (a) Use the value of n from Example 7.11 to show that M nµ = L(a), where L(a) = coth a −1 a is the Langevin function and a = µH kT . (b) Show that fo... | ElementaryCalculus_Page_248_Chunk3783 |
Parametric Equations • Section 7.6 239 7.6 Parametric Equations Recall that Section 6.5 presented two different ways to “identify” or represent points on the unit circle—by angle and by slope, as in Figure 7.6.1: x y 1 θ 1 0 (x, y) = (cos θ,sin θ) (a) Identify points by angle θ x y slope = t 1 −1 θ (x, y) 0 (b) Identif... | ElementaryCalculus_Page_249_Chunk3784 |
240 Chapter 7 • Analytic Geometry and Plane Curves §7.6 Example 7.13 Show that for any constants ω ̸= 0 and r > 0, and for t measured in radians, x = h + r cos ωt and y = k + r sin ωt for −∞< t < ∞ is a parametrization of the circle (x−h)2 +(y−k)2 = r2 with center (h,k) and radius r. Solution: Since ωt is similar to th... | ElementaryCalculus_Page_250_Chunk3785 |
Parametric Equations • Section 7.6 241 Example 7.16 Bézier curves13 are used in Computer Aided Design (CAD) to join the ends of an open polygonal path of noncollinear control points with a smooth curve that models the “shape” of the path. The curve is created via repeated linear interpolation, illustrated in Figure 7.6... | ElementaryCalculus_Page_251_Chunk3786 |
242 Chapter 7 • Analytic Geometry and Plane Curves §7.6 A curve with parametric equations x = x(t) and y = y(t) might not be the graph of a single function y = f (x), but the derivative dy dx can still be found by using the differentials of x and y as functions of t: dy = y′(t)dt and dx = x′(t)dt, so that dy dx = y′(t)... | ElementaryCalculus_Page_252_Chunk3787 |
Parametric Equations • Section 7.6 243 As in formula (7.14) the derivative dy dx is given by: dy dx = y′(θ) x′(θ) = a sin θ a(1 −cos θ) = sin θ 1 −cos θ = cot 1 2θ Thus, dy dx is undefined when cos θ = 1, namely, when θ = 2πk for all integers k, i.e. when x = a(θ −sin θ) = a(2πk −sin 2πk) = 2πka. Notice from Figure 7.6.... | ElementaryCalculus_Page_253_Chunk3788 |
244 Chapter 7 • Analytic Geometry and Plane Curves §7.6 5. The parametrization C : x = 1−t2 1+t2 , y = 2t 1+t2 , −∞< t < ∞of the unit circle C shown earlier makes the unit circle a rational curve, since x and y are rational functions of the parameter t. Is the ellipse x2 a2 + y2 b2 = 1 a rational curve? Justify your an... | ElementaryCalculus_Page_254_Chunk3789 |
Polar Coordinates • Section 7.7 245 7.7 Polar Coordinates Suppose that you wanted to write the equation of a spiral, like the one in Figure 7.7.1. The curve is clearly not the graph of a function y = f (x) in Cartesian coordinates, as it violates the vertical line test. However, this spiral is simple to express using p... | ElementaryCalculus_Page_255_Chunk3790 |
246 Chapter 7 • Analytic Geometry and Plane Curves §7.7 You might be familiar with graphing paper, for plotting points or functions given in Cartesian coordinates. Such paper consists of a rectangular grid, where the horizontal and vertical lines represent where x and y, respectively, are constants, at regular interval... | ElementaryCalculus_Page_256_Chunk3791 |
Polar Coordinates • Section 7.7 247 In general, polar coordinates are useful in describing plane curves that exhibit symmetry about the origin (though there are other situations), which arise in many physical applications. r x y O θ y x (r,θ) (x, y) Figure 7.7.5 Figure 7.7.5 shows how to convert between polar coordinat... | ElementaryCalculus_Page_257_Chunk3792 |
248 Chapter 7 • Analytic Geometry and Plane Curves §7.7 Derivatives in Polar Coordinates Suppose that the polar coordinates (r,θ) for a plane curve are related by a function: r = r(θ). Then by formula (7.17), x = r(θ) cos θ and y = r(θ) sin θ are now parametric equations for the curve in the parameter θ. Thus, by the P... | ElementaryCalculus_Page_258_Chunk3793 |
Polar Coordinates • Section 7.7 249 Integration in Polar Coordinates In some cases polar coordinates can simplify evaluation of a definite integral or finding an area. To determine the polar form of a definite integral, suppose that r is a function of θ: r = f (θ). A polar region swept out by r = f (θ) between θ = α and θ... | ElementaryCalculus_Page_259_Chunk3794 |
250 Chapter 7 • Analytic Geometry and Plane Curves §7.7 Example 7.22 Use polar coordinates to show that the area of a circle of radius R is πR2. Solution: Let the origin be the center of the circle. Then r = R is the polar equation of the circle, with 0 ≤θ ≤2π sweeping out exactly one full circle. The area A inside the... | ElementaryCalculus_Page_260_Chunk3795 |
Polar Coordinates • Section 7.7 251 For Exercises 13-15, sketch the graph of the given curve and indicate all local maxima and minima. 13. r = 1+sin θ 14. r = 1−cos θ 15. r = sin 2θ 16. Find the area inside r = 1+sin θ. 17. Find the area inside r = sin 2θ. B 18. Sketch a rough graph of the meridian voltage component Eθ... | ElementaryCalculus_Page_261_Chunk3796 |
CHAPTER 8 Applications of Integrals 8.1 Area Between Curves The “area under a curve” was defined in Chapter 5 as the area below some curve y = f (x) and above the x-axis over some interval. That was a special case of the area between curves, where in general one curve y = f1(x) is not necessarily always above another cu... | ElementaryCalculus_Page_262_Chunk3797 |
Area Between Curves • Section 8.1 253 Example 8.1 h(x) y x 0 2 y = ex y = e−x Find the area between y = ex and y = e−x over [0,2]. Solution: Since ex ≥e−x for x in [0,2], the height function h(x) for the region between the curves over [0,2] is h(x) = ¯¯ex −e−x¯¯ = ex −e−x. The area A of the region is thus A = Z2 0 (ex ... | ElementaryCalculus_Page_263_Chunk3798 |
254 Chapter 8 • Applications of Integrals §8.1 Formula (8.1) can be extended to find the area between any number of curves, by splitting the integral over subintervals with different height functions. Example 8.4 Find the area of the region bounded by y = 6−x2, y = x and y = −5x above the x-axis. y = −5x y = x y x 0 −1 ... | ElementaryCalculus_Page_264_Chunk3799 |
Area Between Curves • Section 8.1 255 θ = α θ = β y x r1 r2 O Figure 8.1.2 The area between curves given by polar equations can be found similarly. For example, consider curves r = r1(θ) and r = r2(θ) with r1(θ) ≥r2(θ) when α ≤θ ≤β as in Figure 8.1.2. The area A of the region between the curves and those angles is simp... | ElementaryCalculus_Page_265_Chunk3800 |
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