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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+ Ã 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 shown in the graph above—equals 2. The region is inﬁnite in the y direction. Example 5.35 Evaluate Z3 1 ⌊x⌋dx . y x 0 1 2 3 1 2 y = ⌊x⌋ Solution: Recall from Example 3.22 in Section 3.3 that the ﬂoor function y = ⌊x⌋has jump discontinuities at each integer value of x, as shown in the graph on the right. The integral R3 1 ⌊x⌋dx is thus an improper integral over the interval [1,3), which needs to be split at the point of discontinuity x = 2 within that interval: Z3 1 ⌊x⌋dx = Z2 1 ⌊x⌋dx + Z3 2 ⌊x⌋dx = µ lim b→2− Zb 1 ⌊x⌋dx ¶ + µ lim c→3− Zc 2 ⌊x⌋dx ¶ = µ lim b→2− Zb 1 1 dx ¶ + µ lim c→3− Zc 2 2 dx ¶ = lim b→2− Ã x ¯¯¯¯ b 1 ! + lim c→3− µ 2x ¯¯¯¯ c 2 ¶ = lim b→2−(b−1) + lim c→3−(2c−4) = (2−1) + (6−4) = 3 Similar to some of the above examples, the following result is easy to prove (see the exercises): For any real number a > 0, the improper integral Z∞ a dx xp is convergent if p > 1, and divergent if 0 < p ≤1.	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 divergent, then R∞ a f (x)dx is divergent. The idea behind part (a) is that if −g(x) ≤f (x) ≤g(x) over [a,∞), then—thinking of improper integrals as areas—the integral of f is “squeezed” between the two ﬁnite integrals for ±g. There are, however, some subtle issues to prove about the limit in the integral of f —ﬁnite bounds might not necessarily mean the limit exists.5 Example 5.36 Show that Z∞ 1 sin x x2 dx is convergent. y x 0 1 y = 1 x2 y = −1 x2 sin x x2 Solution: By Example 5.29, the integral R∞ 1 1 x2 dx is convergent. So since \|sin x\| ≤1 for all x, then ¯¯¯¯ sin x x2 ¯¯¯¯ ≤ 1 x2 for all x in [1,∞). Thus, by the Comparison Test, R∞ 1 sin x x2 dx is convergent. The graph on the right shows how the curve y = sin x x2 is bounded between the curves y = ± 1 x2 . The rules and properties from Section 5.3 concerning deﬁnite integrals still apply to im- proper integrals, provided the improper integrals are convergent. For example, suppose a func- tion f has a discontinuity or vertical asymptote at x = c. If both improper integrals Rc a f (x)dx and Rb c f (x)dx are convergent, then the improper integral Rb a f (x)dx is convergent and Zb a f (x) dx = Zc a f (x) dx + Zb c f (x) dx . Likewise, if Rc a f (x)dx and R∞ c f (x)dx are convergent, then so is R∞ a f (x)dx, with Z∞ a f (x) dx = Zc a f (x) dx + Z∞ c f (x) dx . 5See pp.140-141 in BUCK, R.C., Advanced Calculus, 2nd ed., New York: McGraw-Hill Book Co., 1965.	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 14. Z∞ 2 dx x lnx 15. Z∞ 1 dx x p x2 −1 16. In a standby system of two non-identical components, the normal operating component A has a failure rate of λA > 0 failures per unit time, while the standby component B—which takes over when A fails—has a failure rate λB > 0 (with λA ̸= λB). (a) Find the standby system’s reliability R(t) beyond time t ≥0, where R(t) = Z∞ t λAλB λA −λB ³ e−λBx −e−λA x´ dx . (b) Show that the system’s mean time to failure (MTTF) m, where m = R∞ 0 R(t)dt , is m = 1 λA + 1 λB . 17. Show that for all a > 0, Z∞ a dx xp is convergent if p > 1, and divergent if 0 < p ≤1. 18. Show that for all a > 0, Za 0 dx xp is convergent if 0 < p < 1, and divergent if p ≥1. 19. Is Z∞ 1 dx x+ x4 convergent? Explain. 20. Is Z∞ 2 dx x−px convergent? Explain. B 21. Example 5.31 showed that R∞ 0 sinxdx is divergent. What is the ﬂaw in the argument that the integral must be 0 since each “hump” of sin x above the x-axis is canceled by one below the x-axis? 22. This exercise concerns the subtraction rule R∞ a (f (x)−g(x))dx = R∞ a f (x)dx − R∞ a g(x)dx. (a) Show that 1 x(x+1) = 1 x − 1 x+1 for all x except 0 and -1 (b) Show that R∞ 1 dx x(x+1) is convergent. (c) Show that both R∞ 1 dx x and R∞ 1 dx x+1 are divergent. (d) Does part (c) contradict parts(a)-(b) and the subtraction rule? Explain. x y 0 −1 n 1 n n y = Dn(x) 23. The improper integral R∞ −∞δ(x)dx = 1 is one of the notable “improprieties” of the Dirac delta function δ. One way to think of that integral is by approximating δ by triangular “pulse” functions Dn (for n ≥1), as in the picture on the right. (a) Write a formula for each Dn(x) over all of R. (b) Show that R∞ −∞Dn(x)dx = 1 for all integers n ≥1. (c) Show that limn→∞Dn(0) = ∞= δ(0). (d) Do the Dn functions begin to resemble δ as n →∞?	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. Differentiating that function, on the other hand, is easy. By the Product Rule for differentials, d(x e−x) = x d(e−x) + d(x) e−x = −x e−x dx + e−x dx d(x e−x) = −x e−x dx −d(e−x) x e−x dx = −d(x e−x) −d(e−x), so integrate both sides to get Z x e−x dx = − Z d(x e−x) − Z d(e−x) = −x e−x −e−x + C since R dF = F + C. Generalizing this process for functions u and v, d(uv) = u dv + v du u dv = d(uv) −v du so that integrating both sides yields the integration by parts formula: For differentiable functions u and v: Z u dv = uv − Z v du (6.1) 1Created by the Swiss mathematician, physicist and astronomer Leonhard Euler (1707-1783). The use of the Greek capital letter Γ for this function is due to the French mathematician Adrien-Marie Legendre (1752-1833). 159	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: The original integral is always of the form R udv, so you must decide which parts of xe−x dx will represent u and dv. Typically you would choose dv to be a differential that you could integrate easily (since you will need to integrate dv to get v) and choose u to be a function whose derivative is simpler than u (since that derivative will appear in vdu, which you hope to be a simpler integral). In this case, pick u = x and dv = e−x dx. Then du = dx and v = R dv = R e−x dx = −e−x (you can omit the generic constant C for now—include it when you have evaluated R vdu). Thus, Z udv = uv − Z vdu Z x u e−x dx dv = x u (−e−x) v − Z −e−x v dx du Z x e−x dx = −x e−x −e−x + C which agrees with the example at the beginning of this section. Note that R vdu = R −e−x dx, which indeed is simpler than the original integral. The Gamma function value Γ(2) can now be evaluated: Γ(2) = Z∞ 0 x e−x dx = −x e−x −e−x ¯¯¯¯ ∞ 0 = lim x→∞(−x e−x −e−x) −(−0 e0 −e0) = − ³ lim x→∞ x ex ´ − ³ lim x→∞e−x´ −(0−1) = 0 −0 + 1 = 1 Example 6.2 What would happen in Example 6.1 if you let u = e−x and dv = xdx? Solution: In this case du = −e−x dx and v = R dv = R xdx = 1 2 x2, so that Z x e−x dx = Z e−x u x dx dv = e−x u 1 2 x2 v − Z 1 2 x2 v (−e−x)dx du = 1 2 x2 e−x + 1 2 Z x2 e−x dx which leads you in the wrong direction: a more difﬁcult integral than the original.	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 Z lnx dx . Solution: Integration by parts ostensibly requires two functions in the integral, whereas here lnx ap- pears to be the only one. However, the choice for dv is a differential, and one exists here: dx. Choosing dv = dx obliges you to let u = ln x. Then du = 1 x dx and v = R dv = R dx = x. Now integrate by parts: Z udv = uv − Z vdu Z ln x dx = (ln x)(x) − Z x· 1 x dx = x ln x − Z 1 dx = x ln x −x + C Note that choosing dv = ln x dx would be pointless, as integrating dv to get v is the original problem! Example 6.4 Evaluate Z x3 ex2 dx . Solution: One frequently useful guideline for integration by parts is to eliminate the most complicated function in the integral by integrating it—as dv—into something simpler (which becomes v). In this integral, ex2 is somewhat complicated but has no closed form antiderivative. However, x ex2 appears in the integral and can be integrated easily (using a substitution as in Section 5.4). So choose dv = x ex2 dx, which means u = x2. Then du = 2xdx and v = R dv = R x ex2 dx = 1 2 ex2. Now integrate by parts: Z udv = uv − Z vdu Z x3 ex2 dx = x2 · 1 2 ex2 − Z 1 2 ex2 ·2x dx = x2 2 ex2 − Z x ex2 dx = x2 2 ex2 −1 2 ex2 + C = (x2 −1) 2 ex2 + C Sometimes multiple rounds of integration by parts are needed, as in the following example.	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 x e−x dx (integrate by parts again) Z x2 e−x dx = −x2 e−x + 2(−x e−x −e−x) + C (by Example 6.1) = −x2 e−x −2x e−x −2 e−x + C In the above example, notice that the u = 2x in the second integral came from the derivative of the u = x2 in the ﬁrst integral. Likewise, the dv = −e−x dx in the second integral came from integrating the dv = e−x dx in the ﬁrst integral. In general, if n rounds of integration by parts were needed, with ui and vi representing the u and v, respectively, for round i = 1, 2, ..., n, then the repeated integration by parts would look like this: Z u1 dv1 = u1v1 − Z v1 du1 = u1v1 − Z u2 dv2 = u1v1 − µ u2v2 − Z v2 du2 ¶ = u1v1 −u2v2 + Z u3 dv3 = u1v1 −u2v2 + µ u3v3 − Z u4 dv4 ¶ = u1v1 −u2v2 + u3v3 − µ u4v4 − Z u5 dv5 ¶ = ··· = u1v1 −u2v2 + u3v3 −u4v4 + u5v5 −··· Z un dvn The last integral R un dvn is one you could presumably integrate easily.	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 column and integrate down the dv column. If the u in the original integral is a polynomial of degree n, then you know from Section 1.6 that its (n + 1)-st derivative will be 0, at which point the tabular method terminates. The integral is then the sum of the indicated products with alternating signs. For example, the tabular method on the integral from Example 6.5 looks like this: u dv x2 e−x dx 2x −e−x (+) + (x2)(−e−x) 2 e−x (–) −(2x)(e−x) 0 −e−x (+) + (2)(−e−x) STOP The integral is the sum of the products, and agrees with the result in Example 6.5: Z x2 e−x dx = + (x2)(−e−x) −(2x)(e−x) + (2)(−e−x) + C = −x2 e−x −2x e−x −2 e−x + C Example 6.6 Evaluate Z x3 e−x dx . Solution: Use the tabular method with u = x3 and dv = e−x dx: u dv x3 e−x dx 3x2 −e−x (+) + (x3)(−e−x) 6x e−x (–) −(3x2)(e−x) 6 −e−x (+) + (6x)(−e−x) 0 e−x (–) −(6)(e−x) STOP Z x3 e−x dx = −x3 e−x −3x2 e−x −6x e−x −6 e−x + 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. Then Z udv = uv − Z vdu Z sec3 x dx = sec x tan x − Z sec x tan2 x dx Z sec3 x dx = sec x tan x − Z sec x (sec2 x −1) dx Z sec3 x dx = sec x tan x + Z sec x dx − Z sec3 x dx 2 Z sec3 x dx = sec x tan x + ln \| sec x + tan x \| + C Z sec3 x dx = 1 2 (sec x tan x + ln \| sec x + tan x \|) + C Example 6.8 Evaluate Z ex sin x dx . Solution: Let u = ex and dv = sin xdx, so that du = ex dx and v = R dv = R sin xdx = −cos x. Then Z udv = uv − Z vdu Z ex sin x dx = −ex cos x + Z ex cos x dx and so integration by parts is needed again, for the integral on the right: let u = ex and dv = cos xdx, so that du = ex dx and v = R dv = R cos xdx = sin x. Then Z ex sin x dx = −ex cos x + µ uv − Z vdu ¶ Z ex sin x dx = −ex cos x + µ ex sin x − Z ex sin x dx ¶ 2 Z ex sin x dx = −ex cos x + ex sin x Z ex sin x dx = ex 2 (sin x −cos x) + C	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−x2, and x p 1−x2 is easy to integrate (via a substitution), let u = x2 and dv = x p 1−x2 dx. Then du = 2xdx and v = R dv = R x p 1−x2 dx = −1 3(1−x2)3/2, and so: Zb a udv = uv ¯¯¯¯ b a − Zb a vdu Z1 0 x3 p 1−x2 dx = −x2 3 (1−x2)3/2 ¯¯¯¯ 1 0 + Z1 0 2x 3 (1−x2)3/2 dx = (0−0) + Ã −2 15 (1−x2)5/2 ¯¯¯¯ 1 0 ! = 0 + 2 15 = 2 15 Exercises A For Exercises 1-25, evaluate the given integral. 1. Z x ln x dx 2. Z x2 ex dx 3. Z x cos x dx 4. Z x3x dx 5. Z x2 ax dx (a > 0) 6. Z ln 4x dx 7. Z ln x2 dx 8. Z x2 sin x dx 9. Z x cos2 x dx 10. Z sin x cos 2x dx 11. Z sin−1 x dx 12. Z cos−1 2x dx 13. Z tan−1 3x dx 14. Z x sec2 x dx 15. Z sin x sin 3x dx 16. Z ln x x3 dx 17. Z x3 ln2 x dx 18. Z x5 ex dx 19. Z2 0 x3 dx p 4−x2 20. Z1 0 x3 p 1+ x2 dx 21. Z sin(ln x) dx 22. Z ln(1+ x2)dx 23. Z x tan−1 x dx 24. Z cot−1 p x dx 25. Z e px dx 26. Evaluate the integral R ex sin xdx from Example 6.8 by using two rounds of the tabular method and the formula R u1 dv1 = u1v1 −u2v2 + R v2 du2 from p.162. B 27. Show that for all constants a and b ̸= 0 : Z eax cos bx dx = eax (a cos bx + b sin bx) a2 + b2 + C and Z eax sin bx dx = eax (a sin bx −b cos bx) a2 + b2 + C	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 extension of the factorial operation to all positive real numbers. In fact, the Gamma function was created for that purpose. 29. Use Exercise 28 to prove for all integers n ≥1: Z∞ 0 rn e−r ln r dr = (n−1)! + n Z∞ 0 rn−1 e−r ln r dr 30. By the Maxwell speed distribution for gas molecules, the average speed 〈ν〉of molecules of mass m in a gas at temperature T is 〈ν〉= 4π ³ m 2πkT ´3/2 Z∞ 0 ν3 e−mν2/2kT dν , where k ≈1.38056×10−23 J/K is the Boltzmann constant. Show that 〈ν〉= s 8kT πm . 31. Some physics texts write integrals in a form like this energy integral from statistical mechanics, Z∞ 0 ln ³ 1−αe−x2´ d(x3) which uses the differential of a function—in this case d(x3)—instead of a variable (e.g. not just plain dx). This often signals that integration by parts is on the way, with the added beneﬁt of having the v = R dv calculation done for you—in the above integral d(x3) means that v = x3, with dv = d(x3) not really being needed for anything else. With that understanding, show that for 0 < α < 1, Z∞ 0 ln ³ 1−αe−x2´ d(x3) = −2α Z∞ 0 x4 e−x2 dx 1−αe−x2 . 32. Find the ﬂaw in the following “proof” that 0 = 1: Evaluating the integral Z dx x using integration by parts with u = 1 x and dv = dx, so that du = −dx x2 and v = x, shows that Z udv = uv − Z vdu Z dx x = µ1 x ¶ · x − Z x · µ −dx x2 ¶ Z dx x = 1 + Z dx x 0 = 1 ✓	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 involving products of sines and cosines with “mixed” angles can be simpliﬁed with the useful product-to-sum formulas:2 sin A cos B = 1 2 (sin (A +B) + sin (A −B)) (6.2) cos A sin B = 1 2 (sin (A +B) −sin (A −B)) (6.3) cos A cos B = 1 2 (cos (A +B) + cos (A −B)) (6.4) sin A sin B = −1 2 (cos (A +B) −cos (A −B)) (6.5) Example 6.10 Evaluate Z 0.5 sin x sin 12x dx . Solution: Using the product-to-sum formula (6.5) with A = x and B = 12x, sin A sin B = −1 2 (cos (A +B) −cos (A −B)) sin x sin 12x = −1 2 (cos (x+12x) −cos (x−12x)) sin x sin 12x = −1 2 (cos 13x −cos 11x) since cos(−11x) = cos 11x. Then Z 0.5 sin x sin 12x dx = −1 4 Z (cos 13x −cos 11x) dx = −1 52 sin 13x + 1 44 sin 11x + C Notice how the product-to-sum formula turned an integral of products of sines into integrals of individ- ual cosines, which are easily integrated. The integrand is an example of a modulated wave, commonly used in electronic communications (e.g. radio broadcasting). The graph is shown below: y x 0 0.5 −0.5 y = 0.5sinx sin12x π 2π The curves y = ±0.5sin x (shown in dashed lines) form an amplitude envelope for the modulated wave. 2See Section 3.4 in CORRAL, M., Trigonometry, http://mecmath.net/trig/, 2009.	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 sin x dx = Z p(u) du where p(u) is a polynomial in the variable u = cos x, and the remaining single sin x is now part of du = −sin x dx. You can then use the Power Formula to integrate that polynomial. Example 6.11 Evaluate Z sin3 x dx . Solution: Let u = cos x so that du = −sin x dx: Z sin3 x dx = Z (sin2 x) sin x dx = Z (1−cos2 x) sin x dx = Z (1−u2) (−du) = Z (u2 −1) du = 1 3 u3 −u + C = 1 3 cos3 x −cos x + C In general R sin2n+1 x dx will be a polynomial of degree 2n +1 in terms of cos x. Similarly, use cos2 x = 1−sin2 x to integrate odd powers of cos x, with the substitution u = sin x: Z cos2n+1 x dx = Z (cos2 x)n cos x dx = Z (1−sin2 x)n p(u) cos x dx du	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 x cos3 x dx = Z sin2 x (cos2 x) cos x dx = Z sin2 x (1−sin2 x) p(u) cos x dx du = Z (u2 −u4) du = 1 3 u3 −1 5 u5 + C = 1 3 sin3 x −1 5 sin5 x + C For even powers of sin x or cos x. You would replace sin2 x or cos2 x with either sin2 x = 1 −cos 2x 2 or cos2 x = 1 + cos 2x 2 , respectively, as often as necessary, then proceed as before if odd powers occur. Example 6.13 Evaluate Z sin4 x dx . Solution: Replace sin2 x by 1−cos 2x 2 : Z sin4 x dx = Zµ1 −cos 2x 2 ¶2 dx = 1 4 Z (1 −2 cos 2x + cos22x) dx = 1 4 Zµ 1 −2 cos 2x + 1 + cos 4x 2 ¶ dx = 1 8 Z (3 −4 cos 2x + cos 4x) dx = 3x 8 −1 4 sin 2x + 1 32 sin 4x + C	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 in an integral of a polynomial p(u) in terms of u = tan x: Z sec2k+2 x tann x dx = Z (sec2 x)k sec2 x tann x dx = Z (1+tan2 x)k tann x p(u) sec2 x dx du Likewise for an odd power n = 2k +1, use tan2 x = sec2 x −1 for all but one of the n powers of tan x, then use the substitution u = sec x, so that du = sec x tan x dx. This results in an integral of a polynomial p(u) in terms of u = sec x: Z secm x tan2k+1 x dx = Z secm−1 x sec x (tan2 x)k tan x dx = Z secm−1 x (sec2 x−1)k p(u) sec x tan xdx du Mimic the above procedure for integrals of the form R cscm x cotn x dx when either m is even or n is odd, using the identity csc2 x = 1+cot2 x in a similar manner. Example 6.14 Evaluate Z sec4 x tan x dx . Solution: Use sec2 x = 1+tan2 x for one sec2 x term, then substitute u = tan x, so that du = sec2 x dx: Z sec4 x tan x dx = Z sec2 x sec2 x tan x dx = Z (1+tan2 x) tan x sec2 x dx = Z (1+ u2)u du = Z (u+ u3) du = 1 2 u2 + 1 4 u4 + C = 1 2 tan2 x + 1 4 tan4 x + C	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 = −sin x dx) = − Z u4 dx = −1 5 cos5 x + C Exercises A For Exercises 1-12, evaluate the given integral. 1. Z sin 2x cos 5x dx 2. Z cos 2x cos 5x dx 3. Z sin 2x sin 5x dx 4. Z cos 2πx sin 3πx dx 5. Z sin3 x cos3/2 x dx 6. Z cos4 x dx 7. Z sin6 x dx 8. Z sin x sin 2x sin 3x dx 9. Z sec4 x dx 10. Z sec2 x tan3 x dx 11. Z tan3 x sec4 x dx 12. Z dx csc2 x cot x B 13. Evaluate R sin3 x cos3 x dx in two different ways: (a) Use sin3 x = (1−cos2 x) sin x and the substitution u = cos x. (b) Use cos3 x = (1−sin2 x) cos x and the substitution u = sin x. Are the answers from parts(a) and (b) equivalent? Explain. 14. Evaluate R sec4 x tan x dx by using sec4 x tan x = sec3 x (sec x tan x) and the substitution u = sec x. Is your answer equivalent to the answer in Example 6.14? Explain. 15. Show that Z 4 tan x (1−tan2 x) (1+tan2 x)2 dx = −1 4 cos 4x + C. 16. The autocorrelation function Rx(τ) of the periodic function x(t) = A cos(ωt+θ) is given by Rx(τ) = ω 2π Z2π/ω 0 x(t) x(t−τ) dt where A, ω and θ are constants. Show that Rx(τ) = A2 2 cos ωτ .	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. Use the circle of radius r > 0 centered at the origin (0,0) in the xy-plane, whose equation is x2 + y2 = r2 (see Figure 6.3.1(a) below). x y r θ r 0 (x, y) = (rcos θ,rsin θ) x2 + y2 = r2 (a) Full circle x y r −r θ = 0 θ = π 0 y = p r2 −x2 (b) Upper hemisphere Figure 6.3.1 Circle of radius r By symmetry about the x-axis, the area A of the circle is twice the area of its upper hemi- sphere (see Figure 6.3.1(b) above), which is the area under the curve y = p r2 −x2: A = 2 Zr −r p r2 −x2 dx To evaluate this integral, recall from trigonometry that any point (x, y) on the circle can be written as (x, y) = (rcos θ,rsin θ), where 0 ≤θ < 2π (in radians) is the angle shown in Figure 6.3.1(a). Figure 6.3.1(b) shows that as x goes from x = −r to x = r, the angle θ goes from θ = π to θ = 0. Now substitute x = r cos θ and dx = −r sin θ dθ into the integral and change the limits of integration from x = −r and x = r to θ = π and θ = 0, respectively: A = 2 Z0 π p r2 −r2 cos2 θ (−r sin θ) dθ = −2 Zπ 0 p r2(1−cos2 θ) (−r sin θ) dθ = 2 Zπ 0 r p sin2 θ r sin θ dθ = 2r2 Zπ 0 sin2 θ dθ = ✁2r2 Zπ 0 1−cos 2θ ✁2 dθ = r2 µ θ −1 2 sin 2θ ¶ ¯¯¯¯ π 0 = r2 µ π −1 2 sin 2π − µ 0 −1 2 sin 0 ¶¶ = πr2 ✓	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 is still in terms of θ. To put this back in terms of u, use θ = cos−1( u a), the double-angle formula sin 2θ = 2 sin θ cos θ, and p a2 −u2 = p a2 sin2 θ = a sin θ. Then Zp a2 −u2 du = −a2 2 cos−1 ³u a ´ + 2a2 sin θ cos θ 4 = −a2 2 cos−1 ³u a ´ + (a cos θ)(a sin θ) 2 + C which results in the following formula: Zp a2 −u2 du = −a2 2 cos−1³u a ´ + 1 2 u p a2 −u2 + C (6.6) It is left as an exercise to show that the substitution u = asin θ gives: Zp a2 −u2 du = a2 2 sin−1³u a ´ + 1 2 u p a2 −u2 + C (6.7) That these two seemingly different antiderivatives are equivalent follows immediately from the identity sin−1 x + cos−1 x = π 2 for all −1 ≤x ≤1, which shows that the antiderivatives differ by the constant πa2 4 (absorbed in the generic constant C): a2 2 sin−1³ u a ´ + 1 2 u p a2 −u2 + C = a2 2 ³π 2 −cos−1³u a ´´ + 1 2 u p a2 −u2 + C = −a2 2 cos−1³ u a ´ + 1 2 u p a2 −u2 + C + πa2 4 C Thus, either substitution—u = acos θ or u = asin θ—can be used when evaluating the integral Rp a2 −u2 du. The latter choice is sometimes preferred, to avoid the negative sign in du and the resulting formula.	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 general, when other methods fail, use the table below as a guide for certain types of inte- grals, making use of the speciﬁed substitution and trigonometric identity: Integral contains Substitution Identity p a2 −u2 u = a sin θ 1 −sin2 θ = cos2 θ p a2 + u2 u = a tan θ 1 + tan2 θ = sec2 θ p u2 −a2 u = a sec θ sec2 θ −1 = tan2 θ For example, the substitution u = a tan θ leads to the following formula: Zp a2 + u2 du = 1 2 u p a2 + u2 + a2 2 ln ¯¯u + p a2 + u2 ¯¯ + C (6.8) Similarly, the substitution u = a sec θ yields this formula: Zp u2 −a2 du = 1 2 u p u2 −a2 −a2 2 ln ¯¯u + p u2 −a2 ¯¯ + C (6.9) The proof of each formula requires this result from Example 6.7 in Section 6.1: Z sec3 θ dθ = 1 2 (sec θ tan θ + ln \| sec θ + tan θ \|) + C (6.10) The above substitutions can be used even if no square roots are present.	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 contains a term of the form a2 + u2 (with a = 1 and u = x), so use the substitution x = tan θ. Then dx = sec2 θ dθ and so Z dx (1+ x2)2 = Z sec2 θ dθ (1+tan2 θ)2 = Z sec2 θ dθ (sec2 θ)2 = Z dθ sec2 θ2 = Z cos2 θ dθ = Z 1+cos 2θ 2 dθ = θ 2 + 1 4 sin 2θ + C = θ 2 + 1 2 sin θ cos θ + C by the trigonometric double-angle identity sin 2θ = 2 sin θ cos θ. p 1+ x2 x 1 θ The simplest way to get expressions for sin θ and cos θ in terms of x is to draw a right triangle with an angle θ such that tan θ = x = x 1, as in the drawing on the right. The hypotenuse must then be p 1+ x2 (by the Pythagorean Theorem), which makes it easy to read off the values of sin θ and cos θ: sin θ = x p 1+ x2 and cos θ = 1 p 1+ x2 Since θ = tan−1 x, putting the integral back in terms of x yields: Z dx (1+ x2)2 = 1 2 tan−1 x + 1 2 x p 1+ x2 1 p 1+ x2 + C = 1 2 tan−1 x + x 2(1+ x2) + C Note: An alternative method for getting sin θ and cos θ in terms of x would be to put tan θ = x in the identity sec2 θ = 1+tan2 θ to solve for cos θ, then use the identity sin2 θ = 1−cos2 θ to solve for sin θ. By completing the square, quadratic expressions in x can be put in one of the forms a2 ± u2 or u2 −a2, enabling the use of the corresponding trigonometric substitution.	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 expression is now of the form u2 −a2 for u = 2(x +1) and a = 3. Use the substitution u = a sec θ, which means 2(x+1) = 3 sec θ. Then 2dx = 3 sec θ tan θ dθ and so: Z dx (4x2 +8x−5)3/2 = Z dx (4(x+1)2 −9)3/2 = 3 2 Z sec θ tan θ dθ (9 sec2 θ −9)3/2 = 3 2 Z sec θ tan θ dθ (9(sec2 θ −1))3/2 = 1 18 Z sec θ tan θ dθ tan3 θ = 1 18 Z sec θ dθ tan2 θ = 1 18 Z cos θ dθ sin2 θ = 1 18 Z csc θ cot θ dθ = −1 18 csc θ + C 2(x+1) p 4x2 +8x−5 3 θ To get an expression for csc θ in terms of x, draw a right triangle with an angle θ such that sec θ = 2(x+1) 3 , as in the drawing on the right. The side opposite θ must then be p 4x2 +8x−5 (by the Pythagorean Theorem), and hence: csc θ = 2(x+1) p 4x2 +8x−5 Putting the integral back in terms of x yields: Z dx (4x2 +8x−5)3/2 = −1 18 2(x+1) p 4x2 +8x−5 + C = − x+1 9 p 4x2 +8x−5 + C Note: Trigonometric identities could have been used to obtain csc θ by knowing sec θ. The following integrals from Section 5.4 might be helpful for the exercises: Z tan u du = ln \|sec u\| + C (6.11) Z sec u du = ln \|sec u + tan u\| + C (6.12) Z csc u du = −ln \|csc u + cot u\| + C (6.13)	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 x2 dx p a2 −x2 (a > 0) 13. Z x3 dx p 9−x2 14. Z p 4−x2 x dx 15. Z (x−4) dx p −9x2 +36x−32 16. Z dx (4x2 +16x+15)3/2 17. Prove formula (6.7) directly by using the substitution u = asin θ. 18. Prove formula (6.8). 19. Prove formula (6.9). 20. Show that using the substitution u = acot θ to evaluate the integral Rp a2 + u2 dx leads to an antiderivative equivalent to the one in formula (6.8). 21. Show that using the substitution u = acsc θ to evaluate the integral Rp u2 −a2 dx leads to an antiderivative equivalent to the one in formula (6.9). B 22. The integrals Z dx p x2 ± a2 can be evaluated without the use of trigonometric substitutions, by using differentials: (a) For u2 = x2 ± a2, show that dx u = d(x+ u) x+ u . (b) Integrate both sides of the result from part (a). Note: In general, many integrals involving p x2 ± a2 can be handled with a similar manipulation of differentials, with varying complexity. 23. According to Newtonian physics the path of a photon grazing the surface of the Sun should be deﬂected by the Sun’s gravitational ﬁeld by an angle θ, given approximately by θ = ¯¯¯¯ 2GMR c2 Z0 ∞ dy (R2 + y2)3/2 ¯¯¯¯ where c = 2.998×108 m/s is the speed of light, G = 6.67×10−11 N/m2/kg2 is the gravitational constant, M = 1.99×1030 kg is the mass of the Sun, and R = 6.96×108 m is the radius of the Sun. Show that θ = 2GM c2R = 4.24×10−6 radians = 2.43×10−4 degrees ≈0.875 seconds of arc, where 1 second of arc = 1/3600 of 1 degree.3 3Albert Einstein published this result in 1911, then showed in 1915 that the true angle should be double that amount, due to the curvature of space. Experiments veriﬁed Einstein’s prediction. See pp.69-71 in SERWAY, R.A., C.J. MOSES AND C.A. MOYER, Modern Physics, Orlando, FL: Harcourt Brace Jovanovich Publishers, 1989.	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 method of partial fractions. The idea behind this method is simple: replace a complicated rational function with simpler ones that are easy to integrate. For example, there is no formula for evaluating the integral Z dx x2 + x but notice that you can write 1 x2 + x = 1 x(x+1) = 1 x − 1 x+1 , called the partial fraction decomposition of 1 x2+x, so that Z dx x2 + x = Zµ1 x − 1 x+1 ¶ dx = ln\|x\| −ln\|x+1\| + C . There is a systematic way to ﬁnd this decomposition. First, assume that 1 x(x+1) = A x + B x+1 for some constants A and B. Get a common denominator on the right side: 1 x(x+1) = A (x+1) + Bx x(x+1) 0x + 1 x(x+1) = (A +B) x + A x(x+1) Now equate coefﬁcients in the numerators of both sides to solve for A and B: constant term : A = 1 coefﬁcient of x : A + B = 0 ⇒ B = −A = −1 Thus, 1 x(x+1) = A x + B x+1 = 1 x + −1 x+1 = 1 x − 1 x+1 as before.	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 than the degree of the polynomial in the denominator. Start with the most basic case, similar to the example above: Case 1 - Distinct linear factors: A rational function p(x) q(x) such that degree(p(x)) < degree(q(x)), where q(x) is a product of n > 1 distinct linear factors q(x) = (a1x+ b1)(a2x+ b2) ··· (anx+ bn) , can be written as a sum of partial fractions p(x) q(x) = A1 a1x+ b1 + A2 a2x+ b2 + ··· + An anx+ bn for some constants A1, A2, ... , An. Those constants can be solved for by getting a common denominator on the right side of the equation and then equating the coefﬁcients of the numerators of both sides. Example 6.19 Evaluate Z dx x2 −7x+10 . Solution: Since x2 −7x+10 = (x−2)(x−5), then 1 x2 −7x+10 = 1 (x−2)(x−5) = A x−2 + B x−5 = (A +B)x + (−5A −2B) (x−2)(x−5) so that coefﬁcient of x : A + B = 0 ⇒ B = −A constant term : −5A −2B = 1 ⇒ −5A + 2A = 1 ⇒ A = −1 3 and B = 1 3 Thus, Z dx x2 −7x+10 = ZÃ −1 3 x−2 + 1 3 x−5 ! dx = −1 3 ln\|x−2\| + 1 3 ln\|x−5\| + C 4For example, see Section 5.10 in HILLMAN, A.P., AND G.L. ALEXANDERSON, A First Undergraduate Course in Abstract Algebra, 3rd ed., Belmont, CA: Wadsworth Publishing Co., 1983.	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)), where q(x) is a product of n > 1 distinct linear factors and one linear factor repeated m > 1 times q(x) = (ax+ b)m (a1x+ b1)(a2x+ b2) ··· (anx+ bn) , can be written as a sum of partial fractions p(x) q(x) = A1 ax+ b + A2 (ax+ b)2 + ··· + Am (ax+ b)m + B1 a1x+ b1 + ··· + Bn anx+ bn for some constants A1, A2, ... , Am and B1, ... , Bn. Those constants can be solved for by the same method as in Case 1. Example 6.20 Evaluate Z x2 + x−1 x3 + x2 dx . Solution: Since x3 + x2 = x2 (x+1), then x2 + x−1 x3 + x2 = x2 + x−1 x2 (x+1) = A x + B x2 + C x+1 = Ax(x+1) + B(x+1) + Cx2 x2 (x+1) = (A +C)x2 + (A +B)x + B x2 (x+1) so that constant term : B = −1 coefﬁcient of x : A + B = 1 ⇒ A = 1 −B = 2 coefﬁcient of x2 : A + C = 1 ⇒ C = 1 −A = −1 Thus, Z x2 + x−1 x3 + x2 dx = Zµ2 x + −1 x2 + −1 x+1 ¶ dx = 2 ln\|x\| + 1 x −ln\|x+1\| + C0 (C0 = generic constant) Case 2 can be extended to more than one repeated factor—the partial fraction decomposition would then have more terms similar to those for the ﬁrst repeated factor.	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 + 2B = 0 ⇒ A = −2B = −2 coefﬁcient of x3 : A + C = 0 ⇒ C = −A = 2 coefﬁcient of x2 : 2A + B + C + D = 0 ⇒ D = −2A −B −C = 1 Thus, Z dx x2(x+1)2 = Zµ−2 x + 1 x2 + 2 x+1 + 1 (x+1)2 ¶ dx = −2 ln\|x\| −1 x + 2 ln\|x+1\| − 1 x+1 + C0 (C0 = generic constant) The partial fraction decompositions for quadratic factors are similar to those for linear fac- tors, except the numerators in each partial fraction can now contain linear terms. A factor of the form ax2+bx+c is considered quadratic only if it cannot be factored into a product of linear terms (i.e. has no real roots) and a ̸= 0. Case 3 - Distinct quadratic factors: A rational function p(x) q(x) such that degree(p(x)) < degree(q(x)), and q(x) is a product of n > 1 distinct quadratic factors q(x) = (a1x2 + b1x+ c1)(a2x2 + b2x+ c2) ··· (anx2 + bnx+ cn) , can be written as a sum of partial fractions p(x) q(x) = A1x+B1 a1x2 + b1x+ c1 + A2x+B2 a2x2 + b2x+ c2 + ··· + Anx+Bn anx2 + bnx+ cn for some constants A1, A2, ... , An and B1, B2, ... , Bn. Those constants can be solved for by the same method as in Case 1.	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 coefﬁcient of x3 : A + C = 0 ⇒ C = −A coefﬁcient of x2 : B + D = 0 ⇒ D = −B coefﬁcient of x : 4A + C = 0 ⇒ 4A −A = 0 ⇒ A = 0 constant term : 4B + D = 1 ⇒ 4B −B = 1 ⇒ B = 1 3 and D = −1 3 Thus, Z dx (x2 +1)(x2 +4) = ZÃ 1 3 x2 +1 + −1 3 x2 +4 ! dx = 1 3 tan−1 x −1 6 tan−1 ³ x 2 ´ + C0 by formula (5.4) in Section 5.4. A repeated quadratic factor is handled in the same way as a repeated linear factor: Case 4 - One repeated quadratic factor + distinct quadratic factors: A rational function p(x) q(x) such that degree(p(x)) < degree(q(x)), where q(x) is a product of n > 1 distinct quadratic factors and one quadratic factor repeated m > 1 times q(x) = (ax2 + bx+ c)m (a1x2 + b1x+ c1)(a2x2 + b2x+ c2) ··· (anx2 + bnx+ cn) , can be written as a sum of partial fractions p(x) q(x) = A1x+B1 ax2 + bx+ c + ··· + Amx+Bm (ax2 + bx+ c)m + C1x+ D1 a1x2 + b1x+ c1 + ··· + Cnx+ Dn anx2 + bnx+ cn for some constants A1, ... , Am, B1, ... , Bm, C1, ... , Cn, and D1, ... , Dn. Those constants can be solved for by the same method as in Case 1.	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 side of the equation expanded as (A + E)x5 + (B + F)x4 + (5A +C +2E)x3 + (5B + D +2F)x2 + (4A +4C + E)x + (4B +4D + F) (x2 +1)2(x2 +4) so that equating coefﬁcients of both sides gives coefﬁcient of x5 : A + E = 0 ⇒ E = −A coefﬁcient of x4 : B + F = 0 ⇒ F = −B coefﬁcient of x3 : 5A + C + 2E = 0 ⇒ 5A + C −2A = 0 ⇒ C = −3A coefﬁcient of x2 : 5B + D + 2F = 0 ⇒ 5B + D −2F = 0 ⇒ D = −3B coefﬁcient of x : 4A + 4C + E = 0 ⇒ 4A −12A −A = 0 ⇒A = 0 ⇒C = 0 and E = 0 constant term : 4B + 4D + F = 1 ⇒ 4B −12B −B = 1 ⇒B = −1 9 ⇒D = 1 3 and F = 1 9 Thus, Z dx (x2 +1)2(x2 +4) = ZÃ −1 9 x2 +1 + 1 3 (x2 +1)2 + 1 9 x2 +4 ! dx = −1 9 tan−1 x + 1 3 µ1 2 tan−1 x + x 2(x2 +1) ¶ + 1 18 tan−1 ³ x 2 ´ + C0 = 1 18 tan−1 x + x 6(x2 +1) + 1 18 tan−1³ x 2 ´ + C0 where the middle integral on the right is from Example 6.17 in Section 6.3. When a rational function has a numerator with degree larger than its denominator, dividing the numerator by the denominator leaves the sum of a polynomial and a new rational function perhaps satisfying the conditions for Cases 1-4. When the numerator and denominator have the same degree, a trick like this might be easier. x2 +2 (x+1)(x+2) = (x2 +3x+2)−3x (x+1)(x+2) = x2 +3x+2 (x+1)(x+2) − 3x (x+1)(x+2) = 1 − 3x (x+1)(x+2) The last rational function on the right can be integrated using partial fractions.	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)2 dx B 13. For all numbers a ̸= b show that Z dx (x−a)(x−b) = 1 a−b ln ¯¯¯¯ x−a x−b ¯¯¯¯ + C . 14. Let q(x) = (x−a1)(x−a2), where a1 ̸= a2. Show that q′(x) q(x) = 1 x−a1 + 1 x−a2 . 15. For q(x) as in Exercise 14, show that 1 q(x) = 1 q′(a1)(x−a1) + 1 q′(a2)(x−a2) . 16. Extend Exercise 15 to three distinct linear factors: if q(x) = (x−a1)(x−a2)(x−a3) then 1 q(x) = 1 q′(a1)(x−a1) + 1 q′(a2)(x−a2) + 1 q′(a3)(x−a3) . This result can be extended to any n ≥2 distinct factors, though you do not need to prove that. 17. Find d100 dx100 µ 1 x2 −9x+20 ¶ . 18. Find d2020 dx2020 ¡ (x2 −1)−1¢ . 19. It is possible to use a form of partial fractions to evaluate integrals that are not rational functions. For example, evaluate the integral Z dx x+ x4/3 by ﬁnding constants A, B and C such that 1 x+ x4/3 = 1 x(1+ x1/3) = A x + B x−2/3 +C 1+ x1/3 . Notice how in the second partial fraction the highest power of x in the numerator is one less than in the denominator, similar to a partial fraction for a quadratic factor in a rational function. C 20. Find a different way to evaluate the integral in Exercise 19.	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 integral rule for “differentiation under the integral sign.”5 This powerful and useful method is best explained with a simple example. Recall from Section 5.4 that Z eαx dx = 1 α eαx + C (6.14) for any constant α ̸= 0. This antiderivative involves the variable x and the constant α. The idea behind the Leibniz rule is to reverse those roles: view α as the variable and x as a constant. The antiderivative is then seen as a function of α, and so its derivative can be taken with respect to α. The big leap that this method makes is to move the differentiation operation inside the integral:6 d dα Z eαx dx = Z d dα (eαx) dx = Z x eαx dx However, differentiating the right side of formula (6.14) shows that d dα Z eαx dx = d dα ¡ 1 α eαx + C ¢ = α (x eαx) −1 · eαx α2 = 1 α x eαx − 1 α2 eαx Thus, Z x eαx dx = 1 α x eαx − 1 α2 eαx + C which can be veriﬁed via integration by parts with the tabular method: u dv x eαx dx 1 1 α eαx (+) + (x)( 1 α eαx) 0 1 α2 eαx (–) −(1)( 1 α2 eαx) STOP Z x eαx dx = 1 α x eαx − 1 α2 eαx + C ✓ 5The renowned physicist Richard Feynman (1918-1988) famously lamented that the technique was no longer being taught. See p.72 in FEYNMAN, R.P., Surely You’re Joking, Mr. Feynman!, New York: Bantam Books, 1986. 6It can be proved that this is valid when the derivative of the integrand is a continuous function of α, which will always be the case in this book. See pp.121-122 in SOKOLNIKOFF, I.S., Advanced Calculus, New York: McGraw- Hill Book Company, Inc., 1939.	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 differentiation—as a variable. In general, that is how the Leibniz rule is used. Typically this means if you want to evaluate a certain integral with the Leibniz rule, then you “work backwards” to ﬁgure out which integral you need to differentiate with respect to some constant (e.g. α) in the integrand. Example 6.24 Use the Leibniz rule to evaluate Z dx (1+ x2)2 . Solution: By formula (5.4) in Section 5.4, Z dx a2 + x2 = 1 a tan−1 ¡ x a ¢ + C for any constant a > 0. So differentiate both sides with respect to a: d da Z dx a2 + x2 = d da ¡ 1 a tan−1 ¡ x a ¢ + C ¢ Z d da µ 1 a2 + x2 ¶ dx = −1 a2 tan−1 ¡ x a ¢ + 1 a · 1 1+ ¡ x a ¢2 · −x a2 Z − 2a (a2 + x2)2 dx = −1 a2 tan−1 ¡ x a ¢ − x a(a2 + x2) Z dx (a2 + x2)2 = 1 2a3 tan−1 ¡ x a ¢ + x 2a2 (a2 + x2) + C That general formula is useful in itself. In particular, for a = 1, Z dx (1+ x2)2 = 1 2 tan−1 x + x 2(1+ x2) + C , which agrees with the result from Example 6.17 in Section 6.3. Notice that there was no generic constant (e.g. a or α) in the statement of the problem. When that happens, you will need to ﬁgure out where the constant should be in order to use the Leibniz rule. You can also use differentiation under the integral sign to evaluate deﬁnite integrals.	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 convergent (and equals 1 2π) by Example 5.32 in Section 5.5. For α ≥0, deﬁne φ(α) = Z∞ 0 α e−α2x2 1+ x2 dx . Then clearly φ(0) = 0, and differentiating under the integral sign shows φ′(α) = Z∞ 0 −2α2e−α2x2 + e−α2x2 1+ x2 dx ⇒ φ′(0) = Z∞ 0 dx 1+ x2 = 1 2π . The substitution y = αx, so that dy = αdx, shows φ(α) can be written as φ(α) = Z∞ 0 e−y2 1+ ¡ y α ¢2 dy ⇒ 0 ≤lim α→∞φ(α) ≤I < ∞. Also, for α > 0, d dα µ 1 α e−α2 φ(α) ¶ = d dα Z∞ 0 e−α2(1+x2) 1+ x2 dx = Z∞ 0 −2α(1+ x2) e−α2(1+x2) 1+ x2 dx = −2α e−α2 Z∞ 0 e−α2x2 dx , now substitute u = αx and du = αdx to get = −2α e−α2 1 α Z∞ 0 e−u2 du = −2 e−α2 I , and so integrating both sides yields Z∞ 0 d dα µ 1 α e−α2 φ(α) ¶ dα = −2I Z∞ 0 e−α2 dα = −2I2 . However, by the Fundamental Theorem of Calculus. Z∞ 0 d dα µ 1 α e−α2 φ(α) ¶ dα = 1 α e−α2 φ(α) ¯¯¯¯ ∞ 0 = µ lim α→∞ φ(α) α eα2 ¶ − µ lim α→0 φ(α) α eα2 ¶ = 0 − µ lim α→0 φ(α) α eα2 ¶ →0 0 , so by L’Hôpital’s Rule = −lim α→0 φ′(α) eα2 +2α2 eα2 = −φ′(0) 1+0 = −1 2π . Thus, −2I2 = −1 2π ⇒ I = 1 2 p π which is the desired result.	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 function Γ(t) can be written as Γ(t) = 2 Z∞ 0 y2t−1 e−y2 dy for all t > 0, and that Γ ¡ 1 2 ¢ = pπ. Solution: Let x = y2, so that dx = 2y dy. Then x = 0 ⇒y = 0 and x = ∞⇒y = ∞, so Γ(t) = Z∞ 0 xt−1 e−x dx = Z∞ 0 (y2)t−1 e−y2 2y dy = 2 Z∞ 0 y2t−1 e−y2 dy . In this form, with the help of Example 6.25 it is now easy to evaluate Γ ¡ 1 2 ¢ : Γ ¡ 1 2 ¢ = 2 Z∞ 0 y1−1 e−y2 dy = 2 Z∞ 0 e−y2 dy = 2 ¡ 1 2 p π ¢ = p π A function closely related to the Gamma function is the Beta function B(x, y), deﬁned by: B(x, y) = Z1 0 tx−1(1−t)y−1 dt for all x > 0 and y > 0 (6.15) It can be shown that7 B(x, y) = Γ(x) Γ(y) Γ(x+ y) for all x > 0 and y > 0. (6.16) Example 6.27 Show that the Beta function B(x, y) can be written as B(x, y) = Z∞ 0 ux−1 (1+ u)x+y du . Solution: Let u = t 1−t, so that t = u 1+u, 1−t = 1 1+u, and dt = du (1+u)2 . Then t = 0 ⇒u = 0 and t = 1 ⇒u = ∞, so B(x, y) = Z1 0 tx−1 (1−t)y−1 dt = Z∞ 0 ³ u 1+ u ´x−1 µ 1 1+ u ¶y−1 du (1+ u)2 = Z∞ 0 ux−1 (1+ u)x+y du . 7See p.18-19 in RAINVILLE, E.D., Special Functions, New York: Chelsea Publishing Company, 1971.	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. It turns out that derivatives of fractional orders (e.g. n = 1 2) can be deﬁned, with the Riemann-Louiville deﬁnition being the most common: For all 0 < α < 1, the fractional derivative of order α of a function f (x) is dα dxα f (x) = 1 Γ(1−α) d dx Zx 0 f (t) (x−t)α dt . (6.17) Example 6.28 Calculate d1/2 dx1/2 (x) . Solution: Here α = 1 2 and f (x) = x, so that d1/2 dx1/2 (x) = 1 Γ(1−1/2) d dx Zx 0 t (x−t)1/2 dt = 1 pπ d dx Zx 0 t dt p x−t since Γ ¡ 1 2 ¢ = pπ by Example 6.26. Use the substitution u = p x−t, so that t = x−u2 and dt = −2udu. Then t = 0 ⇒u = px and t = x ⇒u = 0, so d1/2 dx1/2 (x) = 1 pπ d dx Z0 px (x−u2)(−2u du) u = 2 pπ d dx Zpx 0 (x −u2) du = 2 pπ d dx ¡ xu −1 3 u3¢ ¯¯¯¯ u=px u=0 = 2 pπ d dx ³ x3/2 −1 3 x3/2´ = 2 pπ d dx ³ 2 3 x3/2´ = 2 pπ px Part of the motivation for creating fractional derivatives was to ﬁnd if it were possible to take two “half” derivatives to form a “whole” derivative: d1/2 dx1/2 µ d1/2 dx1/2 f (x) ¶ = d dx f (x) It is left as an exercise to show that the above relation does hold for the function f (x) = x. Derivatives with fractional order 0 < α < 1 and integer order n ≥1 can be combined by taking the derivative of integer order ﬁrst:8 dn+α dxn+α f (x) = dα dxα µ dn dxn f (x) ¶ 8For more details about fractional derivatives, as well as examples of their applications in physics and engineering, see OLDHAM, K.B. AND J. SPANIER, The Fractional Calculus, New York: Academic Press, 1974.	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 via that substitution, with θ as shown in Figure 6.5.1(a) below. 1 x y θ 1 0 (x, y) = (cos θ,sin θ) (a) Identify points by angle x y slope = 1 2 slope = −1 2 slope = 1 slope = -1 slope = 2 slope = −2 1 −1 slope = 0 0 (b) Identify points by slope Figure 6.5.1 Points on the unit circle x2 + y2 = 1 Figure 6.5.1(b) shows a different identiﬁcation of points on the unit circle—by slope. This will be the basis for a half-angle substitution for evaluating certain integrals. Let A be the point (−1,0), then for any other point P on the unit circle draw a line from A through P until it intersects the line x = 1, as shown in Figure 6.5.2 below: x y x = 1 0 1 A P θ/2 θ 1 t p 1+ t2 (a) 0 ≤θ ≤π 2 x y x = 1 1 A 1 t 0 θ/2 θ p 1+ t2 (b) π 2 < θ < π Figure 6.5.2 Half-angle substitution: t = tan ¡ θ 2 ¢ = slope of AP From geometry you know that the inscribed angle that the line AP makes with the x-axis is half the measure of the central angle θ. So the slope of AP is the tangent of that angle: tan 1 2θ = t 1 = t, which is measured along the y-axis and can take any real value. Each point on the unit circle—except A—can be identiﬁed with that slope t.	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 p 1+ t2 = 2t 1+ t2 and cos θ = cos2 1 2θ −sin2 1 2θ = 1 1+ t2 − t2 1+ t2 = 1−t2 1+ t2 . Since θ = 2 tan−1 t, then dθ = d ¡ 2 tan−1 t ¢ = 2dt 1+ t2 . Below is a summary of the substitution: Half-angle substitution: The substitution t = tan 1 2θ yields: sin θ = 2t 1+ t2 cos θ = 1−t2 1+ t2 dθ = 2dt 1+ t2 The half-angle substitution thus turns rational functions of sin θ and cos θ into rational func- tions of t, which can be integrated using partial fractions or another method. Example 6.29 Evaluate Z dθ 1 + sin θ + cos θ . Solution: Using t = tan 1 2θ, the denominator of the integrand is 1 + sin θ + cos θ = 1+ t2 1+ t2 + 2t 1+ t2 + 1−t2 1+ t2 = 2t+2 1+ t2 so that Z dθ 1 + sin θ + cos θ = Z 2dt 1+t2 2t+2 1+t2 = Z dt t+1 = ln\|t+1\| + C = ln ¯¯tan 1 2θ + 1 ¯¯ + C	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 constant term : −2A + B = −1 ⇒ 4B + B = −1 ⇒ B = −1 5 and A = 2 5 Thus, Z dθ 3 sin θ + 4 cos θ = ZÃ 2 5 2t+1 + −1 5 t−2 ! dt = 1 5 ln\|2t+1\| −1 5 ln\|t−2\| + C = 1 5 ln ¯¯2 tan 1 2θ + 1 ¯¯ −1 5 ln ¯¯tan 1 2θ −2 ¯¯ + C By the half-angle substitution t = tan 1 2θ, sin θ 1 + cos θ = 2t 1+ t2 1+ t2 1+ t2 + 1−t2 1+ t2 = 2t 1+ t2 2 1+ t2 = t which yields the useful half-angle identities:9 tan 1 2θ = sin θ 1 + cos θ = 1 −cos θ sin θ (6.18) Example 6.31 Evaluate Z sin θ 1 + cos θ dθ . Solution: Though you could use the half-angle substitution t = tan 1 2θ, it is easier to use the half-angle identity (6.18) directly, since Z sin θ 1 + cos θ dθ = Z tan 1 2θ dθ = 2 ln ¯¯sec 1 2θ ¯¯ + C by formula (6.11) in Section 6.3. 9For a different derivation, see pp.79-80 in CORRAL, M., Trigonometry, http://mecmath.net/trig/, 2009.	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 θ 3 sin θ dθ 11. Z∞ −∞ e−x2/2 dx 12. Z∞ −∞ x2 e−x6 dx 13. Consider the integral Z sin θ 1 + cos θ dθ from Example 6.31. (a) Evaluate the integral using the substitution u = 1+cos θ. (b) Evaluate the integral using the half-angle substitution t = tan 1 2θ. (c) Show that the answers from parts (a) and (b) are equivalent to the result from Example 6.31. B φ 3 4 5 14. Evaluate the integral Z dθ 3 sin θ + 4 cos θ from Example 6.30 by noting that Z dθ 3 sin θ + 4 cos θ = Z dθ 5 ¡ 3 5 sin θ + 4 5 cos θ ¢ = Z dθ 5 ¡ cos φ sin θ + sin φ cos θ ¢ = Z dθ 5 sin(θ +φ) = 1 5 Z csc(θ +φ) dθ by the sine addition formula, where φ is the angle in the right triangle shown above. Complete the integration and show that your answer is equivalent to the result from Example 6.30. 15. Show directly from the deﬁnition of the Beta function that B(x, y) = B(y,x) for all x > 0 and y > 0. 16. Show that the Beta function B(x, y) can be written as B(x, y) = Zπ/2 0 2 sin2x−1(θ) cos2y−1(θ) dθ for all x > 0 and y > 0. 17. Use Exercise 16 and formula (6.16) to show that Zπ/2 0 sinm θ cosn θ dθ = Γ µ m+1 2 ¶ Γ µ n+1 2 ¶ 2Γ ³m+ n 2 +1 ´ for all m > −1 and n > −1. 18. Use Exercise 28 from Section 6.1, as well as Exercise 17 above, to show that for m = 1, 2, 3, ..., Zπ/2 0 sin2m θ dθ = pπ Γ ¡ m+ 1 2 ¢ 2(m!) and Zπ/2 0 sin2m+1 θ dθ = pπ (m!) 2Γ ¡ m+ 3 2 ¢ .	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. Show that Z1 0 1 p 1−xn dx = 1 n B ¡ 1 n, 1 2 ¢ for n ≥1. 25. Show that the Gamma function Γ(t) can be written as Γ(t) = pt Z∞ 0 ut−1 e−pu du for all t > 0 and p > 0. 26. Show that the Gamma function Γ(t) can be written as Γ(t) = Z1 0 µ ln µ 1 u ¶¶t−1 du for all t > 0. 27. Using the result from Exercise 27 in Section 6.1 that Z eax cos bx dx = eax (a cos bx + b sin bx) a2 + b2 for all constants a and b ̸= 0, differentiate under the integral sign to show that for all α > 0 Z∞ 0 x e−x sin αx dx = 2α (1+α2)2 . C 28. Use the Leibniz rule and formula (6.8) from Section 6.3 to show that for all a > 0, Z dx p a2 + x2 = ln ¯¯x+ p a2 + x2 ¯¯ + C . 29. Use Example 6.27 to show that the Beta function satisﬁes the relation B(x,1−x) = Z1 0 t−x + tx−1 1+ t dt for all 0 < x < 1. (Hint: First use a substitution to show that Z∞ 0 ux−1 1+ u du = Z∞ 0 t−x 1+ t dt.) 30. Show that for all a > −1, Zπ/2 0 dθ 1 + a sin2 θ = π 2 p 1+ a .	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 produces an approximate value of the deﬁnite integral—the more rectangles, the better the approximation. y = sin(x2) y x 0 pπ 1 For example, suppose you wanted to evaluate Zpπ 0 sin(x2) dx with the rectangle method. This means ﬁnding the area of the shaded region in the ﬁgure on the right. Computers have eliminated the need to do these sorts of calculations by hand. Though the rectangle method is simple to implement in a traditional programming language (e.g. via a looping construct), there are easier ways in a domain-speciﬁc language (DSL) geared toward scientiﬁc computing. One such DSL is MATLAB®, or its free open-source clone Octave.10 Implementing the rectangle method from scratch in Octave is a one-liner. For example, sup- pose you divide the interval [0,pπ] into 100,000 (105) subintervals of equal length, producing 100,001 (1 + 1e5 in scientiﬁc notation) equally spaced points in [0,pπ] (including 0 and pπ). First use the left endpoints of the 105 subintervals: octave> sum(sin(linspace(0,sqrt(pi),1+1e5)(1:end-1).^2)sqrt(pi)/1e5) ans = 0.8948314693913354 Now use the right endpoints: octave> sum(sin(linspace(0,sqrt(pi),1+1e5)(2:end).^2)sqrt(pi)/1e5) ans = 0.8948314693913354 Finally, use the midpoints of the subintervals: octave> sum(sin((linspace(0,sqrt(pi),1+1e5)(1:end-1)+sqrt(pi)/2e5).^2)* sqrt(pi)/1e5) ans = 0.8948314695305527 The true value of the integral up to 15 decimal places is 0.894831469484145, so all three approximations are accurate to 9 decimal places. 10Octave is freely available at https://www.gnu.org/software/octave/	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 5 7 Get all but the last number in the above list:11 octave> linspace(1,7,4)(1:end-1) ans = 1 3 5 Now square each number in that list of numbers (the dot before the exponentiation operator ^ applies the squaring operation ^2 element-wise in the list): octave> linspace(1,7,4)(1:end-1).^2 ans = 1 9 25 Now take the sine of each of those squared numbers (measured in radians): octave> sin(linspace(1,7,4)(1:end-1).^2) ans = 0.8414709848078965 0.4121184852417566 -0.132351750097773 Now multiply each of those numbers (the heights of the rectangles) by the width (2) of each rectangle, then add up those areas: octave> sum(sin(linspace(1,7,4)(1:end-1).^2)*2) ans = 2.24247543990376 11MATLAB would require you to use two steps: MATLAB>> x = linspace(1,7,4); MATLAB>> x(1:end-1)	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 slope by using nonrectangular regions. For the trapezoid rule those regions are trape- zoids, while Simpson’s rule uses quasi-rectangular regions whose top edges are parabolas, as shown in Figure 6.6.1: y x a b xi xi+1 y = f (x) (a) rectangle method y x a b xi xi+1 y = f (x) (b) trapezoid rule y x a b xi xi+1 xi+2 y = f (x) (c) Simpson’s rule Figure 6.6.1 Comparison of numerical integration methods for Rb a f (x)dx For a partition P = {a = x0 < x1 < ··· < xn−1 < xn = b} of an interval [a,b] into n ≥1 subinter- vals of equal width h = (b−a)/n, let yi = f (xi) for i = 0, 1, ..., n. The trapezoid rule adds up the areas of trapezoids on each subinterval [xi, xi+1], with the top edge being the line segment join- ing the points (xi, yi) and (xi+1, yi+1). The approximation formula is straightforward to derive, based on areas of trapezoids: Trapezoid rule: Zb a f (x) dx ≈h 2 (y0 + 2y1 + 2y2 + ··· + 2yn−1 + yn) Simpson’s rule depends on pairs of neighboring subintervals: [x0, x1] and [x1, x2], [x2, x3] and [x3, x4], ... , [xn−2, xn−1] and [xn−1, xn]. Thus, n ≥2 must be even. The top edge of the region over each pair [xi, xi+1] and [xi+1, xi+2] is the unique parabola joining the 3 points (xi, yi), (xi+1, yi+1), and (xi+2, yi+2). The approximation formula is then:12 Simpson’s rule: Zb a f (x) dx ≈h 3 (y0 + 4y1 + 2y2 + 4y3 + 2y4 + ··· + + 2yn−2 + 4yn−1 + yn) 12For a full derivation of both formulas, see pp.144-149 in HORNBECK, R.W., Numerical Methods, New York: Quantum Publishers, Inc., 1975.	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 summation formulas for both rules. In particular the trapezoid rule approximation becomes Zpπ 0 sin(x2) dx ≈h 2 (0 + 2y1 + 2y2 + ··· + 2yn−1 + 0) = h · Ã n−1 X k=1 yk ! which is simple to implement in Octave: octave> x = linspace(0,sqrt(pi),1+1e5); octave> h = sqrt(pi)/1e5; octave> hsum(sin(x(2:end-1).^2)) ans = 0.8948314693913405 Likewise, the Simpson’s rule approximation becomes Zpπ 0 sin(x2) dx ≈h 3 (4y1 + 2y2 + 4y3 + 2y4 + ··· + + 2yn−2 + 4yn−1) which can be implemented easily by using Octave’s powerful indexing features: octave> (h/3)(4sum(sin(x(2:2:end-1).^2)) + 2sum(sin(x(3:2:end-1).^2))) ans = 0.8948314694841457 In the above command, the statement x(3:2:end-1) allows you to skip every other element in the list x after position 3, by moving up the list in increments of 2 positions all the way to the next-to-last position in the list (end-1). Similarly for x(2:2:end-1), which starts at position 2 and then moves up in increments of 2. Note that Simpson’s rule gives essentially the true value in this case, and the value from the trapezoid rule is virtually the same as the value produced by the built-in trapz function in Octave/MATLAB: octave> trapz(x,sin(x.^2)) ans = 0.8948314693913402 In general you are better off using these sorts of built-in functions instead of implementing your own. Typically Simpson’s rule is slightly more efﬁcient than the trapezoid rule, which is slightly more efﬁcient than the rectangle method. However, in the above examples all the approxima- tions were accurate to at least 9 decimal places (equivalent to getting the distance between Detroit and Chicago correct within the thickness of a toothpick). The running time of each calculation was only a few thousandths of a second. Modern computing has generally made the efﬁciency differences negligible.	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 wi = h 3 , 2h 3 , or 4h 3 , depending on the points ai in the interval [a,b]. The method of Gaussian quadrature transforms an integral over any interval [a,b] into an integral over the speciﬁc interval [−1,1] and then uses a standard set of points in [−1,1] and known weights for those points:13 Gaussian quadrature: Transform the integral Rb a f (x)dx into an integral over [−1,1] by means of the substitution u = 1 b−a(2x−a−b), so that x = b−a 2 u + a+b 2 and dx = b−a 2 du. Then Zb a f (x) dx = b −a 2 Z1 −1 g(u) du ≈b −a 2 nX i=1 wi g(ai) where g(u) = f ³ b−a 2 u + a+b 2 ´ , with the points a1, ..., an and weights w1, ..., wn given in Table 6.1 below for any choice of 2 ≤n ≤10 points in [−1,1]. Table 6.1 Table of Gaussian quadrature points and weights n a1, ..., an w1, ..., wn 2 ±0.577350 1 3 0 8/9 ±0.774597 5/9 4 ±0.339981 0.652145 ±0.861136 0.347855 5 0 0.568889 ±0.538469 0.478629 ±0.906180 0.236927 6 ±0.238619 0.467914 ±0.661209 0.360762 ±0.932470 0.171324 7 0 0.417959 ±0.405845 0.381830 ±0.741531 0.279705 ±0.949108 0.129485 n a1, ..., an w1, ..., wn 8 ±0.183435 0.362684 ±0.525532 0.313707 ±0.796666 0.222381 ±0.960290 0.101229 9 0 0.330239 ±0.324253 0.312347 ±0.613371 0.260611 ±0.836031 0.180648 ±0.968160 0.081274 10 ±0.148874 0.295524 ±0.433395 0.269267 ±0.679410 0.219086 ±0.865063 0.149451 ±0.973907 0.066671 13The details are beyond the scope of this book. See Chapter 4 in RALSTON, A. AND P. RABINOWITZ, A First Course in Numerical Analysis, 2nd ed., New York: McGraw-Hill, Inc., 1978. See also Table 1 in STROUD, A.H. AND D. SECREST, Gaussian Quadrature Formulas, Englewood Cliffs, NJ: Prentice-Hall, Inc., 1966.	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 points ai and weights wi are a1 = −0.339981 w1 = 0.652145 a2 = 0.339981 w2 = 0.652145 a3 = −0.861136 w3 = 0.347855 a4 = 0.861136 w4 = 0.347855 and so Z2 0 dx 1+ x3 = 2−0 2 Z1 −1 g(u) du = Z1 −1 du 1+(u+1)3 ≈ 4X i=1 wi g(ai) = 4X i=1 wi · 1 1+(ai +1)3 ≈ 0.652145 1+(−0.339981+1)3 + 0.652145 1+(0.339981+1)3 + 0.347855 1+(−0.861136+1)3 + 0.347855 1+(0.861136+1)3 ≈1.091621 The true value of the integral to six decimal places is 1.090002. Using more points (e.g. n = 7) is easy to implement in Octave, using element-wise operations on arrays: octave> a = [ 0 -0.405845 0.405845 -0.741531 0.741531 -0.949108 0.949108 ]; octave> w = [ 0.417959 0.381830 0.381830 0.279705 0.279705 0.129485 0.129485 ]; octave> sum(w./(1 + (a+1).^3)) ans = 1.090016688064804 Gaussian quadrature can be applied to improper integrals. For example, Z∞ 0 f (x) e−x dx ≈ nX i=1 wi f (ai) using the points ai and weights wi in Table 6.214 for n = 3,4, or 5 points in [0,∞): Table 6.2 Table of Gaussian quadrature points and weights for R∞ 0 f (x) e−x dx n a1, ..., an w1, ..., wn 3 0.415775 0.711093 2.294280 0.278518 6.289945 0.010389 n a1, ..., an w1, ..., wn 4 0.322548 0.603154 1.745761 0.357419 4.536620 0.038888 9.395071 0.000539 n a1, ..., an w1, ..., wn 5 0.263560 0.521756 1.413403 0.398667 3.596426 0.075942 7.085810 0.003612 12.640801 0.000023 14See Table 6 in STROUD, A.H. AND D. SECREST, Gaussian Quadrature Formulas.	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 x5 e−x dx ≈ nX i=1 wi f (ai) = nX i=1 wi a5 i ≈0.711093(0.415775)5 + 0.278518(2.294280)5 + 0.010389(6.289945)5 ≈119.9974709727211 The true value is Γ(6) = 5! = 120. Note: The points ai in Table 6.2 are the roots of the Laguerre polynomials of degree n. Exercises A 1. A simple pendulum of length l swings through an angle of 90◦on either side of the vertical with period P, given by P = 4 s l g Zπ/2 0 dθ p 1 −0.5 sin2 θ where g = 9.8 m/s2 is the acceleration due to gravity. Use the rectangle method (with left endpoints), the trapezoid rule, and Simpson’s rule to write P as a constant multiple of p l/g. Preferably, use a computer and n = 105 subintervals of equal width (or n = 10 subintervals if calculating by hand). 2. Repeat Exercise 1 using Gaussian quadrature with n = 5 points. 3. Approximate the value of Zpπ 0 sin(x2) dx by using Gaussian quadrature with n = 7 points. 4. Repeat Exercise 3 with n = 9 points. 5. Repeat Exercise 3 with n = 10 points. 6. The points ai in Table 6.1 for Gaussian quadrature are the roots of the Legendre polynomials Pn(x), with P0(x) = 1, P1(x) = x, and Pn(x) deﬁned for integers n ≥2 by the recursion formula nPn(x) = (2n−1)xPn−1(x) −(n−1)Pn−2 (x) (a) Write out Pn(x) explicitly in standard polynomial form for n = 2,3,4,5. (b) Verify that the roots of Pn(x) match the n points a1, ..., an in Table 6.1 for n = 2,3,4,5. (c) With no calculations, explain why R1 −1 P2(x)P3(x)dx = R1 −1 P3(x)P4(x)dx = R1 −1 P4(x)P5(x)dx = 0. (d) For n = 0,1,2,3, verify that Z1 −1 P2 n(x) dx = 2 2n+1 . 7. Repeat Example 6.34 with n = 4 points. 8. Repeat Example 6.34 with n = 5 points. 9. Use Table 6.2 to approximate the value of Z∞ 0 ln(1+ x) e−x dx with n = 5 points.	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 that plane. The ﬁxed point is the center of the circle, while the ﬁxed distance from the center is the circle’s radius. Similarly, the question “What is an ellipse?” would likely be answered with “an oval,” “some- thing egg-shaped,” or “a squished circle.” A precise deﬁnition would be: An ellipse is the set of all points in a plane such that the sum of the distances from each of those points to two ﬁxed points in the plane is the same constant. The two ﬁxed points are the foci—plural of focus—of the ellipse. Figure 7.1.1 illustrates the above deﬁnitions, with a point P moving along each curve. r P C (a) Circle: center C, radius r = constant P d1 d2 F1 F2 (b) Ellipse: foci F1 and F2, d1 + d2 = constant Figure 7.1.1 The circle and its “squished” sibling the ellipse Along the ellipse the sum d1 + d2 of the distances from P to the foci remains constant. 202	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, using only the deﬁnition. Stick two pins in a board and tie the ends of a piece of string to the pins, with the string long enough so that there is some slack (see Figure 7.1.2(a)). The pins will be the foci of the ellipse. (a) Two pins and string on a board P (b) Pull the string taut with a pencil Figure 7.1.2 Construction of an ellipse Pull the string taut with a pencil whose point is touching the board, then move the pencil around as far as possible on all sides of the pins. The drawn ﬁgure will be an ellipse, as in Figure 7.1.2(b). The length of the string is the constant sum d1 + d2 of distances from points on the ellipse to the foci. The symmetry of the ellipse is obvious. There is some terminology connected with ellipses. The principal axis is the line containing the foci, and the center is halfway between the foci, as in Figure 7.1.3: principal axis major axis minor axis F1 (focus) F2 (focus) C (center) V1 (vertex) V2 (vertex) Figure 7.1.3 The parts of an ellipse The vertexes are the points where the ellipse intersects the principal axis. The major axis is the chord joining the vertexes, and the minor axis is the chord through the center that is perpendicular to the major axis. The two semi-major axes are the halves of the major axis joining the center to the vertexes (CV1 and CV2 in Figure 7.1.3). Likewise the semi-minor axes are the two halves of the minor axis. A chord through the center is a diameter. Notice that a circle is the special case of an ellipse with identical foci (i.e. the foci and center are all the same point).	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 methods from analytic geometry—the study of geometric objects in the context of coordinate systems.2 You have already seen the equation of an ellipse in the xy-plane centered at the origin: x2 a2 + y2 b2 = 1, where a > b > 0, with the x-axis as the principal axis. The equation is straightforward to derive from the deﬁnition of the ellipse. x y (x, y) d1 d2 (−c,0) (c,0) In the xy-plane, let the foci of an ellipse be the points (±c,0) for some c > 0, so that the center is the origin (0,0) and the x-axis is the principal axis, as in the ﬁgure on the right. Denote by 2a the constant sum d1 + d2 of the distances from points (x, y) on the ellipse to the foci, with a > 0. Notice that a > c, since the distance 2c between the foci must be less than d1 + d2 = 2a. Then by the distance formula, d1 + d2 = 2a q (x+ c)2 + y2 + q (x−c)2 + y2 = 2a µq (x−c)2 + y2 ¶2 = µ 2a − q (x+ c)2 + y2 ¶2 (x−c)2 + ✓✓ y2 = 4a2 −4a q (x+ c)2 + y2 + (x+ c)2 + ✓✓ y2 4a q (x+ c)2 + y2 = 4a2 + (x+ c)2 −(x−c)2 ✁4a q (x+ c)2 + y2 = ✁4a2 + ✁4xc q (x+ c)2 + y2 = a + c a x x2 + ✟✟ 2cx + c2 + y2 = a2 + ✟✟ 2cx + c2 a2 x2 ³ 1−c2 a2 ´ x2 + y2 = a2 −c2 x2 a2 + y2 a2 −c2 = 1 x2 a2 + y2 b2 = 1 where b2 = a2 −c2 (and so a > b > 0) ✓ 1The word ellipse is in fact due to the Greek astronomer and geometer Apollonius of Perga (ca. 262-190 B.C.), which seems an improvement over the name “thyreos” that Euclid (ca. 360-300 B.C.) had given the shape. 2Pioneered by the French mathematician and philosopher René Descartes (1596-1650), for whom the Cartesian coordinate system is named. The proposition “I think, therefore I am” (Cogito, ergo sum) is due to Descartes.	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 the major axis goes from (−a,0) to (a,0) and has length 2a (i.e. the semi-major axis has length a). Similarly, setting x = 0 shows the minor axis goes from (0,−b) to (0,b), (i.e. the semi-minor axis has length b). Since a > b > 0 and b2 = a2 −c2, then c = p a2 −b2. Thus, for any ellipse of the form x2 a2 + y2 b2 = 1 with a > b > 0, the foci will be at (±c,0), where c = p a2 −b2. The foci can be used to deﬁne an important geometric property of the ellipse: The eccentricity of an ellipse is the ratio of the distance between the foci to the length of the major axis. In the case of an ellipse x2 a2 + y2 b2 = 1 with a > b > 0, the eccentricity e is given by e = c a = p a2 −b2 a . The eccentricity e is a measure of how “oval” an ellipse is, with 0 < e < 1. The boundary case e = 0 is a circle, while e = 1 is a line segment; an ellipse is somewhere in between—the closer e gets to 1, the more “squished” the ellipse. See Figure 7.1.5. C (a) Circle: e = 0 F1 F2 (b) Ellipse: e = 0.5 F1 F2 (c) Ellipse: e = 0.75 F1 F2 (d) Segment: e = 1 Figure 7.1.5 Eccentricity e Earth’s elliptical orbit around the Sun is almost circular: the eccentricity is 0.017. Only the orbits of Venus and Neptune (both at 0.007) have a lower eccentricity among the nine planets in the solar system, while Pluto’s (0.252) has the highest. It is left as an exercise to show that the ellipse x2 a2 + y2 b2 = 1 with a > b > 0 can be written in terms of the eccentricity e: y2 = (1−e2)(a2 −x2) (7.1)	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 hemisphere. Thus, Area = 4 Za 0 y dx = 4b a Za 0 p a2 −x2 dx = 4b a µ a2 2 sin−1 ³ x a ´ + 1 2 x p a2 −x2 ¯¯¯¯ a 0 ¶ (by formula (6.7) in Section 6.3) = 4b a µ a2 2 π 2 + 0 −(0 + 0) ¶ = πab P F1 F2 Figure 7.1.6 A remarkable feature of the ellipse is the reﬂection property: light shone from one focus to any point on the ellipse will reﬂect to the other focus. Figure 7.1.6 shows light emanating from the focus F1 and reﬂecting off a point P on the ellipse to the other focus F2. Fermat’s Principle from Example 4.4 in Section 4.1 showed that the incoming angle θ1 (angle of incidence) of light from point A will equal the outgoing angle θ2 (angle of reﬂection) to point B for light reﬂecting off a ﬂat reﬂective surface at point P, as in Figure 7.1.7(a). Fermat’s Principle also applies to curved surfaces—e.g. an ellipse—with the angles measured relative to the tangent line to the curve at the point of reﬂection, as in Figure 7.1.7(b). A B P θ1 θ2 (a) Flat surface tangent line A B P θ1 θ2 (b) Curved surface Figure 7.1.7 Fermat’s Principle for reﬂection: θ1 = θ2	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 F2 c 0 −a a b N P(x0, y0) n = normal line tangent line α1 α2 (b) Ellipse: show that α1 = α2 Figure 7.1.8 Fermat’s Principle with normal line Thus, to prove the reﬂection property, it sufﬁces to prove that the normal line n to the ellipse at P bisects the angle ∠F1PF2 in Figure 7.1.8(b)—this would make α1 = α2, so that the indi- cated path from F1 to P to F2 satisﬁes Fermat’s Principle. First, let P = (x0, y0) be a point on the ellipse x2 a2 + y2 b2 = 1, with a > b > 0. Assume that P is not a vertex (i.e. y0 ̸= 0), otherwise the reﬂection property holds trivially. By Exercise 15 in Section 3.4, the equation of the tangent line to the ellipse at P = (x0, y0) is xx0 a2 + yy0 b2 = 1 , (7.2) so that its slope is −b2x0 a2 y0 . Hence, the negative reciprocal a2 y0 b2x0 is the slope of the normal line n, whose equation—valid even when x0 = 0 (i.e. when y0 = ±b)— is then b2x0 (y−y0) = a2y0 (x−x0) . (7.3) Setting y = 0 and solving for x shows the x-intercept of n is at x = (a2 −b2) x0y0 a2y0 = c2 a2 x0 = e2 x0 Let N = (e2x0,0) be that x-intercept, as in Figure 7.1.8(b). The distance F1N from the focus F1 = (−c,0) = (−ea,0) to N is then F1N = e2 x0 −(−ea) = e(a+ ex0) while the distance F2N from the focus F2 = (c,0) = (ea,0) to N is F2N = ea −e2 x0 = e(a−ex0) . Therefore, F1N F2N = e(a+ ex0) e(a−ex0) = a+ ex0 a−ex0 .	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 F2 = (ea,0) to P = (x0, y0) is given by (F2P)2 = (x0 −ea)2 + y2 0 = x2 0 −2eax0 + e2a2 + (1−e2)(a2 −x2 0) (F2P)2 = a2 −2eax0 + e2x2 0 = (a−ex0)2 F2P = a−ex0 . F1 F2 N P α1 α2 θ 180◦−θ Thus, F1P F2P = a+ ex0 a−ex0 = F1N F2N , which means that α1 = α2:3 by the Law of Sines, and with θ = ∠F2NP as in the ﬁgure on the right, sin α2 F2N = sin θ F2P = sin(180◦−θ) F2P = sin(180◦−θ) F1P · F1P F2P = sin α1 F1N · F1N F2N = sin α1 F2N and thus sin α2 = sin α1, so that α2 = α1 (since 0◦< α1, α2 < 90◦). ✓ x y −c c −b b 0 a −a x2 b2 + y2 a2 = 1 Figure 7.1.9 An ellipse of the form x2 b2 + y2 a2 = 1 with a > b > 0 simply switches the roles of x and y in the previ- ous examples: the principal axis is now the y-axis, the foci are at (0,±c), where c = p a2 −b2, and the vertexes are at (0,±a), as in Figure 7.1.9. Thus, the largest denominator on the left side of an equation of the form x2 □2 + y2 □2 = 1 tells you which axis is the principal axis. For example, the principal axis of the ellipse x2 25 + y2 16 = 1 is the x-axis (since 25 > 16), while the ellipse x2 4 + y2 9 = 1 has the y-axis as its principal axis (since 9 > 4). 3This follows directly from Proposition 3 in Book VI of Euclid’s Elements. See the purely geometric proof on pp.125-126 in EUCLID, Elements, (Thomas L. Heath translation), Santa Fe, NM: Green Lion Press, 2002.	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 eccentricity e. 2. x2 25 + y2 16 = 1 3. x2 4 + y2 9 = 1 4. 4x2 25 + y2 4 = 1 5. x2 +4y2 = 1 6. 25x2 +9y2 = 225 7. Show that for a > b > 0 the ellipse x2 a2 + y2 b2 = 1 with eccentricity e can be written as y2 = (1−e2)(a2−x2). 8. Use Example 7.1 to show the area inside the ellipse x2 a2 + y2 b2 = 1 with eccentricity e is πa2 p 1−e2. 9. For all a > b > 0, ﬁnd the points of intersection of the ellipses x2 a2 + y2 b2 = 1 and x2 b2 + y2 a2 = 1. 10. Show that the vertexes are the closest and farthest points on an ellipse to either focus. B 11. Show that any line of slope m that is tangent to the ellipse x2 a2 + y2 b2 = 1 must be of the form y = mx ± p a2m2 + b2 . 12. A 10ft ladder with a mark 3ft from the top rests against a wall. If the top of the ladder slides down the wall, with the foot of the ladder sliding away from the wall on the ground, as in Figure 7.1.10, then show the mark moves along part of an ellipse. y x 3 7 Figure 7.1.10 Exercise 12 F P D G Figure 7.1.11 Exercise 13 T1 T2 Figure 7.1.12 Exercise 14 13. Another deﬁnition of an ellipse is the set of points P for which the ratio of the distance from P to a ﬁxed point F (a focus) to the distance from P to a ﬁxed line D (the directrix) is a constant e < 1 (the eccentricity): PF PG = e, as in Figure 7.1.11. Use this deﬁnition to show that the equation of an ellipse with focus (c,0) can be written as x2 a2 + y2 b2 = 1 for some a > b > 0. Find the equation of the directrix. 14. Show that the set of intersection points of all perpendicular tangent lines to an ellipse form a circle, as in Figure 7.1.12 (showing two such tangent lines T1 ⊥T2). 15. A chord of an ellipse that passes through a focus and is perpendicular to the major axis is a latus rectum. Show that for the ellipse x2 a2 + y2 b2 = 1 with a > b > 0 the length of each latus rectum is 2b2 a . 16. Suppose that the normal line at one end of a latus rectum of an ellipse passes through an end of the minor axis. Show that the eccentricity e is a root of the equation e4 + e2 −1 = 0, then ﬁnd e. 17. Show that the set of all midpoints of a family of parallel chords in an ellipse lie on a diameter. (Hint: Use symmetry with chords of slope m ̸= 0.)	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 Section 7.1 is, in fact, similar to the deﬁnition of the parabola: A parabola is the set of all points in a plane that are equidistant from a ﬁxed point (the focus) and a ﬁxed line (the directrix). D F P G Figure 7.2.1 Parabola: PF = PG Figure 7.2.1 illustrates the above deﬁnition, with a point P moving along the parabola so that the distance from P to the focus F equals the distance from P to the directrix D. Note that the point halfway between the focus and directrix must be on the parabola—that point is the vertex, which is the point on the parabola closest to the directrix. The axis of the parabola is the line that passes through the focus and is perpendicular to the directrix. Notice that the ratio PF PG equals 1, whereas that ratio for an ellipse—by the alternative deﬁnition—was the eccentricity e < 1. The eccentricity of the parabola, therefore, is always 1.4 To construct a parabola from the deﬁnition, cut a piece of string to have the same length AB as one side of a drafting triangle, as in Figure 7.2.2. D F P B A C P Figure 7.2.2 Construction of a parabola Fasten one end of the string to the vertex A of the triangle and the other end to a pin somewhere between A and B—the pin will be the focus F of the parabola. Hold the string taut against the edge AB of the triangle at a point P on either side of the pin, then move the edge BC of the triangle along the directrix D. The drawn ﬁgure will be a parabola, as the lengths PF and PB will be equal (since the length of the string being AB = AP +PF means PF = PB). 4The eccentricity e of the parabola being 1 means there is no second vertex, unlike the ellipse (where e < 1 forced the existence of two vertexes in the alternative deﬁnition).	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 point (x, y) whose distances d1 and d2 from the focus (0, p) and directrix y = −p, respectively, are equal. Then d2 1 = d2 2 (x−0)2 + (y−p)2 = (x−x)2 + (y+ p)2 x2 + ✓✓ y2 −2py +	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 and y, the tangent line to the parabola 4px = y2 at a point (x0, y0) is: 2p(x+ x0) = y0 y (7.5) x y F = (p,0) P (x0, y0) (−x0,0) Q 0 β β Figure 7.2.4 4px = y2 Formula (7.5) simpliﬁes the proof of the reﬂection property for parabolas: light shone from the focus to any point on the parabola will reﬂect in a path parallel to the axis of the parabola. Figure 7.2.4 shows light emanating from the focus F = (p,0) and reﬂecting off a point P = (x0, y0) on the parabola 4px = y2. If that line of reﬂection is parallel to the x-axis—the axis of the parabola—then the tangent line to the parabola at (x0, y0) should make the same angle β with the line of reﬂection as it does with the x-axis. So extend the tangent line to intersect the x-axis and use formula (7.5) to ﬁnd the x-intercept: 2p(x+ x0) = y0 y = y0 ·0 = 0 ⇒ x = −x0 Let Q = (−x0,0), so that the distance FQ equals p + x0. The goal is to show that the angle of incidence ∠FPQ equals the angle of reﬂection β. The focal radius FP has length FP = q (p −x0)2 +(0−y0)2 = q p2 −2px0 + x2 0 +4px0 = q p2 +2px0 + x2 0 = p + x0 . Thus, FQ = FP in the triangle △FPQ, so that ∠FPQ = ∠FQP = β, i.e. the light’s path does indeed satisfy Fermat’s Principle for curved surfaces. ✓ The parabola’s reﬂection property shows up in some engineering applications, typically by revolving part of a parabola around its axis, producing a parabolic surface in three dimen- sions called a paraboloid. For example, it used to be common for vehicle headlights to use paraboloids for their inner reﬂective surface, with a bulb at the focus, so that—by the reﬂection property—the light would shine straight ahead in a solid beam. Many ﬂashlights still operate on that principle. The reﬂection property also works in the opposite direction, which is why satellite dishes and radio telescopes are often wide paraboloids with a signal receiver at the focus, to maximize reception of incoming reﬂected signals.	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 parabola. v2 0 2g 0 v2 0 g Solution: Recall from Example 4.3 in Section 4.1 that if the object is launched at an angle 0 < θ < π 2 with the ground, then the height y attained by the object as a function of the hori- zontal distance x that it travels is given by y = − gx2 2v2 0 cos2 θ + xtan θ . The curve is a parabola, with the ﬁgure on the right showing these parabolic trajectories for 500 values of the angle θ. Clearly each parabola in- tersects at least one other.The maximum hori- zontal distance v2 0 g occurs only for θ = π 4 , as was shown in Example 4.3. The maximum vertical height v2 0 2g is attained when the object is launched straight up (i.e. θ = π 2 ), as was shown in Exercise 18 in Section 5.1. By symmetry only the angles 0 < θ ≤π 2 in the same vertical plane need be considered. So in the above ﬁgure, imagine if the trajectories for all possible angles were included, ﬁlling up a region that does appear to have a parabolic boundary. This will now be shown to be true. First, it turns out that all the parabolas for 0 < θ < π 2 have the same directrix y = v2 0 2g. To see why, recall from Exercise 11 in Section 4.1 that the maximum height reached by the object is v2 0 sin2 θ 2g , which is thus the y-coordinate of the vertex of the parabola. That vertex is midway between the focus and the directrix. The parabola is of the form 4py = x2 + bx, where b is a constant that does not affect the distance between the vertex and directrix5, and 4p is a constant with p < 0 such that the directrix is −p units above the vertex (since p < 0), just as in the case 4py = x2. The equation of the parabola then shows that 1 4p = − g 2v2 0 cos2 θ ⇒ p = − v2 0 cos2 θ 2g so that the directrix is at y = y-coordinate of the vertex + (−p) = v2 0 sin2 θ 2g + − Ã − v2 0 cos2 θ 2g ! = v2 0 2g(sin2 θ + cos2 θ) y = v2 0 2g It is perhaps surprising that all the parabolic trajectories share the same directrix y = v2 0 2g, which is independent of the angle θ. Note that the heights of each vertex µ v2 0 sin2 θ 2g ¶ and focus µ v2 0 2g(sin2 θ −cos2 θ) ¶ do depend on θ. The common directrix is the key to the remainder of the proof. 5This will be discussed further in Section 7.4.	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 (b) Outside the envelope D y x P O v2 0 2g v2 0/g (c) On the envelope Figure 7.2.5 Trajectory envelope and a point P The origin O = (0,0) is on each trajectory, so by deﬁnition of a parabola the foci for all the trajectories must be a distance v2 0 2g from O, i.e. the distance from O to D. That is, the foci of all the trajectories must lie on the circle C0 of radius v2 0 2g centered at O. If P is any other point inside the envelope, so that it lies on at least one trajectory, then it must be a distance r > 0 below the line D. By deﬁnition of a parabola, P must be the same distance from the foci of any trajectories it belongs to. That is, the foci must be on a circle C of radius r centered at P and touching the directrix D, as in Figure 7.2.6: D y x P O v2 0 2g v2 0/g C F1 F2 C0 r (a) Two intersections F1, F2 D y x P O v2 0 2g v2 0/g C0 C r (b) No intersections D y x P F O v2 0 2g v2 0/g C0 C r (c) One intersection F Figure 7.2.6 Circles C and C0 intersect at foci of trajectories L D y x P O v2 0 2g v2 0 g v2 0/g r v2 0 2g In Figure 7.2.6(a) C and C0 intersect at two points F1 and F2, so P belongs to two trajectories; P must then be inside the envelope. In Figure 7.2.6(b) C and C0 do not intersect, so P must be outside the envelope (since it is not on a parabola with a focus on C0). If C and C0 intersect at only one point F, as in Figure 7.2.6(c), then P must be on the envelope. In that case, P is a distance r + v2 0 2g from O, which is also the distance from P to the line y = v2 0 g (denoted by L). Thus, by deﬁnition of a parabola, P is on a parabola with focus O and directrix L. The vertex is at µ 0, v2 0 2g ¶ . Therefore, the envelope is a parabola: the boundary of the shaded region in the ﬁgure on the right. ✓	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 suspension bridges: the suspension cables supporting a horizontal bridge (via vertical suspenders, as in the ﬁgure on the right) have to be parabolas if the weight of the bridge is uniformly distributed.6 Exercises A 1. Construct a parabola using the procedure shown in Figure 7.2.2. For Exercises 2-6, sketch the graph of the given parabola and indicate the exact locations of the focus, vertex, and directrix. 2. 8y = x2 3. y = 8x2 4. x = y2 5. x = −3y2 6. −1000y = x2 7. Find the points of intersection of the parabolas 4py = x2 and 4px = y2 when p > 0. What is the equation of the line through those points? 8. A vehicle headlight in the shape of a paraboloid is 3in deep and has an open edge with diameter 8in. Where should the center of the bulb be placed in order to be at the focus, measured in inches relative to the vertex? 9. The latus rectum of a parabola is the chord that passes through the focus and is parallel to the directrix. Find the length of the latus rectum for the parabola 4py = x2. 10. Show that the circle whose diameter is the latus rectum of a parabola touches the parabola’s direc- trix at one point. 11. Find the points on the parabola 4px = y2 such that the focal radii to those points have the same length as the latus rectum. 12. From each end of the latus rectum of a parabola draw a line to the point where the directrix and axis intersect. Show that the two drawn lines are perpendicular. B 13. Show that any point not on a parabola is on either zero or two tangent lines to the parabola. 14. Show that y = mx−2mp−m3 p is the normal line of slope m to the parabola 4px = y2. 15. From a point P on a parabola with vertex V let PQ be the line segment perpendicular to the axis at a point Q. Show that PQ2 equals the product of QV and the length of the latus rectum. 16. Show that the curve y = ax2 + bx+ c is a parabola for a ̸= 0, using only the deﬁnition of a parabola. Find the focus, vertex and directrix. 17. Show that the set of all midpoints of a family of parallel chords in a parabola lie on a line parallel to the parabola’s axis. 6See pp.159-161 in SMITH, C.E., Applied Mechanics: Statics, New York: John Wiley & Sons, Inc., 1976.	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 hyperbola is the set of all points in a plane such that the ratio of the distance from a ﬁxed point (a focus) to the distance from a ﬁxed line (a directrix) is a constant e > 1, called the eccentricity of the hyperbola. x y 0 y = 1 x Figure 7.3.1 It will be shown in Section 7.4 that the curve y = 1 x is a hyperbola, which has two branches (see Figure 7.3.1). In general a hyperbola resembles a “wider” or less “cupped” parabola, and it has two symmet- ric branches (and hence two foci and two directrices) as well as two asymptotes. The ratio of distances referred to in the deﬁnition of the hyperbola also appears in the second deﬁnition of the ellipse (where the ratio is smaller than 1) and in the deﬁnition of the parabola (where the ratio equals 1). In all three cases that ratio is the eccentricity. See Figure 7.3.2(a) for the comparisons. F e < 1 e = 1 e > 1 D (a) Eccentricity e Sun v < vE v = vE v > vE D (b) Orbit velocity v and escape velocity vE Figure 7.3.2 Hyperbola, parabola and ellipse with focus F and directrix D There is an analogue for Figure 7.3.2(a) in terms of orbits. For example, an object approach- ing the Sun must meet or exceed the escape velocity to overcome the Sun’s gravitational pull and avoid returning to orbit. Figure 7.3.2(b) shows the three possible trajectories—hyperbola, parabola and ellipse—in terms of the object’s velocity v and the escape velocity vE. Note the apparent correlation between the eccentricity of the object’s path and its speed as a fraction of the escape velocity (i.e. v vE ).	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- tricity). x y (ea,0) x = a e d2 d1 0 (x, y) Figure 7.3.4 To derive the equation of a hyperbola with eccentricity e > 1, assume the focus is on the x-axis at (ea,0), with a > 0, and the line x = a e is the directrix, as in Figure 7.3.4. Pick a point (x, y) whose distances d1 and d2 from the focus (ea,0) and directrix x = a e , respectively, satisfy the condition for a hyperbola: d1 d2 = e > 1. Then d2 1 = e2d2 2 (x−ea)2 + (y−0)2 = e2 µ³ x−a e ´2 + (y−y)2 ¶ x2 −✘✘✘ 2eax + e2a2 + y2 = e2x2 −✘✘✘ 2eax + a2 (e2 −1) x2 −y2 = (e2 −1)a2 x2 a2 −y2 b2 = 1 where b2 = (e2−1)a2 > 0. The hyperbola thus has two branches (x = ± a b p y2 + b2), as in Figure 7.3.5. Let c = ea, so that c > a and b2 = c2 −a2. By symmetry the hyperbola has two foci (±c,0) and two directrices x = ± a2 c , and the lines y = ± b a x are asymptotes. The vertexes are the points on the hyperbola closest to the directrices. x y (c,0) focus vertex (−c,0) focus vertex 0 a2 c −a2 c a −a asymptote y = b a x asymptote y = −b a x conjugate axis transverse axis directrix directrix Figure 7.3.5 Parts of the hyperbola x2 a2 −y2 b2 = 1, with b2 = c2 −a2 The center is the point midway between the foci. The transverse axis and conjugate axis are the perpendicular lines through the foci and center, respectively.	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 directrix—is simply a consequence of the symmetry about both axes imposed by the equation x2 a2 −y2 b2 = 1. A parabola, in comparison, is symmetric about only one axis. To see why the lines y = ± b a x are asymptotes, consider the upper half y = b a p x2 −a2 of both branches of the hyperbola. The difference between the line y = b a x and the upper right branch approaches zero as x approaches inﬁnity: lim x→∞ µb a x −b a p x2 −a2 ¶ = b a lim x→∞ ³ x − p x2 −a2 ´ · x + p x2 −a2 x + p x2 −a2 = b a lim x→∞ x2 −(x2 −a2) x + p x2 −a2 = lim x→∞ ab x + p x2 −a2 = 0 Thus the line y = b a x is an (oblique) asymptote for the upper half y = b a p x2 −a2 of the right branch in the ﬁrst quadrant of the xy-plane. So by symmetry the lines y = ± b a x are asymptotes for both branches, i.e. for the entire hyperbola. Conversely, given a hyperbola in the form x2 a2 −y2 b2 = 1, let c2 = a2 + b2 to get the foci (±c,0) and the directrices x = ± a2 c , while the eccentricity is e = c a. Example 7.3 For the hyperbola x2 −y2 = 1 ﬁnd the vertexes, foci, directrices, asymptotes and eccentricity. Solution: Here a = b = 1, so that c2 = a2 + b2 = 2, i.e. c = p 2. The vertexes are thus (±1,0), the asymptotes are y = ±x, the foci are (± p 2,0), the directrices are x = ± 1 p 2, and the eccentricity is e = p 2. x y c a y = a2 c −c −a y = −a2 c y = a b x y = −a b x Figure 7.3.6 y2 a2 −x2 b2 = 1 Switching the roles of x and y produces the rotated hyper- bola y2 a2 −x2 b2 = 1, shown in Figure 7.3.6. The transverse axis is the y-axis, the conjugate axis is the x-axis, the vertexes are at (0,±a), and the foci are at (0,±c), where c2 = a2+b2. The directrices are y = ± a2 c , the asymptotes are y = ± a b x, and the eccentricity is e = c a. For example, the hyperbola y2 −x2 = 1 has a = b = 1, so that c = p 2. The vertexes are (0,±1), the asymptotes are y = ±x, the foci are (0,± p 2), the directrices are y = ± 1 p 2, and the eccentricity is e = p 2. The hyperbola y2 −x2 = 1 is just the hyperbola x2 −y2 = 1 rotated 90◦.	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 and F2. The difference d1 −d2 of the distances d1 and d2 can be positive or negative depending on which branch the point (x, y) is on, which is why the absolute value \|d1 −d2\| is used. x y F2 F1 d1 d2 0 (x, y) Figure 7.3.7 Hyperbola: \|d1 −d2\| = constant > 0 F2 F1 P A P L Figure 7.3.8 Construction of a hyperbola It is left as an exercise to show that this second deﬁnition yields the same equation of the hyperbola as from the ﬁrst deﬁnition. The second deﬁnition is often used as the primary def- inition in many textbooks, perhaps because it provides a simple way to construct a hyperbola by hand. Figure 7.3.8 shows the procedure for foci F1 and F2: at the focus F1 fasten one end of a ruler of length L, and at the other end A of the ruler fasten one end of a string of length L−d for some number 0 < d < F1F2. Fasten the other end of the string to the focus F2 and hold the string taut with a pencil against the ruler at a point P, while rotating the ruler about F1. The drawn ﬁgure will be one branch of a hyperbola, since the difference PF1 −PF2 will always be the positive constant d: PF1 −PF2 = (L −AP) −PF2 = L −(AP + PF2) = L −(L −d) = d Reverse the roles of F1 and F2 to draw the other branch of the hyperbola. By Exercise 16 in Section 3.4, the tangent line to the hyperbola x2 a2 −y2 b2 = 1 at a point (x0, y0) is xx0 a2 −yy0 b2 = 1 , (7.6) so that its slope is b2x0 a2 y0 when y0 ̸= 0. Note that by the above equation, when y0 = 0 (so that x0 = ±a) the two tangent lines are the vertical lines x = ±a.	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 along the line through P and the other focus F1. F2 F1 P Figure 7.3.9 Reﬂection property x y F2 (c,0) F1 (−c,0) P (x0, y0) α1 α2 L θ1 θ2 θ2 θ A Figure 7.3.10 Hyperbola x2 a2 −y2 b2 = 1 with foci (±c,0) By Fermat’s Principle for curved surfaces, the reﬂection property is equivalent to saying the tangent line L to the hyperbola at the point P = (x0, y0) bisects the angle ∠F1PF2, i.e. θ1 = θ2 as in Figure 7.3.10. The reﬂection property holds trivially when y0 = 0 (the light reﬂects straight back along the x-axis), so to show that θ1 = θ2 assume y0 ̸= 0. By symmetry only x0 > 0 need be considered. For the case x0 ̸= c, let A be the x-intercept of the tangent line L. Then by Figure 7.3.10, since the sum of the angles in the triangle △F1PA equals 180◦, α1 +θ2 +(180◦−θ) = 180◦ ⇒ θ2 = θ −α1 ⇒ tan θ2 = tan(θ −α1) . Thus, since tan θ is the slope of the tangent line L (i.e. b2x0 a2y0 ), and tan α1 = y0 x0+c, then by the subtraction formula for the tangent function tan θ2 = tan θ −tan α1 1 + tan θ tan α1 = b2x0 a2 y0 − y0 x0 + c 1 + b2x0 a2 y0 · y0 x0 + c = b2x2 0 −a2y2 0 + b2x0c ✭✭✭✭✭ a2 y0 (x0 + c) (a2 + b2) x0 y0 + a2 y0c ✭✭✭✭✭ a2 y0 (x0 + c) = a2b2 + b2x0c c2x0y0 + a2y0c (since c2 = a2 + b2, and x2 0 a2 − y2 0 b2 = 1 ⇒b2x2 0 −a2y2 0 = a2b2) = b2✘✘✘✘✘ (a2 + x0c) cy0✘✘✘✘✘ (a2 + x0c) = b2 cy0 Similarly, since the sum of the angles in the triangle △F2PA equals 180◦, α2 +θ1 +θ = 180◦ ⇒ θ1 = 180◦−(θ +α2) ⇒ tan θ1 = −tan(θ +α2) .	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 −x0c) = b2 cy0 = tan θ2 , i.e. θ1 = θ2 (since 0◦< θ1, θ2 < 90◦). ✓ Note: The case x0 = c is left as an exercise. Ellipses, parabolas and hyperbolas are sometimes called conic sections, due to being formed by intersections of planes with a double circular cone of unlimited extent: (a) Ellipse (b) Parabola (c) Hyperbola Figure 7.3.11 Conic sections Each double cone in Figure 7.3.11 has two nappes—a cone extending upward and one ex- tending downward. When a plane intersects only one nappe in a closed noncircular curve, as in Figure 7.3.11(a), that curve is an ellipse. A plane that is parallel to a line on one nappe, as in Figure 7.3.11(b), intersects only that nappe in a parabola. The intersection of a plane with both nappes, as in Figure 7.3.11(c), is a hyperbola. O P1 P2 Q Figure 7.3.12 To prove that the ellipse, parabola and hyperbola really are repre- sented by the indicated conic sections, ﬁrst a minor result is needed from three-dimensional geometry: tangent line segments to a sphere from the same point have equal lengths, as in Figure 7.3.12. Since the right triangles △QOP1 and △QOP2 share the same hypotenuse QO and have legs OP1 and OP2 of equal length (the radius of the sphere), the result QP1 = QP2 follows by the Pythagorean Theorem.	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 7.3.13. So β is the angle the cone makes with any circular base of the cone. This constant an- gle β, with 0◦< β < 90◦, is an intrinsic prop- erty of the cone. Let Pc be a plane that inter- sects the lower nappe of the cone in a curve C, such that Pc makes an angle α with any base circle of the cone. By symmetry, only the angles 0◦< α ≤90◦need be considered.8 Inscribe a sphere in the cone so that it touches Pc at a point F, as in Figure 7.3.13 (Pc is the tangent plane to the sphere at F). Let P0 be the plane through the circle where the inscribed sphere touches the cone, and let D be the line of intersection of the planes Pc and P0. It will be shown that F and D are the focus and directrix, respectively, of the curve C. Let P be any point on the curve C, then let Q be the point on the plane P0 that lies on a line through P and the cone’s vertex. Drop a perpendicular line segment from P to the point A in the plane P0. From A draw a perpendicular line segment to the point G on the line D. Then as Figure 7.3.13 shows, △QAP and △PAG are right triangles, with sin α = PA PG and sin β = PA PQ . However, since PF and PQ are both tangent line segments to the inscribed sphere from the same point P, the result proved earlier shows that PQ = PF. Thus, PG sin α = PA = PQ sin β = PF sin β ⇒ PF PG = sin α sin β . Let e = PF PG . Then e is the same constant sin α sin β for any point P on the curve C. Thus, by deﬁnition, e is the eccentricity of the curve C with focus F and directrix D. If 0◦< α < β then 0◦< sin α < sin β so that 0 < sin α sin β < 1, which means that 0 < e < 1, and hence C is an ellipse (by the second deﬁnition of an ellipse). Likewise, if α = β then e = 1, so that C is a parabola. Finally, if β < α ≤90◦then e > 1, so that C is a hyperbola (and will intersect both nappes of the cone). Thus, the ellipse, parabola and hyperbola truly are conic sections. ✓ 7The proof can be extended to oblique double cones. See §364 in SALMON, G.S., A Treatise on Conic Sections, London: Longmans, Green and Co., 1929. 8The case where α = 0◦results in a circle, which is typically not considered a conic section.	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, indicate each directrix and asymptote, and ﬁnd the eccentricity e. 2. x2 16 −y2 9 = 1 3. x2 8 −y2 15 = 1 4. 4x2 25 −y2 4 = 1 5. x2 −4y2 = 1 6. 25y2 −9x2 = 225 7. Find the equation of the hyperbola with foci (±5,0) and vertexes (±3,0). B 8. In the second deﬁnition of the hyperbola on p.219, let 2a be the constant absolute value of the difference of distances from points on the hyperbola to the foci, for some a > 0. Let the foci be at (±c,0) for some c > 0. Show that this second deﬁnition yields a hyperbola having an equation of the form x2 a2 −y2 b2 = 1, with b > 0. 9. Prove the reﬂection property for the hyperbola (see pp.220-221) when x0 = c for the point P = (x0, y0) on the hyperbola with foci at (±c,0). 10. Show that a straight line parallel to an asymptote of a hyperbola intersects the hyperbola at exactly one point. 11. A latus rectum (plural: latus recta) of a hyperbola is a chord through either focus perpendicular to the transverse axis. Show that the latus recta of the hyperbola x2 a2 −y2 b2 = 1 have length 2b2 a . 12. Show that the segment of an asymptote of a hyperbola between the two directrices has the same length as the line segment between the vertexes. 13. Show that the segment of a tangent line to a hyperbola between the hyperbola’s asymptotes has its midpoint at the point of tangency. 14. The focal radii of a hyperbola are the line segments from the foci to points on the hyperbola. Show that the lengths of the focal radii to points (x, y) on the hyperbola x2 a2 −y2 b2 = 1 are a + ex and ex −a, where e is the eccentricity. 15. Show that a tangent line to a hyperbola together with the hyperbola’s asymptotes bounds a triangle of constant area (i.e. the area is independent of the point of tangency on the hyperbola). 16. Show that the product of the perpendicular distances from any point on the hyperbola x2 a2 −y2 b2 = 1 to the asymptotes y = ± b a x is a constant. 17. A person at a point P = (x, y) hears the crack of a riﬂe located at the point F1 = (−1000,0) and the sound of the ﬁred bullet hitting its target located at the point F2 = (1000,0) at the same time. The bullet’s speed is 2000 ft/sec and the speed of sound is 1100 ft/sec. Find an equation relating x and y. 18. Show that the set of all midpoints of a family of parallel chords either in one branch or between the two branches of a hyperbola lie on a line through the center of the hyperbola. 19. Prove a second reﬂection property of hyperbolas: a light shone between the two branches and directed toward one focus will reﬂect toward the other focus.	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 means of coordinate transformations. x y x′ y′ k h P = (x, y) P′ = (x′, y′) x−h y−k O O′ Figure 7.4.1 Translation You have probably seen in earlier courses how the graph of a function y = f (x) can be shifted horizontally by an amount h and vertically by an amount k by replacing x and y by x−h and y−k, respectively: y−k = f (x−h) This coordinate transformation is called translation, and can be applied to any curve in the xy-plane. The origin O = (0,0) is shifted to the point O′ = (h,k), which serves as the origin of the x′y′-plane,9 as in Figure 7.4.1. Let P = (x, y) be a point in the xy-plane. Consider P as a point P′ = (x′, y′) in the x′y′-plane, so that relative to the origin O′, the x′ and y′ coordinates of P′ are: x′ = x−h y′ = y−k Using these translation equations (i.e. the substitutions x 7→x−h and y 7→y−k), the graphs of the conic sections can be translated to any center (h,k): Ellipse: For a > b > 0, an equation of the form (x−h)2 a2 + (y−k)2 b2 = 1 describes an ellipse with center (h,k), vertexes (h±a,k), and foci (h±c,k), where c2 = a2−b2. The eccentricity is e = c a, and the principal axis is the line y = k. Likewise, an equation of the form (y−k)2 a2 + (x−h)2 b2 = 1 describes an ellipse with center (h,k), vertexes (h,k±a), and foci (h,k±c), where c2 = a2−b2. The eccentricity is e = c a, and the principal axis is the line x = h. 9The prime symbol (′) does not indicate differentiation—it acts merely to distinguish the new axes.	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,k) and focus (h+ p,k). The directrix is the line x = h−p and the axis is the line y = k Hyperbola: For a ̸= 0 and b ̸= 0, an equation of the form (x−h)2 a2 −(y−k)2 b2 = 1 describes a hyperbola with center (h,k), vertexes (h ± a,k), and foci (h ± c,k), where c2 = a2+b2. The eccentricity is e = c a, the directrices are the lines x = h± a2 c , the asymptotes are the lines y−k = ± b a(x−h), and the transverse axis is the line y = k. Likewise, an equation of the form (y−k)2 a2 −(x−h)2 b2 = 1 describes a hyperbola with center (h,k), vertexes (h,k ± a), and foci (h,k ± c), where c2 = a2+b2. The eccentricity is e = c a, the directrices are the lines y = k± a2 c , the asymptotes are the lines y−k = ± a b(x−h), and the transverse axis is the line x = h. Example 7.4 x y y = 1 x = 2 0 2 4 (2− p 3,1) (2+ p 3,1) Figure 7.4.2 Ellipse (x−2)2 4 +(y−1)2 = 1 Find the vertexes and foci of the ellipse (x−2)2 4 +(y−1)2 = 1. Solution: The translation coordinates are (h,k) = (2,1). Also, a = 2 and b = 1. Thus, the vertexes are (h± a,k) = (2±2,1) = (0,1) and (4,1). Since c2 = a2 −b2 = 3 then c = p 3, so the foci are (h± c,k) = (2± p 3,1), as shown in Figure 7.4.2. Note that the ellipse (x−2)2 4 +(y−1)2 = 1 in the xy-plane is the ellipse x′2 4 + y′2 = 1 in the x′y′-plane, for the translation equations x′ = x −2 and y′ = y−1 (i.e. the x′-axis is the line y = 1 and the y′-axis is the line x = 2).	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 x2 + b a x ¶ = y −c a µ x2 + b a x + b2 4a2 ¶ = y −c + a b2 4a2 µ x + b 2a ¶2 = 1 a µ y + b2 −4ac 4a ¶ So h = −b 2a, k = 4ac−b2 4a , and 4p = 1 a means p = 1 4a. Thus, the vertex is (h,k) = ³ −b 2a, 4ac−b2 4a ´ , the focus is (h,k+ p) = ³ −b 2a, 4ac−b2+1 4a ´ , and the directrix is the line y = k−p = 4ac−b2−1 4a . r x y x′ y′ P = (x, y) P′ = (x′, y′) θ α−θ α O Figure 7.4.3 Rotation Rotation is another common coordinate transformation. Consider the case of rotating the xy-plane about the origin by an angle θ, as in Figure 7.4.3. The origin of the result- ing x′y′-plane is unchanged from the origin O = (0,0) of the xy-plane. To ﬁnd the rotation equations for x′ and y′, let P = (x, y) be a point in the xy-plane away from the origin. From trigonometry you know that for r = p x2 + y2 ̸= 0 and the angle 0◦≤α < 360◦measured from the positive x-axis to OP, x = r cos α and y = r sin α . Considering P as a point P′ = (x′, y′) in the x′y′-plane, OP′ makes an angle α −θ with the positive x′-axis, so that by the sine and cosine subtraction identities, x′ = r cos(α−θ) = r cos α cos θ + r sin α sin θ = x cos θ + y sin θ (7.7) and y′ = r sin(α−θ) = r sin α cos θ −r cos α sin θ = y cos θ −x sin θ . (7.8) Similar to the translation substitutions, the above rotation equations allow any curve in the xy-plane to be rotated: To rotate a curve in the xy-plane about the origin by an angle θ, make the following sub- stitutions: x 7→x cos θ + y sin θ and y 7→−x sin θ + y cos θ (7.9)	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→−x sin θ + y cos θ = −x sin 45◦+ y cos 45◦= −x+ y p 2 The equation of the rotated ellipse (shown in Figure 7.4.4) is then: 1 4 µ x+ y p 2 ¶2 + µ−x+ y p 2 ¶2 = 1 x2 + 2xy + y2 + 4(x2 −2xy + y2) = 8 5x2 −6xy + 5y2 −8 = 0 In the above example note the presence of the 6xy term in the equation of the rotated ellipse. In general if a conic section has a second-degree equation of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 (7.10) then B ̸= 0 indicates rotation, and either D ̸= 0 or E ̸= 0 indicates translation. Example 7.7 Find the value of a such that rotating the hyperbola x2 a2 −y2 a2 = 1 by 45◦counterclockwise about the origin results in the curve xy = 1. Solution: Since θ = 45◦then as in Example 7.6 the substitutions are again: x 7→x+ y p 2 and y 7→−x+ y p 2 The equation of the rotated hyperbola is then: 1 a2 µ x+ y p 2 ¶2 − 1 a2 µ−x+ y p 2 ¶2 = 1 x2 +2xy+ y2 −(x2 −2xy+ y2) 2a2 = 1 2xy a2 = 1 Thus, when a = p 2 the rotated hyperbola has the equation xy = 1, which shows that the curve y = 1 x is a hyperbola. In general any curve of the form Bxy = 1 is a hyperbola for B ̸= 0.	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 ellipse (or a circle, point, or no curve) (b) B2 −4AC = 0: a parabola (or a line, two parallel lines, or no curve) (c) B2 −4AC > 0: a hyperbola (or two intersecting lines) If B ̸= 0 and the curve is a conic section, then the rotation angle θ is given by: (a) θ = 45◦if A = C (b) tan 2θ = B A−C if A ̸= C, with 0◦< θ < 90◦ Recall that the rotation equations (7.7) and (7.8) for x′ and y′ in terms of x and y provided substitutions that allowed the second-degree equation of the rotated conic section to be found. Conversely, to transform a second-degree equation in x and y into a “standard” conic section equation in terms of x′ and y′ (to simplify sketching the graph), the “reverse” rotation equations for x and y in terms of x′ and y′ are needed: x = x′ cos θ −y′ sin θ (7.11) y = x′ sin θ + y′ cos θ (7.12) Example 7.8 Determine the type of curve whose equation is 5x2 +4xy+8y2 −36 = 0, and sketch its graph. Solution: Since A = 5, B = 4, and C = 8, then B2 −4AC = −144 < 0, so the curve is an ellipse if it is a conic section. Since B ̸= 0 and A ̸= C, the rotation angle θ would be given by tan 2θ = B A−C = 4 −3, with 0◦< θ < 90◦. Then 2θ is in the second quadrant, so that sin 2θ = 4 5 and cos 2θ = −3 5 . Use a half-angle identity to ﬁnd θ: tan θ = sin 2θ 1+cos 2θ = 4 5 1+ −3 5 = 2 Hence, sin θ = 2 p 5 and cos θ = 1 p 5. Now use equations (7.11) and (7.12) to get expressions for x and y in terms of x′ and y′: x = x′ cos θ −y′ sin θ = x′ −2y′ p 5 y = x′ sin θ + y′ cos θ = 2x′ + y′ p 5 10For a proof see Section 6.8 in PROTTER, M.H. AND C.B. MORREY, Analytic Geometry, 2nd ed., Reading, MA: Addison-Wesley Publishing Company, Inc., 1975.	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 is just the ellipse x2 4 + y2 9 = 1 rotated counterclockwise by the angle θ = tan−1 2 ≈63.4◦, as shown in the ﬁgure on the right. Exercises A For Exercises 1-3, sketch the graph of the given ellipse with the exact locations of the foci and vertexes. 1. (x−3)2 25 + (y−2)2 16 = 1 2. (x+1)2 9 + (y+3)2 4 = 1 3. 4x2 + y2 +24x−2y+21 = 0 For Exercises 4-7, sketch the graph of the given parabola with the exact locations of the focus, vertex and directrix. 4. 4(x−1)2 = 3(y+2) 5. y = 4x2 +24x+21 6. 3(y+2)2 = 5(x−2) 7. 4x2 −4x+4y−5 = 0 For Exercises 8-10, sketch the graph of the given hyperbola with the exact locations of the foci, vertexes, directrices, and asymptotes. 8. (x−3)2 25 −(y−2)2 16 = 1 9. (x+1)2 9 −(y+3)2 4 = 1 10. 4x2 −y2 +24x−2y+21 = 0 11. Find the foci, vertexes, directrices and asymptotes for the hyperbola y = 1 x in Example 7.7. 12. Find the equation of the parabola 4y = x2 when rotated 60◦counterclockwise about the origin. 13. Prove equations (7.11) and (7.12). (Hint: Use equations (7.7) and (7.8).) B 14. Determine the type of curve whose equation is 8x2 −12xy+17y2 −20 = 0, and sketch its graph. 15. Determine the type of curve whose equation is 7x2 +6xy−y2 −32 = 0, and sketch its graph. 16. Let A′x′2+B′x′y′ +C′y′2+D′x′+E′y′ +F′ = 0 be the equation obtained by substituting the “reverse” translation equations x = x′ + h and y = y′ + k into equation (7.10). Show that A = A′, B = B′ and C = C′ (i.e. A, B and C are invariant under translation). 17. Similar to Exercise 16, substitute the “reverse” rotation equations (7.11) and (7.12) into equation (7.10) to show that A +C and B2 −4AC are invariant under rotation. 18. Sketch the graph of the conic section whose equation is x2 −2xy+ y2 +4x+4y+10 = 0. (Hint: First handle the rotation (since A = C use θ = 45◦even though B2 −4AC = 0), then the translation (by completing the square).)	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 for the unit hyperbola x2 −y2 = 1, whose points can be deﬁned in terms of hyperbolic functions. x y O P = (x, y) 1 θ x2 + y2 = 1 a 2 = sector area = 1 2 r2θ = θ 2 x = cos θ = cos a y = sin θ = sin a Figure 7.5.1 Circular y x x O P = (x, y) 1 x2 −y2 = 1 a 2 = shaded area Figure 7.5.2 Hyperbolic For a point P = (x, y) on the unit hyperbola x2 −y2 = 1, the hyperbolic angle a is twice the area of the shaded hyperbolic sector in Figure 7.5.2).11 The area a 2 thus equals the area of the right triangle with hypotenuse OP and legs of length x and y (so that the triangle’s area is 1 2 xy) minus the area under the hyperbola over the interval [1, x]. So since the upper half of the hyperbola x2 −y2 = 1 is the function y = p x2 −1, a 2 = (area of triangle) −(area under the hyperbola from 1 to x) = 1 2 xy − Zx 1 p u2 −1 du a 2 = 1 2 x p x2 −1 − µ1 2 x p x2 −1 −1 2 ln ³ x+ p x2 −1 ´¶ (by formula (6.9)) a = ln ³ x+ p x2 −1 ´ ea = x+ p x2 −1 (ea −x)2 = ³p x2 −1 ´2 e2a −2xea +	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, analogous to the trigonometric (circular) functions: The hyperbolic sine, hyperbolic cosine, hyperbolic tangent, hyperbolic cotan- gent, hyperbolic secant and hyperbolic cosecant, denoted by sinh, cosh, tanh, coth, sech and csch, respectively, are: sinh x = ex −e−x 2 for all x cosh x = ex + e−x 2 for all x tanh x = sinh x cosh x for all x coth x = 1 tanh x for all x ̸= 0 sech x = 1 cosh x for all x csch x = 1 sinh x for all x ̸= 0 The graphs of the hyperbolic functions are shown below: x y 1 0 y = cosh x y = sinh x (a) sinh x and cosh x x y 1 −1 0 y = tanh x y = coth x (b) tanh x and coth x x y 1 0 y = sech x y = csch x (c) sech x and csch x Figure 7.5.3 Graphs of the six hyperbolic functions The graph of y = cosh x in Figure 7.5.3(a) might look familiar: a catenary—a uniform cable hanging from two ﬁxed points—has the shape of a hyperbolic cosine function.	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 ± cosh u sinh v sinh 2x = 2 sinh x cosh x cosh(u ± v) = cosh u cosh v ± sinh u sinh v cosh 2x = cosh2 x + sinh2 x tanh(u ± v) = tanh u ± tanh v 1 ± tanh u tanh v tanh 2x = 2 tanh x 1 + tanh2 x The identity cosh2 x−sinh2 x = 1 was proved when deriving the coordinates of points on the unit hyperbola x2−y2 = 1 in terms of the hyperbolic angle (since such a point (x, y) = (cosh a,sinh a) must satisfy x2 −y2 = 1). The addition identities can be proved similarly using hyperbolic angles (i.e. areas).12 However, it is simpler to use the deﬁnitions of sinh and cosh in terms of exponential functions. For example: sinh u cosh v+cosh u sinh v = eu −e−u 2 · ev + e−v 2 + eu + e−u 2 · ev −e−v 2 = eu+v + eu−v −e−u+v −e−u−v + eu+v −eu−v + e−u+v −e−u−v 4 = 2eu+v −2e−u−v 4 = eu+v −e−(u+v) 2 = sinh(u + v) ✓ The identity for sinh 2x is then easy to prove by letting u = v = x in the above identity: sinh 2x = sinh(x+ x) = sinh x cosh x+cosh x sinh x = 2 sinh x cosh x ✓ Note that the identities cosh(−x) = cosh x and sinh(−x) = −sinh x mean that cosh is an even function and sinh is an odd function. Those two functions thus (sort of) serve as even and odd versions of the exponential function (which is neither even nor odd). Both cosh x and sinh x grow exponentially (the e−x term for both functions becomes negligible as x →∞), while sinh x decreases exponentially to −∞as x →−∞. 12See pp.25-29 in SHERVATOV, V.G., Hyperbolic Functions, Boston: D.C. Heath and Company, 1963.	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 = −csch x + C Z sinh x dx = cosh x + C Z sech x tanh x dx = −sech x + C Z sech2 x dx = tanh x + C Z csch2 x dx = −coth x + C For example, by deﬁnition of cosh x: d dx (cosh x) = d dx µ ex + e−x 2 ¶ = ex −e−x 2 = sinh x ✓ Example 7.9 Find the derivative of y = sinh x3. Solution: By the Chain Rule, dy dx = 3x2 cosh x3. Example 7.10 Evaluate Z tanh x dx. Solution: Use the deﬁnition of tanh x and the substitutions u = cosh x, du = sinh x dx: Z tanh x dx = Z sinh x cosh x dx = Z du u = ln\|u\| + C = ln(cosh x) + C For any constant a > 0, both y = cosh at and y = sinh at satisfy the differential equation y′′(t) = a2y(t) , which models rectilinear motion of a particle under a repulsive force proportional to the dis- placement. This is just one of the many reasons why hyperbolic functions appear in so many physical applications.	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, µ is the magnetic moment, k is Boltzmann’s constant, and T is the temperature of the gas. Show that n = 4πNkT µH sinh µµH kT ¶ . Solution: Let a = µH kT , and let u = cos θ so that du = −sin θ dθ: n = −2πN Z−1 1 eau du = 2πN Z1 −1 eau du = 2πN a eau ¯¯¯¯ 1 −1 = 2πN a µ 2· ea −e−a 2 ¶ = 4πNkT µH sinh µµH kT ¶ Since d dx (sinh x) = cosh x = ex+e−x 2 > 0 for all x, then y = sinh x is an increasing function and thus its inverse function x = sinh−1 y is deﬁned. The remaining inverse hyperbolic functions can be deﬁned similarly, with the following domains and ranges (switching the roles of x and y, as usual) plus their graphs: function sinh−1 x cosh−1 x tanh−1 x csch−1 x sech−1 x coth−1 x domain all x x ≥1 \|x\| < 1 all x ̸= 0 0 < x ≤1 \|x\| > 1 range all y y ≥0 all y all y ̸= 0 y ≥0 all y ̸= 0 x y 1 y = cosh−1 x y = sinh−1 x (a) sinh−1 x and cosh−1 x x y 1 −1 y = coth−1 x y = tanh−1 x (b) tanh−1 x and coth−1 x y 1 y = sech−1 x y = csch−1 x (c) sech−1 x and csch−1 x Figure 7.5.4 Graphs of the six inverse hyperbolic functions	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 Ã 1 x + s 1 x2 +1 ! for x ̸= 0 Notice that the above formula for cosh−1 x was actually proved at the beginning of this section. The formula for sinh−1 x follows from the deﬁnition of an inverse function: y = sinh−1 x ⇒ x = sinh y = ey −e−y 2 ⇒ ey −2x−e−y = 0 ⇒ e2y −2xey −1 = 0 ⇒ u2 −2xu −1 = 0 for u = ey ⇒ u = 2x± p 4x2 −4(1)(−1) 2 = x± p x2 +1 ⇒ ey = u = x+ p x2 +1 since ey > 0 ⇒ y = ln(x+ p x2 +1) ✓ The remaining formulas can be proved similarly. Example 7.12 y x O P (x, y) = (cosh a,sinh a) A = (1,0) x2 −y2 = 1 a 2 = shaded area of OAP Show that for the hyperbolic angle a of a point P = (x, y) on the unit hyperbola x2 −y2 = 1, the area a 2 of the hyperbolic sector OAP (the shaded region in the ﬁgure on the right) is a 2 = 1 2 cosh−1 x . Solution: It was shown earlier that the area of OAP is a 2 = 1 2 ln(x+ p x2 −1) so by the formula cosh−1 x = ln(x+ p x2 −1) the result follows. This makes sense, since x = cosh a and so cosh−1 x = cosh−1(cosh a) = a, by deﬁnition of an inverse. You can use either the general formula for the derivative of an inverse function or the above formulas to ﬁnd the derivatives of the inverse hyperbolic functions:	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 the derivative of tanh−1 x: d dx (tanh−1 x) = d dx µ1 2 ln 1+ x 1−x ¶ = 1 2 d dx (ln(1+ x) −ln(1−x)) = 1 2 µ 1 1+ x −−1 1−x ¶ = = 1−x+(1+ x) 2(1+ x)(1−x) = 1 1−x2 ✓ Exercises A 1. Prove the identities for sinh(−x), cosh(−x), tanh(−x), coth(−x), sech(−x), and csch(−x) on p.232. 2. Prove the identities for cosh(u± v), tanh(u± v), and tanh 2x on p.232. For Exercises 3-15 prove the given identity. Your proofs can use other identities. 3. (cosh x+sinh x)r = cosh rx+sinh rx for all r 4. sinh 3x = 3 sinh x+4 sinh3 x 5. sinh A cosh B = 1 2 (sinh(A +B) + sinh(A −B)) 6. tanh2 x+sech2 x = 1 7. sinh A sinh B = 1 2 (cosh(A +B) −cosh(A −B)) 8. coth2 x−csch2 x = 1 9. cosh A cosh B = 1 2 (cosh(A +B) + cosh(A −B)) 10. coth−1 ¡ 1 x ¢ = tanh−1 x 11. cosh 2x = cosh2 x+sinh2 x = 2 cosh2 x−1 = 1+2 sinh2 x 12. sinh2 x 2 = cosh x −1 2 13. cosh2 x 2 = cosh x + 1 2 14. tanh2 x 2 = cosh x −1 cosh x + 1 15. tanh x 2 = cosh x −1 sinh x = sinh x cosh x + 1 16. Prove the identities for tanh−1 x, coth−1 x, sech−1 x, and csch−1 x on p.235. 17. Prove the derivative formulas for sinh x, tanh x, coth x, sech x, and csch x on p.233. 18. Prove the derivative formulas for sinh−1 x, cosh−1 x, coth−1 x, sech−1 x, and csch−1 x on p.236. 19. Show that cosh x = O(ex) and sinh x = O(ex). 20. Show that d dx (tan−1(sinh x)) = sech x . 21. Verify that the curve y = tanh x has asymptotes y = ±1. 22. Show that sinh dx = dx and cosh dx = 1 for any inﬁnitesimal dx. (Hint: Use 1 1+dx = 1 −dx from Exercise 3 in Section 1.3 along with edx = 1+ dx from Exercise 29 in Section 2.3.)	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 transformations for two inertial frames S and S′ are x′ = x −vt p 1 −(v/c)2 and t′ = t −vx/c2 p 1 −(v/c)2 where S′ moves with speed v > 0 parallel to the x-axis of S. Let v/c = tanh ξ. Show that x′ = −ct sinh ξ + x cosh ξ and t′ = t cosh ξ −(x/c) sinh ξ . 26. Evaluate the integral I = R3 2 dx 1−x2 in two different ways: (a) Use d dx (coth−1 x) = 1 1−x2 for \|x\| > 1 to show that I = coth−1 3 −coth−12. (b) Use the substitution u = 1 x in the integral, then use d dx (tanh−1 x) = 1 1−x2 for \|x\| < 1 to show that I = tanh−1 1 3 −tanh−1 1 2. Is this answer equivalent to the answer from part (a)? Explain. B 27. The general solution of the differential equation y′′ = a2 y is y(t) = y1(t) = c1eat + c2e−at, where a is a positive constant, and c1 and c2 are arbitrary constants. (a) Verify that y(t) = y2(t) = k1 cosh at+ k2 sinh at is also a solution of y′′ = a2 y. (b) Show that for any c1 and c2, y1(t) = c1eat+c2e−at can be written as y1(t) = k1 cosh at+k2 sinh at for some constants k1 and k2 in terms of c1 and c2. 28. Verify that for positive constants β, p, l and c the function η(x) = βx p −βc sinh(px/c) p2 cosh(pl/c) is a solution of the differential equation (related to water displacement in a canal of length 2l) d2η dx2 −p2 c2 η = −βxp c2 . 29. For any constant a and for s > \|a\| the Laplace transform L (s) of the function f (t) = sinh at is L (s) = Z∞ 0 e−st sinh at dt . Show that L (s) = a s2−a2 . C 30. In quantum mechanics the scaling factor eπ/s0 of the three-particle Eﬁmov trimer is the solution s = s0 > 0 of the equation s cosh ¡ πs 2 ¢ = 8 sinh ¡ πs 6 ¢ p 3 , Use a numerical method to approximate s0, then calculate eπ/s0.	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 for a = µH kT , µmax > 0, and x = µmaxH kT , Zµmax 0 L(a) dµ = 1 x ln µsinh x x ¶ . 32. The age t0 of the universe (in years) is given by t0 = τ0 Z1 0 dx q 2q0 x +1−2q0 , where τ0 = 2 × 1010 years is the Hubble time and q0 ≥0 is the deceleration parameter. For the cosmological model with 0 < q0 < 1 2, use the substitution x = 2q0 1−2q0 sinh θ to show that t0 = τ0 Ã 1 1−2q0 − 2q0 (1−2q0)3/2 cosh−1 Ã 1 p 2q0 !! . What fraction of τ0 is t0 (i.e. what is t0 τ0 ) when q0 = 1 4? (Hint: Use Exercise 11 or 12.) 33. Circular rotations preserve the area of circular sectors (see Figure 7.5.5(a)). For any constant c > 0 the hyperbolic rotation φ : (x, y) 7→(cx, y/c) moves points along the hyperbola xy = k (for k > 0), as shown in Figure 7.5.5(b) for c > 1. This hyperbolic rotation preserves the area of hyperbolic sectors. x y O 1 −1 θ θ rotate OBQ by α to OB′Q′ B Q B′ Q′ α (a) Unit circle x2 + y2 = 1 y x 0 (x, y) (cx, y/c) (b) Hyperbola xy = k y x O P′ P = A′ (cosha,sinh a) A = (1,0) (c) Unit hyperbola x2 −y2 = 1 Figure 7.5.5 Circular and hyperbolic rotations As an example of why this is true, let a > 0 and consider the unit hyperbola in Figure 7.5.5(c). Then: (a) Let c > 0. When c ̸= 1 the mapping φ : (x, y) 7→(cx, y/c) does not move points along the unit hyperbola. Find the formula for φ on the unit hyperbola as follows: use the rotation equations from Section 7.4 to rotate the unit hyperbola 45◦to a hyperbola of the form xy = k, apply φ to a generic point on that hyperbola, then rotate that hyperbola by −45◦back to the unit hyperbola. (b) Use part (a) to ﬁnd the value of c such that φ maps A = (1,0) to P = (cosha,sinh a). (c) Let P′ be the point that φ maps P to, and let A′ = P. Use part (b) to ﬁnd the coordinates of P′, then show that the hyperbolic sectors OAP and OA′P′ have the same area a/2.	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) Identify points by slope t Figure 7.6.1 Points on the unit circle x2 + y2 = 1 When identifying by the angle θ, all points (x, y) on the unit circle can be written as x = cos θ and y = sin θ for any angle θ. When identifying by the slope t of lines through the point (−1,0), recall from the derivation of the half-angle substitution that sin θ = 2t 1+t2 and cos θ = 1−t2 1+t2. So all points (x, y) on the unit circle except (−1,0) can be written as x = 1−t2 1+ t2 and y = 2t 1+ t2 for any slope t. These two distinct “identiﬁcations” are called parametrizations of the unit circle, with parameter θ in the ﬁrst case and parameter t in the second. In general, one way to describe a plane curve C (i.e. a curve in the xy-plane) is to write its x and y-coordinates as functions of a variable t: x = x(t) and y = y(t) These are parametric equations of C, which consists of all points (x, y) such that x = x(t) and y = y(t) for the parameter t in some interval I. The shorthand for this is: C : x = x(t), y = y(t), t in I Notice the ﬂexibility that parametric equations provide, since plane curves can take any shape, not limited to the graph of a single function y = f (x). In fact, a curve y = f (x) is the special case where the parametric equations are x = t and y = f (t). In physical settings the parameter t often denotes time, but it can represent anything and any symbol can be used in its place. A curve can have many parametrizations.	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 the angle θ in Figure 7.6.1(a), it sufﬁces to show that (x−h)2+(y−k)2 = r2: (x−h)2 + (y−k)2 = r2 cos2 ωt + r2 sin2 ωt = r2 ✓ The constant ω determines how fast and in which direction the circle is traced as the parameter t varies. For example, for ω = 2 the circle C is traced counterclockwise at twice the speed of the parametrization C : x = h + rcos t, y = k + rsin t. In all cases the circle is re-traced every 2π/ω radians. For that reason the interval for t is often restricted to the interval [0,2π/ω], so that the circle is traced only once. Example 7.14 x y (x, y) = (a cos t,b sin t) −a a b −b Show that for a > 0 and b > 0 the parametric equations x = a cos t and y = b sin t for 0 ≤t ≤2π describe an ellipse. Solution: Since x2 a2 + y2 b2 = a2 cos2 t a2 + b2 sin2 t b2 = cos2 t + sin2 t = 1 for all t, then the points (x, y) = (a cos t,b sin t) lie on the ellipse x2 a2 + y2 b2 = 1. It should be obvious that the entire ellipse is traced, but it is left as an exercise to show what the parameter t represents. Example 7.15 y x O (x, y) = (cosh t,sinh t) 1 x2 −y2 = 1 t 2 = shaded area Show that the parametric equations x = cosh t and y = sinh t for −∞< t < ∞describe one branch of a hyperbola. Solution: Since x2 −y2 = cosh2 t −sinh2 t = 1 for all t, then the points (x, y) = (cosh t,sinh t) lie on the unit hyperbola x2 −y2 = 1. This was in fact shown in Section 7.5, where t is half the area of the shaded region in the above ﬁgure for t > 0. For t < 0 the shaded region is reﬂected below the x-axis. Since cosh t ≥1 and sinh t can take any value, then the entire right branch of the hyperbola is traced as t varies. Likewise the left branch has parametric equations x = −cosh t and y = sinh t. The hyperbola is thus parametrized by area (or negative area for t < 0). In general, for a > 0 and b > 0 the hyperbola x2 a2 −y2 b2 = 1 has parametric equations x = ±a cosh t and y = b sinh t for all t.	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 7.6.2 and described below for n = 3 points: B0 B1 B2 A0 A1 P 40% 40% 40% (a) t = 0.4 = 40% B0 B1 B2 B (b) Many values of t in [0,1] Figure 7.6.2 Bézier curve B with 3 control points B0, B1, B2 For three control points B0, B1, B2, pick 0 ≤t ≤1. Say t = 0.4, which you can think of as a percentage: 0.4 = 40%. Let A0 and A1 be the points 40% of the way from B0 to B1 and from B1 to B2, respectively, as in Figure 7.6.2(a). Then the point P that is 40% of the way from A0 to A1 is on the Bézier curve B joining B0 and B2. Do this for every t in [0,1] to ﬁll out the curve B, as in Figure 7.6.2(b). It can be shown via de Casteljau’s algorithm that the Bézier curve B for any three control points B0 = (x0, y0), B1 = (x1, y1) and B2 = (x2, y2) in the xy-plane has parametric equations x = (1−t)2x0 + 2t(1−t)x1 + t2x2 and y = (1−t)2 y0 + 2t(1−t) y1 + t2 y2 (7.13) for 0 ≤t ≤1. Write out and simplify the explicit parametric equations for the Bézier curve B with control points B0 = (1,2), B1 = (2,4) and B2 = (4,1). Solution: The parametric equations for B0 = (x0, y0) = (1,2), B1 = (x1, y1) = (2,4), B2 = (x2, y2) = (4,1) are: x = (1−t)2(1) + 2t(1−t)(2) + t2(4) = 1−2t+ t2 +4t−4t2 +4t2 = t2 +2t+1 y = (1−t)2(2) + 2t(1−t)(4) + t2(1) = 2−4t+2t2 +8t−8t2 + t2 = −5t2 +4t+2 y x B0 B1 B2 1 2 3 4 1 2 3 0 B The Bézier curve B : x = t2 + 2t + 1, y = −5t2 + 4t + 2, 0 ≤t ≤1 for B0, B1, B2 is shown in the ﬁgure on the right. It is left as an exercise to show that this curve is part of a parabola. In general Bézier curves can be created for n ≥3 control points in the plane, with the parametric equations being polynomials of degree n−1 in the parameter t. In the exercises you will be guided in how to derive the parametric equations in the cases n = 3 and n = 4. Bézier curves can also be constructed for control points in three-dimensional space. A similar construct—a Bézier surface—is used in three dimensions to model the boundary of a polyhedron (i.e. a solid whose faces are polygons). 13Developed and popularized in the 1960s by two engineers, Pierre Bézier and Paul de Casteljau, for vehicle body modeling at the French automotive manufacturers Renault and Citroën, respectively.	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)dt x′(t)dt = y′(t) x′(t) = dy dt dx dt (7.14) when dx dt ̸= 0. The second derivative d2 y dx2 can then be found via the Chain Rule: d dt µ dy dx ¶ = µ d dx µdy dx ¶¶ · dx dt = d2y dx2 · dx dt ⇒ d2y dx2 = d dt µdy dx ¶ dx dt (7.15) For t in [a,b] with x1 = x(a) and x2 = x(b), the integral Rx2 x1 ydx is given by: Zx2 x1 y dx = Zb a y(t) x′(t) dt (7.16) Example 7.17 A cycloid is the path of a point P on a circle rolling along a straight line. Figure 7.6.3 shows the cycloid C traced by a circle of radius a rolling along the x-axis so that P touches the origin during the roll: P y a y x 0 x aθ 2πa t θ C Figure 7.6.3 Cycloid C for a circle of radius a Find parametric equations for C and ﬁnd dy dx . Solution: For the angles t and θ—measured in radians—shown in Figure 7.6.3, t + θ + π/2 = 2π, so t = 3π/2 −θ. The point P touches the origin as the circle rolls, so the horizontal distance from the circle’s center to the y-axis is the length of the circular arc with central angle θ, namely aθ. So by the parametrization of the circle as in Example 7.13, but with center (h,k) = (aθ,a), radius r = a, and ω = 1, x = aθ + a cos t = a ¡ θ + cos ¡ 3π 2 −θ ¢¢ = a(θ −sin θ) y = a + a sin t = a ¡ 1 + sin ¡ 3π 2 −θ ¢¢ = a(1 −cos θ) Thus, C : x = a(θ −sin θ), y = a(1 −cos θ), −∞< θ < ∞is a parametrization of the cycloid C.	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 7.6.3 that the cycloid has cusps at those values of x. B A A cycloid appears in the solution of the famous brachistochrone problem:14 ﬁnd the plane curve joining two points A and B—where B is at a lower height than A but not directly under it—along which an object slides frictionless under the force of gravity alone from A to B in the shortest time. It turns out that the optimal path is not a straight line, but part of an inverted (upside- down) cycloid with a cusp at A, as in the ﬁgure on the right.15 Exercises A 1. For a > b > 0, in the ellipse x2 a2 + y2 b2 = 1 inscribe a circle of radius b centered at the origin, called the minor auxiliary circle. This circle is parametrized by x = bcos t and y = bsin t for 0 ≤t ≤2π, with the angle t shown in Figure 7.6.4. From each point on that circle draw a horizontal line segment to the point P on the ellipse in the same quadrant, as shown. Show that P = (acos t,bsin t). Note: The angle t is called the eccentric angle of the ellipse, and it is the parameter for the parametriza- tion of the ellipse in Example 7.14. x y P = (x, y) −a a b −b t (bcost,bsin t) Figure 7.6.4 Exercise 1 x y P = (x, y) −a a b −b t (acost,asin t) Figure 7.6.5 Exercise 2 2. Similar to Exercise 1, circumscribe a circle of radius a (called the major auxiliary circle) around the ellipse x2 a2 + y2 b2 = 1. This circle is parametrized by x = acos t and y = asin t for 0 ≤t ≤2π, with the eccentric angle t shown in Figure 7.6.5. From each point on that circle draw a vertical line segment to the point P on the ellipse in the same quadrant, as shown. Show again that P = (acos t,bsin t). 3. Show that the cycloid in Example 7.17 has global maxima at x = (2k + 1)πa for all integers k, and that the cycloid is always concave down. 4. Show that the area under the cycloid in Example 7.17 over the interval [0,2πa] is 3πa2. 14First solved in 1696 by the Swiss physicist and mathematician Johann Bernoulli (1667-1748). 15See pp.60-62 in CLEGG, J.C., Calculus of Variations, Edinburgh: Oliver & Boyd, Ltd., 1968. For Bernoulli’s proof see pp.644-655 in SMITH, D.E., A Source Book in Mathematics, New York: Dover Publications, Inc., 1959.	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 answer. 6. Let P0 = (x0, y0) and P1 = (x1, y1) be distinct points in the xy-plane. For t in [0,1] deﬁne x(t) = (1−t)x0 + tx1 and y(t) = (1−t) y0 + t y1 and let P(t) = (x(t), y(t)). (a) Show that C : x = x(t), y = y(t), 0 ≤t ≤1 is a parametrization of the line segment from P0 to P1. (b) Show that the parameter t is the proportion of the length P0P(t) to the length P0P1: P0P(t) P0P1 = t. 7. In an ellipsograph a rod of length a has a peg at one end and another peg a distance b from the other end, so that the rod can slide along two thin perpendicular rails, with the pegs in the rails as in Figure 7.6.6. Show that a marker at the endpoint P traces an ellipse as the rod moves. (Hint: Treat the rails as the x and y-axes. Find parametric equations, with a different angle than in Exercise 1.) a−b b P Figure 7.6.6 Exercise 7 a x y O A B P = (x, y) θ C Figure 7.6.7 Exercise 8 B 8. The end P of a thread is held taut as it is unwound from a circle of radius a starting from a point A, as in Figure 7.6.7. Show that the path C that the end traces—called the involute of the circle—has parametric equations x = a(cosθ +θsinθ), y = a(sinθ −θcosθ). 9. Recall that a parabola of the form y = ax2+bx+c has a constant second derivative d2 y dx2 = 2a. Consider the Bézier curve B in Example 7.16. (a) Find d2 y dx2 for B. Is it a constant? (b) Show that B is part of a parabola. Does this contradict part (a)? Explain. (Hint: Show the curve has a second-degree equation of the form (7.10).) (c) Find the point where the curve B has a global maximum. 10. Use Exercise 6 to derive the formulas (7.13) for the parametric equations of the general Bézier curve for n = 3 control points. 11. To form the Bézier curve for n = 4 control points B0 = (x0, y0), B1 = (x1, y1), B2 = (x2, y2), B3 = (x3, y3), for 0 ≤t ≤1 use the three points that are 100t% of the way along the line segments B0B1, B1B2, B2B3 as the control points in the n = 3 case. Show that the resulting parametrization is: x = (1−t)3x0 +3t(1−t)2x1 +3t2(1−t)x2 + t3x3 and y = (1−t)3 y0 +3t(1−t)2 y1 +3t2(1−t)y2 + t3 y3 12. Each point on a plane curve lies on some line through the origin. Use that fact to show that the equation y2 = x2 + x3 deﬁnes a rational curve (see Exercise 5). (Hint: For all real t, ﬁnd the intersections of the lines y = tx with the curve. Consider also the special case of the line x = 0.) 13. Sketch the graph of the curve C : x = 2t−4t3, y = t3 −3t4, −∞< t < ∞.	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 polar coordinates.16 Recall that any point P distinct from the origin (denoted by O) in the xy-plane is a distance r > 0 from the origin, and the ray −−→ OP makes an angle θ with the positive x-axis, as in Figure 7.7.2. Imagine −−→ OP swings around the “pole” at the origin. x y 0 1 2 3 Figure 7.7.1 Spiral r x y O θ P = (r,θ) Figure 7.7.2 Polar coordinates (r,θ) −r x y O θ P = (−r,θ) Figure 7.7.3 Negative r: (−r,θ) The pair (r,θ) contains the polar coordinates of P, and the positive x-axis is called the polar axis of this coordinate system. For the angle θ measured in radians, (r,θ) = (r,θ + 2πk) for all integers k. Thus, the polar coordinates of a point are not unique. By convention r can be negative, by deﬁning (−r,θ) = (r,θ + π) for any angle θ: the ray −−→ OP is drawn in the opposite direction from the angle θ, as in Figure 7.7.3. When r = 0, the point (r,θ) = (0,θ) is the origin O, regardless of the value of θ. Example 7.18 Express the spiral in Figure 7.7.1 in polar coordinates, such that the distance between any two points separated by 2π radians is always 1. Solution: The goal is to ﬁnd some equation involving r and θ (measured in radians) that describes the spiral. The distance between any two points separated by 2π radians is always 1, for example: θ = 0 ⇒ r = 1 θ = 2π ⇒ r = 2 θ = 4π ⇒ r = 3 ... θ = 2πk ⇒ r = 1+ k for all integers k ≥0. In fact r = 1 + k when θ = 2πk for all real k ≥0, by the assumption about the distance. So solving for k in terms of θ yields the polar equation r = 1+ θ 2π for all θ ≥0. 16Created by the Flemish mathematician Grégoire de Saint-Vincent (1584-1667) and Italian mathematician Bonaventura Cavalieri (1598-1647) in the 17th century, later used by Newton in his Method of Fluxions (1671).	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 intervals. Similar graphing paper exists for polar coordinates, as in Figure 7.7.4. 0◦ 15◦ 30◦ 45◦ 60◦ 75◦ 90◦ 105◦ 120◦ 135◦ 150◦ 165◦ 180◦ 195◦ 210◦ 225◦ 240◦ 255◦ 270◦ 285◦ 300◦ 315◦ 330◦ 345◦ O Figure 7.7.4 Polar coordinate graphing paper This polar grid is radial, not rectangular. The concentric circles around the origin O are where r is constant (e.g. r = 1, r = 2), while the lines through the origin are where θ is constant, at regular intervals for each. The angle θ shown here is in degrees, though radians are often preferred for their “unitless” nature.	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 coordinates and Carte- sian coordinates. Namely, for a point with polar coordinates (r,θ) and Cartesian coordinates (x, y): Polar to Cartesian: x = r cos θ y = r sin θ (7.17) Cartesian to Polar: r = ± q x2 + y2 tan θ = y x if x ̸= 0 (7.18) In formula (7.18), if x = 0 then θ = π/2 or θ = 3π/2. If x ̸= 0 and y ̸= 0 then the two possible solutions for θ in the equation tanθ = y x are in opposite quadrants (for 0 ≤θ < 2π). If the angle θ is in the same quadrant as the point (x, y), then r = p x2 + y2 (i.e. r is positive); otherwise r = − p x2 + y2 (i.e. r is negative). Example 7.19 Write the equation of the unit circle x2 + y2 = 1 in polar coordinates. Solution: By formula (7.18), r2 = x2 + y2 = 1, so in polar coordinates the equation is simply r = 1. In general the circle x2 + y2 = a2 of radius a > 0 has the simpler expression r = a in polar coordinates. Example 7.20 Write the equation x2 +(y−4)2 = 16 in polar coordinates. Solution: This is the equation of a circle of radius 4 centered at the point (0,4). Expand the equation: x2 + (y−4)2 = 16 x2 + y2 −8y + 16 = 16 x2 + y2 = 8y r2 = 8rsin θ (by formulas (7.17) and (7.18)) r = 8 sin θ Is it valid to cancel r from both sides in the last step? Yes. The point (0,0) is on the circle, so canceling r does not eliminate r = 0 as a potential solution of the equation (e.g. θ = 0 would make r = 8sin θ = 0). Thus, the polar equation is r = 8sin θ. Notice that this polar equation is actually less intuitive than its Cartesian equivalent. From the equa- tion x2 +(y−4)2 = 16 it is easy to identify the curve as a circle and read off its radius and center; these properties are not so obvious from the polar equation. Since the circle’s center is not the origin, there is no symmetry about the origin, which is when polar coordinates are often better suited.	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 Product Rule and formulas (7.14) and (7.15) from Section 7.6, with dx = x′(θ)dθ and dy = y′(θ)dθ: For a plane curve with polar equation r = r(θ), dy dx = r′(θ) sin θ + r(θ) cos θ r′(θ) cos θ −r(θ) sin θ and d2y dx2 = d dθ µdy dx ¶ r′(θ) cos θ −r(θ) sin θ . (7.19) Example 7.21 Sketch the graph of r = 1+cos θ. Solution: First sketch the graph treating (r,θ) as Cartesian coordinates, for 0 ≤θ ≤2π as in Figure 7.7.6(a). Then use that graph to trace out a rough graph in polar coordinates, as in Figure 7.7.6(b).17 r θ 0 π 2π 2 (a) θr-plane y x 2 0 1 −1 ¡ 3 2, π 3 ¢ ¡ 3 2, 5π 3 ¢ ¡ 1 2, 2π 3 ¢ ¡ 1 2, 4π 3 ¢ (b) Polar coordinates (r,θ) Figure 7.7.6 Graph of r = 1+cos θ To ﬁnd the maxima, minima, and inﬂection points it is still necessary to ﬁnd dy dx and d2 y dx2 . It is left as an exercise to use formula (7.19) and double-angle identities to show that dy dx = −cos θ +cos 2θ sin θ +sin 2θ and d2 y dx2 = − 3(1+cos θ) (sin θ +sin 2θ)3 and that for θ in [0,2π], dy dx = 0 only when θ = π 3 and 5π 3 , while dy dx is undeﬁned for θ = 0, 2π 3 , π, 4π 3 , and 2π. Since d2 y dx2 ¯¯¯ θ= π 3 < 0 and d2 y dx2 ¯¯¯ θ= 5π 3 > 0 then the curve has a local maximum when θ = π 3 and a local minimum when θ = 5π 3 . It can also be shown that d2 y dx2 changes sign around θ = 0, 2π 3 , π, 4π 3 , so that the inﬂection points occur at those values of θ, as shown (with the correct concavity) in Figure 7.7.6(b). 17There are far too many interesting plane curves to cover here. For an extensive collection, see LAWRENCE, J.D., A Catalog of Special Plane Curves, New York: Dover Publications, Inc., 1972. See also SEGGERN, D.H. VON, CRC Handbook of Mathematical Curves and Surfaces, Boca Raton, FL: CRC Press, Inc., 1990.	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 = β would look like the shaded region in Figure 7.7.7 with area A: θ = α θ = β y x r = f (θ) O Figure 7.7.7 Area A r r + dr ds y x dθ (r,θ) (r + dr,θ + dθ) Figure 7.7.8 Area dA c a b A C B Figure 7.7.9 Area = 1 2 bc sin A A typical inﬁnitesimal wedge of that region—shown in Figure 7.7.8—is produced by an in- ﬁnitesimal change dθ in the angle θ, which results in an inﬁnitesimal change dr in r. By the Microstraightness Property, the curve r = f (θ) is a straight line with inﬁnitesimal length ds over that inﬁnitesimal angle dθ. The area dA of the wedge thus equals the area of the trian- gle with sides of lengths r and r + dr with included angle dθ and a side of length ds. Since sin dθ = dθ and dr = f ′(θ)dθ, then by the formula for the area of a triangle (as shown in Figure 7.7.9),18 dA = 1 2 (r)(r + dr) sin dθ = 1 2 r2dθ + 1 2 r(dr)dθ = 1 2 r2dθ + 1 2 r(f ′(θ)dθ)dθ = 1 2 r2dθ + 0 = 1 2 r2dθ since (dθ)2 = 0. The area A of the region is the sum of these inﬁnitesimal areas dA: For a plane curve with polar equation r = f (θ), the area A of the region swept out between θ = α and θ = β is: A = Zθ=β θ=α dA = Zβ α 1 2 r2dθ = Zβ α 1 2 (f (θ))2dθ (7.20) If f is periodic then choose the angle interval [α,β] so that the area is swept out only once. 18The formula 1 2 bc sin A for the area of a triangle △ABC is derived in most trigonometry texts. For example, see p.54 in CORRAL, M., Trigonometry, http://mecmath.net/trig/, 2009.	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 circle is then A = Z2π 0 1 2 r2dθ = Z2π 0 1 2 R2dθ = 1 2 R2θ ¯¯¯¯ 2π 0 = 1 2 R2 (2π−0) = πR2 . ✓ Note the simplicity of this integral compared to the trigonometric substitution required when using Cartesian coordinates, as in Section 6.3. Notice also that using a larger interval—say [0,4π]—for θ would result in an incorrect area (4πR2) even though the region inside the curve is the same. Example 7.23 Find the area inside the curve r = 1+cos θ. Solution: Choose 0 ≤θ ≤2π as in Example 7.21, so that the area A is: A = Z2π 0 1 2 r2dθ = Z2π 0 1 2 (1+cos θ)2dθ = Z2π 0 µ1 2 + cos θ + cos2 θ 2 ¶ dθ = Z2π 0 µ1 2 + cos θ + 1+cos 2θ 4 ¶ dθ = Z2π 0 µ3 4 + cos θ + cos 2θ 4 ¶ dθ = 3 4 θ + sin θ + 1 8 sin 2θ ¯¯¯¯ 2π 0 = 3π 2 Exercises A For Exercises 1-8 write the given equation in polar coordinates. 1. (x−3)2 + y2 = 9 2. y = x 3. x2 −y2 = 1 4. 3x2 +4y2 −6x = 9 5. y = −x 6. y = x+1 7. y = x2 8. y = x3 9. Write the polar equation r2 = 4 cos 2θ in Cartesian coordinates. 10. Find the tangent line to r = cos 2θ at ¡ 1 2, π 6 ¢ . 11. Find the tangent line to r = 8 sin2 θ at ¡ 2, 5π 6 ¢ . 12. Recall the curve r = 1+cos θ from Example 7.21. (a) Verify that for this curve, dy dx = −cos θ +cos 2θ sin θ +sin 2θ and d2 y dx2 = − 3(1+cos θ) (sin θ +sin 2θ)3 . (b) Verify that for θ in [0,2π], dy dx = 0 only when θ = π 3 and 5π 3 , and is undeﬁned for θ = 0, 2π 3 , π, 4π 3 , and 2π. (c) Verify that d2 y dx2 ¯¯¯¯ θ= π 3 < 0 and d2 y dx2 ¯¯¯¯ θ= 5π 3 > 0. (d) Verify that d2 y dx2 changes sign around θ = 0, 2π 3 , π, and 4π 3 .	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 for a half-wavelength linear antenna: Eθ = r(θ) = cos( π 2 cos θ) sin θ 19. Show that the distance d between two points (r1,θ1) and (r2,θ2) in polar coordinates is d = q r2 1 + r2 2 −2r1r2 cos(θ1 −θ2) . 20. For a point P = (r,θ) on the curve r = r(θ), let α be the angle that the tangent line through P makes with the positive x-axis. Let ψ = α−θ be the angle between the tangent line and the line through P and the origin. Show that tan ψ = r(θ) r′(θ). (Hint: Use formulas (7.19) and (3.2).) 21. For the parabola with focus at (0,0), vertex at (0,−p), and directrix y = −2p (with p > 0), show that the polar equation is r = 2p 1 −sin θ . 22. For the ellipse with foci (0,0) and (2c,0), vertexes (c ± a,0), and eccentricity e = c a (with 0 < c < a), show that the polar equation is r = a(1 −e2) 1 −e cos θ . 23. For the hyperbola with foci (0,0) and (2c,0), vertexes (c±a,0), and eccentricity e = c a (with 0 < a < c), show that the polar equation is r = a(e2 −1) 1 + e cos θ . F1 F2 y x −a 0 a d1 d2 P = (x, y) η 24. The bipolar coordinates19 (ξ,η) of a point P = (x, y) relative to two poles F1 = (−a,0) and F2 = (a,0) are given by ξ = ln d1 d2 and η = ∠F1PF2, where d1 = F1P and d2 = F2P, as in the ﬁgure on the right . Then −∞< ξ < ∞except at F1 (which corresponds to ξ = −∞) and F2 (ξ = ∞), while 0 ≤η ≤π for points on or above the x-axis, and π < η < 2π below the x-axis. (a) Show that x = a sinh ξ cosh ξ −cos η and y = a sin η cosh ξ −cos η . (Hint: Use the distance formula, the Law of Cosines, and the Law of Sines.) (b) Show that for constants τ ̸= 0 and σ ̸= 0 or π, the curves ξ = τ and η = σ are the respective circles (x −a coth τ)2 + y2 = a2 sinh2 τ and x2 + (y −a cot σ)2 = a2 sin2 σ . 19Inspired by the lines of force and equipotential lines for an electric dipole. See pp.55-56 in STRATTON, J.A., Electromagnetic Theory, New York: McGraw-Hill Book Company, Inc., 1941.	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 curve y = f2(x) over the entire interval, as in Figure 8.1.1 for an interval [a,b]. y x a b x x+ dx 0 y = f2(x) y = f1(x) dA h(x) = \| f1(x)−f2(x)\| dx Figure 8.1.1 The area A between two curves y = f1(x) and y = f2(x) over [a,b] The area A of the region between the curves in Figure 8.1.1 cannot be negative. Thus, a typical inﬁnitesimal area element dA of the region is of the form h(x)dx, where the height function h(x) is the nonnegative difference in the y-coordinates of the curves at each x in [a,b]: h(x) = ¯¯ f1(x)−f2(x) ¯¯. Hence: The area A between two curves y = f1(x) and y = f2(x) over an interval [a,b] is: A = Zb a dA = Zb a ¯¯ f1(x)−f2(x) ¯¯ dx (8.1) The interval [a,b] can be replaced by any interval—ﬁnite or inﬁnite—over which the inte- gral is deﬁned. Neither curve is required to be above the x-axis. 252	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 −e−x) dx = ex + e−x ¯¯¯¯ 2 0 = e2 + e−2 −(1+1) = 2(cosh 2 −1) . Example 8.2 y x 0 1 y = x y = x2 Find the area of the region bounded by y = x2 and y = x. Solution: A “bounded” region will always mean a region of ﬁnite area, as op- posed to unbounded regions. The curves y = x2 and y = x intersect at x = 0 and x = 1, so the region the curves bound is the shaded region shown in the ﬁgure on the right. Since x ≥x2 for 0 ≤x ≤1, then the height function for the region is h(x) = \|x2 −x\| = x−x2. The region’s area A is then A = Z1 0 (x−x2) dx = 1 2 x−1 3 x2 ¯¯¯¯ 1 0 = 1 2 −1 3 = 1 6 . Example 8.3 y x 0 π 4 π 3 y = sin x y = cos x 1 Find the area of the region bounded by y = sin x and y = cos x over [0,π/3]. Solution: As shown in the ﬁgure on the right, the curves intersect at x = π 4 , and cos x ≥sin x for 0 ≤x ≤π 4 , while sin x ≥cos x for π 4 ≤x ≤π 3 . The area A of the region thus needs to be split into two integrals: A = Zπ/3 0 \|sin x −cos x\| dx = Zπ/4 0 (cos x −sin x) dx + Zπ/3 π/4 (sin x −cos x) dx = Ã sin x + cos x ¯¯¯¯ π/4 0 ! + Ã −cos x −sin x ¯¯¯¯ π/3 π/4 ! = µ 1 p 2 + 1 p 2 ¶ −(0−1) + Ã −1 2 − p 3 2 ! − µ −1 p 2 −1 p 2 ¶ = 4 p 2−3− p 3 2	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 2 y = 6−x2 Solution: As shown in the ﬁgure on the right, above the x-axis the curve y = 6−x2 intersects the line y = x at x = 2 and intersects the line y = −5x at x = −1. Since 6−x2 ≥−5x over [−1,0] and 6−x2 ≥x over [0,2], the area A of the shaded region needs to be split into two integrals: A = Z0 −1 ¯¯6−x2 −(−5x) ¯¯ dx + Z2 0 ¯¯6−x2 −x ¯¯ dx = Z0 −1 (6−x2 +5x) dx + Z2 0 (6−x2 −x) dx = Ã 6x−1 3 x3 + 5 2 x2 ¯¯¯¯ 0 −1 ! + Ã 6x−1 3 x3 −1 2 x2 ¯¯¯¯ 2 0 ! = 0 − µ −6+ 1 3 + 5 2 ¶ + µ 12−8 3 −2 ¶ −0 = 21 2 For some areas between curves it might be easier to switch the roles of x and y, so that instead of a vertical height function you would use a horizontal width function. Example 8.5 Find the area of the region bounded by x = y2 −2 and y = x. y x 0 −2 −1 2 w(y) y = x x = y2 −2 Solution: As shown in the ﬁgure on the right, the parabola x = y2 −2 intersects the line y = x at x = −1 and x = 2. The region has differ- ent height functions h(x) for −2 ≤x ≤−1 and −1 ≤x ≤2, so that two integrals would be required for the area A. However, notice that the width function w(y) has one deﬁnition over the entire region between the curves x = y2 −2 and x = y: w(y) = ¯¯y −(y2 −2) ¯¯ = y −(y2 −2). Thus, instead of integrating the vertical strips dA = h(x)dx along the x-axis, integrate the horizontal strips dA = w(y)dy along the y-axis, from y = −1 to y = 2: A = Z2 −1 w(y) dy = Z2 −1 ¯¯y−(y2 −2) ¯¯ dy = Z2 −1 (y−(y2 −2)) dy = 1 2 y2 −1 3 y3 +2y ¯¯¯¯ 2 −1 = µ 2−8 3 +4 ¶ − µ1 2 + 1 3 −2 ¶ = 9 2	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 simply the difference between the “outer” and “inner” areas, each given by formula (7.20): A = Zβ α 1 2 r2 1dθ − Zβ α 1 2 r2 2dθ = Zβ α 1 2 (r2 1 −r2 2) dθ In general, to include cases where the “outer” and “inner” curves switch positions, take the absolute value of the difference: The area A between two polar curves r = r1(θ) and r = r2(θ) for α ≤θ ≤β is: A = Zβ α 1 2 ¯¯r2 1 −r2 2 ¯¯ dθ (8.2) Example 8.6 θ = π 6 θ = π 3 y x r = 1+cosθ O r = 1−cosθ Find the area between r = 1+cos θ and r = 1−cos θ for π 6 ≤θ ≤π 3 . Solution: Let r1(θ) = 1 + cos θ and r2(θ) = 1 −cos θ. Since r1(θ) > r2(θ) for π 6 ≤θ ≤π 3 , the area A of the region (shown in the ﬁgure on the right) is A = Zπ/3 π/6 1 2 ¯¯r2 1 −r2 2 ¯¯ dθ = Zπ/3 π/6 1 2 ((1+cos θ)2 −(1−cos θ)2) dθ = Zπ/3 π/6 2 cos θ dθ = 2 sin θ ¯¯¯¯ π/3 π/6 = 2 Ãp 3 2 −1 2 ! = p 3 −1 y x Figure 8.1.3 Monte Carlo integration is a technique for approximating the area of a region by taking a large number of random points in a rectangle that encloses the region (see Figure 8.1.3). The idea is simple: # of points in the region # of points in the rectangle ≈ area of the region area of the rectangle For example, if 20% of the random points in the rectangle fall inside the region, then—by randomness—you would expect the area of the region to be about 20% of the area of the rectangle. The more random points you take, the better the approximation. Since the area of the rectangle is known, as well as the number of random points inside the region and the rectangle, the area of the region is easy to approximate.	ElementaryCalculus_Page_265_Chunk3800

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