Problem Set 1 Solutions: Differentiation

SOLUTIONS TO 18.01 EXERCISES Unit 1. Differentiation 1A. Graphing 1A-1,2 a) y = (x − 1)2 − 2 b) y = 3(x2 + 2x) + 2 = 3(x + 1)2 − 1 2 2 1 1 -2 1b 1a 1A-3 a) f (−x) = 1 -2 -1 2a 2b (−x)3 − 3x −x3 − 3x = = −f (x), so it is odd. 1 − (−x)4 1 − x4 b) (sin(−x))2 = (sin x)2 , so it is even. c) odd , so it is odd even � ±(1 + x)4 : neither. d) (1 − x)4 = e) J0 ((−x)2 ) = J0 (x2 ), so it is even. 1A-4 a) p(x) = pe (x) + po (x), where pe (x) is the sum of the even powers and po (x) is the sum of the odd powers b) f (x) = F (x) = f (x) + f (−x) f (x) − f (−x) + 2 2 f (x) + f (−x) f (x) − f (−x) is even and G(x) = is odd because 2 2 COPYRIGHT DAVID JERISON AND MIT 1996, 2003 1 E. Solutions to 18.01 Exercises F (−x) = 1. Differentiation f (−x) + f (−(−x)) = F (x); 2 G(−x) = f (x) − f (−x) = −G(−x). 2 c) Use part b: 1 1 2a 2a + = = 2 x + a −x + a (x + a)(−x + a) a − x2 even 1 1 −2x −2x − = = 2 x + a −x + a (x + a)(−x + a) a − x2 odd =⇒ 1 a x = 2 − 2 2 x+a a −x a − x2 x−1 3y + 1 . Crossmultiply and solve for x, getting x = , so the 2x + 3 1 − 2y 3x + 1 inverse function is . 1 − 2x 1A-5 a) y = b) y = x2 + 2x = (x + 1)2 − 1 (Restrict domain to x ≤ −1, so when it’s flipped about the diagonal y = x, √ you’ll still get the graph of√a function.) Solving for x, we get x = y + 1 − 1, so the inverse function is y = x + 1 − 1 . g(x) g(x) f(x) f(x) 5b 5a 1A-6 a) A = b) √ √ 1 + 3 = 2, tan c = 3 ,c= 1 π 3. So sin x + √ 3 cos x = 2 sin(x + π3 ) . √ π 2 sin(x − ) 4 1A-7 a) 3 sin(2x − π) = 3 sin 2(x − b) −4 cos(x + π ), amplitude 3, period π, phase angle π/2. 2 π ) = 4 sin x amplitude 4, period 2π, phase angle 0. 2 2 1. Differentiation E. Solutions to 18.01 Exercises 3 4 π π 2π 2π -4 -3 7a 7b 1A-8 f (x) odd =⇒ f (0) = −f (0) =⇒ f (0) = 0. So f (c) = f (2c) = · · · = 0, also (by periodicity, where c is the period). 1A-9 3 2 -8 -7 -5 -3 -1 1 3 -4 4 8 12 5 -1 9c 9ab period = 4 -6 c) The graph is made up of segments joining (0, −6) to (4, 3) to (8, −6). It repeats in a zigzag with period 8. * This can be derived using: (1) x/2 − 1 = −1 =⇒ x = 0 and g(0) = 3f (−1) − 3 = −6 (2) x/2 − 1 = 1 =⇒ x = 4 and g(4) = 3f (1) − 3 = 3 (3) x/2 − 1 = 3 =⇒ x = 8 and g(8) = 3f (3) − 3 = −6 (4) 1B. Velocity and rates of change 1B-1 a) h = height of tube = 400 − 16t2 . average speed h(2) − h(0) (400 − 16 · 22 ) − 400 = = −32ft/sec 2 2 (The minus sign means the test tube is going down. You can also do this whole problem using the function s(t) = 16t2 , representing the distance down measured from the top. Then all the speeds are positive instead of negative.) b) Solve h(t) = 0 (or s(t) = 400) to find landing time t = 5. Hence the average speed for the last two seconds is h(5) − h(3) 0 − (400 − 16 · 32 ) = = −128ft/sec 2 2 3 E. Solutions to 18.01 Exercises 1. Differentiation c) h(t) − h(5) 400 − 16t2 − 0 16(5 − t)(5 + t) = = t−5 t−5 t−5 = −16(5 + t) → −160ft/sec as t → 5 (5) (6) 1B-2 A tennis ball bounces so that its initial speed straight upwards is b feet per second. Its height s in feet at time t seconds is s = bt − 16t2 a) (7) (8) (9) (10) s(t + h) − s(t) b(t + h) − 16(t + h)2 − (bt − 16t2 ) = h h bt + bh − 16t2 − 32th − 16h2 − bt + 16t2 = h bh − 32th − 16h2 = h = b − 32t − 16h → b − 32t as h → 0 Therefore, v = b − 32t. b) The ball reaches its maximum height exactly when the ball has finished going up. This is time at which v(t) = 0, namely, t = b/32. c) The maximum height is s(b/32) = b2 /64. d) The graph of v is a straight line with slope −32. The graph of s is a parabola with maximum at place where v = 0 at t = b/32 and landing time at t = b/16. v b s b/32 t graph of velocity b/32 b/16 graph of position t e) If the initial velocity on the first bounce was b1 = b, and the √velocity of the second bounce is b2 , then b22 /64 = (1/2)b21 /64. Therefore, b2 = b1 / 2. The second (continued →) bounce is at b1 /16 + b2 /16. 4 1. Differentiation E. Solutions to 18.01 Exercises f) If the ball continues to bounce then the landing times form a geometric series √ √ (11) b1 /16 + b2 /16 + b3 /16 + · · · = b/16 + b/16 2 + b/16( 2)2 + · · · √ √ (12) = (b/16)(1 + (1/ 2) + (1/ 2)2 + · · · ) (13) = b/16 √ 1 − (1/ 2) √ Put another way, the ball stops bouncing after 1/(1 − (1/ 2)) ≈ 3.4 times the length of time the first bounce. 1C. Slope and derivative. 1C-1 a) π(r + h)2 − πr2 π(r2 + 2rh + h2 ) − πr2 π(2rh + h2 ) = = h h h = π(2r + h) (14) (15) → 2πr as h → 0 (16) b) (19) (4π/3)(r + h)3 − (4π/3)r3 (4π/3)(r3 + 3r2 h + 3rh2 + h3 ) − (4π/3)r3 = h h (4π/3)(3r2 h + 3rh2 + h3 ) = h 2 = (4π/3)(3r + 3rh + h2 ) (20) → 4πr2 as h → 0 (17) (18) 1C-2 f (x) − f (a) (x − a)g(x) − 0 = = g(x) → g(a) as x → a. x−a x−a 1C-3 a) (21) (22) (23) (24) � � � � 1 1 1 1 2x + 1 − (2(x + h) + 1) − = h 2(x + h) + 1 2x + 1 h (2(x + h) + 1)(2x + 1) � � 1 −2h = h (2(x + h) + 1)(2x + 1) −2 = (2(x + h) + 1)(2x + 1) −2 as h → 0 −→ (2x + 1)2 5 E. Solutions to 18.01 Exercises 1. Differentiation b) (25) 2(x + h)2 + 5(x + h) + 4 − (2x2 + 5x + 4) 2x2 + 4xh + 2h2 + 5x + 5h − 2x2 − 5x = h h 4xh + 2h2 + 5h (26) = = 4x + 2h + 5 h (27) −→ 4x + 5 as h → 0 c) (28) (29) (30) (31) (32) � � � � 1 1 1 1 (x2 + 1) − ((x + h)2 + 1) − = h (x + h)2 + 1 x2 + 1 h ((x + h)2 + 1)(x2 + 1) � 2 � 1 x + 1 − x2 − 2xh − h2 − 1 = h ((x + h)2 + 1)(x2 + 1) � � 1 −2xh − h2 = h ((x + h)2 + 1)(x2 + 1) −2x − h = ((x + h)2 + 1)(x2 + 1) −2x −→ 2 as h → 0 (x + 1)2 d) Common denominator: √ � � �√ � 1 1 1 x− x+h 1 √ √ −√ = √ h h x x+h x+h x Now numerator and denominator by √ simplify the numerator by multiplying x + h, and using (a − b)(a + b) = a2 − b2 : √ � � � � √ 1 ( x)2 − ( x + h)2 1 x − (x + h) √ √ √ √ (33) = √ √ √ √ h h x + h x( x + x + h) x + h x( x + x + h) � � 1 −h √ √ (34) = √ √ h x + h x( x + x + h) � � −1 √ (35) = √ √ √ x + h x( x + x + h) −1 1 (36) −→ √ 3 = − x−3/2 as h → 0 2 2( x) √ x+ e) For part (a), −2/(2x + 1)2 < 0, so there are no points where the slope is 1 or 0. For slope −1, √ √ −2/(2x + 1)2 = −1 =⇒ (2x + 1)2 = 2 =⇒ 2x + 1 = ± 2 =⇒ x = −1/2 ± 2/2 For part (b), the slope is 0 at x = −5/4, 1 at x = −1 and −1 at x = −3/2. 6 1. Differentiation E. Solutions to 18.01 Exercises 1C-4 Using Problem 3, a) f � (1) = −2/9 and f (1) = 1/3, so y = −(2/9)(x − 1) + 1/3 = (−2x + 5)/9 b) f (a) = 2a2 + 5a + 4 and f � (a) = 4a + 5, so y = (4a + 5)(x − a) + 2a2 + 5a + 4 = (4a + 5)x − 2a2 + 4 c) f (0) = 1 and f � (0) = 0, so y = 0(x − 0) + 1, or y = 1. √ d) f (a) = 1/ a and f � (a) = −(1/2)a−3/2 , so √ y = −(1/2)a3/2 (x − a) + 1/ a = −a−3/2 x + (3/2)a−1/2 1C-5 Method 1. y � (x) = 2(x − 1), so the tangent line through (a, 1 + (a − 1)2 ) is y = 2(a − 1)(x − a) + 1 + (a − 1)2 In order to see if the origin is on this line, plug in x = 0 and y = 0, to get the following equation for a. 0 = 2(a − 1)(−a) + 1 + (a − 1)2 = −2a2 + 2a + 1 + a2 − 2a + 1 = −a2 + 2 √ Therefore a = ± 2 and the two tangent lines through the origin are √ √ y = 2( 2 − 1)x and y = −2( 2 + 1)x (Because these are lines throught the origin, the constant terms must cancel: this is a good check of your algebra!) Method 2. Seek tangent lines of the form y = mx. Suppose that y = mx meets y = 1 + (x − 1)2 , at x = a, then ma = 1 + (a − 1)2 . In addition we want the slope y � (a) = 2(a − 1) to be equal to m, so m = 2(a − 1). Substituting for m we find 2(a − 1)a = 1 + (a − 1)2 √ This√is the same equation as in method 1: a2 − 2 = 0, so a = ± 2 and m = 2(± 2 − 1), and the two tangent lines through the origin are as above, √ √ y = 2( 2 − 1)x and y = −2( 2 + 1)x 1D. Limits and continuity 1D-1 Calculate the following limits if they exist. If they do not exist, then indicate whether they are +∞, −∞ or undefined. a) −4 b) 8/3 7 E. Solutions to 18.01 Exercises 1. Differentiation c) undefined (both ±∞ are possible) d) Note that 2 − x is negative when x > 2, so the limit is −∞ e) Note that 2 − x is positive when x < 2, so the limit is +∞ (can also be written ∞) f) 4x2 4x ∞ = → = ∞ as x → ∞ x−2 1 − (2/x) 1 g) 4x2 4x2 − 4x(x − 2) 8x 8 − 4x = = = → 8 as x → ∞ x−2 x−2 x−2 1 − (2/x) i) 1 + (2/x) + (3/x2 ) 1 x2 + 2x + 3 → as x → ∞ = 3x2 − 2x + 4 3 − (2/x) + 4/x2 ) 3 j) x−2 x−2 1 1 = = → as x → 2 2 x −4 (x − 2)(x + 2) x+2 4 1D-2 a) lim x→ 0 √ x=0 b) lim x→1+ 1 =∞ x−1 lim x→1− 1 = −∞ x−1 c) lim (x − 1)−4 = ∞ (left and right hand limits are same) x→ 1 d) lim | sin x| = 0 (left and right hand limits are same) x→0 e) lim x→0+ |x| =1 x lim x→0− 1D-3 a) x = 2 removable infinite c) x = 0 removable removable |x| = −1 x x = −2 infinite d) x = 0 removable b) x = 0, ±π, ±2π, ... e) x = 0 jump (−1,1) (0,.5) 2 4b 4a 8 f) x = 0 1. Differentiation E. Solutions to 18.01 Exercises 1D-5 a) for continuity, want ax + b = 1 when x = 1. a+b=1 Ans.: all a, b such that dy d(x2 ) d(ax + b) = = 2x = 2 when x = 1 . We have also = a. dx dx� dx Therefore, to make f (x) continuous, we want a = 2. b) Combining this with the condition a + b=1 from part (a), we get finally b = −1, a = 2. 1D-6 a) f (0) = 02 + 4 · 0 + 1 = 1. Match the function values: f (0− ) = lim ax + b = b, x→0 so b = 1 by continuity. Next match the slopes: f � (0+ ) = lim 2x + 4 = 4 � x→ 0 � − and f (0 ) = a. Therefore, a = 4, since f (0) exists. b) f (1) = 12 + 4 · 1 + 1 = 6 and f (1− ) = lim ax + b = a + b x→1 Therefore continuity implies a + b = 6. The slope from the right is f � (1+ ) = lim 2x + 4 = 6 x→1 Therefore, this must equal the slope from the left, which is a. Thus, a = 6 and b = 0. 1D-7 f (1) = c12 + 4 · 1 + 1 = c + 5 and f (1− ) = lim ax + b = a + b x→ 1 Therefore, by continuity, c + 5 = a + b. Next, match the slopes from left and right: f � (1+ ) = lim 2cx + 4 = 2c + 4 and f � (1− ) = lim a = a x→ 1 x→ 1 Therefore, a = 2c + 4 and b = −c + 1. 1D-8 a) f (0) = sin(2 · 0) = 0 and f (0+ ) = lim ax + b = b x→ 0 Therefore, continuity implies b = 0. The slope from each side is f � (0− ) = lim 2 cos(2x) = 2 and f � (0+ ) = lim a = a x→0 x→0 Therefore, we need a = � 2 in order that f not be differentiable. 9 E. Solutions to 18.01 Exercises 1. Differentiation b) f (0) = cos(2 · 0) = 1 and f (0+ ) = lim ax + b = b x→0 Therefore, continuity implies b = 1. The slope from each side is f � (0− ) = lim −2 sin(2x) = 0 and f � (0+ ) = lim a = a x→ 0 x→ 0 Therefore, we need a = � 0 in order that f not be differentiable. 1D-9 There cannot be any such values because every differentiable function is continuous. 1E: Differentiation formulas: polynomials, products, quotients 1E-1 Find the derivative of the following polynomials a) 10x9 + 15x4 + 6x2 b) 0 (e2 + 1 ≈ 8.4 is a constant and the derivative of a constant is zero.) c) 1/2 d) By the product rule: (3x2 + 1)(x5 + x2 ) + (x3 + x)(5x4 + 2x) = 8x7 + 6x5 + 5x + 3x2 . Alternatively, multiply out the polynomial first to get x8 + x6 + x5 + x3 and then differentiate. 4 1E-2 Find the antiderivative of the following polynomials a) ax2 /2 + bx + c, where a and b are the given constants and c is a third constant. b) x7 /7 + (5/6)x6 + x4 + c c) The only way to get at this is to multiply it out: x6 + 2x3 + 1. Now you can take the antiderivative of each separate term to get x7 x4 + +x+c 7 2 Warning: The answer is not (1/3)(x3 + 1)3 . (The derivative does not match if you apply the chain rule, the rule to be treated below in E4.) 1E-3 y � = 3x2 + 2x − 1 = 0 =⇒ (3x − 1)(x + 1) = 0. Hence x = 1/3 or x = −1 and the points are (1/3, 49/27) and (−1, 3) 10 1. Differentiation E. Solutions to 18.01 Exercises 1E-4 a) f (0) = 4, and f (0− ) = lim 5x5 + 3x4 + 7x2 + 8x + 4 = 4. Therefore the x→ 0 function is continuous for all values of the parameters. f � (0+ ) = lim 2ax + b = b and f � (0− ) = lim 25x4 + 12x3 + 14x + 8 = 8 x→ 0 x→ 0 Therefore, b = 8 and a can have any value. b) f (1) = a + b + 4 and f (1+ ) = 5 + 3 + 7 + 8 + 4 = 27. So by continuity, a + b = 23 f � (1− ) = lim 2ax + b = 2a + b; f � (1+ ) = lim 25x4 + 12x3 + 14x + 8 = 59. x→ 1 x→1 Therefore, differentiability implies 2a + b = 59 Subtracting the first equation, a = 59 − 23 = 36 and hence b = −13. 1E-5 a) 1 (1 + x)2 b) 1 − 2ax − x2 (x2 + 1)2 c) −x2 − 4x − 1 (x2 − 1)2 d) 3x2 − 1/x2 1F. Chain rule, implicit differentiation 1F-1 a) Let u = (x2 + 2) d 2 du d 2 u = u = (2x)(2u) = 4x(x2 + 2) = 4x3 + 8x dx dx du Alternatively, d 2 d 4 (x + 2)2 = (x + 4x2 + 4) = 4x3 + 8x dx dx d 100 du d 100 b) Let u = (x2 + 2); then u = u = (2x)(100u99 ) = (200x)(x2 + 2)99 . dx dx du 1F-2 Product rule and chain rule: 10x9 (x2 + 1)10 + x10 [10(x2 + 1)9 (2x)] = 10(3x2 + 1)x9 (x2 + 1)9 1F-3 y = x1/n =⇒ y n = x =⇒ ny n−1 y � = 1. Therefore, y� = 1 1 1 1 = y 1−n = x n −1 ny n−1 n n 1F-4 (1/3)x−2/3 + (1/3)y −2/3 y � = 0 implies y � = −x−2/3 y 2/3 Put u = 1 − x1/3 . Then y = u3 , and the chain rule implies dy du = 3u2 = 3(1 − x1/3 )2 (−(1/3)x−2/3 ) = −x−2/3 (1 − x1/3 )2 dx dx 11 E. Solutions to 18.01 Exercises 1. Differentiation The chain rule answer is the same as the one using implicit differentiation because y = (1 − x1/3 )3 =⇒ y 2/3 = (1 − x1/3 )2 1F-5 Implicit differentiation gives cos x + y � cos y = 0. Horizontal slope means y � = 0, so that cos x = 0. These are the points x = π/2 + kπ for every integer k. Recall that sin(π/2 + kπ) = (−1)k , i.e., 1 if k is even and −1 if k is odd. Thus at x = π/2 + kπ, ±1 + sin y = 1/2, or sin y = �1 + 1/2. But sin y = 3/2 has no solution, so the only solutions are when k is even and in that case sin y = −1 + 1/2, so that y = −π/6 + 2nπ or y = 7π/6 + 2nπ. In all there are two grids of points at the vertices of squares of side 2π, namely the points (π/2 + 2kπ, −π/6 + 2nπ) and (π/2 + 2kπ, 7π/6 + 2nπ); k, n any integers. 1F-6 Following the hint, let z = −x. If f is even, then f (x) = f (z) Differentiating and using the chain rule: f � (x) = f � (z)(dz/dx) = −f � (z) because dz/dx = −1 But this means that f � is odd. Similarly, if g is odd, then g(x = −g(z). Differenti­ ating and using the chain rule: g � (x) = −g � (z)(dz/dx) = g � (z) 1F-7 a) because dz/dx = −1 dD 1 x−a = ((x − a)2 + y0 2 )−1/2 (2(x − a)) = � dx 2 (x − a)2 + y0 2 b) dm −1 −2v m0 v = m0 · (1 − v 2 /c2 )−3/2 · 2 = 2 dv 2 c c (1 − v 2 /c2 )3/2 c) dF 3 −3mgr = mg · (− )(1 + r2 )−5/2 · 2r = dr 2 (1 + r2 )5/2 d) −6bt a a(1 − 5bt2 ) dQ = at · + = dt (1 + bt2 )4 (1 + bt2 )3 (1 + bt2 )4 1F-8 a) V = 1 2 1 −r2 −r πr h =⇒ 0 = π(2rr� h + r2 ) =⇒ r� = = 3 3 2rh 2h b) P V c = nRT =⇒ P � V c + P · cV c−1 = 0 =⇒ P � = − cP V c−1 cP =− c V V c) c2 = a2 + b2 − 2ab cos θ implies 0 = 2aa� + 2b − 2(cos θ(a� b + a)) =⇒ a� = 1G. Higher derivatives 12 −2b + 2 cos θ · a a cos θ − b = 2a − 2 cos θ · b a − b cos θ 1. Differentiation E. Solutions to 18.01 Exercises 1G-1 a) 6 − x−3/2 b) −10 (x + 5)3 c) −10 (x + 5)3 d) 0 1G-2 If y ��� = 0, then y �� = c0 , a constant. Hence y � = c0 x + c1 , where c1 is some other constant. Next, y = c0 x2 /2 + c1 x + c2 , where c2 is yet another constant. Thus, y must be a quadratic polynomial, and any quadratic polynomial will have the property that its third derivative is identically zero. 1G-3 x2 y2 2x 2yy � + = 1 = ⇒ + 2 = 0 =⇒ y � = −(b2 /a2 )(x/y) a2 b2 a2 b Thus, (37) (38) �� � � 2 �� � b2 y − xy � b y + x(b2 /a2 )(x/y) y =− =− y2 y2 a2 a2 � 4 � 4 b b =− (y 2 /b2 + x2 /a2 ) = − 2 3 3 2 y a a y �� � 1G-4 y = (x + 1)−1 , so y (1) = −(x + 1)−2 , y (2) = (−1)(−2)(x + 1)−3 , and y (3) = (−1)(−2)(−3)(x + 1)−4 . The pattern is y (n) = (−1)n (n!)(x + 1)−n−1 13 E. Solutions to 18.01 Exercises 1G-5 a) y � = u� v + uv � =⇒ 1. Differentiation y �� = u�� v + 2u� v � + uv �� b) Formulas above do coincide with Leibniz’s formula for n = 1 and n = 2. To calculate y (p+q) where y = xp (1 + x)q , use�u �= xp and v = (1 + x)q . The only n (p) (q) term in the Leibniz formula that is not 0 is u v , since in all other terms k either one factor or the other is 0. If u = xp , u(p) = p!, so y (p+q) = � � n n! p!q! = · p!q! = n! p p!q! 1H. Exponentials and Logarithms: Algebra 1H-1 a) To see when y = y0 /2, we must solve the equation 1 2 = e−kt . Take ln of both sides: − ln 2 = −kt, from which t = b) y1 = y0 ekt1 by assumption, λ = 1 e− ln 2 = y1 · 2 y0 = y0 e−kt , or 2 ln 2 . k − ln 2 y0 ek(t1 +λ) = y0 ekt1 · ekλ = y1 · k 1H-2 pH = − log10 [H + ]; by assumption, [H + ]dil = 12 [H + ]orig . Take − log10 of both sides (note that log 2 ≈ .3): − log [H + ]dil = log 2 − log [H + ]orig =⇒ pHdil = pHorig + log2 . 1H-3 a) ln(y + 1) + ln(y − 1) = 2x + ln x; exponentiating both sides and solving for y: (y + 1) · (y − 1) = e2x · x =⇒ y 2 − 1 = xe2x =⇒ y = √ xe2x + 1, since y > 0. 2 y+1 = 10−x . Solve for y−1 2 y; to simplify the algebra, let A = 10−x . Crossmultiplying, y + 1 = Ay − A =⇒ 2 A+1 10−x + 1 y= = −x2 A−1 10 −1 b) log(y + 1) − log(y − 1) = −x2 ; exponentiating, c) 2 ln y − ln(y + 1) = x; exponentiating both sides and solving for y: 14 1. Differentiation E. Solutions to 18.01 Exercises √ y2 ex e2x + 4ex = ex =⇒ y 2 − ex y − ex = 0 =⇒ y = , since y − 1 > 0. y+1 2 1H-4 c ln a = c ⇒ ln a = c ln b ⇒ a = ec ln b = eln b = bc . Similarly, ln b log a = c ⇒ a = bc . log b u2 + 1 = y; this u2 − 1 1 y+1 x = ln( ) 2 y−1 1H-5 a) Put u = ex (multiply top and bottom by ex first): gives u2 = y+1 y+1 = e2x ; taking ln: 2x = ln( ), y−1 y−1 1 b) ex + e−x = y; putting u = ex gives u + = y ; solving for u gives u � � y ± y2 − 4 y ± y2 − 4 x 2 u −yu+1 = 0 so that u = = e ; taking ln: x = ln( ) 2 2 1H-6 A = log e · ln 10 = ln(10log e ) = ln(e) = 1 ; similarly, logb a · loga b = 1 1H-7 a) If I1 is the intensity of the jet and I2 is the intensity of the conversation, then � � I1 /I0 log10 (I1 /I2 ) = log10 = log10 (I1 /I0 ) − log10 (I2 /I0 ) = 13 − 6 = 7 I2 /I0 Therefore, I1 /I2 = 107 . b) I = C/r2 and I = I1 when r = 50 implies I1 = C/502 =⇒ C = I1 502 =⇒ I = I1 502 /r2 This shows that when r = 100, we have I = I1 502 /1002 = I1 /4 . It follows that 10 log10 (I/I0 ) = 10 log10 (I1 /4I0 ) = 10 log10 (I1 /I0 ) − 10 log10 4 ≈ 130 − 6.0 ≈ 124 The sound at 100 meters is 124 decibels. The sound at 1 km has 1/100 the intensity of the sound at 100 meters, because 100m/1km = 1/10. 10 log10 (1/100) = 10(−2) = −20 so the decibel level is 124 − 20 = 104. 1I. Exponentials and Logarithms: Calculus 1I-1 a) (x + 1)ex 2 4xe2x b) 4xe2x c) (−2x)e−x 15 2 d) ln x e) 2/x f) 2(ln x)/x g) E. Solutions to 18.01 Exercises 1. Differentiation � �� h) (xx )� = ex ln x = (x ln x)� ex ln x = (ln x + 1)ex ln x = (1 + ln x)xx i) (ex − e−x )/2 j) (ex + e−x )/2 k) −1/x l) −1/x(ln x)2 m) −2ex /(1 + x 2 e ) 1 (even) 1I-3 a) As n → ∞, h = 1/n → 0. � � 1 ln(1 + h) ln(1 + h) − ln(1) d n ln(1 + ) = = → −→ ln(1 + x)�� =1 h→0 n h h dx x=0 Therefore, lim n ln(1 + n→∞ 1 )=1 n b) Take the logarithm of both sides. We need to show lim ln(1 + n→∞ 1 n ) = ln e = 1 n But 1 n 1 ) = n ln(1 + ) n n so the limit is the same as the one in part (a). ln(1 + 1I-4 a) � 1+ 1 n �3n = �� �n �3 1 −→ e3 as n → ∞, 1+ n b) Put m = n/2. Then � �5n � �10m �� �m �10 2 1 1 −→ e10 as m → ∞ 1+ = 1+ = 1+ n m m c) Put m = 2n. Then � �5n � �5m/2 �� �m �5/2 1 1 1 −→ e5/2 as m → ∞ 1+ = 1+ = 1+ 2n m m 1J. Trigonometric functions 1J-1 a) 10x cos(5x2 ) b) 6 sin(3x) cos(3x) 16 c) −2 sin(2x)/ cos(2x) = −2 tan(2x) 1. Differentiation E. Solutions to 18.01 Exercises d) −2 sin x/(2 cos x) = − tan x. (Why did the factor 2 disappear? Because ln(2 cos x) = ln 2 + ln(cos x), and the derivative of the constant ln 2 is zero.) e) x cos x − sin x x2 i) (x2 sin x)� 2x sin x + x2 cos x 2 = = + cot x. Alternatively, 2 2 x sin x x sin x x f) −(1+y � ) sin(x+y) g) − sin(x+y) 2 h) 2 sin x cos xesin x ln(x2 sin x) = ln(x2 ) + ln(sin x) = 2 ln x + ln sin x Differentiating gives 2 cos x 2 + = + cot x x sin x x j) 2e2x sin(10x) + 10e2x cos(10x) k) 6 tan(3x) sec2 (3x) = 6 sin x/ cos3 x √ √ l) −x(1 − x2 )−1/2 sec( 1 − x2 ) tan( 1 − x2 ) m) Using the chain rule repeatedly and the trigonometric double angle formu­ las, (39) (cos2 x − sin2 x)� = −2 cos x sin x − 2 sin x cos x = −4 cos x sin x; (40) (2 cos2 x)� = −4 cos x sin x; (41) (cos(2x))� = −2 sin(2x) = −2(2 sin x cos x). The three functions have the same derivative, so they differ by constants. And indeed, cos(2x) = cos2 x − sin2 x = 2 cos2 x − 1, (using sin2 x = 1 − cos2 x). n) 5(sec(5x) tan(5x)) tan(5x) + 5(sec(5x)(sec2 (5x)) = 5 sec(5x)(sec2 (5x) + tan2 (5x)) Other forms: 5 sec(5x)(2 sec2 (5x) − 1); 10 sec3 (5x) − 5 sec(5x) o) 0 because sec2 (3x)−tan2 (3x) = 1, a constant — or carry it out for practice. p) Successive use of the chain rule: (42) (43) � � 1 (sin ( x2 + 1))� = cos ( x2 + 1) · (x2 + 1)−1/2 · 2x 2 � x =√ cos ( x2 + 1) x2 + 1 17 E. Solutions to 18.01 Exercises 1. Differentiation q) Chain rule several times in succession: � � � −x (44) (cos2 1 − x2 )� = 2 cos 1 − x2 · (− sin 1 − x2 ) · √ 1 − x2 � x (45) =√ sin(2 1 − x2 ) 1 − x2 r) Chain rule again: � � x x x x+1−x (46) tan2 ( ) = 2 tan( ) · sec2 ( )· x+1 x+1 x+1 (x + 1)2 2 x x (47) = tan( ) sec2 ( ) (x + 1)2 x+1 x+1 1J-2 Because cos(π/2) = 0, cos x cos x − cos(π/2) d = lim = cos x|x=π/2 = − sin x|x=π/2 = −1 x − π/2 dx x→π/2 x − π/2 x→π/2 lim 1J-3 a) (sin(kx))� = k cos(kx). Hence (sin(kx))�� = (k cos(kx))� = −k 2 sin(kx). Similarly, differentiating cosine twice switches from sine and then back to cosine with only one sign change, so (cos(kx)�� = −k 2 cos(kx) Therefore, sin(kx)�� + k 2 sin(kx) = 0 and cos(kx)�� + k 2 cos(kx) = 0 √ Since we are assuming k > 0, k = a. b) This follows from the linearity of the operation of differentiation. With k 2 = a, (48) (c1 sin(kx) + c2 cos(kx))�� + k 2 (c1 sin(kx) + c2 cos(kx)) (49) = c1 (sin(kx))�� + c2 (cos(kx))�� + k 2 c1 sin(kx) + k 2 c2 cos(kx) (50) = c1 [(sin(kx))�� + k 2 sin(kx)] + c2 [(cos(kx))�� + k 2 cos(kx)] (51) = c1 · 0 + c2 · 0 = 0 c) Since φ is a constant, d(kx + φ)/dx = k, and (sin(kx + φ)� = k cos(kx + φ), (sin(kx + φ)�� = (k cos(kx + φ))� = −k 2 sin(kx + φ) Therefore, if a = k 2 , (sin(kx + φ)�� + a sin(kx + φ) = 0 d) The sum formula for the sine function says sin(kx + φ) = sin(kx) cos(φ) + cos(kx) sin(φ) In other words sin(kx + φ) = c1 sin(kx) + c2 cos(kx) 18 1. Differentiation E. Solutions to 18.01 Exercises with c1 = cos(φ) and c2 = sin(φ). 1J-4 a) The Pythagorean theorem implies that c2 = sin2 θ + (1 − cos θ)2 = sin2 θ + 1 − 2 cos θ + cos2 θ = 2 − 2 cos θ Thus, c= √ � 2 − 2 cos θ = 2 1 − cos θ = 2 sin(θ/2) 2 b) Each angle is θ = 2π/n, so the perimeter of the n-gon is n sin(2π/n) As n → ∞, h = 2π/n tends to 0, so 2π sin h − sin 0 d n sin(2π/n) = sin h = 2π → 2π sin x|x=0 = 2π cos x|x=0 = 2π h h dx 19