Problem Set 2 Solutions: Applications of Differentiation

SOLUTIONS TO 18.01 EXERCISES 2. Applications of Differentiation 2A. Approximation 2A-1 √ d √ b b ⇒ f (x) ≈ a + √ x by formula. a + bx = √ dx 2 a 2 a + bx By algebra: 2A-2 D( √ � √ bx √ bx a + bx = a 1 + ≈ a(1 + ), same as above. a 2a 1 −b 1 b 1 1/a ≈ )= ⇒ f (x) ≈ − 2 x; OR: = a + bx (a + bx)2 a a a + bx 1 + b/ax 1 b (1 − x). a a 2A-3 D( (1 + 2x) · (1 + x)3/2 )= 1 + 2x 3 2 · (1 + x)3/2 − (1 + x)3/2 · 2 1 ⇒ f � (0) = − 2 (1 + 2x) 2 1 ⇒ f (x) ≈ 1− x; OR, by algebra, 2 2A-4 Put (1 + x)3/2 3 1 ≈ (1+ x)(1−2x) ≈ 1− x. 1 + 2x 2 2 h g 2h = �; then w = ≈ g(1 − �)2 ≈ g(1 − 2�) = g(1 − ). R (1 + �)2 R 2A-5 A reasonable assumption is that w is propotional to volume v, which is in turn proportional to the cube of a linear dimension, i.e., a given person remains similar to him/herself, for small weight changes.) Thus w = Ch3 ; since 5 feet = 60 inches, we get w(60 + �) C(60 + �)3 � 3� 1 = = (1+ )3 ⇒ w(60+�) ≈ w(60)·(1+ ) ≈ 120·(1+ ) ≈ 126. w(60) C(60)3 60 60 20 [Or you can calculate the linearization of w(h) arround h = 60 using derivatives, and using the value w(60) to determine C. getting w(h) ≈ 120 + 6(h − 60) COPYRIGHT DAVID JERISON AND MIT 1996, 2003 1 E. Solutions to 18.01 Exercises 2A-6 tan θ = 2. Applications of Differentiation sin θ θ ≈ θ(1 + θ2 //2) ≈ θ ≈ cos θ 1 − θ2 /2 sec x 1 1 1 2 √ 2A-7 √ = ≈ 1 2 1 2 ≈ 1 − x2 ≈ 1 + x 2 2 (1 − x )(1 − x ) 1−x cos x 1 − x 2 2 2A-8 1 1 = = 1 1−x 1 − ( 2 + Δx) 1 2 1 2 = 1 − 2Δx − Δx ≈ 2(1 + 2Δx + 4(Δx)2 ) ≈ 2 + 4(x − 12 ) + 8(x − 21 )2 2A-10 y = (1 + x)r , y � = r(1 + x)r−1 , y �� = r(r − 1)(1 + x)r−2 Therefore y(0) = 1, y � (0) = r, y �� (0) = r(r − 1), giving (1 + x)r ≈ 1 + rx + r(r − 1) 2 x . 2 2A-11 pv k = c ⇒ p = cv −k = c((v0 + Δv)−k = cv0 −k (1 + ≈ 2A-12 a) b) c v0 k (1 − k Δv −k ) v0 Δv k(k + 1) Δv 2 + ( ) ) v0 2 v0 ex x2 5 ≈ (1 + x + )(1 + x + x2 ) ≈ 1 + 2x + x2 1−x 2 2 ln(1 + x) x ≈ ≈1−x x xe x(1 + x) 2 c) e−x ≈ 1 − x2 [Substitute into ex ≈ 1 + x] d) ln(cos x) ≈ ln(1 − x2 x2 )≈− [since ln(1 + h) ≈ h] 2 2 e) x ln x = (1 + h) ln(1 + h) ≈ (1 + h)(h − 2 (x − 1) + (x − 1) 2 h2 h2 ) ≈ h+ 2 2 2A-13 Finding the linear and quadratic approximation 2 ⇒ x ln x ≈ 2. Applications of Differentiation E. Solutions to 18.01 Exercises a) 2x (both linear and quadratic) b) 1, 1 − 2x2 c) 1, 1 + x2 /2 (Use (1 + u)−1 ≈ 1 − u with u = x2 /2: sec x = 1/ cos x ≈ 1/(1 − x2 /2) = (1 − x2 /2)−1 ≈ 1 + x2 /2 d) 1, 1 + x2 e) Use (1 + u)−1 ≈ 1 − u + u2 : (a + bx)−1 = a−1 (1 + (bx/a))−1 ≈ a−1 (1 − bx/a + (bx/a)2 ) Linear approximation: (1/a) − (b/a2 )x Quadratic approximation: (1/a) − (b/a2 )x + (b2 /a3 )x2 f) f (x) = 1/(a + bx) so that f � (1) = −b(a + b)−2 and f �� (1) = 2b2 /(a + b)−3 . � 0. We need to assume that these numbers are defined, in other words that a + b = Then the linear approximation is 1/(a + b) − (b/(a + b)2 )(x − 1) and the quadratic approximation is 1/(a + b) − (b/(a + b)2 )(x − 1) + (b/(a + b)3 )(x − 1)2 Method 2: Write 1/(a + bx) = 1/(a + b + b(x − 1)) Then use the expansion of problem (e) with a+b in place of a and b in place of b and (x − 1) in place of x. The requirement a �= 0 in (e) corresponds to the restriction a + b �= 0 in (f). f � (x) = −3 sin(3x), 2A-15 f (x) = cos(3x), (1) f (0) = 1, � f (π/6) = cos(π/2) = 0, (2) f (0) = −3 sin 0 = 0, (3) f �� (x) = −9 cos(3x). Thus, f (π/3) = cos π = −1 � f (π/6) = −3 sin(π/2) = −3, f �� (0) = −9, f �� (π/6) = 0, f � (π/3) = −3 sin π = 0 f �� (π/3) = 9 Using these values, the linear and quadratic approximations are respectively: (4) for x ≈ 0 : f (x) ≈ 1 and f (x) ≈ 1 − (9/2)x2 (5) for x ≈ π/6 : both are f (x) ≈ −3(x − π/6) (6) for x ≈ π/3 : f (x) ≈ −1 and f (x) ≈ −1 + (9/2)(x − π/3)2 3 E. Solutions to 18.01 Exercises 2. Applications of Differentiation 2A-16 a) The law of cosines says that for a triangle with sides a, b, and c, with θ opposite the side of length c, c2 = a2 + b2 − 2ab cos θ Apply it to one of the n triangles with vertex at the origin: a = b = 1 and θ = 2π/n. So the formula is � c = 2 − 2 cos(2π/n) b) The perimeter is n near 0 is � 2 − 2 cos(2π/n). The quadratic approximation to cos θ cos θ ≈ 1 − θ2 /2 Therefore, as n → ∞ and θ = 2π/n → 0, � � � n 2 − 2 cos(2π/n) ≈ n 2 − 2(1 − (1/2)(2π/n)2 ) = n (2π/n)2 = n(2π/n) = 2π In other words, lim n n→∞ � 2 − 2 cos(2π/n) = 2π, the circumference of the circle of radius 1. 2B. Curve Sketching 2B-1 a) y = x3 − 3x + 1, y � = 3x2 − 3 = 3(x − 1)(x + 1). y � = 0 =⇒ x = ±1. Endpoint values: y → −∞ as x → −∞, and y → ∞ as x → ∞. Critical values: y(−1) = 3, y(1) = −1. Increasing on: −∞ < x < −1, 1 < x < ∞. Decreasing on: −1 < x < 1. Graph: (−∞, −∞) � (−1, 3) � (1, −1) � (∞, ∞), crossing the x-axis three times. (-1, 3) 1a (1,-1) b) y = x4 − 4x + 1, y � = x3 − 4. y � = 0 =⇒ x = 41/3 . 4 2. Applications of Differentiation E. Solutions to 18.01 Exercises Increasing on: 41/3 < x < ∞; decreasing on: −∞ < x < 41/3 . Endpoint values: y → ∞ as x → ±∞; critical value: y(41/3 ) = 1. Graph: (−∞, ∞) � (41//3 , 1) � (∞, ∞), never crossing the x-axis. (See below.) c) y � (x) = 1/(1 + x2 ) and y(0) = 0. By inspection, y � > 0 for all x, hence always increasing. Endpoint values: y → c as x → ∞ and by symmetry y → −c as x → −∞. (But it is not clear at this point in the course whether c = ∞ or some finite value. It turns out (in Lecture 26) that y → c = π/2. Graph: (−∞, −c) � (∞, c), crossing the x-axis once (at x = 0). (See below.) (2,4) (4/3, 1) (odd) 1 1 -4 1b 1c 1d 1e d) y = x2 /(x − 1), y � = (2x(x − 1) − x2 )/(x − 1)2 = (x2 − 2x)/(x − 1)2 = (x − 2)x/(x − 1)2 . Endpoint values: y → ∞ as x → ∞ and y → −∞ as x → −∞. Singular values: y(1+ ) = +∞ and y(1− ) = −∞. Critical values: y(0) = 0 and y(2) = 4. New feature: Pay attention to sign changes in the denominator of y � . Increasing on: −∞ < x < 0 and 2 < x < ∞ Decreasing on: 0 < x < 1 and 1 < x < 2 Graph: (−∞, −∞) � (0, 0) � (1, −∞) ↑ (1, ∞) � (2, 4) � (∞, ∞), crossing the x-axis once (at x = 0). 5 E. Solutions to 18.01 Exercises 2. Applications of Differentiation Commentary on singularities: Look out for sign changes both where y � is zero and also where y � is undefined: y � = 0 indicates a possible sign change in the numerator and y � undefined indicates a possible sign change in the denominator. In this case there was no sign change in y � at x = 1, but there would have been a sign change, if there had been an odd power of (x − 1) in the denominator. e) y = x/(x + 4), y � = ((x + 4) − x)/(x + 4)2 = 4/(x + 4)2 . No critical points. Endpoint values: y → 1 as x → ±∞. Increasing on: −4 < x < ∞. Decreasing on: −∞ < x < −4. Singular values: y(−4+ ) = −∞, y(−4− ) = +∞. Graph: (−∞, 1) � (−4, ∞) ↓ (−4, −∞) � (∞, 1), crossing the x-axis once (at x = 0). √ f) y = x + 1/(x − 3), y � = −(1/2)(x + 5)(x + 1)−1/2 (x − 3)−2 No critical points because x = −5 is outside of the domain of definition, x ≥ −1. Endpoint values: y(−1) = 0, and as x → ∞, 1 + x1 1 y=√ 3 → ∞ =0 √ x− x Singular values: y(3+ ) = +∞, y(3− ) = −∞. Increasing on: nowhere Decreasing on: −1 < x < 3 and 3 < x < ∞. Graph: (−1, 0) � (3, −∞) ↑ (3, ∞) � (∞, 0), crossing the x-axis once (at x = −1). g) y = 3x4 − 16x3 + 18x2 + 1, y � = 12x3 − 48x2 + 36x = 12x(x − 1)(x − 3). y � = 0 =⇒ x = 0, 1, 3. Endpoint values: y → ∞ as x → ±∞. 6 2. Applications of Differentiation E. Solutions to 18.01 Exercises Critical values: y(0) = 1, y(1) = 6, and y(3) = −188. Increasing on: 0 < x < 1 and 3 < x < ∞. Decreasing on: −∞ < x < 0 and 1 < x < 3. Graph: (−∞, ∞) � (0, 1) � (1, 6) � (3, −188) � (∞, ∞), crossing the x-axis once. (1,6) 1 -1 1 3 1g 1f 1h 1i (3,-26) 2 2 h) y = e−x , y � = −2xe−x . y � = 0 =⇒ x = 0. Endpoint values: y → 0 as x → ±∞. Critical value: y(0) = 1. Increasing on: −∞ < x < 0 Decreasing on: 0 < x < ∞ Graph: (−∞, 0) � (0, 1) � (∞, 0), never crossing the x-axis. (The function is even.) 2 i) y � = e−x and y(0) = 0. Because y � is even and y(0) = 0, y is odd. No critical points. Endpoint values: y → c as x → ∞ and by symmetry y → −c as x → −∞. It is not clear at this point in the course whether c is finite or infinite. But we will be able to show that c is finite when we discuss improper integrals in Unit 6. √ (Using a trick with iterated integrals, a subject in 18.02, one can show that c = π/2.) Graph: (−∞, −c) � (∞, c), crossing the x-axis once (at x = 0). 2B-2 a) One inflection point at x = 0. (y �� = 6x) 7 E. Solutions to 18.01 Exercises 2. Applications of Differentiation b) No inflection points. y �� = 3x2 , so the function is convex. x = 0 is not a point of inflection because y �� > 0 on both sides of x = 0. c) Inflection point at x = 0. (y �� = −2x/(1 + x2 )2 ) d) No inflection points. Reasoning: y �� = 2/(x − 1)3 . Thus y �� > 0 and the function is concave up when x > 1, and y �� < 0 and the function is concave down when x < 1. But x = 1 is not called an inflection point because the function is not continuous there. In fact, x = 1 is a singular point. e) No inflection points. y �� = −8/(x+1)3 . As in part (d) there is a sign change in y �� , but at a singular point not an inflection point. 1) f) y �� = −(1/2)[(x + 1)(x − 3) − (1/2)(x + 5)(x − 3) − 2(x + 5)(x + 1)](x + (x − 3)3 −3/2 = −(1/2)[−(3/2)x2 − 15x − 11/2](x + 1)−3/2 (x − 3)3 Therefore there are two inflection points, x = (−30 ± √ 768)/6, ≈ 9.6, .38. g) y �� = 12(3x2 − 8x + 36). Therefore there are no inflection points. The quadratic equation has no real roots. 2 �� 2 −x . Therefore there are two inflection points at x = √h) y = (−2 + 4x )e ±1/ 2. 2 i) One inflection point at x = 0. (y �� = −2xe−x ) 2B-3 a) y � = 3x2 + 2ax + b. The roots of the quadratic polynomial are distinct real numbers if the discriminant is positive. (The discriminant is defined as the number under the square root in the quadratic formula.) Therefore there are distinct real roots if and only if (2a)2 − 4(3)b > 0, or a2 − 3b > 0. From the picture, since y → ∞ as x → ∞ and y → −∞ as x → −∞, the larger root of 3x2 + 2ax + b = 0 (with the plus sign in the quadratic formula) must be the local min, and the smaller root must be the local max. Since y <<-1 when x << -1 and y >> 1 when x >> 1, the local max. is to the left of the local min. 8 2. Applications of Differentiation E. Solutions to 18.01 Exercises b) y �� = 6x + 2a, so the inflection point is at −a/3. Therefore the condition y < 0 at the inflection point is � y � (−a/3) = 3(−a/3)2 + 2a(−a/3) + b = −a2 /3 + b < 0, which is the same as a2 − 3b > 0. If y � < 0 at some point x0 , then the function is decreasing at that point. But y → ∞ as x → ∞, so there must be a local minimum at a point x > x0 . Similarly, since y → −∞ as x → −∞, there must be a local maximum at a point x < x0 . Comment: We evaluate y � at the inflection point of y (x = −a/3) since we are trying to decide (cf. part (b)) whether y � is ever negative. To do this, we find the minimum of y � (which occurs where y �� = 0). 4 5 7 8 10 Max is at x = 5 or x = 10; Min is at x = 0 or x = 8. 2B−4 Graph of function −7 −3 −1 5 Graph of derivative; note that local maximum point above corresponds to zero below; point of inflection above corresponds to local minimum below. 2B−5 2B-6 a) Try y � = (x + 1)(x − 1) = x2 − 1. Then y = x3 /3 − x + c. The constant c won’t matter so set c = 0. It’s also more convenient to multiply by 3: y = x3 − 3x b) This is an odd function with local min and max: y(1) = −2 and y(−1) = 2. The endpoints values are y(3) = 18 and y(−3) = −18. It is very steep: y � (3) = 8 . 9 E. Solutions to 18.01 Exercises 2. Applications of Differentiation c) -3 -1 1 3 Δy Δx→0 Δx 2B-7 a) f � (a) = lim If y increasing, then Therefore, lim Δx→0 Δy > 0 ⇒ Δx > 0 Δy < 0 ⇒ Δx < 0 � . So in both cases Δy > 0. Δx Δy ≥ 0. Δx Δy b) Proof breaks down at the last step. Namely, > 0 doesn’t imply Δx Δx lim >0 Δx→0 Δy [Limits don’t preserve strict inequalities, only weak ones. For example, u2 > 0 for u �= 0, but lim u2 = 0 ≥ 0, not > 0.] u→0 Counterexample: f (x) = x3 is increasing for all x, but f � (0) = 0. c) Use f (a) ≥ f (x) to show that lim Δy/Δx ≤ 0 and Δx→0+ lim Δy/Δx ≥ 0. Δx→0− Since the left and right limits are equal, the derivative must be zero. 2C. Max-min problems 2C-1 The base of the box has sidelength 12 − 2x and the height is x, so the volume is V = x(12 − 2x)2 . At the endpoints x = 0 and x = 6, the volume is 0, so the maximum must occur in between at a critical point. V � = (12 − 2x)2 + x(2)(12 − 2x)(−2) = (12 − 2x)(12 − 2x − 4x) = (12 − 2x)(12 − 6x). 10 2. Applications of Differentiation E. Solutions to 18.01 Exercises x 12-2x x It follows that V � = 0 when x = 6 or x = 2. At the endpoints x = 0 and x = 6 the volume is 0, so the maximum occurs when x = 2. 2C-2 We want to minimize the fence length L = 2x + y, where the variables x and y are related by xy = A = 20, 000. y x x Choosing x as the independent variable, we have y = A/x, so that L = 2x+A/x. At the endpoints x = 0 and x = ∞ (it’s a long barn), we get L = ∞, so the minimumof L must occur at a critical point. A A L� = 2 − 2 ; L = 0 =⇒ x2 = = 10, 000 =⇒ x = 100 feet x 2 2C-3 We have y = (a − x)/2, so xy = x(a − x)/2. At the endpoints x = 0 and x = a, the product xy is zero (and beyond it is negative). Therefore, the maximum occurs at a critical point. Taking the derivative, a − 2x d x(a − x) = ; this is 0 when x = a/2. dx 2 2 2C-4 If the length is y and the cross-section is a square with sidelength x, then 4x + y = 108. Therxefore the volume is V = x2 y = 108x2 − 4x3 . Find the critical points: (108x2 − 4x3 )� = 216x − 12x2 = 0 =⇒ x = 18 or x = 0. The critical point x = 18 (3/2 ft.) corresponds to the length y = 36 (3 ft.), giving therefore a volume of (3/2)2 (3) = 27/4 = 6.75 cubic feet. The endpoints are x = 0, which gives zero volume, and when x = y, i.e., x = 9/5 feet, which gives a volume of (9/5)3 cubic feet, which is less than 6 cubic feet. So the critical point gives the maximum volume. 2C-5 We let r = radius of bottom and h = height, then the volume is V = πr2 h, and the area is A = πr2 + 2πrh. Using r as the independent variable, we have using the above formulas, A − πr2 h= , 2πr 2 V = πr h = � � A π 3 r− r ; 2 2 11 dV A 3π 2 = − r . dr 2 2 E. Solutions to 18.01 Exercises 2. Applications of Differentiation h r Therefore, dV /dr = 0 implies A = 3πr2 , from which h = A − πr2 = r. 2πr Checking the endpoints, at one h = 0 and V = 0; at the other, lim V = 0 r →0+ (using the expression above for V in terms of r); thus the critical point must occur at a maximum. (Another way to do this problem is to use implicit differentiation with respect to r. Briefly, since A is fixed, dA/dr = 0, and therefore (7) (8) dA r+h = 2πr + 2πh + 2πrh� = 0 =⇒ h� = − ; dr r dV = 2πrh + πr2 h� = 2πrh − πr(r + h) = πr(h − r). dr It follows that V � = 0 when r = h or r = 0, and the latter is a rejected endpoint. 2C-6 To get max and min of y = x(x + 1)(x − 1) = x3 − x, first 1 points: y � = 3x2 − 1 = 0 if x = ± √ ; 3 1 1 2 −2 1 1 2 y( √ ) = √ (− ) = √ , rel. min. y(− √ ) = − √ (− ) = 3 3 3 3 3 3 3 3 Check endpoints: absolute min. y(2) = 6 ⇒ 2 is absolute max.; find the critical 2 √ , rel. max. 3 3 y(−2) = −6 ⇒ −2 is (This is an endpoint problem. The endpoints should be tested unless the physical or geometric picture already makes clear whether the max or min occurs at an endpoint.) 2C-7 Let r be the radius, which is √ fixed. Then the height a of the rectangle is in the interval 0 ≤ a ≤ r. Since b = 2 r2 − a2 , the area A is given in terms of a by � A = 2a r2 − a2 . The value of A at the endpoints a = 0 and a = r is zero, so the maximum occurs at a critical point in between. 12 2. Applications of Differentiation E. Solutions to 18.01 Exercises r a b � dA 2a2 2(r2 − a2 ) − 2a2 √ = 2 r2 − a2 − √ = da r2 − a2 r2 − a2 √ r 2 2 Thus dA/da = 0 implies 2r = a , from which we get a = √ , b = r 2. 2 (We use the positive square root since a ≥ 0. Note that b = 2a and A = r2 .) 2C-8 a) Letting a and b be the two legs and x and y the sides of the rectangle, we have y = −(b/a)(x − a) and the area A = xy = (b/a)x(a − x). The area is zero at the two ends x = 0 and x = a, so the maximum occurs in between at a critical point: A� = (b/a)((a − x) − x); = 0 if x = a/2. Thus y = (b/a)(a − x) = b/2 and A = ab/4. b) This time let x be the point shown on the accompanying figure; using similar triangles, the sides of the rectangle are b y x a �1 = x� 2 b a + b2 and �2 = √ (a − x) 2 a a + b2 Therefore the area is l b x 1 l2 a-x A = �1 �2 = (b/a)x(a − x) This is the same formula for area as in part (a), so the largest area is the same, occurring when x = a/2, and the two maximal rectangles both have the same area; they have different dimensions though, since in the present case, one side length is half the hypotenuse: � � �1 = a2 + b2 /2 and �2 = ab/2 a2 + b2 . 13 E. Solutions to 18.01 Exercises 2. Applications of Differentiation 2C-9 The distance is � L= x2 + 1 + � (a − x)2 + b2 The endpoint values are x → ±∞, for L → ∞, so the minimum value is at a critical point. b 1 θ x � x θ2 1 a-x a−x x L =√ = −� x2 + 1 (a − x)2 + b2 � √ (a − x)2 + b2 − (a − x) x2 + 1 � √ x2 + 1 (a − x)2 + b2 Thus L� = 0 implies (after squaring both sides), x2 ((a − x)2 + b2 ) = (a − x)2 (x2 + 1), or x2 b2 = (a − x)2 or bx = (a − x); we used the positive square roots since both sides must be positive. Rewriting the above, 1 b = , or tan θ1 = tan θ2 . a−x x Thus θ1 = θ2 : the angle of incidence equals the angle of reflection. 2C-10 The total time is � √ 1002 + (a − x)2 1002 + x2 T = + 5 2 As x → ±∞, T → ∞, so the minimum value will be at a critical point. B 100 P x α β a x Q 100 A x (a − x) sin α sin β √ − − � = . 2 2 2 2 5 2 5 100 + x 2 100 + (a − x) Therefore, if T � = 0 , it follows that sin α sin β sin α 5 = or = . 5 2 sin β 2 T� = 2C-11 Use implicit differentiation: 14 2. Applications of Differentiation E. Solutions to 18.01 Exercises x2 + y 2 = d2 =⇒ 2x + 2yy � = 0 =⇒ y � = −x/y. We want to maximize xy 3 . At the endpoints x = 0 and y = 0, the strength is zero, so there is a maximum at a critical point. Differentiating, d y x 0 = (xy 3 )� = y 3 + 3xy 2 y � = y 3 + 3xy 2 (−x/y) = y 3 − 3x2 y Dividing by x3 , (y/x)3 − 3(y/x) = 0 =⇒ (y/x)2 = 3 =⇒ y/x = √ 3. 2C-12 The intensity is proportional to √ sin θ x/ 1 + x2 y= = = x(1 + x2 )−3/2 1 + x2 1 + x2 Endpoints: y(0) = 0 and y → 0 as x → ∞, so the maximum will be at a critical point. Critical points satisfy S θ P 1 √ y � = (1 − 2x2 )(1 + x2 )−5/2 = 0 =⇒ 1 − 2x2 = 0 =⇒ x = 1/ 2 √ The best height is 1/ 2 feet above the desk. (It’s not worth it. Use a desk lamp.) 2C-13 a) Let p denote the price in dollars. Then there will be 100 + (2/5)(200 − p) passengers. Therefore the total revenue is R = p(100 + (2/5)(200 − p) = p(180 − (2/5)p) At the “ends” zero price p = 0, and no passengers p = (5/2)180 = 450, the revenue is zero. So the maximum occurs in between at a critical point. R� = (180 − (2/5)p) − (2/5)p = 180 − (4/5)p = 0 =⇒ p = (5/4)180 = $225 b) P = xp − x(10 − x/105 ) with x = 105 (10 − p/2) Therefore, the profit is cents is P = 105 (10 − p/2)(p − 10 + (10 − p/2)) = 105 (10 − p/2)(p/2) = (105 /4)p(20 − p) dP = (105 /2)(10 − p) dp 15 E. Solutions to 18.01 Exercises 2. Applications of Differentiation The critical point at p = 10. This is x = 105 (10 − 5) = 5 × 105 kilowatt hours, which is within the range available to the utility company. The function P has second derivative −105 /2, so it is concave down and the critical point must be the maximum. (This is one of those cases where checking the second derivative is easier than checking the endpoints.) Alternatively, the endpoint values are: x = 0 =⇒ 105 (10 − p/2) = 0 =⇒ p = 20 =⇒ P = 0. (9) x = 8 × 105 =⇒ 8 × 105 = 105 (10 − p/2) (10) =⇒ 10 − p/2 = 8 =⇒ p = 4 (11) =⇒ P = (105 /4)4(20 − 4) = 16 × 105 cents = $160, 000 The profit at the crit. pt. was (105 /4)10(20 − 10) = 2.5 × 106 cents = $250, 000 2C-14 a) Endpoints: y = −x2 ln(x) → 0 as x → 0+ and y → −∞ as x → ∞. √ Critical points: y � = −2x ln x − x = 0 =⇒ ln x = −1/2 =⇒ x = 1/ e. √ Critical value: y(1/ e) = 1/2e. √ Maximum value: 1/2e, attained when x = 1/ e. (min is not attained) b) Endpoints: y = −x ln(2x) → 0 as x → 0+ and y → −∞ as x → ∞. Critical points: y � = − ln(2x) − 1 = 0 =⇒ x = 1/2e. Critical value: y(1/2e) = −(1/2e) ln(1/e) = 1/2e. Maximum value: 1/2e, attained at x = 1/2e. (min is not attained) 2C-15 No minimum. The derivative is −xe−x < 0, so the function decreases. (Not needed here, but it will follows from 2G-8 or from L’Hospital’s rule in E31 that xe−x → 0 as x → ∞.) 16 2. Applications of Differentiation E. Solutions to 18.01 Exercises (1//e, 1/2e) 1 14a (1/2e, 1/2e) 1/2 14b 2D. More Max-min Problems 2D-3 The milk will be added at some time t1 , such that 0 ≤ t1 ≤ 10. In the interval 0 ≤ t < t1 the temperature is y(t) = (100 − 20)e−(t−0)/10 + 20 = 80e−t/10 + 20 Therefore, −t1 /10 T1 = y(t− + 20 1 ) = 80e We are adding milk at a temperature T2 = 5, so the temperature as we start the second interval of cooling is 9 1 1 T1 + T2 = 72e−t1 /10 + 18 + 10 10 2 Let Y (t) be the coffee temperature in the interval t1 ≤ t ≤ 10. We have just calculated Y (t1 ), so 5Y (t) = (Y (t1 ) − 20)e−(t−t1 )/10 + 20 = (72e−t1 /10 − 1.5)e−(t−t1 )/10 + 20 The final temperature is � � T = Y (10) = (72e−t1 /10 − 1.5)e−(10−t1 )/10 + 20 = e−1 72 − (1.5)et1 /10 + 20 We want to maximize this temperature, so we look for critical points: dT = −(1.5/10e)et1 /10 < 0 dt1 Therefore the function T (t1 ) is decreasing and its maximum occurs at the left endpoint: t1 = 0. Conclusion: The coffee will be hottest if you put the milk in as soon as possible. 2E. Related Rates 17 E. Solutions to 18.01 Exercises 2. Applications of Differentiation 2E-1 The distance from robot to the point on the ground directly below the street lamp is x = 20t. Therefore, x� = 20. x+y y = (similar triangles) 30 5 Therefore, 30 5 x y (x� + y � )/30 = y � /5 =⇒ y � = 4 and (x + y)� = 24 The tip of the shadow is moving at 24 feet per second and the length of the shadow is increasing at 4 feet per second. 18 2. Applications of Differentiation E. Solutions to 18.01 Exercises 2E-2 x and dθ/dt = 3(2π) = 6π 4 with t is measured in minutes and θ measured in radians. The light makes an angle of 60◦ with the shore when θ is 30◦ or θ = π/6. Differentiate with respect to t to get (sec2 θ)(dθ/dt) = (1/4)(dx/dt) Since sec2 (π/6) = 4/3, we get dx/dt = 32π miles per minute. tan θ = θ x y shoreline 2E-3 The distance is x = 10, y = 15, x� = 30 and y � = 30. Therefore, � �� (12) (x2 + y 2 )1/2 = (1/2)(2xx� + 2yy � )(x2 + y 2 )−1/2 � (13) = (10(30) + 15(30))/ 102 + 152 √ (14) = 150/ 13miles per hour 2E-4 V = (π/3)r2 h and 2r = d = (3/2)h implies h = (4/3)r. /x 2+ y 2 y x Therefore, V = (π/3)r2 h = (4π/9)r3 Moreover, dV /dt = 12, hence � � dV dV dr d dr 3 (4π/9)r ) = = = (4π/3)r2 (12) = 16πr2 . dt dr dt dr dt dV = 36πm3 /minute. 2E-5 The information is dt When h = 2, r = 3/2, so that h = (4/3)r r r 19 E. Solutions to 18.01 Exercises 2. Applications of Differentiation z� = 4 x2 + 102 = z 2 , We want to evaluate x� at x = 20. (Derivatives are with respect to time.) Thus 2xx� = 2zz � and z 2 = 202 + 102 = 500 Therefore, z 10 x weight √ √ x� = (zz � )/x = 4 500/20 = 2 5 2E-6 x� = 50 and y � = 400 and z 2 = x2 + y 2 + 22 The problem is to evaluate z � when x = 50 and y = 400. Thus 2 z y x 2 x+y boat 2 2zz � = 2xx� + 2yy � =⇒ z � = (xx� + yy � )/z √ √ √ and z = 502 + 4002 + 4 = 162504. So z � = 162500/ 162504 ≈ 403mph. (The fact that the plane is 2 miles up rather than at sea level changes the answer by only about 4/1000. Even the boat speed only affects the answer by about 3 miles per hour.) 2E-7 V = 4(h2 + h/2), V � = 1. To evaluate h� at h = 1/2, 1 = V � = 8hh� + 2h� = 8(1/2)h� + 2h� = 6h� Therefore, h 1/2 h h h Cross-sectional area = h 2+ h/2 20 2. Applications of Differentiation E. Solutions to 18.01 Exercises h� = 1/6 meters per second 2E-8 x� = 60, y � = 50 and x = 60 +� 60t, y = 50t. Noon is t = 0 and t is measured in hours. To find the time when z = x2 + y 2 is smallest, we may as well minimize z 2 = x2 + y 2 . We know that there will be a minimum at a critical point because when t → ±∞ the distance tends to infinity. Taking the derivative with respect to t, the critical points satisfy z x y 2xx� + 2yy � = 0 This equation says 2((60 + 60t)60 + (50t)50) = 0 =⇒ (602 + 502 )t = −602 Hence t = −36/61 ≈ −35min The ships were closest at around 11 : 25 am. 2E-9 dy/dt = 2(x − 1)dx/dt. Notice that in the range x < 1, x − 1 is negative and √ so (x − 1) = − y. Therefore, √ √ √ dx/dt = (1/2(x − 1))(dy/dt) = −(1/2 y)(dy/dt) = +( y)(1 − y)/2 = 1/4 2 Method 2: Doing this directly turns out to be faster: √ x = 1 − y =⇒ dx/dt = 1 − (1/2)y −1/2 dy/dt and the rest is as before. 2E-10 r = Ct1/2 . The implicit assumption is that the volume of oil is constant: πr2 T = V or r2 T = (V /π) = const Therefore, differentiating with respect to time t, (r2 T )� = 2rr� T + r2 T � = 0 =⇒ T � = −2r� /r But r� = (1/2)Ct−1/2 , so that r� /r = 1/2t. Therefore T � = −1/t (Although we only know the rate of change of r up to a constant of proportionality, we can compute the absolute rate of change of T .) 21 E. Solutions to 18.01 Exercises 2. Applications of Differentiation 2F. Locating zeros; Newton’s method 2F-1 a) y � = − sin x − 1 ≤ 0. Also, y � < 0 except at a discrete list of points (where sin x = −1). Therefore y is strictly decreasing, that is, x1 < x2 =⇒ y(x1 ) < y(x2 ). Thus y crosses zero only once. Upper and lower bounds for z such that y(z) = 0: y(0) = 1 and y(π/2) = −π/2. Therefore, 0 < z < π/2. b) xn+1 = xn − (cos xn − xn )/(sin xn + 1) (15) x1 = 1, x2 = .750363868, (16) x3 = .739112891, x4 = .739085133 Accurate to three decimals at x3 , the second step. Answer .739. 5π/2 Slopes at the points are 0 or -2. 3π/2 π/2 −5π/2 −3π/2 3π/2 5π/2 −π/2 −3π/2 −5π/2 c) Fixed point method takes 53 steps to stabilize at .739085133. Newton’s method takes only three steps to get to 9 digits of accuracy. (See x4 .) 2F-2 y = 2x − 4 + y� = 2 − 1 (x − 1)2 −∞ 0. For 1 < x < 2, the 22 2. Applications of Differentiation E. Solutions to 18.01 Exercises numerator is still negative, but the denominator is positive, so y � < 0. For 2 < x, both numerator and denominator are positive, so y � > 0. 6 y �� = (x − 1)4 Therefore, y �� > 0 for x �= 1. Critical value: y(2) = 1 Singular values: y(1− ) = y(1+ ) = ∞ Endpoint values: y → ∞ as x → ∞ and y → −∞ as x → −∞. (2,1) 1 -3 Conclusion: The function increases from −∞ to ∞ on the interval (−∞, 1). Therefore, the function vanishes exactly once in this interval. The function de­ creases from ∞ to 1 on the interval (1, 2) and increases from 1 to ∞ on the interval (2, ∞). Therefore, the function does not vanish at all in the interval (1, ∞). Finally, the function is concave up on the intervals (−∞, 1) and (1, ∞) 23 E. Solutions to 18.01 Exercises 2. Applications of Differentiation 2F-3 2x3 − 1 x2 Therefore y � = 0 implies x3 = 1/2 or x = 2−1/3 . Moreover, y � > 0 when x > 2−1/3 , � 0. The sign does not change across the singular and y � < 0 when x < 2−1/3 and x = point x = 0 because the power in the denominator is even. (continued →) y � = 2x − x−2 = 2(x3 + 1) x3 �� 3 Therefore y = 0 implies x = −1, or x = −1. Keeping track of the sign change in the denominator as well as the numerator we have that y �� > 0 when x > 0 and y �� < 0 when −1 < x < 0. Finally, y �� > 0 when x < −1, and both numerator and denominator are negative. y �� = 2 + 2x−3 = Critical value: y(2−1/3 ) = 2−2/3 + 21/3 ≈ 1.9 Singular value: y(0+ ) = +∞ and y(0− ) = −∞ Endpoint values: y → ∞ as x → ±∞ P P is the critical point Conclusions: The function decreases from ∞ to −∞ in the interval (−∞, 0). Therefore it vanishes exactly once in this interval. It jumps to ∞ at 0 and decreases from ∞ to 2−2/3 + 21/3 in the interval (0, 2−1/3 ). Finally it increases from 2−2/3 + 21/3 to ∞ in the interval (2−1/3 , ∞). Thus it does not vanish on the interval (0, ∞). The function is concave up in the intervals (−∞, −1) and (0, ∞) and concave down in the interval (−1, 0), with an inflection point at −1. 2F-4 From the graph, x5 − x − c = 0 has three roots for any small value of c. The value of c gets too large if it exceeds the local maximum of x5 − x labelled. To calculate that local maximum, consider y � = 5x4 −1 = 0, with solutions x = ±5−1/4 . The local maximum is at x = −5−1/4 and the value is (−5−1/4 )5 − (−5−1/4 ) = 5−1/4 − 5−5/4 ≈ .535 Since .535 > 1/2, there are three roots. 24 2. Applications of Differentiation E. Solutions to 18.01 Exercises P y=c √ 2F-5 a) Answer: x1 = ±1/ 3. f (x) = x − x3 , so f � (x) = 1 − 3x2 and xn+1 = xn − f (xn )/f � (xn ) √ So x2 is undefined if f � (x1 ) = 0, that is x1 = ±1/ 3. √ b) Answer: x1 = ±1/ 5. (This value can be found by experimentation. It can be also be found by iterating the inverse of the Newton method function.) Here is an explanation: Using the fact that f is odd and that x3 = x1 suggests that x2 = −x1 . This greatly simplifies the equation. xn+1 = xn − (xn − x3n )/(1 − 3xn2 ) = −2x3n 1 − 3x2n Therefore we want to find x satisfying −2x3 1 − 3x2 2 3 2 This equation is the same √ as x(1 − 3x ) = 2x , which implies √ x = 0 or 5x = 1.√In other words, √ x = ±1/ 5. Now one can check that if x1 = 1/ 5, then x2 = −1/ 5, x3 = +1/ 5, etc. −x = √ √ √c) Answers: √If x1 < −1/ 3, then xn → −1. If x1 > 1/ 3, then xn → 1. If −1/ 5 < x1 < 1/ 5, then xn → 0. This can be found experimentally, numerically. For a complete analysis and proof one needs the methods of an upper level course like 18.100. 2F-6 a) To simplify this problem to its essence, let V = π. (We are looking for ratio r/h and this will be the same no matter what value we pick for V .) Thus r2 h = 1 and A = πr2 + 2π/r Minimize B = A/π instead. B = r2 + 2r−1 =⇒ B � = 2r − 2r−2 and B � = 0 implies r = 1. Endpoints: B → ∞ as r → 0 and as r → ∞, so we have found the minimum at r = 1. (The constraint r2 h = 1 shows that this minimum is achieved when r = h = 1. As a doublecheck, the fact that the minimum area is achieved for r/h = 1 follows from 2C/5; see part (b).) The minimum of B is 3 attained at r = 1. Ten percent more than the 25 E. Solutions to 18.01 Exercises 2. Applications of Differentiation minimum is 3.3, so we need to find all r such that B(r) ≤ 3.3 Use Newton’s method with F (r) = B(r) − 3.3. (It is unwise to start Newton’s method at r = 1. Why?) The roots of F are approximately r = 1.35 and r = .72. B B=3.3 (1,3) allowable r-interval r Since r2 h = 1, h = 1/r2 and the ratio, r/h = r3 Compute (1.35)3 ≈ 2.5 and (.72)3 ≈ .37. Therefore, the proportions with at most 10 percent extra glass are approximately .37 < r/h < 2.5 b) The connection with Problem 2C-5 is that the minimum area r = h is not entirely obvious, and not just because we are dealing with glass beakers instead of tin cans. In 2C/5 the area is fixed whereas here the volume is held fixed. But because one needs a larger surface area to hold a larger volume, maximizing volume with fixed area is the same problem as minimizing surface area with fixed volume. This is an important so-called duality principle often used in optimization problems. In Problem 2C-5 the answer was r = h, which is the proportion with minimum surface area as confirmed in part (a). 26 2. Applications of Differentiation E. Solutions to 18.01 Exercises 2F-7 Minimize the distance squared, x2 + y 2 . The critical points satisfy 2x + 2yy � = 0 The constraint y = cos x implies y � = − sin x. Therefore, 0 = x + yy � = x − cos x sin x There is one obvious solution x = 0. The reason why this problem is in this section is that one needs the tools of inequalities to make sure that there are no other solutions. Indeed, if f (x) = x − cos x sin x, then f � (x) = 1 − cos2 x + sin2 x = 2 sin2 x ≥ 0 Furthermore, f � (x)0 is strictly positive except at the points x = kπ, so f is increas­ ing and crosses zero exactly once. There is only one critical point and the distance tends to infinity at the endpoints x → ±∞, so this point is the minimum. The point on the graph closest to the origin is (0, 1). y = cos x Alternative method: To show that (0, 1) is closest it suffices to show that for −1 ≤ x < 0 and 0 < x ≤ 1, � 1 − x2 < cos x 2 2 Squaring gives 1 − x < cos x. This can be proved using the principles of problems 6 and 7. The derivative of cos2 x − (1 − x2 ) is twice the function f above, so the methods are very similar. 27 E. Solutions to 18.01 Exercises 2. Applications of Differentiation 2G. Mean-value Theorem 2G-1 a) slope chord = 1; f � (x) = 2x ⇒ f � (c) = 1 if c = b) slope chord = ln 2; f � (x) = c) for x3 − x: slope chord = 1 x 1 . 2 ⇒ f � (c) = ln 2 if c = 1 . ln 2 f (2) − f (−2) 6 − (−6) = = 3; 2 − (−2) 4 f � (x) = 3x2 − 1 ⇒ f � (c) = 3c2 − 1 = 3 ⇒ c = ± √23 From the graph, it is clear you should get two values for c. (The axes are not drawn to the same scale.) (2,6) -c c (-2,-6) 2G-2 a) f (x) = f (a) + f � (c)(x − a) ; Take a = 0; f (ξ) = sin x, f � (x) = cos x ⇒ f (x) = 0 + cos c · x ⇒ sin x < x (since cos c < 1 for 0 < c < 2π) Thus the inequality is valid for 0 < x ≤ 2π; since the function is periodic, it is also valid for all x > 0. b) √ d √ 1 1 1 1+x= √ ⇒ 1+x=1+ √ x < 1 + x, since dx 2 2 1+x 2 1+c c > 0. 2G-3. Let s(t) = distance; then average velocity = slope of chord = 121 = 66. 11/6 Therefore, by MVT, there is some time t = c such that s� (c) = 66 > 65. (An application of the mean-value theorem to traffic enforcement...) 28 2. Applications of Differentiation E. Solutions to 18.01 Exercises 2G-4 According to Rolle’s Theorem (Thm.1 p.800 : an important special case of the M.V.T, and a step in its proof), between two roots of p(x) lies at least one root of p� (x). Therefore, between the n roots a1 , . . . , an of p(x), lie at least n − 1 roots of p� (x). There are no more than n − 1 roots, since degree of p� (x) = n − 1; thus p� (x) has exactly n − 1 roots. 2G-5 Assume f (x) = 0 at a, b, c. By Rolle’s theorem (as in MVT-4), there are two points q1 , q2 where f � (q1 ) = 0, f � (q2 ) = 0. By Rolle’s theorem again, applied to q1 and q2 and f � (x), there is a point p where f (p) = 0. Since p is between q1 and q2 , it is also between a and c. �� 2G-6 a) Given two points xi such that a ≤ x1 < x2 ≤ b, we have f (x2 ) = f (x1 ) + f � (c)(x2 − x1 ), where x1 < c < x2 . Since f � (x) > 0 on [a, b], f � (c) > 0; also x2 − x1 > 0. Therefore f (x2 ) > f (x1 ), which shows f (x) is increasing . b) We have f (x) = f (a) + f � (c)(x − a) where a < c < x. Since f � (c) = 0, f (x) = f (a) for a ≤ x ≤ b, which shows f (x) is constant on [a, b]. 29