7. (3), (5) d(x) = x 2 x 3 (4) maximize d (6) 0 x 1 (7) d (x) = 2x 3x 2, d (x) = 0 when x = 0 and x = 2 3. d(0) = 0, d( 2 3 ) = 4 9.

Modeling Tools

Optimization

Answers to exercises from 4.22, page 353

1. (1), (2)

Barn

H W

(3) A = HW , 50 = H + 2W (4) maximize A (5) A = (50−2W )W = 50W −2W² (6) W can’t be less than 0 nor greater than 50 yards. (7) A⁰= 50 − 4W . A⁰is 0 when W =⁵⁰₄ = 12.5. When W is, respectively, 0, 12.5, and 50 yards, A is 0, 312.5, and 0 sq.yds. Clearly, the best dimensions are W = 12.5 yards, H = 50 − 2W = 25 yards.

3. (1), (2)

H W

(3) A = 3HW , 100 = 4H + 6W (4) maximize A (5) A = 3H^100−4H₆ = 50W − 2W² = 50H − 2H² sq.yds.

(6)H can’t be less than 0 nor greater than ¹⁰⁰₄ = 25 yards. (7) A⁰= 50 − 4H. A⁰is 0 when H = ⁵⁰₄ = 12.5.

When H is, respectively, 0, 12.5, and 25 yards, A is 0, 312.5, and 0 sq.yds. Clearly, the best dimensions are H = 12.5, W =⁵⁰₆ = 8¹₃ yards.

5. (3), (5) h(t) = −5t²+ 50t + 30 meters (4) maximize h (6) t ≥ 0, no upper bound. (7) h⁰(t) = −10t + 50. h⁰ is 0 when t = 5. h is 30 when t = 0 seconds, 155 meters when t = 5 seconds. From the shape of the curve (a parabola opening down) the ball will reach a height of 155 meters.

7. (3), (5) d(x) = x²− x³ (4) maximize d (6) 0 ≤ x ≤ 1 (7) d⁰(x) = 2x − 3x², d⁰(x) = 0 when x = 0 and x = ²₃. d(0) = 0, d(²₃) = ₂₇⁴, d(1) = 0. The answer is²₃. 9. (1), (2)

d t (3, −4) (cos(t), sin(t))

(3), (5) d(t) = p(3 − cos(t))²+ (−4 − sin(t))². (4) minimize d (6) 0 ≤ t ≤ 2π (one possible answer) (7) d⁰(t) = .5((3 − cos(t))² + (−4 − sin(t))²)⁻.5((3 − cos(t))²+ (−4 − sin(t))²)⁰ = .5((3 − cos(t))² + (−4 − sin(t))²)⁻.5(−6 sin(t) + 8 cos(t)). This is 0 only when

−6 sin(t) + 8 cos(t) is 0, which is when _cos(t)^sin(t) = tan(t) =

−⁴₃, which is when t is approximately −.9272952 radi- ans. At this point, cos(t) = _sec(t)¹ = √ ¹

1+tan²(t) =

1 q

1+(−⁴₃)² = ³₅, while in a similar way, sin(t) = −⁴₅, so d(t) = 4 (after some calculation). At t = 0 and t = 2π, d =√

20. Clearly, the distance is a minimum at ³₅, −⁴₅
.

11. (1), (2)

h w

559

(3) A = hw, h = _w¹ (4) maximize A (5) A = _w¹w = 1 (6) and (7) don’t matter! The area is 1 no matter what shape we choose!

13. (1), (2)

h

1 2a

a

(3) a²h = 1, A = (4a)(h + a) (4) minimize area, A (5) A = (4a)(_a¹2 + a) (6) a > 0 (7) A⁰ = _a⁴+ 4a²
0

=

−_a⁴2 + 8a. This is 0 when a = √3¹

2. While a < √3¹ 2, A⁰ < 0, A is descending. Similarly, when a > √3¹

2, A⁰> 0, so A is ascending. The minimum possible area at a = √3¹

2 is approximately 7.56 sq.meters.

15. (1), (2)

h

.25

.25 .5 b .5 b

(3) .5hb = 1, A = (2b + 1)(h + .5) (4) minimize area, A (5) A = (2b + 1)(²_b + .5) (6) b > 0 (7) A⁰= 4.5 +²_b+ b
0

= −_b²2+ 1. This is 0 when b =√ 2.

As in problem 13, this must be the minimum possible area, because when b < √

2, A⁰ < 0, and A descends until b =√

2 and when b >√

2, A⁰> 0, so A ascends after b =√

2.

17. Max: 4, min: 0 19. Max: 0, no min

21. Max: 1, min: 0 23. No max, min: −3 25. No max, no min

27. No max, min: 1

29. (1), (2)

y x b

Problem 16 deals with the same question but with right triangles. The answer to problem 16 is that the great- est rectangle has height half that of the triangle. Now suppose we split this triangle in two, creating two right triangles

and ask the same question about the two right triangles.

The answer is that in both pieces, the largest rectangle is half the height of the triangle. Putting them together, the largest rectangle in the whole triangle has height (and area) half the triangle.

31. The slope of the line tangent to f (x) = 1/x is f⁰(a) =

−_a¹2. The equation of the line is y = −_a¹2(x − a) +¹_a. From this, we compute the x- and y-intercepts as, re- spectively, 2a and ²_a.

a 2a

2 a

Our problem reduces to the problem of finding the largest rectangle that can be constructed inside a right triangle with one side on the hypotenuse. This is a spe- cial case of problem 29. The answer to that problem is that the largest rectangle has half the base of the triangle and half the area. The area of this triangle is 2a²_a= 2, so the largest rectangle here has area 1 (compare this with the answer to problem 11)—no matter what a we choose!

33. To get the angle that yields the farthest distance, we first split the initial velocity, v, into two components, the ver- tical, vy, and the horizontal, vx. Initially, vy= 100 sin α and vx = 100 cos α. We also have to consider the ef- fect that gravity has on the vertical height of the can- non ball, given here as a constant, −g. We want to know the amount of time, t, the cannon ball takes to return to its original height, h = 0 meters. By inte- grating the acceleration due to gravity with respect to time, a(t) = −g, we get an expression for the vertical velocity at any time, namely v1(t) = −gt + 100 sin α.

We still don’t know the amount of time the cannon ball takes to reach its original height, h, so, we must inte- grate this once more to get h = −¹₂gt²+ (100 sin α)t.

The height of the cannon ball is 0 when t = 0 and t = −^{200 sin α}g . Finding the horizontal distance, d, that the cannon ball travels is much easier. We know that gravity does not affect this component, and that initially, vx= 100 cos α. Then, d = (100 cos α)t. If we substitute t = −^{200 sin α}g and maximize d, then we get an angle for which the cannon ball travels the farthest distance.

d⁰ = ²⁰⁰⁰⁰_g (cos²α)(sin²α) is 0 when cos²α = sin²α, α = ^π₄. At this value, the cannon ball travels ¹⁰⁰⁰⁰_g meters.

35. (3) A = (2b + 2a)(h + a), 1 = hab (4) minimize A (5) A = (2b + 2a)(_ab¹ + a) (6) There are no end- points (7) (Take the derivative of A with respect to b) A⁰= −2a + 2a²b². This is 0 when b =√¹

a meters.

37. Imagine a stretchable and shrinkable line moving around the corner, always keeping contact with the two walls and the corner, as in the picture. We are interested in how short that shrinkable line becomes. That shortest length is the largest piece of plywood that can make the turn.

(1), (2)

6 9 y

x

d

d

(3) ^x₉ = _y⁶ = ^d_d²

1, d1 =p81 + y², d2 =√

36 + x² (4) minimize d = d1+ d2 (5) d =p81 + y²+_y⁶p81 + y² (6) y > 0 (7) d⁰= (p81 + y²)(−_y⁶2)+(1+⁶_y)(√ ^y

81+y²).

This is 0 when y³= 486, so y = 3√³

18. One can show al- gebraically that for y < 3√³

18, y⁰< 0 and for y > 3√³ 18, y⁰> 0, so that we have indeed found the minimum for d. The largest piece of plywood that can make the turn is approximately 21.07 feet long.

39. From problem 31, Melba knows that if she chooses m, Hosni’s best move is to choose 2m − 1. In that case, Melba can assume that is Hosni’s move and maximize P . That will be the best payoff she can guarantee. Then P = (2m − 1)²− 28m − 4m(2m − 1) + 2(2m − 1) − 35 =

−4m²− 24m − 36. P⁰ = −8m − 24, which is 0 when m = −3. From the expression for P , this is a maximum.

The value of P at this point is 0. If Melba plays the number −3, she can be sure of not losing money. On the other hand, if she chooses any other number, Hosni can win.

41. We know two ways to attack this. One is the standard method of wrapping. (1), (2)

a 2

a

c 2

b a

a

b 2

b 2

(3) For the box with dimensions a, b, and c, shown above, the area of the wrapping paper, A = (2a + 2b)(c + a).

Also, 12 = 2a+2b = c+a, so b = 6−a and c = 12−a. (4) maximize volume, V (5) V = abc = a(6 − a)(12 − a) = 72a − 18a²+ a³ (here, you could also maximize the vol- ume in terms of b or c) (6) a > 0 (7) V⁰= 72−36a+3a². This is 0 when a = 6 ±√

12. If a = 6 −√ 12, then b =√

12, and c = 6 +√

12. V = 24√

12, a little over 83 cubic inches.

The other method is to place a box with a square base in the middle of the wrapping paper diagonally. It turns out that you can wrap a slightly larger box this way. Try it. When you’re ready, the answer is on page 581.

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% The real and legitimate goal of the sciences is

the endowment of human life with new inven- tions and riches.

– Francis Bacon

Newton’s Method

Answers to exercises from 4.23, page 365

1. Starting with 1 as a guess, we get successively 2, 1.8, and 1.79129. Two steps are shown below (graph not to scale):

1

1.79129

Another root (start with a guess of −3) is approximately

−2.79129.

3. Starting with 2 as a guess we get successively 1.67668, 1.61165, 1.60967, and 1.60944. Two steps are shown be- low:

1.60944

1 2

5. Starting with a guess of .5 we get successively .80903,.82426, and.82413. Two steps are shown below:

1 .82413

Another root is approximately −.82413

7. Starting with a guess of 3 we get successively 2.68750, 2.63270, and 2.63110. Two steps are shown below:

2 3

2.63110

9. Starting with 1 as a guess we get successively .8, .65196, .56178, .52835, .524208, and .52415. Two steps are shown below:

-1

.52415 1

The other roots of this function are approximately

−1.1015, and 0.

11. Starting with 0 as a guess we get successively 1, .40487, .43595, .4352. Two steps are shown below:

.4352

1

The other roots of this function are approximately

−1.4738, −.80295, 1.8519, 2.40727, 3.20135, 3.41073.

13. Starting with 0 as a guess we get successively −.33333 and −.32899. The two steps are shown below:

-.32899

-1

The other roots of this function are approximately

−1.71415 and 1.51946.

15. Starting with 1 as a guess we get successively 2.62945, 1.88071, 2.00106, and 2. Two steps are shown below:

1

2

17. Thep(5) is a root of the function f(x) = x²− 5. Using Newton Rapheson method with a guess of 2.5 we get successively 2.25, 2.23611, and 2.23607. Two steps are shown below:

2 3

2.23607

19. Thep(17) is a root of the function f(x) = x²−17 Start- ing with a guess of 4.2 we get successively 4.12381 and 4.12312. Two steps are shown below:

5 4.12312

21. cos(^π₄) = ^√₂² is a root of the function f (x) = 4x²− 2.

Starting with 1 as a guess we get successively .75, .70833, and .70711. Two steps are shown below:

.70711 1

23. sec(^π₃) =√²

3 is a root of the function f (x) = .75x²− 1.

Starting with a guess of 1 we get successively ⁷₆ and ⁹⁷₈₄. Two steps are shown below:

1

97 84

25. Starting with 1.2 as a guess we get successively 1.49444, 1.44405 and 1.44225. Two steps are shown below:

2 1

1.44225

27. Starting with 2 as a guess we get successively 2.75, 2.58264, 2.57133, and 2.57128. Two steps are shown below:

3 2

2.57128

29. √⁹

200 is a root of the function f (x) = x⁹− 200. Starting with a guess of of 2 we get successively 1.86458, 1.80951, 1.80178, and 1.80165. Two steps are shown below:

1

1.80165

31. Starting with 1 as a guess we get successively .73576, .69404, and .69315. Two steps are shown below:

.69315

33. Starting with 0 as a guess we get successively 1.5, 1.05783, .92585, .91634, and .91629. Two steps are shown below:

.91629

35. −6, −5, and 7 37. −27, −4, and 4 39. −9, −6, and 8









The “Newton’s method” used today is an improvement by Joseph Raphson of Newton’s original technique. It is some- times called the Newton-Raphson method.

41. −2, 3, and 6

43. −2, 2, and 9 45. −3, −2, 2, and 3 47. g −^g2_2g⁻ⁿ =^g2_2g⁺ⁿ

49. g −_kg^gkk−1⁻ⁿ =^g(kgk−1_kgk−1^)−gk−n

51. A guess of 0 should help you find the root of the problem:

.322349. A guess of 2 gives you an undefined quotient, since the denominator will equal 0. At this point x = 2, the derivative of the function is equal to 0, meaning this point is a relative minimum or maximum, so the tangent lines at that point will never cross the x-axis.

53. Using the Newton-Raphson method on any nonzero guess g will give you an answer of −2g, thus both bounc- ing you back and forth across the y-axis and doubling the value of g with each iteration:

-2g g 4g

This is because for this function g −

|g|

4 3 g 1 3g2

3

= g −

|g|⁴³3g²³

g = g −^3gg² = g − 3g = −2g.

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% A science is said to be useful if its develop-

ment tends to accentuate the existing inequal- ities in the distribution of wealth, or more di- rectly promotes the destruction of human life.

– G.H. Hardy

Implicit Differentiation

Answers to exercises from 4.24, page 376

1. Remember that we are treating y as a composite func- tion. This function is a circle of radius 5 because x² + y² = r². Differentiating both sides gives us (x²+ y²)⁰= (25)⁰. (x²)⁰= 2x and (y²)⁰= 2yy⁰. Thus, y⁰= ^−x_y . So, at the point (−3, 4), y⁰=³₄.

3. This is also a circle. Here (2x²+ 3y²)⁰= (11)⁰. (2x²)⁰= 4x and (3y²)⁰= 6yy⁰, so y⁰= ^−4x_6y . Thus, at the point (2, −1), y⁰=⁴₃.

5. x²+ 3x + 4y²+ 3y = 14 at (1, −2). (x²+ 3x + 4y²+ 3y)⁰ = (14)⁰. We take the derivative of the function:

(x²) + 3x + (4y²)⁰+ 3y⁰= 14⁰. 2x + 3 + 8yy⁰+ 3y⁰= 0 and y⁰= ^−2x−3_8y+3 . Thus y⁰= ₁₃⁵ at the point (1, −2).

7. x³y − y³x = 30 at (2, −3).

(x³y − y³x)⁰= (30)⁰ (x³y)⁰− (y³x)⁰= 0

x³y⁰+ 3yx²−y³−3xy²y⁰= 0 and y⁰=^−3yx_x3−3xy^2+y23. Thus, y⁰= ⁻⁹₄₆ at the point (2, −3).

9. y³+ yx² = 3y − x²at (1, 1).

3y²y⁰+ 2xy + y⁰x²= 3y⁰− 2x y⁰(3y²+ x²− 3) = −2x − 2xy y⁰= _3y^−2x−2xy2+x²−3

At the point (1, 1), y⁰=₃₊₁₋₃⁻²⁻² = −4.

11. x²y²+ x⁴= y²at (⁻¹√ 2,√¹

2).

(x²y²)⁰+ (x⁴)⁰= (y²)⁰. x²yy⁰− yy⁰= (−xy²− 2x³)⁰ y⁰= ^−xy_x2²y^−2x−y³

At the point (⁻¹√ 2,√¹

2), y⁰= −3.

13. x⁴+ 2x²y²+ y⁴= 4x²− 2 at (^√₂³,¹₂).

(x⁴)⁰+ (2x²y²)⁰+ (y⁴)⁰= (4x²)⁰− (2)⁰. y⁰(4x²y + 4y³) = 8x − 4x³− 4xy² y⁰= ^8x−4x_4x3y+4y^3−4xy22

At the point (^√₂³,¹₂), y⁰=√ 3.

15. x³y − xy⁴= 3x²y²− xy at (1, −1).

3x²y + x³y⁰− yt − 4xy³y⁰= 6xy²+ 6x²yy⁰− y − xy⁰ y⁰(x³− 4xy³− 6x²y + x) = 6xy²− y − 3x²y + y⁴ y⁰= ^6xy_x3−4xy^2−y−3x3−6x^2y+y2y+x⁴

At the point (1, −1), y⁰=¹¹₁₂.

17. yx⁴+2x²y³+y⁵−5x⁴−10x²y²−5y⁴+20x²+20y²= 16 at (−1, −1).

(yx⁴)⁰+ (2x²y³)⁰+ (y⁵)⁰− (5x⁴)⁰− (10x²y²)⁰− (5y⁴)⁰+ (20x²)⁰+ (20y²)⁰= 0.

y⁰(x⁴+ 6x²y²+ 5y⁴− 20x²y − 20y³+ 40y) = −4yx³− 4xy³+ 20x³+ 20xy²− 40x.

y⁰= _x^−4yx4+6x^32−4xyy²+5y^3+20x4−20x^3+20xy2y−20y^2−40x3+40y

At the point (−1, −1), y⁰= −²₃ .

19. This is a tricky one, so we commend you for even try- ing this problem! There are many ways to describe a curve in the xy plane, one way is by polar coordi- nates; x = r cos(θ) and y = r sin(θ). Question 15 can be re-written as 4r⁴cos⁴− 3r²cos²− rcos + 4r⁴sin⁴− 3r²sin²+ 8r⁴cos²sin² = 0. This simplifies further to

−3r − cos + 4r³= 0. At (−0.5, 0), y⁰has two values, be- cause that is where the curve changes direction to form the inner curve.

21. If y = x and y = −x, y − x = 0 and y + x = 0 respec- tively. Since these two equations are equal to zero, we can multiply them together, (y − x)(y + x) = 0.

23. x²y²+ x⁴− y²= 0 at (√¹ 2,√¹

2).

x²y²− y²= −x⁴. y²(x²− 1) = −x⁴ y²= _1−x^x42

y = ±^√_1−x^x2 2

y⁰= ±^2x(1−x2)

1

2+x³(1−x²)

−1 2

1−x² = 3.

25. From question 10, we know y⁰= ^−2xy_2x2y^2−4x3

−2y . y⁰⁰= ^{(2x2y−2y)(−2xy2−4x}_(2x³₂⁾⁰_y^{−(−2xy2−4x3)(2x2y−2y)0}

−2y)²

y⁰⁰= ^(2x2y−2y)(−4xyy⁰−2y²−12x³)−(−2xy²−4x³)(2x²y⁰+4xy−2y⁰) (2x²y−2y)²

 The real end of science is the honor of the

human mind.

– Karl Jacobi

Related Rates

Answers to exercises from 4.25, page 383

1.

a

b d

a⁰= 55, b⁰= 30, a²+ b²= d². When a = 55 and b = 30, d = 62.65. (a²+ b²)⁰ = (d²)⁰

2aa⁰+ 2bb⁰ = 2dd⁰ 2(55)(55) + 2(30)(30) = 2(62.65)d⁰

6050 + 1800 = 125.3d⁰ 62.65 mph = d⁰

3. To find the speed of the car, the patrolman should mul- tiply the number on the radar gun by ^r_c (where r is the distance between the patrolman and the car, and c is the distance of the car down the road.) In this case, speed of car= c⁰=⁵₄r⁰.

5.

h

w l

volume= lhw, v⁰= .3 m³/s. Because the box is rectan- gular, l and w are constants. v = lhw

v⁰ = (lw)h⁰ .3 = 8h⁰ .0375 m/s = h⁰ 7.

h r

We know v⁰ = 1.3 m³/min, and _h^r = ³₇, thus r = ^3h₇. v = ¹₃πr²h

v = ^π₃(^3h₇)²h v = ^3π₄₉h³ v⁰ = ^3π₄₉3h²h⁰ 1.3 = (^3π₄₉)3(7)²h⁰ 1.3 = 9πh⁰

h⁰ = ^1.3_9π ≈ .0460 m/s 9.

boat dock

b

h l

h = 10, l⁰= −2 ft/sec.h²+b²= l², thus when l = 30, b = 28.3. h²+ b² = l²

2hh⁰+ 2bb⁰ = 2ll⁰ 0 + 2(28.3)b⁰ = 2(30)(−2)

b⁰ = −2.12 ft/sec

11. The percentage of alcohol is modeled by the function p = ^a_v where p is the percentage of alcohol and v is the total volume of liquid in the conainer. The total volume v is equal to the sum of the volume of alco- hol, a, and the volume of water, w. Thus, p = _a+w^a . We know that a = 13 and that it is a constant, so p = _13+w¹³ . Now we differentiat: p⁰ = _(13+v)^13w02. We are given that v⁰ = 2 and since we are asked about the moment when a = w, w must be equal to 13, Thus

p⁰ = _(13+13)¹³⁽²⁾2

= ₂₆²⁶2

= .0385 percent per minute.

13.

r

r =length of rope on one side of the pulley. r⁰ = .6 in/sec. h =height of man off the ground. r0=beginning length of rope when even. r = r0+ h

r⁰ = 0 + h⁰ 6 in/sec = h⁰

15. Volume v = ⁴₃πr³. Surface area s = 8πr². v⁰ = .03 m³/s. When v = .35,

r = .437. v = ⁴₃πr³

v⁰ = 4πr²r⁰ .03 = 4π(.191)r⁰ .0125 m/s = r⁰ s = 8πr²

s⁰ = 16πrr⁰

s⁰ = 16π(.437)(.0125) s⁰ ≈ .275 m²/s 17.

l

.5l h

m

e d

l = 12 and d⁰ = .1. From similar triangles, we have ^.5l_. = ^m_h = ^e_d, thus when d = 3, m = 5.81 and e = .025.e⁰ = .5d⁰ = .05.

(.5l)² = m²+ e² .25l² = m²+ e²

.5ll⁰ = 2mm⁰+ 2ee⁰

.5(12)(0) = 2(5.81)m⁰+ 2(1.5)(.05) 0 = 11.6m⁰+ .15

−.013 m/s = m⁰

19. From problem 11, we know y = ³_z and x = 2y. Also, we know z⁰ = 18, y⁰ = −1.5, and x⁰ = −3. z⁰⁰ = −3.

y⁰ = ^−3z_z2⁰

y⁰z² = −3z⁰ y⁰⁰z²+ y⁰2zz⁰ = −3z⁰⁰ y⁰⁰(36) + (−1.5)(2)(6)(18) = (−3)(−3)

36y⁰⁰ = 9 + 324 y⁰⁰ = 9.25 x⁰ = 2y⁰

x⁰⁰ = 2y⁰⁰

x⁰⁰ = 2(9.25)x⁰⁰ = 18.5

21. We label the angle between the ladder and the horizon- tal α, the distance of the woman from the center r, and the vertical position of the woman on the ladder h. We know α⁰=₃₀₀^2π radians/sec, r = 30 and r⁰= −1 ft/sec.

a.

α

α =^3π₂ h = r sin α

h⁰ = r⁰sin α + r cos α(α⁰)

h⁰ = −1(sin^3π₂ ) + 30(cos^3π₂)(₃₀₀^2π) h⁰ = 1 + 0

h⁰ = 1 ft/sec

b.

α

α =^π₂ h = r sin α

h⁰ = r⁰sin α + r cos α(α⁰) h⁰ = −1(sin^π₂) + 30(cos^π₂)(₃₀₀^2π) h⁰ = −1 + 0

h⁰ = −1 ft/sec

c.

α = 0 h = r sin α

h⁰ = r⁰sin α + r cos α(α⁰) h⁰ = −1(sin 0) + 30(cos 0)(₃₀₀^2π) h⁰ = 0 + 30(1)₃₀₀^2π

h⁰ = .2π ft/sec

d.

α

α =^π₄ h = r sin α

h⁰ = r⁰sin α + r cos α(α⁰) h⁰ = −1(sin^π₄) + 30(cos^π₄)(₃₀₀^2π) h⁰ = ⁻¹√

2+√³⁰ 2(₃₀₀^2π) h⁰ ≈ −.263 ft/sec

23. s wall

c 20

10

From similar triangles, we know ₁₀^c = ₂₀^s.

2c = s

2c⁰ = s⁰ 2(−60) = s⁰

−120 mph = s⁰

The shadow is shrinking at 120 mph.

25.

l h

g

p is the position of the plane, and c is the position of the car. h = 1 mile, l⁰= −136 mph, p⁰ = −120 mph, and g⁰= p⁰− c⁰. When l = 1.5 miles, g = 1.118 miles.

l² = h²+ g² 2ll⁰ = 0 + 2gg⁰ 2(1.5)(−136) = 2(1.118)g⁰

−408 = = 2.236g⁰

−182.46 mph = g⁰

g⁰ = p⁰− c⁰

−182.46 = −120 − c⁰ 62.46 mph = c⁰

The car is going only 62.46 mph, which is under the speed limit of 65 mph. Thus the car should not get a ticket.

27. v = ¹₃πr²h, surface area exposed s = πr², and v⁰= ks, where k is a constant. Because the ratio of radius to height does not change, we know _h^r = q, where q is some constant. v = ¹₃πr²h

v = ¹₃π(qh)²h v⁰ = (q²π)h²h⁰ kπr² = (q²π)h²h⁰ kπ(qh)² = (q²π)h²h⁰

k = h⁰

'

&

$

%

Karl Marx and Friedrich Engels, philosophical founders of communism, struggled with the calculus and its place in their system. Engels once wrote to Marx:

“Last week in a dream I gave a fellow my shirt but- tons to differentiate and this fellow ran away with them.”

Curve-sketching and the Mean Value Theorem

Answers to exercises from 4.26, page 392

1. f⁰(x) = 4 − 2x

f⁰(x) is positive when x is less than 2, and negative when x is greater than 2. Thus f (x) is increasing when x < 2 and decreasing when x > 2.

3. f⁰(x) = 3x²− 3

f⁰(x) > 0 when x > 1 and when x < −1. f⁰(x) < 0 when

−1 < x < 1. Thus f(x) is decreasing when −1 < x < 1 and increasing elsewhere.

5. f⁰(x) = 3x²+ 4x + 1

f⁰(x) > 0 when x < −1 or when x > −¹₃ and f⁰(x) < 0 when −1 < x < −¹₃. f (x) is decreasing when −1 < x <

−¹₃ and increasing elsewhere.

7. f⁰(x) =^−2x_(x2+2x+3)^2+6x+122

f⁰(x) > 0 when ^3+√₂³³ < x < ^2−√₂³³ and f⁰(x) < 0 otherwise. f (x) is increasing when ³⁺

√33

2 < x <³⁻

√33

and decreasing elsewhere. 2

9. f⁰(x) =√ ^x+1

x²+2x+3

f⁰(x) > 0 when x > −1 and f⁰(x) < 0 when x < −1.

f (x) is increasing when x > −1 and decreasing when x < −1.

11. f⁰(x) =₂^3x−4√ x−2

f⁰(x) = 0 when x < ⁴₃, but f⁰(x) is undefined when x < 2. f⁰(x) > 0 when x > 2. f (x) is increasing when x > 2 and undefined elsewhere.









The Mean Value Theorem was probably discovered inde- pendently by several mathematicians. Lagrange certainly understood it. Cauchy proved it.

13. f⁰(x) = xe^x+ e^x

f⁰(x) > 0 when x > −1. f⁰(x) < 0when x < −1. f⁰(x) is increasing when x > −1 and decreasing when x < −1.

15. f⁰(x) = 1 − e^x

f⁰(x) > 0 when x < 0 and f⁰(x) < 0 when x > 0. f (x) is increasing when x < 0 and decreasing when x > 0.

17. f⁰(x) = − sin x cos (cos x)

f⁰(x) > 0 when π < x < 2π, when 3π < x < 4π, when 5π < x < 6π etc. f⁰(x) < 0 when 0 < x < π, when 2π < x < 3π etc. f (x) is increasing when π < x < 2π, when 3π < x < 4π, when 5π < x < 6π etc. and decreas- ing when 0 < x < π, when 2π < x < 3π etc.

19. (3x²− 3) sec²(x³− 3x)

f⁰(x) < 0 when −1 < x < 1 and positive otherwise. f(x) is decreasing when −1 < x < 1 and increasing elsewhere.

21. f⁰(x) = (300x²− 300)(x³− 3x)⁹⁹ f⁰(x) > 0 when −√

3 < x < −1, when 0 < x < 1 and when x > √

3. f⁰(x) < 0 when x < −√ 3, when

−1 < x < 0 and when 1 < x < √

3. f (x) is in- creasing when −√

3 < x < −1, when 0 < x < 1 and when x >√

3. f (x) is decreasing when x < −√ 3, when

−1 < x < 0 and when 1 < x <√ 3.

23. f⁰(x) = _|x^3x3²−3x|⁻³

f⁰(x) > 0 when −√

3 < x < −1 and when x > √ 3.

f⁰(x) < 0 when −1 < x < 0. f⁰9x) is undefined when 0 < x <√

3 and when x < −√

3. f (x) is increasing when

−√

3 < x < −1 and when x >√

3, f (x) is decreasing when −1 < x < 0 and is undefined elsewhere.

25. f⁰(x) = 4 − 2x 0 = 4 − 2x x = 2

2

There is a relative maximum at x = 2. f (x) is differen- tiable at that point.

27. f⁰(x) = 3x²− 3 0 = 3x²− 3 x = ±1

-1 1

There is a relative minimum at 1 and a relative maxi- mum at −1. f(x) is differentiable at both points.

29. f⁰(x) = 3x²+ 4x + 1 0 = 3x²+ 4x + 1 x = −1 or x = −¹₃

−¹₃ -1

There is a relative minimum at x = −¹₃ and a rela- tive maximum at x = −1. f(x) is differentiable at both points.

31. f⁰(x) =^−2x_(x2+2x+3)^2+6x+122

0 =^−2x_(x2+2x+3)^2+6x+122

x =^3±

√33 2

3−√ 33

2 3−√

33 2

There is a relative minimum at x = ^3+√₂³³ and a rela- tive maximum at x = ^3−√₂²². f (x) is differentiable at both points.

33. f⁰(x) = √ ^x+1

x²+2x+3

0 =√ ^x+1

x²+2x+3

x = −1

-1

There is a relative minimum at x = −1. f(x) is differ- entiable at that point.

35. f⁰(x) = ₂^3x−4√ x−2

0 =₂^3x−4√ x−2

x =⁴₃

4 3

2

f (x) is undefined when x < 2. There is a minimum at x = 2. f (x) is not differentiable at that point.

37. f⁰(x) = xe^x+ e^x 0 = xe^x+ e^x x = −1

-1

There is a relative minimum at x = −1. f(x) is differ- entiable at this point.

39. f⁰(x) = 1 − e^x 0 = 1 − e^x x = 0

There is a relative maximum at x = 0. f (x) is differen- tiable at that point.

41. f⁰(x) = − sin x cos (cos x) 0 = − sin x cos (cos x) x = 0, x = π, x = 2π etc.

π 2π

3π

There are relative minima at x = π, x = 3π, x = 5π etc.

and relative maxima at x = 0, x = 2π, x = 4π etc. f (x) is differentiable at each of these points.

43. f⁰(x) = (3x²− 3) sec²(x³− 3x) 0 = (3x²− 3) sec²(x³− 3x) x = ±1

-1 1

There is a relative minimum at x = 1 and a relative maximum at x = −1. f(x) is differentiable at both of these points.

45. f⁰(x) = (300x²− 300)(x³− 3x)⁹⁹ 0 = (300x²− 300)(x³− 3x)⁹⁹ x = 0, x = ±1 or x = ±√

3

−√ 3

-1 1

√3

There are relative minima at x = ±√

3 and x = 0 and relative maxima at x = ±1. f(x) is differentiable at each of these points.

(Graph is not to scale.)

47. f⁰(x) =_|x^3x3²−3x|⁻³

0 =_|x^3x3−3x|²⁻³

x = ±1

-1 1

There are relative maxima at x = ±1. f(x) is differen- tiable at both of these points.

51. f⁰(x) = 2|x|

-1 1

There are no relative extrema.

53. For x < 1, f⁰(x) = −3x²+ 4x − 1.

For x = 1, f⁰(x) = 0.

For x > 1, f⁰(x) = 3x²− 4x + 1.

f⁰(x) = 0 x =¹₃ or x = 1.

1 1

3

There is a relative minimum at x = ¹₃ and a relative maximum at x = 1. f (x) is differentiable at both points.

55. Suppose for a function f , f⁰(x), 0 for all x in an interval [a, b]. If f were increasing on the interval [a, b] then for some r and some s in [a, b] with r < s, f (r) < f (s).

Then ^{f(s)−f (r)}_s−r > 0. By the Mean Value Theorem, f⁰(c) =^{f(s)−f (r)}_s−r , but f⁰(c) < 0 since c is in the interval [a, b]. Thus f is not increasing, but rather decreasing.

57. The converse of Proposition 6.5 is false. The function f (x) = x³is increasing on [−1, 1], but f⁰(0) is not posi- tive.

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&

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% The closely held consolation of some rational-

izing mathematicians, that dy and dx are in fact only infinitely small and [that their ratio]

only approaches

is a chimera.

– Karl Marx

Curve-sketching and the Second Derivative

Answers to exercises from 4.27, page 402

1. x < 0, y < 0, y⁰< 0, y⁰⁰< 0

3. x < 0, y < 0, y⁰< 0, y⁰⁰> 0

5. x < 0, y > 0, y⁰> 0, y⁰⁰> 0

7. x > 0, y < 0, y⁰> 0, y⁰⁰> 0

9. f (x) = 4x + 5 − x² f⁰(x) = 4 − 2x f⁰⁰(x) = −2

There are no inflection points for this function.

11. f (x) = x³− 3x f⁰(x) = 3x²− 3 f⁰⁰(x) = 6x

There is an inflection point at (0, 0).

13. f (x) = x³− x²− x + 1 f⁰(x) = 3x²− 2x − 1 f⁰⁰(x) = 6x − 2

There is an inflection point at (¹₃,¹⁶₂₇).

15. f (x) = x⁴− 5x² f⁰(x) = 4x³− 10x f⁰⁰(x) = 12x²− 10

There are inflection points at (−q

5

6, −¹²⁵₃₆) and (q

5 6, −¹²⁵₃₆).

17. f (x) = (x³+ 2)² f⁰(x) = 6x⁵+ 12x² f⁰⁰(x) = 2x(15x³+ 12)

There are inflection points at (0, 4) and (q³

−⁴₅,³⁶₂₅).

19. f (x) =_1+x¹2

f⁰(x) = _(1+x^−2x2)²

f⁰⁰(x) = _(1+x^6x2−22)³

There are inflection points at (q

1

3,³₄) and (−q

1 3,³₄).

21. f (x) =√

x²+ 2x + 3 f⁰(x) = √ ^x+1

x²+2x+3

f⁰⁰(x) = ²

(x²+2x+3)³2

There are no inflection points.

23. f (x) = x√ x − 2 f⁰(x) = ₂^3x−4√

x−2

f⁰⁰(x) = ^3x−8

4(x−2)³²

There is an inflection point at (⁸₃,^8√₉⁶).

25. f (x) = sin²(x) f⁰(x) = 2 sin(x) cos(x) f⁰⁰(x) = 4 cos²(x) − 2

There are inflection points at (−^π₄, 1), (^π₄, 1), (^3π₄, 1), (^5π₄ , 1) etc.

27. f (x) = sin(x) − cos(x) f⁰(x) = cos(x) + sin(x) f⁰⁰(x) = cos(x) − sin(x)

There are inflection points at (−^7π₄,^√₂²), (−^3π₄ , −^√₂²), (^π₄,^√₂²), (^5π₄, −^√₂²), (^9π₄ ,^√₂²) etc.

29. f (x) = x − e^x f⁰(x) = 1 − e^x f⁰⁰(x) = −e^x

There are no inflection points.

31. f (x) = x ln(x) f⁰(x) = ln(x) + 1 f⁰⁰(x) = ¹_x

There are no inflection points.

33. f (x) =√ 5x³− 3x f⁰(x) = 3√

5x²− 3 f⁰⁰(x) = 6√

5x

There is an inflection point at (0, 0).

37. f (x) = e^x3−3x

f⁰(x) = (3x62 − 3)e^x3−3x

f⁰⁰(x) = (9x⁴− 18x²+ 6x + 9)e^x3−3x

There are inflection points at (−.60436, 4.91522) and (−1.40248, 4.23915).

39. For x < 0, f (x) = −2x + 1, f⁰(x) = −2, and f⁰⁰(x) = 0.

For 0 < x < 1, f (x) = 1, f⁰(x) = 0, and f⁰⁰(x) = 0.

For x > 1, f⁰(x) = 2x − 1, f⁰(x) = 2, and f⁰⁰(x) = 0.

There are no inflection points.

41. There are no inflection points.

43. False

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% . . . the mathematics of variables, whose most

important part is the infinitesimal calculus, is in essence nothing other than the application of dialectics to mathematical relations.

– Friedrich Engels

The Second Derivative Test

Answers to exercises from 4.28, page 406

1. f⁰(x) = 4 − 2x 0 = 4 − 2x x = 2 f⁰⁰(x) = −2 f⁰⁰(2) = −2

There is a relative maximum at x = 2.

3. f⁰(x) = 3x²− 3 0 = 3x²− 3 x = ±1 f⁰⁰(x) = 6x f⁰⁰(−1) = −6 f⁰⁰(1) = 6

There is a relative minimum at x = 1 and a relative maximum at x = −1.

5. f⁰(x) = 3x²+ 4x + 1 0 = 3x²+ 4x + 1 x = −1 or x = −¹₃ f⁰⁰(x) = 6x + 4 f⁰⁰(−1) = −2 f⁰⁰(−¹₃) = 2

There is a relative minimum at x = −¹₃ and a relative maximum at x = −1.

7. f⁰(x) = _(1+x^−2x2)²

0 =_(1+x^−2x2)²

x = 0

f⁰⁰(x) = _(1+x^6x2−22)²

f⁰⁰(0) = −2

There is a relative maximum at x = 0.

9. f⁰(x) = √^x

1+x²

0 =√^x

1+x²

x = 0 f⁰⁰(x) = ¹

(1+x²)³²

f⁰⁰(0) = 1

There is a relative minimum at x = 0.

11. f⁰(x) = ^3√₂^x− 4 0 =^3√₂^x− 4 x =⁶⁴₉ f⁰⁰(x) = ₄√³

x

f⁰⁰(⁶⁴₉) = ¹⁶₃

There is a relative minimum at x =⁶⁴₉.

13. f⁰(x) = e^x+ xe^x 0 = e^x+ xe^x x = −1

f⁰⁰(x) = 2e^x+ xe^x f⁰⁰(−1) = ¹e

There is a relative minimum at x = −1.

15. f⁰(x) = 1 − e^x 0 = 1 − e^x x = 0 f⁰⁰(x) = −e^x f⁰⁰(0) = −1

There is a relative maximum at x = 0.

17. f⁰(x) = cos(x) + sin(x) 0 = cos(x) + sin(x) x =^3π₄ ,^7π₄ ,^11π₄ ,^15π₄ etc.

f⁰⁰(x) = cos(x) − sin(x) f⁰⁰(^3π₄ ) = −√

2 f⁰⁰(^7π₄ ) =√

2 f⁰⁰(^11π₄ ) = −√

2 f⁰⁰(^15π₄ ) =√

2

There are relative minima at x = ^7π₄ , ^15π₄ etc. and relative maxima at x = ^3π₄ , ^15π₄ etc.

19. f (x) = x² f⁰(x) = 2x f⁰⁰(x) = 2

a. f⁰(x) > 0 when x > 0.

b. f⁰(x) < 0 when x < 0.

c. f⁰(x) = 0 when x = 0.

d. f⁰⁰(x) > 0 for all x.

e. f⁰⁰(x) < 0 for no x.

f. f⁰⁰(x) = 0 for no x.

g.

21. f (x) = x⁴ f⁰(x) = 4x³ f⁰⁰(x) = 12x²

a. f⁰(x) > 0 when x > 0.

b. f⁰(x) < 0 when x < 0.

c. f⁰(x) = 0 when x = 0.

d. f⁰⁰(x) ≥ 0 for all x.

e. f⁰⁰(x) < 0 for no x.

f. f⁰⁰(x) = 0 for x = 0.

g.

23. f (x) = x³− x² f⁰(x) = 3x²− 2x

f⁰⁰(x) = 6x − 2

a. f⁰(x) > 0 when 0 < x < ²₃.

b. f⁰(x) < 0 when x < 0 and when x >²₃. c. f⁰(x) = 0 when x = 0 and when x =²₃. d. f⁰⁰(x) > 0 when x > ¹₃.

e. f⁰⁰(x) < 0 when x < ¹₃. f. f⁰⁰(x) = 0 when x =¹₃. g.

25. f (x) =_x¹ f⁰(x) = ⁻¹_x2

f⁰⁰(x) = _x²3

a. f⁰(x) > 0 for no x.

b. f⁰(x) < 0 for all x.

c. f⁰(x) = 0 for no x.

d. f⁰⁰(x) > 0 when x > 0.

e. f⁰⁰(x) < 0 when x < 0.

f. f⁰⁰(x) = 0 for no x.

g.

27. f (x) =_1+x^x2

f⁰(x) = _(1+x^1−x2²)²

f⁰⁰(x) = _(1+x^2x3−6x2)³

a. f⁰(x) > 0 when −1 < x < 1.

b. f⁰(x) < 0 when x < −1 and when x > 1.

c. f⁰(x) = 0 when x = ±1.

d. f⁰⁰(x) > 0 when x < −√

3 and when 0 < x <√ 3.

e. f⁰⁰(x) < 0 when −√

3 < x < 0 and when x >√ 3.

f. f⁰⁰(x) = 0 when x = ±√

3 and when x = 0.

g.

29. f (x) = x²− x f⁰(x) = 3x²− 1 f⁰⁰(x) = 6x

a. f⁰(x) > 0 when x < −^√₃³ and when x > ^√₃³..

b. f⁰(x) < 0 when −^√₃³ < x <^√₃³. c. f⁰(x) = 0 when x = ±^√₃³. d. f⁰⁰(x) > 0 when x > 0.

e. f⁰⁰(x) < 0 when x < 0.

f. f⁰⁰(x) = 0 when x = 0.