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Active Calculus

Matthew Boelkins

Appendix B Answers to Activities

This appendix contains answers to all activities in the text. Answers for preview activities are not included.
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1 Understanding the Derivative
1.1 How do we measure velocity?
1.1.1 Position and average velocity

Activity 1.1.2.

Answer.
  1. AV[0.4,0.8]=12.8 ft/sec; AV[0.7,0.8]=8 ft/sec; the other average velocities are, respectively, 6.56, 6.416, 0, 4.8, 6.24, 6.384, all in ft/sec.
  2. m=12.8 is the average velocity of the ball between t=0.4 and t=0.8.
  3. Like a straight line with slope about 6.4.
  4. About 6.4 feet per second.

1.1.2 Instantaneous Velocity

Activity 1.1.3.

Answer.
  1. AV[1.5,2]=−24 ft/sec, which is negative.
  2. The instantaneous velocity at t=1.5 is approximately −16 ft/sec; at t=2, the instantaneous velocity is about −32 ft/sec, and −16>−32.
  3. When the ball is rising, its instantaneous velocity is positive, while when the ball is falling, its instantaneous velocity is negative.
  4. Zero.

Activity 1.1.4.

Answer.
AV[2,2+h]=−32−16h

1.2 The notion of limit
1.2.1 The Notion of Limit

Activity 1.2.2.

Answer.
  1. 2.
  2. 12.
  3. 12.

1.2.2 Instantaneous Velocity

Activity 1.2.3.

Answer.
  1. 6+h.
  2. 6.2 meters/min.
  3. 6 meters per minute.

Activity 1.2.4.

Answer.
  1. AV[0.5,1]=1−11−0.5=0, AV[1.5,2.5]=3−12.5−1.5=2, and AV[0,5]=5−05−0=1.
  2. Take shorter and shorter time intervals and draw the lines whose slopes represent average velocity. If those lines’ slopes are approaching a single number, that number represents the instantaneous velocity.
  3. The instantaneous velocity at t=2 is greater than the average velocity on [1.5,2.5].

1.3 The derivative of a function at a point
1.3.1 The Derivative of a Function at a Point

Activity 1.3.2.

Answer.
  1. f is linear.
  2. The average rate of change on [1,4], [3,7], and [5,5+h] is −2.
  3. f′(1)=−2.
  4. f′(2)=−2, f′(π)=−2, and f′(−2)=−2, since the slope of a linear function is the same at every point.

Activity 1.3.3.

Answer.
  1. The vertex is (12,36).
  2. s(2)−s(1)2−1=−32 feet per second.
  3. s′(1)=−16.
  5. s′(a) is positive whenever 0≤a<12; s′(a) to be negative whenever 12<a<2; s′(12)=0.

Activity 1.3.4.

Answer.
  2. AV[2,4]≈9171 people per decade is expected to be the average rate of change of the city’s population over the two decades from 2030 to 2050.
  3. P′(2)=limh→0P(2+h)−P(2)h=limh→025000e2/5(eh/5−1h)
    Because there is no way to remove a factor of h from the numerator, we cannot eliminate the h that is making the denominator go to zero.
  4. P′(2)=limh→0P(2+h)−P(2)h≈7458.5
    which is measured in people per decade.
  5. See the graph provided in (a) above. The magenta line has slope equal to the average rate of change of P on [2,4], while the green line is the tangent line at (2,P(2)) with slope P′(2).
  6. It appears that the tangent line’s slope at the point (a,P(a)) will increase as a increases.

1.4 The derivative function
1.4.1 How the derivative is itself a function

Activity 1.4.2.

Answer.

Activity 1.4.3.

Answer.
  1. f′(x)=0.
  2. g′(t)=1.
  3. p′(z)=2z.
  4. q′(s)=3s2.
  5. F′(t)=−1t2.
  6. G′(y)=12y.

1.5 Interpreting, estimating, and using the derivative
1.5.2 Toward more accurate derivative estimates

Activity 1.5.2.

Answer.
  1. F′(30)≈=3.85 degrees per minute.
  2. F′(60)≈=1.56 degrees per minute.
  3. F′(75)>F′(90).
  4. The value F(64)=330.28 is the temperature of the potato in degrees Fahrenheit at time 64, while F′(64)=1.341 measures the instantaneous rate of change of the potato’s temperature with respect to time at the instant t=64, and its units are degrees per minute. Because at time t=64 the potato’s temperature is increasing at 1.341 degrees per minute, we expect that at t=65, the temperature will be about 1.341 degrees greater than at t=64, or in other words F(65)≈330.28+1.341=331.621. Similarly, at t=66, two minutes have elapsed from t=64, so we expect an increase of 2⋅1.341 degrees: F(66)≈330.28+2⋅1.341=332.962.
  5. Throughout the time interval [0,90], the temperature F of the potato is increasing. But as time goes on, the rate at which the temperature is rising appears to be decreasing. That is, while the values of F continue to get larger as time progresses, the values of F′ are getting smaller (while still remaining positive). We thus might say that “the temperature of the potato is increasing, but at a decreasing rate.”

Activity 1.5.3.

Answer.
  1. It costs $800 to make 2000 feet of rope.
  2. “dollars per foot.”
  3. C(2100)≈=835,.
  4. Either C′(2000)=C′(3000) or C′(2000)>C′(3000).
  5. Impossible. The total cost function C(r) can never decrease.

Activity 1.5.4.

Answer.
  1. f′(90)≈0.0006 liters per kilometer per kilometer per hour.
  2. At 80 kilometers per hour, the car is using fuel at a rate of 0.015 liters per kilometer.
  3. When the car is traveling at 90 kilometers per hour, its rate of fuel consumption per kilometer is increasing at a rate of 0.0006 liters per kilometer per kilometer per hour.

1.6 The second derivative
1.6.3 Concavity

Activity 1.6.2.

Answer.
  1. Increasing: 0<t<2, 3<t<5, 7<t<9, and 10<t<12. Decreasing: never.
  2. Velocity is increasing on 0<t<1, 3<t<4, 7<t<8, and 10<t<11; y=v(t) is decreasing on 1<t<2, 4<t<5, 8<t<9, and 11<t<12. Velocity is constant on 2<t<3, 5<t<7, and 9<t<10.
  3. a(t)=v′(t) and a(t)=s″(t).
  4. s″(t) is positive since s′(t) is increasing.
    • increasing.
    • decreasing.
    • constant.
    • increasing.
    • decreasing.
    • constant.
    • concave up.
    • concave down.
    • linear.

Activity 1.6.3.

Answer.
  1. Degrees Fahrenheit per minute.
  2. F″(30)≈−0.119.
  3. At the moment t=30, the temperature of the potato is 251 degrees; its temperature is rising at a rate of 3.85 degrees per minute; and the rate at which the temperature is rising is falling at a rate of 0.119 degrees per minute per minute.
  4. Increasing at a decreasing rate.

Activity 1.6.4.

Answer.

1.7 Limits, Continuity, and Differentiability
1.7.1 Having a limit at a point

Activity 1.7.2.

Answer.
  1. f(−2)=1; f(−1) is not defined; f(0)=73; f(1)=2; f(2)=2.
  2. limx→−2−f(x)=2 andlimx→−2+f(x)=1
    limx→−1−f(x)=53 andlimx→−1+f(x)=53
    limx→0−f(x)=73 andlimx→0+f(x)=73
    limx→1−f(x)=3 andlimx→1+f(x)=3
    limx→2−f(x)=2 andlimx→2+f(x)=2
  3. limx→−2f(x) does not exist. The values of the limits as x→a for a=−1,0,1,2 are 53,73,3,2.
  4. a=−2, a=−1, and a=1.

1.7.2 Being continuous at a point

Activity 1.7.3.

Answer.
  1. a=−2; a=+2.
  2. a=3.
  3. a=−1; a=3.
  4. a=−2; a=2; a=3; a=−1.
  5. “If f is continuous at x=a, then f has a limit at x=a.”

1.7.3 Being differentiable at a point

Activity 1.7.4.

Answer.
  1. g is piecewise linear.
  2. g′(0)=(limh→0g(0+h)−g(0)h=(limh→0|0+h|−|0|h=(limh→0|h|h
  3. limh→0+|h|h=1, but limh→0−|h|h=−1.
  4. a=−3,−2,−1,1,2,3.
  5. True.

1.8 The Tangent Line Approximation
1.8.2 The local linearization

Activity 1.8.2.

Answer.
  1. L(−1)=−2; L′(−1)=3.
  2. g(−1)=−2; g′(−1)=3.
  3. Less.
  4. g(−1.03)≈L(−1.03)=−2.09.
  5. Concave up.
  6. The illustration below shows a possible graph of y=g(x) near x=−1, along with the tangent line y=L(x) through (−1,g(−1)).

Activity 1.8.3.

Answer.
  1. L(x)=−1+2(x−2).
  2. f(2.07)≈L(2.07)=−0.86.
  3. See the image in part e.
  4. Neither.
  5. See the image below, which shows, at left, a possible graph of y=f(x) near x=2, along with the tangent line y=L(x) through (2,f(2)).
  6. Too large.

2 Computing Derivatives
2.1 Elementary derivative rules
2.1.2 Constant, Power, and Exponential Functions

Activity 2.1.2.

Answer.
  1. f′(t)=0.
  2. g′(z)=7zln⁡(7).
  3. h′(w)=34w−1/4.
  4. dpdx=0.
  5. r′(t)=(2)tln⁡(2).
  6. ddq[q−1]=−q−2.
  7. dmdt=−3t−4=−3t4.

2.1.3 Constant Multiples and Sums of Functions

Activity 2.1.3.

Answer.
  1. f′(x)=53x2/3−4x3+2xln⁡(2).
  2. g′(x)=14ex+3⋅5x4−1.
  3. h′(z)=12z−1/2−4z−5+5zln⁡(5).
  4. drdt=53⋅7t6−πet.
  5. dsdy=4y3.
  6. q′(x)=2x−2x−2.
  7. p′(a)=12a3−6a2+14a−1.

Activity 2.1.4.

Answer.
  1. h′(4)=316.
  2. (i.)P′(4)=2(1.37)4ln⁡(1.37)≈2.218 million cells per day; (ii.) the population is growing at an increasing rate.
  3. y−25=−33(a+1).
  4. The slope is a number, while the equation is, well, an equation.

2.2 The sine and cosine functions
2.2.1 The sine and cosine functions

Activity 2.2.2.

Answer.
  1. Figure 2.2.2.
  2. 1,0,−1,0,1,0,−1,0,1.
  3. f′(0)=f′(−2π)=f′(2π)=1.
  4. Figure 2.2.2.
  5. ddx[sin⁡(x)]=cos⁡(x).
Figure 2.2.2. At left, the graph of y=f(x)=sin⁡(x). At right, the graph of y=f′(x).

Activity 2.2.3.

Answer.
  1. Figure 2.2.5.
  2. 0,−1,0,1,0,−1,0,1,0.
  3. g′(π2)=g′(−3π2)=−1.
  4. Figure 2.2.5.
  5. ddx[cos⁡(x)]=−sin⁡(x).
Figure 2.2.5. At left, the graph of y=g(x)=cos⁡(x). At right, the graph of y=g′(x)

Activity 2.2.4.

Answer.
  1. dhdt=−3sin⁡(t)−4cos⁡(t).
  2. f′(π6)=2+34.
  3. y−π24=(π−2)(x−π2).
  4. p′(z)=4z3+4zln⁡(4)−4sin⁡(z).
  5. P′(2)=8cos⁡(2)≈−3.329 hundred animals per decade.

2.3 The product and quotient rules
2.3.1 The product rule

Activity 2.3.2.

Answer.
  1. m′(w)=3w17⋅4wln⁡(4)+4w⋅51w16.
  2. h′(t)=(sin⁡(t)+cos⁡(t))⋅4t3+t4⋅(cos⁡(t)−sin⁡(t)).
  3. f′(1)=e(cos⁡(1)+sin⁡(1))≈3.756.
  4. L(x)=−12(x+1).

2.3.2 The quotient rule

Activity 2.3.3.

Answer.
  1. r′(z)=(z4+1)3zln⁡(3)−3z(4z3)(z4+1)2.
  2. v′(t)=(cos⁡(t)+t2)cos⁡(t)−sin⁡(t)(−sin⁡(t)+2t)(cos⁡(t)+t2)2.
  3. R′(0)=29.
  4. I′(0.5)=50e0.5≈30.327, I′(2)=−100e2≈−13.534, and I′(5)=−400e5≈−2.695, each in candles per millisecond.

2.3.3 Combining rules

Activity 2.3.4.

Answer.
  1. f′(r)=(5r3+sin⁡(r))[4rln⁡(4)+2sin⁡(r)]+(4r−2cos⁡(r))[15r2+cos⁡(r)].
  2. p′(t)=t6⋅6t[−sin⁡(t)]−cos⁡(t)[t6⋅6tln⁡(6)+6t⋅6t5](t6⋅6t)2.
  3. g′(z)=3[z7ez+7z6ez]−2[z2cos⁡(z)+2zsin⁡(z)]+(z2+1)1−z(2z)(z2+1)2.
  4. s′(1)=−2sin⁡(1)−4cos⁡(1)e1≈−1.414 feet per second.
  5. p′(3)=30 and q′(3)=138.

2.4 Derivatives of other trigonometric functions
2.4.1 Derivatives of the cotangent, secant, and cosecant functions

Activity 2.4.2.

Answer.
  1. All real numbers x such that x≠π2+kπ, where k=±1,±2,….
  2. h′(x)=sin⁡(x)cos2⁡(x).
  3. h′(x)=sec⁡(x)tan⁡(x).
  4. h and h′ have the same domain: all real numbers x such that x≠π2+kπ, where k=0,±1,±2,….

Activity 2.4.3.

Answer.
  1. All real numbers x such that x≠kπ, where k=0,±1,±2,….
  2. h′(x)=−cos⁡(x)sin2⁡(x).
  3. h′(x)=−csc⁡(x)cot⁡(x).
  4. p and p′ have the same domain: all real numbers x such that x≠kπ, where k=0,±1,±2,….

Activity 2.4.4.

Answer.
  1. m=f′(π3)=103+43.
  2. p′(π4)=π2162+2π2+π2−1.
  3. h′(t)=(t2+1)sec2⁡(t)−2ttan⁡(t)(t2+1)2+2etsin⁡(t)−2etcos⁡(t).
  4. g′(r)=rsec⁡(r)tan⁡(r)+sec⁡(r)−rln(5)sec⁡(r)5r.
  5. s′(2)=15cos⁡(2)−15sin⁡(2)e2≈−2.69 inches per second.

2.5 The chain rule
2.5.1 The chain rule

Activity 2.5.2.

Answer.
  1. h′(x)=−4x3sin⁡(x4).
  2. h′(x)=sec2⁡(x)2tan⁡(x).
  3. h′(x)=2sin⁡(x)ln⁡(2)cos⁡(x).
  4. h′(x)=−5cot4⁡(x)csc2⁡(x).
  5. h′(x)=9(sec⁡(x)+ex)8(sec⁡(x)tan⁡(x)+ex).

2.5.2 Using multiple rules simultaneously

Activity 2.5.3.

Answer.
  1. p′(r)=4(6r5+2er)2r6+2er.
  2. m′(v)=−3v2sin⁡(v2)sin⁡(v3)+2vcos⁡(v3)cos⁡(v2).
  3. h′(y)=(e4y+1)[−10sin⁡(10y)]−cos⁡(10y)[4e4y](e4y+1)2.
  4. s′(z)=2z2sec⁡(z)ln⁡(2)[z2sec⁡(z)tan⁡(z)+sec⁡(z)⋅2z].
  5. c′(x)=cos⁡(ex2)[ex2⋅2x].

Activity 2.5.4.

Answer.
  1. y−2=14(x−0).
  2. v(1)=s′(1)=−38 inches per second; the particle is moving left at the instant t=1.
  3. P′(1000)=30e−0.0323(−0.0000323)≈−0.000938 inches of mercury per foot.
  4. C′(2)=−10; D′(−1)=−20.

2.6 Derivatives of Inverse Functions
2.6.2 The derivative of the natural logarithm function

Activity 2.6.2.

Answer.
  1. h′(x)=x+2xln⁡(x).
  2. p′(t)=(et+1)1t−ln⁡(t)⋅et(et+1)2.
  3. s′(y)=1cos⁡(y)+2⋅(−sin⁡(y)).
  4. z′(x)=sec2⁡(ln⁡(x))⋅1x.
  5. m′(z)=1ln⁡(z)⋅1z.

2.6.3 Inverse trigonometric functions and their derivatives

Activity 2.6.3.

Answer.
  1. tan⁡(r(x))=x.
  2. r′(x)=cos2⁡(r(x)).
  3. r′(x)=cos2⁡(arctan⁡(x)).
  4. With θ=arctan⁡(x),
  5. cos⁡(arctan⁡(x))=11+x2.
  6. r′(x)=11+x2.

Activity 2.6.4.

Answer.
  1. f′(x)=[x3⋅11+x2+arctan⁡(x)⋅3x2]+[ex⋅1x+ln⁡(x)⋅ex].
  2. p′(t)=2tarcsin⁡(t)ln⁡(2)[t⋅11−t2+arcsin⁡(t)⋅1].
  3. h′(z)=27(arcsin⁡(5z)+arctan⁡(4−z))26[11−(5z)2⋅5+11+(4−z)2⋅(−1)].
  4. s′(y)=−1y2.
  5. m′(v)=1sin2⁡(v)+1⋅[2sin⁡(v)cos⁡(v)].
  6. g′(w)=11+(ln⁡(w)1+w2)2⋅[(1+w2)1w−ln⁡(w)⋅2w(1+w2)2]

2.7 Derivatives of Functions Given Implicitly
2.7.1 Implicit Differentiation

Activity 2.7.2.

Answer.
  1. The graph of the curve fails the vertical line test.
  2. dydx=15y4−15y2+4.
  3. y=−16x+1.
  4. (1.418697,0.543912), (−1.418697,−0.543912), (−3.63143,1.64443), and (3.63143,−1.64443).

Activity 2.7.3.

Answer.
  1. Horizontal at x≈0.42265, thus (0.42265,−1.05782);(0.42265,0.229478);(0.42265,0.770522);(0.42265,2.05782). There are four more points where x≈1.57735.
  2. When y=12,1±52, so one point is (2.21028,12).
  3. y−1=12(x−1).

Activity 2.7.4.

Answer.
  1. dydx(−3y2−6x)=6y−3x2 and the tangent line has equation y−3=1(x+3).
  2. dydx=3x2+1cos⁡(y)+1 and the tangent line has equation y=12x.
  3. dydx=3e−xy−3xye−xy3x2e−xy+2y and the tangent line is y−1=0.234950(x−0.619061).

2.8 Using Derivatives to Evaluate Limits
2.8.1 Using derivatives to evaluate indeterminate limits of the form 00.

Activity 2.8.2.

Answer.
  1. limx→0ln⁡(1+x)x=1.
  2. limx→πcos⁡(x)x=−1π.
  3. limx→12ln⁡(x)1−ex−1=−2.
  4. limx→0sin⁡(x)−xcos⁡(2x)−1=0.

Activity 2.8.3.

Answer.
  1. limx→2f(x)g(x)=18.
  2. limx→2p(x)q(x)=1.
  3. limx→2r(x)s(x)<0.

2.8.2 Limits involving ∞

Activity 2.8.4.

Answer.
  1. limx→∞xln⁡(x)=∞.
  2. limx→∞ex+x2ex+x2=12.
  3. limx→0+ln⁡(x)1x=0.
  4. limx→π2−tan⁡(x)x−π2=−∞.
  5. limx→∞xe−x=0.

3 Using Derivatives
3.1 Using derivatives to identify extreme values
3.1.1 Critical numbers and the first derivative test

Activity 3.1.2.

Answer.
  1. x=−4 or x=1.
  2. g has a local maximum at x=−4 and neither a max nor min at x=1.
  3. g does not have a global minimum; it is unclear (at this point in our work) if g increases without bound, so we can’t say for certain whether or not g has a global maximum.
  4. limx→∞g′(x)=∞.
  5. A possible graph of g is the following.

3.1.2 The second derivative test

Activity 3.1.3.

Answer.
  1. x=−1 is an inflection point of g.
  2. g is concave up for x<−1, concave down for −1<x<2, and concave down for x>2.
  3. g has a local minimum at x=−1.67857351.
  4. g is a degree 5 polynomial.

Activity 3.1.4.

Answer.
  1. In the graph below, h(x)=x2+cos⁡(3x) is given in dark blue, while h(x)=x2+cos⁡(1.6x) is shown in light blue.
  2. If 2k2>1, then the equation cos⁡(kx)=2k2 has no solution. Hence, whenever k2<2, or k<2≈1.414, it follows that the equation cos⁡(kx)=2k2 has no solutions x, which means that h″(x) is never zero (indeed, for these k-values, h″(x) is always positive so that h is always concave up). On the other hand, if k≥2, then 2k2≤1, which guarantees that cos⁡(kx)=2k2 has infinitely many solutions, due to the periodicity of the cosine function. At each such point, h″(x)=2−k2cos⁡(kx) changes sign, and therefore h has infinitely many inflection points whenever k≥2.
  3. To see why h can only have a finite number of critical numbers regardless of the value of k, consider the equation
    0=h′(x)=2x−ksin⁡(kx),
    which implies that 2x=ksin⁡(kx). Since −1≤sin⁡(kx)≤1, we know that −k≤ksin⁡(kx)≤k. Once |x| is sufficiently large, we are guaranteed that |2x|>k, which means that for large x, 2x and ksin⁡(kx) cannot intersect. Moreover, for relatively small values of x, the functions 2x and ksin⁡(kx) can only intersect finitely many times since ksin⁡(kx) oscillates a finite number of times. This is why h can only have a finite number of critical numbers, regardless of the value of k.

3.2 Using derivatives to describe families of functions
3.2.1 Describing families of functions in terms of parameters

Activity 3.2.2.

Answer.
  1. p has two critical numbers (x=±a3) whenever a>0 and no critical numbers when a<0.
  2. When a<0, p is always increasing and has no relative extreme values. When a>0, p has a relative maximum at x=−a3 and a relative minimum at x=+a3.
  3. p is CCD for x<0 and p is CCU for x>0, making x=0 an inflection point.

Activity 3.2.3.

Answer.
  1. h is an always increasing function.
  2. h is always concave down.
  3. limx→∞a(1−e−bx)=a, and limx→∞a(1−e−bx)=−∞.
  4. If b is large and x is close to zero, h′(x) is relatively large near x=0, and the curve’s slope will quickly approach zero as x increases. If b is small, the graph is less steep near x=0 and its slope goes to zero less quickly as x increases.

Activity 3.2.4.

Answer.
  1. L is an always increasing function.
  2. L is concave up for all t<−1kln⁡(1c) and concave up for all other values of t.
  3. limt→∞A1+ce−kt=A, and
    limt→∞A1+ce−kt=0.
  4. The inflection point on the graph of L is (−1kln⁡(1c),A2).

3.3 Global Optimization
3.3.1 Global Optimization

Activity 3.3.2.

Answer.
  1. x=±2≈±1.414.
  3. On [−2,3], g has a global maximum at x=3 and a global minimum at x=2.
  4. On [−2,2], g has a global maximum at x=−2 and a global minimum at x=2.
  5. On [−2,1], g has a global maximum at x=−2 and a global minimum at x=1.

Activity 3.3.3.

Answer.
  1. Absolute maximum: e−1; absolute minimum: 0.
  2. Absolute maximum: 2; absolute minimum: −1.
  3. Absolute maximum: 9.8; absolute minimum: 8.
  4. Absolute minimum 3; no absolute maximum.
  5. Absolute minimum 0; absolute maximum 1ae−1.
  6. Absolute minimum b−1; no absolute maximum.

3.3.2 Moving toward applications

Activity 3.3.4.

Answer.
  2. V(x)=x(10−2x)(15−2x)=4x3−50x2+150x.
  3. 1≤x≤3.
  4. x=25±576≈6.371459426,1.961873908.
    • V(1.961873908)=132.0382370
    • V(1)=104
    • V(3)=108
  6. Absolute maximum: 132.0382370; absolute minimum: 104.

3.4 Applied Optimization
3.4.1 More applied optimization problems

Activity 3.4.2.

Answer.
  1. Let the can have radius r and height h.
  2. V=πr2h; S=2πr2+2πrh; C=2πr2⋅0.027+2πrh⋅0.015.
  3. C(r)=0.054πr2+0.481r, r>0.
  4. r=0.480.108π3≈1.12259; h≈4.041337; minimum cost C(1.12259)≈0.64137.

Activity 3.4.3.

Answer.
The absolute minimum time the hiker can achieve is 0.99302 hours, which is attained by hiking about 2.2 km from P to Q and then turning into the woods for the remainder of the trip.

Activity 3.4.4.

Answer.
Maximum area: A(53)=50093≈96.225. Maximum perimeter: P(1)=52. At x=82−13 the absolute maximum of combined perimeter and area occurs.

Activity 3.4.5.

Answer.
A(1.19606)≈2.2018 is the absolute maximum cross-sectional area, which leads to the absolute maximum volume.

3.5 Related Rates
3.5.1 Related Rates Problems

Activity 3.5.2.

Answer.
  2. r=34h.
  3. V=316πh3.
  4. dVdt=916πh2dhdt.
  5. dhdt|h=3=6481π≈0.2515 feet per minute.
  6. Most rapidly when h=3.

Activity 3.5.3.

Answer.
  2. dhdt=4000sec2⁡(θ)dθdt.
  3. hdhdt=zdzdt.
  4. dzdt|h=3000=360 feet/sec; dθdt|h=3000=12125 radians per second.
  5. greater.

Activity 3.5.4.

Answer.
  2. 3s=2x.
  3. 3dsdt=2dxdt.
  4. dsdt|x=8=2 feet per second.
  5. at a constant rate.
  6. Let y represent the location of the tip of the shadow; dydt=5 feet/sec.

Activity 3.5.5.

Answer.
Let x denote the position of the ball at time t and z the distance from the ball to first base, as pictured below.
dzdt|x=45=1005≈44.7214 feet/sec.
Let r be the runner’s position at time t and let s be the distance between the runner and the ball, as pictured.
dsdt|x=45=43017≈104.2903 feet/sec.

4 The Definite Integral
4.1 Determining distance traveled from velocity
4.1.1 Area under the graph of the velocity function

Activity 4.1.2.

Answer.
  1. A=(v(0.0)⋅0.5+v(0.5)⋅0.5+v(1.0)⋅0.5+v(1.5)⋅0.5=(1.500⋅0.5+1.9375⋅0.5+2.000⋅0.5+2.0625⋅0.5=(3.75
    Thus, D≈3.75 miles.
  2. Using 8 rectangles of width 0.25, D≈3.875.
  3. s(t)=18t4−12t3+34t2+32t.
  4. s(2)−s(0)=1824−1223+3422+322=4.

4.1.2 Two approaches: area and antidifferentiation

Activity 4.1.3.

Answer.
  1. On (0,1), s is increasing because velocity is positive.
  2. s(t)=32t−16t2.
  3. s(1)−s(12)=4.
  4. A=4 feet is the total distance the ball traveled vertically on [12,1].
  5. s(1)−s(0)=16 is the vertical distance the ball traveled on the interval [0,1]. Equivalently, the area between the velocity curve and the t-axis on [0,1] is A=16 feet.
  6. s(2)−s(0)=0, so the ball has zero change in position on the interval [0,2].

4.1.3 When velocity is negative

Activity 4.1.4.

Answer.
  1. Total distance traveled is 2; change in position is 0.
  2. 0<t<1 and 4<t<8.
  3. s(8)−s(0)=5 m, while the distance traveled on [0,8] is D=13, and thus these two quantities are different.
  4. See the figure below.

4.2 Riemann Sums
4.2.1 Sigma Notation

Activity 4.2.2.

Answer.
  1. 65
  2. 32
  3. 3+7+11+15+⋯+27=∑k=174k−1.
  4. 4+8+16+32+⋯+256=∑i=282i.
  5. ∑i=1612i=6364.

4.2.2 Riemann Sums

Activity 4.2.3.

Answer.
  2. L4=31148≈6.47917, R4=33548≈6.97917, and M4=63796≈6.63542.
  3. L4+M42=64696≠63796=M4.
  4. Ln is an under-estimate; Rn is an over-estimate.

4.2.3 When the function is sometimes negative

Activity 4.2.4.

Answer.
  1. M5=−3625=−1.44
  2. The change in position is approximately −1.44 feet.
  3. D≈2.336.
  4. −43 is the object’s total change in position on [1,5].

4.3 The Definite Integral
4.3.1 The definition of the definite integral

Activity 4.3.2.

Answer.
  1. ∫013xdx=32.
  2. ∫−14(2−2x)dx=−5.
  3. ∫−111−x2dx=π2.
  4. ∫−34g(x)dx=3π4−32.

4.3.2 Some properties of the definite integral

Activity 4.3.3.

Answer.
  1. ∫52f(x)dx=−2.
  2. ∫05g(x)dx=3.
  3. ∫05(f(x)+g(x))dx=2.
  4. ∫25(3x2−4x3)dx=−492.
  5. ∫50(2x3−7g(x))dx=−5832.

4.3.3 How the definite integral is connected to a function’s average value

Activity 4.3.4.

Answer.
  1. y=v(t)=4−(t−2)2 is the top half of the circle (t−2)2+y2=4, which has radius 2 and is centered at (2,0).
  2. ∫04v(t)dt=2π.
  3. The object moved 2π meters in 4 minutes.
  4. vAVG[0,4]=π2, meters per minute,.
  5. The height of the rectangle is the average value of v, vAVG[0,4]=π2≈1.57.
  6. D=2π.

4.4 The Fundamental Theorem of Calculus
4.4.1 The Fundamental Theorem of Calculus

Activity 4.4.2.

Answer.
  1. ∫−14(2−2x)dx=−5.
  2. ∫0π2sin⁡(x)dx=1.
  3. ∫01exdx=e−1.
  4. ∫−11x5dx=0.
  5. ∫02(3x3−2x2−ex)dx=233−e2.

4.4.2 Basic antiderivatives

Activity 4.4.3.

Answer.
given function, f(x) antiderivative, F(x)  
k, (k≠0) kx
xn, n≠−1 1n+1xn+1
1x, x>0 ln⁡(x)
sin⁡(x) −cos⁡(x)
cos⁡(x) sin⁡(x)
sec⁡(x)tan⁡(x) sec⁡(x)
csc⁡(x)cot⁡(x) −csc⁡(x)
sec2⁡(x) tan⁡(x)
csc2⁡(x) −cot⁡(x)
ex ex
ax (a>1) 1ln⁡(a)ax
11+x2 arctan⁡(x)
11−x2 arcsin⁡(x)
  1. ∫01(x3−x−ex+2)dx=114−e.
  2. ∫0π/3(2sin⁡(t)−4cos⁡(t)+sec2⁡(t)−π)dt=1−3−π23.
  3. ∫01(x−x2)dx=13.

4.4.3 The total change theorem

Activity 4.4.4.

Answer.
  1. The person burned exactly 4003 calories in the first 10 minutes of the workout.
  2. C(40)−C(0)=∫040C′(t)dt=∫040c(t)dt is the total calories burned on [0,40].
  3. The exact average rate at which the person burned calories on 0≤t≤40 is
    cAVG[0,40]=140−0∫040c(t)dt=140⋅17003=1700120≈14.17 cal/min.
  4. One time at which the instantaneous rate at which calories are burned equals the average rate on [0,40] is t=53(6−6)≈5.918.

5 Evaluating Integrals
5.1 Constructing Accurate Graphs of Antiderivatives
5.1.1 Constructing the graph of an antiderivative

Activity 5.1.2.

Answer.
  1. F is increasing on (0,2) and (5,7); F is decreasing on (2,5).
  2. F is concave up on (0,1), (4,6); concave down on (1,3), (6,7); neither on (3,4).
  3. A relative maximum at x=2; a relative minimum at x=5.
  4. F(1)=−12; F(2)=π4−12; F(3)=π4−1; F(4)=π4−2; F(5)=π4−52; F(6)=π2−52; F(7)=3π4−52; F(8)=3π4−52; and F(−1)=−1.
  5. Use the function values found in (d) and the earlier information regarding the shape of F.
  6. G(x)=F(x)+1.

5.1.2 Multiple antiderivatives of a single function

Activity 5.1.3.

Answer.
  2. H(x)=−cos⁡(x)+2.

5.1.3 Functions defined by integrals

Activity 5.1.4.

Answer.
  1. A is increasing on (0,1.5), (4,6); A is decreasing on (1.5,4).
  2. A is concave up on (0,1) and (3,5); A is concave down on (1,3) and (5,6).
  3. At x=1.5, A has a relative maximum; A has a relative minimum at x=4.
  4. A(0)=−12; A(1)=0; A(2)=0; A(3)=−2; A(4)=−3.5, A(5)=−2, A(6)=−0.5.
  5. Use your work in (a)-(d) appropriately.
  6. B(x)=A(x)+12.

5.2 The Second Fundamental Theorem of Calculus
5.2.1 The Second Fundamental Theorem of Calculus

Activity 5.2.2.

Answer.
  1. A′(x)=f(x).
  2. A(1)=−π4.
  3. A is increasing wherever f is positive; A is CCU wherever f is increasing. A(2)=0, A(3)=−0.5, A(4)=−1.5, A(5)=−2, A(6)=−2+π4, and A(7)=−2+π2.
  4. F and A differ by the constant π4−12.
  5. B and C have the same shape as A and F, and differ from A by a constant. Observe that B(3)=0 and C(1)=0.

5.2.2 Understanding Integral Functions

Activity 5.2.3.

Answer.
  1. See the plot at below left.
  2. F′=f.
  3. F is increasing for all x>0; F is decreasing for x<0
  4. F is CCU on −1<x<1 and CCD for x<−1 and x>1.
  5. F(5)≈1.64038; F(10)≈2.35973.
  6. See the graph at below right.

5.2.3 Differentiating an Integral Function

Activity 5.2.4.

Answer.
  1. ddx[∫4xet2dt]=ex2.
  2. ∫−2xddt[t41+t4]dt=x41+x4−1617.
  3. ddx[∫x1cos⁡(t3)dt]=−cos⁡(x3).
  4. ∫3xddt[ln⁡(1+t2)]dt=ln⁡(1+x2)−ln⁡(10).
  5. ddx[∫4x3sin⁡(t2)dt]=sin⁡(x6)⋅3x2.

5.3 Integration by Substitution
5.3.1 Reversing the Chain Rule: First Steps

Activity 5.3.2.

Answer.
  1. ∫sin⁡(8−3x)dx=−13(−cos⁡(8−3x))+C.
  2. ∫sec2⁡(4x)dx=14tan⁡(4x)+C.
  3. ∫111x−9dx=111ln⁡|11x−9|+C.
  4. ∫csc⁡(2x+1)cot⁡(2x+1)dx=−12csc⁡(2x+1)+C.
  5. ∫11−16x2dx=14arcsin⁡(4x)+C
  6. ∫5−xdx=−1ln⁡(5)5−x+C.

5.3.2 Reversing the Chain Rule: u-substitution

Activity 5.3.3.

Answer.
  1. ∫x25x3+1dx=115ln⁡(5x3+1)+C.
  2. ∫exsin⁡(ex)dx=−cos⁡(ex)+C.
  3. ∫cos⁡(x)xdx=2sin⁡(x)+C.

5.3.3 Evaluating Definite Integrals via u-substitution

Activity 5.3.4.

Answer.
  1. ∫x=1x=2x1+4x2dx=18(ln⁡(17)−ln⁡(5)).
  2. ∫01e−x(2e−x+3)9dx=−120(2e−1+3)10+120(2e0+3)10.
  3. ∫2/π4/πcos⁡(1x)x2dx=1−22.

5.4 Integration by Parts
5.4.1 Reversing the Product Rule: Integration by Parts

Activity 5.4.2.

Answer.
  1. ∫te−tdt=−te−t−e−t+c.
  2. ∫4xsin⁡(3x)dx=−43xcos⁡(3x)+49sin⁡(3x)+c.
  3. ∫zsec2⁡(z)dz=ztan⁡(z)+ln⁡|cos⁡(z)|+c.
  4. ∫xln⁡(x)dx=12x2ln⁡(x)−14x2+c.

5.4.2 Some Subtleties with Integration by Parts

Activity 5.4.3.

Answer.
  1. ∫arctan⁡(x)dx=xarctan⁡(x)−12ln⁡(|1+x2|)+c.
  2. ∫ln⁡(z)dz=zln⁡(z)−z+c.
  3. ∫t3sin⁡(t2)dt=12(−t2cos⁡(t2)+sin⁡(t2)).
  4. ∫s5es3ds=13(s3es3−es3)+c.
  5. ∫e2tcos⁡(et)dt=etsin⁡(et)+cos⁡(et)+c.

5.4.3 Using Integration by Parts Multiple Times

Activity 5.4.4.

Answer.
  1. ∫x2sin⁡(x)dx=−x2cos⁡(x)+2xsin⁡(x)+2cos⁡(x)+c.
  2. ∫t3ln⁡(t)dt=14t4ln⁡(t)−116t4+c.
  3. ∫ezsin⁡(z)dz=−12ezcos⁡(z)+12ezsin⁡(z)+c.
  4. ∫s2e3sds=13s2e3s−29se3s+227e3s+c.
  5. ∫tarctan⁡(t)dt=12t2arctan⁡(t)−12t−12arctan⁡(t)+c.

5.5 Other Options for Finding Algebraic Antiderivatives
5.5.1 The Method of Partial Fractions

Activity 5.5.2.

Answer.
  1. ∫1x2−2x−3dx=14ln⁡|x−3|−14ln⁡|x+1|+C.
  2. ∫x2+1x3−x2dx=−ln⁡|x|+x−1+2ln⁡|x−1|+C.
  3. ∫x−2x4+x2dx=ln⁡|x|+2x−1−12ln⁡|1+x2|+2arctan⁡(x)+C.

5.5.2 Using an Integral Table

Activity 5.5.3.

Answer.
  1. ∫x2+4dx=x2x2+4+2ln⁡|x+x2+4|+C.
  2. ∫xx2+4dx=x2+4+C.
  3. ∫216+25x2dx=25ln⁡|5x+16+25x2|+C.
  4. ∫1x249−36x2dx=−49−36x249x+C.

5.6 Numerical Integration
5.6.1 The Trapezoid Rule

Activity 5.6.2.

Answer.
  1. ∫121x2dx=12.
  2. The table below gives values of the trapezoid rule and corresponding errors for different n-values.
    n Tn ET,n
    4 0.50899 0.00899
    8 0.50227 0.00227
    16 0.50057 0.00057
  3. The table below gives values of the midpoint rule and corresponding errors for different n-values.
    n Mn EM,n
    4 0.49555 −0.00445
    8 0.49887 −0.00113
    16 0.49972 −0.00028
  4. The trapezoid rule overestimates; the midpoint rule underestimates.
  5. f(x)=1x2 is concave up on [1,2].

5.6.3 Simpson’s Rule

Activity 5.6.3.

Answer.
  1. Plot the data.
  2. ∫01.8v(t)dt.
  3. L3=165.6 ft R3=105.6 ft T3=135.6 ft.
    R3 and T3 are underestimates.
  4. M3=143.4 ft; overestimate.
  5. S6=140.8 ft.
  6. Simpson’s rule gives the best approximation of the distance traveled, ∫01.8v(t)dt≈140.8 ft, which leads to AV[0,1.8]≈140.81.8≈78.22 ft/sec.

5.6.4 Overall observations regarding Ln, Rn, Tn, Mn, and S2n.

Activity 5.6.4.

Answer.
  1. For L1 and T1:
    Table 5.6.10. Left and Trapezoid rules.
    f g h
    L1=2 L1=2 L1=2
    R1=1 R1=1 R1=1
    The values of L1 and R1 are the same for all three.
  2. For the M1,
    Table 5.6.11. Midpoint Rule.
    f g h
    M1=74 M1=158 M1=3116
  3. For T1 and S2,
    Table 5.6.12. Trapezoid and Simpson’s Rule.
    f g h
    T1=32 T1=32 T1=32
    S2=53≈1.6667 S2=74 S2=4324≈1.79167
  4. ∫01f(x)dx=53∫01g(x)dx=74∫01h(x)dx=95
  5. Left endpoint rule results are overestimates; right endpoint rules are underestimates; midpoint rules are overestimates; trapezoid rules are underestimates. Simpson’s rule is exact for both f and g, while a slight underestimate of ∫01h(x)dx.

6 Using Definite Integrals
6.1 Using Definite Integrals to Find Area and Length
6.1.1 The Area Between Two Curves

Activity 6.1.2.

Answer.
  1. A=∫016(x−14x)dx=323.
  2. A=∫−20/320/3((12−2x2)−(x2−8))dx160533≈68.853.
  3. A=∫0π4(cos⁡(x)−sin⁡(x))dx=2−1.
  4. The left-hand region has area
    A1=∫1−520((x3−x)−x2)dx=13−5524≈0.075819.
    The right-hand region has area
    A2=∫01+52(x2−(x3−x))dx=13+5524≈1.007514.

6.1.2 Finding Area with Horizontal Slices

Activity 6.1.3.

Answer.
  1. A=∫y=−2y=2(6−2y2−y2)dy=82≈11.314.
  2. A=∫y=−1y=1(2−2y2−(1−y2))dy=43.
  3. A=∫y=0y=1(2−y−y)dy=56
  4. A=∫03(y−(y2−2y))dy=92.

6.1.3 Finding the length of a curve

Activity 6.1.4.

Answer.
  1. L≈2.95789.
  2. L=∫−2244−x2dx=2π.
  3. L=∫011+e6x(9x2+6x+1)dx≈20.1773.
  4. We will usually have to estimate the value of ∫ab1+f′(x)2dx using computational technology.
  5. Approximately (14.9165,f(14.9165))=(14.9165,23.2502).

6.2 Using Definite Integrals to Find Volume
6.2.1 The Volume of a Solid of Revolution

Activity 6.2.2.

Answer.
  1. V=∫04π(x)2dx=∫04πxdx=8π.
  2. V=∫04π(4−(x)2)dx=∫04π(4−x)dx=8π.
  3. V=∫01π(x−x6)dx=514π.
  4. V=∫−33π((x2+4)2−(2x2+1)2)dx=13635π.
  5. V=∫02πy4dy=325π.

6.2.2 Revolving about the y-axis

Activity 6.2.3.

Answer.
  1. V=∫02πy4Δdy.
  2. V=∫02π(16−y4)dy.
  3. V=int02π(4x2−x6)dx.
  4. V=∫022π(y2/3−y2/4)dy.
  5. V=∫03π((y+1)2−(y−1)4)dy.

6.2.3 Revolving about horizontal and vertical lines other than the coordinate axes

Activity 6.2.4.

Answer.
  1. V=∫02π((2x+2)2−(x3+2)2)dx=421(21+82)π≈19.336.
  2. V=∫02π((4−x3)2−(4−2x)2)dx=(8−32221)π≈18.3626.
  3. V=∫022π((y1/3+1)2−(12y+1)2)dy=215(15+82)π≈11.022.
  4. V=∫022π((5−12y)2−(5−y1/3)2)dy=215(75−82)π≈26.677.

6.3 Density, Mass, and Center of Mass
6.3.1 Density

Activity 6.3.2.

Answer.
  1. M=10−10e−2≈8.64665 grams.
    1. V=∫05π(4−45x)2dx=80π3≈83.7758m3.
    2. M=64000π3≈67020.6433kg.
    3. M=∫05(400+2001+x2)⋅π(4−45x)2dx=128π(2653+24arctan⁡(5)−5ln⁡(26))≈42224.8024kg
  3. b≈3.0652.

6.3.2 Weighted Averages

Activity 6.3.3.

Answer.
  1. x―=x1+x22=3.
  2. x―=x1+x2+x3+x44=3.
  3. x―=x1+x2+x3+x44=2.75.
  4. x―=2x1+3x2+1x3+1x47=167.
  5. x―=2x1+3x2+1x3+1x47=177.
  6. x―=2x1+3x2+2x3+1x48=208.
  7. Answers will vary.
  8. If we have an existing arrangement and balancing point, moving one of the locations to the left will move the balancing point to the left; similarly, moving one of the locations to the right will move the balancing point to the right. If instead we add weight to an existing location, if that location is left of the balancing point, the balancing point will move left; the behavior is similar if on the right.

6.3.3 Center of Mass

Activity 6.3.4.

Answer.
  1. M=∫0204+0.1xdx=100 g.
  2. Greater than 10.
  3. x―=∫020x(4+0.1x))dx∫0204+0.1xdx=323.
  4. 5 g/cm.
  5. Slightly to the right of the center of mass for ρ(x).
  6. x―=∫020x4e0.020732xdx∫0204e0.020732xdx≈10.6891,

6.4 Physics Applications: Work, Force, and Pressure
6.4.1 Work

Activity 6.4.2.

Answer.
  1. W=∫02000.3(200−h)dh=6000 foot-pounds.
  2. W=∫0100(40−0.1h)dh=3500foot-pounds.
  3. BAVG[0,100]≈25.9798 pounds.
  4. For the given spring,
    1. k=15.
    2. W=∫0115xdx=152 foot-pounds.
    3. W=∫11.515xdx=9.375 foot-pounds.

6.4.2 Work: Pumping Liquid from a Tank

Activity 6.4.3.

Answer.
  1. W=∫239.81⋅4000π⋅xdx=308 190newton-meters.
  2. W=∫3862.4π(100−x2)(x+5)dx≈673593foot-pounds.
  3. W=∫1362.4(50−252x)xdx=5720foot-pounds.

6.4.3 Force due to Hydrostatic Pressure

Activity 6.4.4.

Answer.
  1. F=∫x=0x=50(6240x)dx=7 800 000 pounds .
  2. F=∫x=10x=30124.8(x−10)900−x2dx=800 244 pounds .
  3. F=∫x=1x=462.4(x−1)(5−1.25x)dx=351 pounds .

6.5 Improper Integrals
6.5.1 Improper Integrals Involving Unbounded Intervals

Activity 6.5.2.

Answer.
    1. ∫1101xdx=ln⁡(10) ∫110001xdx=ln⁡(1000) ∫11000001xdx=ln⁡(100000)
    2. ∫1b1xdx=ln⁡(b).
    3. limb→∞∫1b1xdx=limb→∞ln⁡(b)=∞
    1. ∫1101x3/2dx=2−210 ∫110001x3/2dx=2−21000 ∫11000001x3/2dx=2−2100000
    2. ∫1b1x3/2dx=2−2b.
    3. limb→∞∫1b1x3/2dx=limb→∞(2−2b)=2
  3. Both graphs have a vertical asymptote at x=0 and for both graphs, the x-axis is a horizontal asymptote. However, the graph of y=1x3/2 will ’’approach the x-axis faster’’ than the graph of y=1x.
  4. The area bounded by the graph of y=1x, the x-axis, and the vertical line x=1 is infinite or unbounded. However, The area bounded by the graph of y=1x3/2, the x-axis, and the vertical line x=1 is equal to 2.

6.5.2 Convergence and Divergence

Activity 6.5.3.

Answer.
  1. ∫1∞1x2dx=1
  2. ∫0∞e−x/4dx=4
  3. ∫2∞9(x+5)2/3dx=∞
  4. ∫4∞3(x+2)5/4dx=1261/4
  5. ∫0∞xe−x/4dx=16
  6. If 0<p<1, ∫1∞1xpdx diverges, while if p>1, the integral converges.

6.5.3 Improper Integrals Involving Unbounded Integrands

Activity 6.5.4.

Answer.
  1. ∫011x1/3dx=32
  2. ∫02e−xdx=1−e−2
  3. ∫1414−xdx=23
  4. ∫−221x2dx diverges.
  5. ∫0π/2tan⁡(x)dx=∞
  6. ∫0111−x2dx=π2

7 Differential Equations
7.1 An Introduction to Differential Equations
7.1.1 What is a differential equation?

Activity 7.1.2.

Answer.
  1. Let P be the population t the time in years; dPdt=0.0125P.
  2. Let m be the mass t the time in days; dmdt=−0.056m.
  3. Let B be the balance t be time in years; dBdt=0.04B−1000.
  4. Let t be time in minutes H the temperature of the hot chocolate; dHdt=−0.1(H−70).
  5. Let t be time in minutes and H the temperature of the soda;
    dHdt=0.1(70−H)=−0.1(H−70).

7.1.2 Differential equations in the world around us

Activity 7.1.3.

Answer.
  1. For the skydiver:
    dvdt|(v=0.5)≈1.5dvdt|(v=1)≈1.2dvdt|(v=1.5)≈0.9dvdt|(v=2)≈0.6dvdt|(v=2.5)≈0.3
  2. For the meteorite:
    dvdt|(v=3.5)≈−0.3dvdt|(v=4)≈−0.6dvdt|(v=4.5)≈−0.9dvdt|(v=5)≈−1.2
    A graph of the points from parts (a) and (b) is shown in the following diagram:
  3. dvdt=−0.6v+1.8.
  4. The rate of change of velocity with respect to time is a linear function of velocity.
  5. 0<v<3.
  6. 3<v<5.
  7. v=3.

7.1.3 Solving a differential equation

Activity 7.1.4.

Answer.
  1. v(t)=1.5t−0.25t2 is not a solution to the given DE.
  2. v(t)=3+2e−0.5t is a solution to the given DE.
  3. v(t)=3 is a solution to the given DE.
  4. v(t)=3+Ce−0.5t is a solution to the given DE for any choice of C.

7.2 Qualitative behavior of solutions to DEs
7.2.1 Slope fields

Activity 7.2.2.

Answer.
  1. When y<4, y is an increasing function of t. When y>4, y is a decreasing function of t.
  4. dydt=2(−12e−t/2)=−e−t/2
    and
    −12(y−4)=−12(4+2e−t/2)=−e−t/2
    In addition, y(0)=4+2e0=6.
  5. A constant function.

7.2.2 Equilibrium solutions and stability

Activity 7.2.3.

Answer.
  1. When y<0 and when y>4, y is a decreasing function of t. When 0<y<4, y is a increasing function of t.
  2. y=0 and y=4.
  5. y=4 is stable; y=0 is unstable.
  6. Tend to 4.
  7. Figure 7.2.11 is for an ustable equilibrium; Figure 7.2.12 is for a stable equilibrium.

7.3 Euler’s method
7.3.1 Euler’s Method

Activity 7.3.2.

Answer.
  1. ti yi dy/dt Δy
    0 0 −1 −0.2
    0.2 −0.2 −0.6 −0.12
    0.4 −0.32 −0.2 −0.04
    0.6 −0.36 0.2 0.04
    0.8 −0.32 0.6 0.12
    1.0 −0.2 1 0.2
  2. y=t2−t, with errors e1=0.04, e2=0.08, e3=0.12, e4=0.16, e5=0.2.
  3. If we first think about how y1 is generated for the initial value problem dydt=f(t)=2t−1, y(0)=0, we see that y1=y0+Δt⋅f(t0). Since y0=0, we have y1=Δt⋅f(t0). From there, we know that y2 is given by y2=y1+Δtf(t1). Substituting our earlier result for y1, we see that y2=Δt⋅f(t0)+Δtf(t1). Continuing this process up to y5, we get
    y5=Δt⋅f(t0)+Δtf(t1)+Δtf(t2)+Δtf(t3)+Δtf(t4)
    This is precisely the left Riemann sum with five subintervals for the definite integral ∫01(2t−1) dt.
  4. Solutions to this differential equation all differ by only a constant.

Activity 7.3.3.

Answer.
  2. y=0 or y=6; y=0 is unstable, y=6 is stable.
  3. The solution will tend to y=6.
  4. ti yi dy/dt Δy
    0.0 1.0000 5.0000 1.0000
    0.2 2.0000 8.0000 1.6000
    0.4 3.6000 8.6400 1.7280
    0.6 5.3280 3.5804 0.7161
    0.8 6.0441 −0.2664 −0.0533
    1.0 5.9908 0.0551 0.0110
  5. The value of yi=6 for every value of i.

7.4 Separable differential equations
7.4.1 Solving separable differential equations

Activity 7.4.2.

Answer.
  1. dPdt=0.03P
  2. P=Ce0.03t.
  3. P=10000e0.03t.
  4. The doubling time is t=ln⁡(2)0.03≈23.105 years.
  5. The doubling time is t=1kln⁡(2).

Activity 7.4.3.

Answer.
  1. k=130
  2. T=75+Ce−t/30
  3. The temperature of the coffee tends to 75 degrees.
  4. T(20)=75+30e−2/3≈90.4∘F.
  5. t=−30ln⁡(16)≈53.75 minutes.

Activity 7.4.4.

Answer.
  1. y=−1+Ce(2t−t22).
  2. y=12ln⁡(et2+C).
  3. y=−1+3e2t.
  4. y=−12t+32=−24t+3.
  5. y=4t2+1.

7.5 Modeling with differential equations
7.5.1 Developing a differential equation

Activity 7.5.2.

Answer.
  1. dAdt=0.05A.
  2. dAdt=0.05A−10000.
  3. The only equilibrium solution is A=200000.
  4. t=20ln⁡(2)≈13.86years.
  5. At least $200000.
  6. Up to $15000 every year.

Activity 7.5.3.

Answer.
  1. dMdt=−kM, where k is a positive constant.
  2. k=−12ln⁡(12)≈0.34657.
  3. dMdt=3−kM, where k is a positive constant.
  4. The equilibrium solution mM=3k is stable.
  5. M=3k(1−e−kt).
  6. About 2.426 milligrams per hour.

7.6 Population Growth and the Logistic Equation
7.6.1 The earth’s population

Activity 7.6.2.

Answer.
  1. P′(0)≈0.0755.
  2. P(0)=6.084.
  3. k≈0.012041.
  4. P(t)=6.084e0.012041t.
  5. P(10)≈6.8878.
  6. t=10.012041ln⁡(126.084)≈56.41, or in the year 2056.
  7. P(500)≈3012.3 billion.

7.6.2 Solving the logistic differential equation

Activity 7.6.3.

Answer.
  1. When P=N2.
  2. When the population is 6.125 billion.
  3. P=12.51.0546e−0.025t+1; P(100)=11.504 billion.
  4. t=1−0.025ln⁡((12.59−1)1.0546)≈39.9049 (so in about year 2040).
  5. limt→∞P(t)=N.

8 Taylor Polynomials and Taylor Series
8.1 Approximating f(x)=ex
8.1.1 Finding a quadratic approximation

Activity 8.1.2.

Answer.
  1. T2″(x)=2c2.
  2. f′(x)=f″(x)=ex.
  3. Table 8.1.4. Formulas and values for f(x) and T2(x).
    f(x)= ex T2(x)= c0+c1x+c2x2
    f′(x)= ex T2′(x)= c1+2c2x
    f″(x)= ex T2″(x)= 2c2
    f(0)= 1 T2(0)= c0
    f′(0)= 1 T2′(0)= c1
    f″(0)= 1 T2″(0)= 2c2
  4. See the bottom half of the table above.
  5. c0=1; c1=1; c2=12. So, T2(x)=1+x+12x2.
  6. T2(x) is a better approximation to f(x)=ex near a=0 than the tangent line; |f(x)−T2(x)|<0.1 for approximately −0.9<x<0.8, and for any x-value in that interval, |f(x)−T2(x)|<|f(x)−T1(x)|.

8.1.2 Over and over again

Activity 8.1.3.

Answer.
  1. Table 8.1.9. Formulas and values for f(x) and T3(x).
    f(x)= ex T3(x)= k0+k1x+k2x2+k3x3
    f′(x)= ex T3′(x)= k1+2k2x+3k3x2
    f″(x)= ex T3″(x)= 2k2+3⋅2k3x
    f‴(x)= ex T3‴(x)= 3⋅2k3
    f(0)= 1 T3(0)= k0
    f′(0)= 1 T3′(0)= k1
    f″(0)= 1 T3″(0)= 2k2
    f‴(0)= 1 T3‴(0)= 6k3
  2. k0=1; k1=1; k2=12; k3=16.
  3. T3(x) appears to be an even better approximation than T2(x) near a=0 and that the quality of the approximation extends further; |f(x)−T3(x)|<0.1 for approximately −1.3<x<1.1.
  4. T4(x)=1+x+12x2+16x3+124x4; T4(x) is an even better approximation to f(x)=ex and on a still wider interval.

8.1.3 As the degree of the approximation increases

Activity 8.1.4.

Answer.
  1. The first seven columns and eleven rows of the spreadsheet are:
    Table 8.1.14. Comparing f(x)=ex and its degree 1, 2, 3, 4 and approximations near a=0.
    Δx x f(x) T1(x) T2(x) T3(x) T4(x)
    0.1 −1.0 0.36787 0.00000 0.50000 0.33333 0.37500
    0.1 −0.9 0.40657 0.10000 0.50500 0.38350 0.41083
    0.1 −0.8 0.44933 0.20000 0.52000 0.43467 0.45173
    0.1 −0.7 0.49659 0.30000 0.54500 0.48783 0.49784
    0.1 −0.6 0.54881 0.40000 0.58000 0.54400 0.54940
    0.1 −0.5 0.60653 0.50000 0.62500 0.60417 0.60677
    0.1 −0.4 0.67032 0.60000 0.68000 0.66933 0.67040
    0.1 −0.3 0.74082 0.70000 0.74500 0.74050 0.74084
    0.1 −0.2 0.81873 0.80000 0.82000 0.81867 0.81873
    0.1 −0.1 0.90484 0.90000 0.90500 0.90483 0.90484
    0.1 0.0 1.00000 1.00000 1.00000 1.00000 1.00000
    The next four columns and four rows of the spreadsheet are:
    Table 8.1.15. The absolute error between f(x)=ex and its degree 1, 2, 3, and 4 approximations at x=−1 and x=−0.9.
    |f(x)−T1(x)| |f(x)−T2(x)| |f(x)−T3(x)| |f(x)−T4(x)|
    0.36787 0.13212 0.03454 0.00712
    0.30657 0.09843 0.02307 0.00426
    0.24933 0.07067 0.01466 0.00240
    0.19659 0.04841 0.00875 0.00125
  2. |f(−1)−T2(−1)|≈0.13212; |f(1)−T2(1)|≈0.21828.
  3. |f(−1)−T3(−1)|≈0.03455; |f(1)−T3(1)|≈0.05162.
  4. |f(−1)−T4(−1)|≈0.00712; |f(1)−T4(1)|≈0.00995.
  5. As the degree of the approximation increases, at each fixed x-value, the approximation gets better, and in addition the interval of values on which the approximation is within a certain tolderance gets wider.
  6. Answers will vary. But, as we widen the interval of x-values, the errors of each polynomial approximation increase near the endpoints of the interval.

8.2 Taylor Polynomials
8.2.1 Taylor polynomials

Activity 8.2.2.

Answer.
  1. Table 8.2.13. Finding the derivatives of f(x)=cos⁡(x) at a=0.
    f(x)= cos⁡(x) f(0)= cos⁡(0)=1
    f′(x)= −sin⁡(x) f′(0)= 0
    f″(x)= −cos⁡(x) f″(0)= −1
    f‴(x)= sin⁡(x) f‴(0)= 0
    f(4)(x)= cos⁡(x) f(4)(0)= 1
    f(5)(x)= −sin⁡(x) f(5)(0)= 0
    f(6)(x)= −cos⁡(x) f(6)(0)= −1
    f(7)(x)= sin⁡(x) f(7)(0)= 0
    f(8)(x)= cos⁡(x) f(8)(0)= 1
  2. T8(x)=1+0x−12!x2+0x3+14!x4+0x5−16!x6+0x7+18!x8.
  3. T10(x)=1−12!x2+14!x4−16!x6+18!x8−110!x10.
  4. Figure 8.2.14. The function f(x)=cos⁡(x) and its degree 2 Taylor approximation T2(x)=1−12x2 near the point (0,f(0)).
    As the degree of the approximation increases, the accuracy of the approximation improves at each fixed x-value and in how large the interval is on which the approximation is accurate.
  5. Table 8.2.15. Comparing f(x)=cos⁡(x) and its degree 2, 4, and 6 approximations near a=0.
    Δx x f(x) T2(x) T4(x) T6(x)
    0.2 −2.0 −0.41615 −1.00000 −0.33333 −0.42222
    0.2 −1.8 −0.22720 −0.62000 −0.18260 −0.22984
    0.2 −1.6 −0.02920 −0.28000 −0.00693 −0.03024
    ⋯ ⋯ ⋯ ⋯ ⋯ ⋯
    0.2 1.6 −0.02920 −0.28000 −0.00693 −0.03024
    0.2 1.8 −0.22720 −0.62000 −0.18260 −0.22984
    0.2 2.0 −0.41615 −1.00000 −0.33333 −0.42222
    Table 8.2.16. The absolute error between f(x)=cos⁡(x) and its degree 2, 4, and 6 approximations.
    |f(x)−T2(x)| |f(x)−T4(x)| |f(x)−T6(x)|
    0.58385 0.08281 0.00608
    0.39280 0.04460 0.00263
    0.25080 0.02227 0.00104
    ⋯ ⋯ ⋯
    0.25080 0.02227 0.00104
    0.39280 0.04460 0.00263
    0.58385 0.08281 0.00608
  6. |f(x)−T2(x)|<0.1 for roughly −1.2<x<1.2; |f(x)−T4(x)|<0.1 for −2<x<2; |f(x)−T6(x)|<0.1 for approximately −2.8<x<2.8.

8.2.2 Taylor polynomial approximations centered at an arbitrary value a

Activity 8.2.3.

Answer.
  1. Table 8.2.24. Finding the derivatives of f(x)=ln⁡(x) at a=1.
    f(x)= ln⁡(x) f(1)= 0
    f′(x)= x−1 f′(1)= 1
    f″(x)= −1⋅x−2 f″(1)= −1
    f‴(x)= (−2)(−1)x−3 f‴(1)= (−2)(−1)
    f(4)(x)= (−3)(−2)(−1)x−4 f(4)(1)= (−3)(−2)(−1)
  2. T4(x)=1(x−1)−12(x−1)2+13(x−1)3−14(x−1)4.
  3. Figure 8.2.25. The function f(x)=ln⁡(x) and its degree 1 Taylor approximation T1(x)=x−1 near the point (1,f(1)).
    T4(x) provides a much more accurate approximation of f(x) than T1(x).
  4. |f(x)−T4(x)|<0.1 for approximately 0.29<x<1.97.
  5. T5(x)=1(x−1)−12(x−1)2+13(x−1)3−14(x−1)4+15(x−1)5; T6(x)=1(x−1)−12(x−1)2+13(x−1)3−14(x−1)4+15(x−1)5−16(x−1)6.
    |f(x)−T5(x)|<0.1 for about 0.24<x<1.999; |f(x)−T6(x)|<0.1 for about 0.21<x<2. While the interval of accuracy gets wider as the degree increases, it seems not to extend past x=2 and doesn’t move much to the left.

Activity 8.2.4.

Answer.
  1. Table 8.2.30. Finding the derivatives of f(x)=ln⁡(x) at a=2.
    f(x)= ln⁡(x) f(2)= ln⁡(2)
    f′(x)= x−1 f′(2)= 12
    f″(x)= −1⋅x−2 f″(2)= −122
    f‴(x)= (−2)(−1)x−3 f‴(2)= (−2)(−1)23
    f(4)(x)= (−3)(−2)(−1)x−4 f(4)(2)= (−3)(−2)(−1)24
  2. T4(x)=ln⁡(2)+12(x−2)−12⋅22(x−2)2+13⋅23(x−2)3−14⋅24(x−2)4.
  3. Figure 8.2.31. The function f(x)=ln⁡(x) and its degree 1 and 4 Taylor approximations T1(x) and T4(x) near the point (2,f(2)).
    T4(x) provides a much better approximation of f(x) near a=2 and on a wider interval.
  4. |f(x)−T4(x)|<0.1 for approximately 0.58<x<3.95.
  5. T5(x)=ln⁡(2)+12(x−2)−12⋅22(x−2)2+13⋅23(x−2)3−14⋅24(x−2)4+15⋅25(x−2)5; T6(x)=ln⁡(2)+12(x−2)−12⋅22(x−2)2+13⋅23(x−2)3−14⋅24(x−2)4+15⋅25(x−2)5−16⋅26(x−2)6.
    |f(x)−T5(x)|<0.1 for roughly 0.48<x<4; |f(x)−T6(x)|<0.1 for about 0.42<xlt4. By moving to a=2, which is further away from the asymptote at x=0, we can get approximations of ln(x) that seem to be good all the way up to x=4.

8.3 Geometric Sums
8.3.1 Finite Geometric Series

Activity 8.3.2.

Answer.
  1. 25⋅Sn=25+(25)2(25)3+⋯+(25)n.
  2. The key observation is that 1+25+(25)2+⋯+(25)n−1−25−(25)2−(25)3−⋯−(25)n=1−(25)n,
  3. Sn=1−(25)n35
  4. Sn=7−7⋅(25)n35

8.3.2 Infinite Geometric Series

Activity 8.3.3.

Answer.
  1. S5=a−ar51−r=1−(1/3)51−(1/3)=12181≈1.4938; S10=a−ar101−r=1−(1/3)101−(1/3)=2952419683≈1.49997; S=a1−r=11−13=123=32.
  2. S5=a−ar51−r=4−4⋅(−1/2)51−(−1/2)=114; S10=a−ar101−r=4−4⋅(−1/2)101−(−1/2)=341128=2.6640625; S=a1−r=41+12=432=83.
  3. S5=a−ar51−r=2−2⋅(4/3)51−(4/3)=156281≈19.284; S10=a−ar101−r=2−2⋅(4/3)101−(4/3)=197905419683≈100.546; the infinite series diverges.
  4. S5=a−ar51−r=5−5⋅(3/4)51−(3/4)=3905256≈15.254; S10=a−ar101−r=5−5⋅(3/4)101−(3/4)=4947635262144≈18.874; S=a1−r=51−34=20.
  5. S5=a−ar51−r=43−43⋅(−2/3)51−(−2/3)=220243≈0.905; S10=a−ar101−r=43−43⋅(−2/3)101−(−2/3)=4642059049≈0.786; S=a1−r=431+23=45=0.8.

8.3.3 How geometric series naturally connect to Taylor polynomials

Activity 8.3.4.

Answer.
  1. Table 8.3.9. Finding the derivatives of f(x)=11−x.
    f(x)= 11−x=(1−x)−1
    f′(x)= (−1)(1−x)−2(−1)
    f″(x)= (−2)(−1)(1−x)−3(−1)(−1)
    f‴(x)= (−3)(−2)(−1)(1−x)−4(−1)(−1)(−1)
    f(4)(x)= (−4)(−3)(−2)(−1)(1−x)−5(−1)(−1)(−1)(−1)
    f(5)(x)= (−5)(−4)(−3)(−2)(−1)(1−x)−6(−1)(−1)(−1)(−1)(−1)
  2. Table 8.3.10. Finding the Taylor polynomial coefficients ck.
    f(0)= 11−0=1 c0= f(0)=1
    f′(0)= (−1)(1−0)−2(−1)=1 c1= f′(0)1!=11!=1
    f″(0)= (−2)(−1)(1)−3(−1)(−1)=2! c2= f″(0)2!=2!2!=1
    f‴(0)= (−3)(−2)(−1)(1)−4(−1)(−1)(−1)=3! c3= f‴(0)3!=3!3!=1
    f(4)(0)= (−4)(−3)(−2)(−1)(1)−5(−1)(−1)(−1)(−1)=4! c4= f(4)(0)4!=4!4!=1
    f(5)(0)= (−5)(−4)(−3)(−2)(−1)(1)−6(−1)(−1)(−1)(−1)(−1)=5! c5= f(5)(0)5!=5!5!=1
  3. T5(x)=1+x+x2+⋯+x5 and Tn(x)=1+x+x2+⋯+xn.
  4. Tn(x) is a finite geometric sum with r=x and a=1.
  5. T(x)=1+x+x2+⋯+xn+⋯ which is an infinite geometric sum with a=1 and r=x whose sum (for |r|=|x|<1) is a1−r=11−x=f(x).

8.4 Taylor Series
8.4.1 Taylor series and the Ratio Test

Activity 8.4.2.

Answer.
  1. rn(x)=nn+1⋅12⋅(x−1)
  2. r(x)=12⋅(x−1)
  3. T(x) is guaranteed to converge whenever −1<x<3.
  4. On 1<x<3, T10 and f are nearly indistinguishable for most of the interval. This suggests that T(x) is the Taylor series of f(x)=ln⁡(2)−ln⁡(3−x).

8.4.2 Taylor series of several important functions

Activity 8.4.3.

Answer.
  1. f(k)(0)=1 for every natural number k because f(k)(x)=ex for every natural number k.
  2. Tf(x)=∑n=0∞1k!xk=1+x+12!x2+13!x3+⋯+1n!xn+⋯
  3. rn(x)=1n+1⋅x.
  4. r(x)=limn→∞rn(x)=0; |r(x)|<1 for every value of x; Tf(x) converges for every real number x.
  5. f(x) and T10(x) are almost indistinguishable on the interval −4<x<4; f(x) and T20(x) are almost indistinguishable on −8<x<8.

8.5 Finding and Using Taylor Series
8.5.1 Using substitution and algebra to find new Taylor series expressions

Activity 8.5.2.

Answer.
  1. g(x)=x3sin⁡(x2)=x5−13!x9+15!x13−⋯; we expect this series to converge for all real numbers x.
  2. h(x)=e−x4=1−x4+12!x8−13!x12+⋯; the series converges for all real numbers x.
  3. p(x)=11+5x=1−5x+(−5x)2+(−5x)3+⋯=1−5x+52x2−53x3+⋯; the series converges for all x such that −15<x<15.
  4. q(x)=x2ln⁡(1+x4)=x6−12x10+13x14−⋯; the series converges when |x|<1.

8.5.2 Differentiating and integrating Taylor series

Activity 8.5.3.

Answer.
  1. For g(x)=arctan⁡(x):
    1. g′(x)=11+x2.
    2. f(x)=g′(x)=11+x2=1−x2+x4−x6+⋯.
    3. By the Second FTC we have g(x)=∫0xf(u)du; integrating, it follows that g(x)=x−13x3+15x5−17x7+⋯.
    4. The series for g(x) is certain to converge on |x|<1.
  2. For h(x)=ln⁡(1+x):
    1. p(u)=11−u=1+u+u2+u3+⋯ and this series converges for |u|<1.
    2. r(x)=11+x=p(−x)=1−x+x2−x3+⋯.
    3. ln⁡(1+x)=x−12x2+13x3−14x4+⋯; and this series is guaranteed to converge on −1<x<1.

Activity 8.5.4.

Answer.
  1. e−t2=1−t2+12!t4−13!t6+⋯
  2. ∫0xe−t2dt=x−13x3+15⋅2!x5−17⋅3!x7+⋯
  3. erf(x)=2π(x−13x3+15⋅2!x5−17⋅3!x7+⋯)
  4. For every real value of x.
  5. Extra \left or missing \right

8.5.3 Alternating series of real numbers

Activity 8.5.5.

Answer.
  1. sin⁡(1)≈1−13!+15!−17!=42415040=0.8414682539682539…, and this approximation has error at most 19!≈0.00000276.
  2. ∫01e−x2dx≈1−13+15⋅2!−17⋅3!+19⋅4!−111⋅5!+113⋅6!=16147792162160=0.7468360343… and this approximation has error at most 115⋅7!=1756600≈0.0000132
  3. ∫01cos⁡(x2)dx≈1−15⋅2!+19⋅4!−113⋅6!=2539928080=0.904522792022792022792… and this approximation has error at most 115⋅8!=1604800=0.0000016534….
  4. The alternating series
    1−12⋅12+13⋅13−⋯+(−1)n−11n⋅1n+⋯
    converges by the Alternating Series Theorem, and its exact sum is ln⁡(2).
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