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Active Calculus

Matthew Boelkins

Appendix C Answers to Selected Exercises

This appendix contains answers to all non-WeBWorK exercises in the text.
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1 Understanding the Derivative
1.1 How do we measure velocity?
1.1.4 Exercises

1.1.4.9.

Answer.
  1. s(15)−s(0)≈−98.75.
  2. AV[0,15]=s(15)−s(0)15−0≈−6.58AV[0,2]=s(2)−s(0)2−0≈−47.63AV[1,6]=s(6)−s(1)6−1≈−13.25AV[8,10]=s(10)−s(8)10−8≈−7.35
  3. Most negative average velocity on [0,4]; most positive average velocity on [4,8].
  4. 21.31+22.252=21.78 feet per second.
  5. The average velocities are negative; the instantaneous velocity was positive. Downward motion corresponds to negative average velocity; upward motion to positive average velocity.

1.1.4.10.

Answer.
  1. Sketch a plot where the diver’s height at time t is on the vertical axis. For instance, h(2.45)=0.
  2. AV[2.45,7]≈−3.5−07−2.45=−3.54.55=−0.7692 m/sec. The average velocity is not the same on every time interval within [2.45,7].
  3. When the diver is going upward, her velocity is positive. When she is going downward, her velocity is negative. At the peak of her dive and when her feet touch the bottom of the pool.
  4. It looks like when the position function is steep, the velocity function’s value is farther away from zero, and that whenever the height/position function is rising/increasing, the velocity function has a positive value. Similarly, whenever the position function is decreasing, the velocity is negative.

1.1.4.11.

Answer.
  1. 15957 people.
  2. In an average year the population grew by about 798 people/year.
  3. The slope of a secant line through the points (a,f(a)) and (b,f(b)).
  4. AV[0,20]≈798 people per year.
  5. AV[5,10]≈734.50AV[5,9]≈733.06AV[5,8]≈731.62AV[5,7]≈730.19AV[5,6]≈728.7535

1.2 The notion of limit
1.2.4 Exercises

1.2.4.8.

Answer.
  1. All real numbers except x=±2.
  2. x f(x)
    2.1 −8.41
    2.01 −8.0401
    2.001 −8.004001
    1.999 −7.996001
    1.99 −7.9601
    1.9 −7.61
    limx→2f(x)=−8.
  3. limx→216−x4x2−4=−8.
  4. False.
  5. False.

1.2.4.9.

Answer.
  1. All real numbers except x=−3.
  2. x g(x)
    −2.9 −1
    −2.99 −1
    −2.999 −1
    −3.001 1
    −3.01 −1
    −3.1 −1
    The limit does not exist.
  3. If x>−3,
    −|x+3|x+3=−x+3x+3=−1;
    if x<−3, it follows that
    −|x+3|x+3=−−(x+3)x+3=+1.
    Hence the limit does not exist.
  4. False.
  5. False.

1.2.4.10.

Answer.

1.2.4.11.

Answer.
  1. AV[1,1+h]=100cos⁡(0.75(1+h))⋅e−0.2(1+h)−100cos⁡(0.75)⋅e−0.2h
  2. limh→0AV1,1+h≈−53.837.
  3. The instantaneous velocity of the bungee jumper at the moment t=1 is approximately −53.837 ft/sec.

1.3 The derivative of a function at a point
1.3.3 Exercises

1.3.3.10.

Answer.
  2. AV[−3,−1]≈1.15; AV[0,2]≈−0.4.
  3. f′(−3)≈3; f′(0)≈−12.

1.3.3.11.

Answer.
  1. For instance, you could let f(−3)=3 and have f pass through the points (−3,3), (−1,−2), (0,−3), (1,−2), and (3,−1) and draw the desired tangent lines accordingly.
  2. For instance, you could draw a function g that passes through the points (−2,3), (−1,2), (1,0), (2,0), and (3,3) in such a way that the tangent line at (−1,2) is horizontal and the tangent line at (2,0) has slope 1.

1.3.3.12.

Answer.
  1. AV[0,7]=0.11757≈0.01679 billion people per year; P′(7)≈0.1762 billion people per year; P′(7)>AV[0,7].
  2. AV[19,29]≈0.02234 billion people/year.
  3. We will say that today’s date is July 1, 2015, which means that t=22.5;
    P′(22.5)=limh→0115(1.014)22.5+h−115(1.014)22.5h;
    P′(22.5)≈0.02186 billions of people per year.
  4. y−1.57236=0.02186(t−22.5).

1.3.3.13.

Answer.
  1. All three approaches show that f′(2)=1.
  2. All three approaches show that f′(1)=−1.
  3. All three approaches show that f′(1)=12.
  4. All three approaches show that f′(1) does not exist.
  5. The first two approaches show that f′(π2)=0.

1.4 The derivative function
1.4.3 Exercises

1.4.3.10.

Answer.
  1. See the figure below.
  2. See the figure below.
  3. One example of a formula for f is f(x)=12x2−1.

1.4.3.11.

Answer.
  1. g′(x)=2x−1.
  3. p′(x)=10x−4.
  4. The constants 3 and 12 don’t seem to affect the results at all. The coefficient −4 on the linear term in p(x) appears to make the ``−4’’ appear in p′(x)=10x−4. The leading coefficient 5 in (x)=5x2−4x+12 leads to the coefficient of ``10’’ in p′(x)=10x−4.

1.4.3.12.

Answer.
  1. g is linear.
  2. On −3.5<x<−2, −2<x<0 and 2<x<3.5.
  3. At x=−2,0,2; g must have sharp corners at these points.

1.4.3.13.

Answer.

1.5 Interpreting, estimating, and using the derivative
1.5.4 Exercises

1.5.4.6.

Answer.
  1. F′(10)≈−3.33592.
  2. The coffee’s temperature is decreasing at about 3.33592 degrees per minute.
  3. F′(20).
  4. We expect F′ to get closer and closer to 0 as time goes on.

1.5.4.7.

Answer.
  1. If a patient takes a dose of 50 ml of a drug, the patient will experience a body temperature change of 0.75 degrees F.
  2. ``degrees Fahrenheit per milliliter.’’
  3. For a patient taking a 50 ml dose, adding one more ml to the dose leads us to expect a temperature change that is about 0.02 degrees less than the temperature change induced by a 50 ml dose.

1.5.4.8.

Answer.
  1. t=0.
  2. v′(1)=−32.
  3. ``feet per second per second’’; v′(1)=−32 tells us that the ball’s velocity is decreasing at a rate of 32 feet per second per second.
  4. The acceleration of the ball.

1.5.4.9.

Answer.
  1. AV[40000,55000]≈−0.153 dollars per mile.
  2. h′(55000)≈−0.147 dollars per mile. During 550001st mile, we expect the car’s value to drop by 0.147 dollars.
  3. h′(30000)<h′(80000).
  4. The graph of h might have the general shape of the graph of y=e−x for positive values of x: always positive, always decreasing, and bending upwards while tending to 0 as x increases.

1.6 The second derivative
1.6.5 Exercises

1.6.5.10.

Answer.
  1. f is increasing and concave down near x=2.
  2. Greater.
  3. Less.

1.6.5.11.

Answer.
  1. g′(2)≈1.4.
  2. At most one.
  3. 9.
  4. g″(2)≈5.5.

1.6.5.12.

Answer.
  1. h′(4.5)≈14.3; h′(5)≈21.2; h′(5.5)≈=23.9; rising most rapidly at t=5.5.
  2. h′(5)≈9.6.
  3. Acceleration of the bungee jumper in feet per second per second.
  4. 0<t<2, 6<t<10.

1.6.5.13.

Answer.

1.7 Limits, Continuity, and Differentiability
1.7.5 Exercises

1.7.5.7.

Answer.
  1. a=0.
  2. a=0,3.
  3. a=−2,0,1,2,3.

1.7.5.8.

Answer.
  1. f(x)=|x−2|.
  2. Impossible.
  3. Let f be the function defined to be f(x)=1 for every value of x≠−2, and such that f(−2)=4.

1.7.5.9.

Answer.
  1. h must be piecewise linear with slope of 1 or −1, depending on the interval.
  2. h′(x) is not defined for x=−2,0,2.
  3. It is possible that h is not continuous at x=−2,0,2.
  4. Two of the many possible graphs for h are shown in the following figure.

1.7.5.10.

Answer.
  1. At x=0.
    g′(0)=limh→0g(0+h)−g(0)h=limh→0|h|−|0|h=limh→0|h|h
  2. h 0.1 0.01 0.001 0.0001 −0.1 −0.01 −0.001 −0.0001
    |h|/h 3.162 10 31.62 100 −3.162 −10 −31.62 −100
    g′(0) does not exist.

1.8 The Tangent Line Approximation
1.8.4 Exercises

1.8.4.7.

Answer.
  1. p(3)=−1 and p′(3)=−2.
  2. p(2.79)≈−0.58.
  3. Too large.

1.8.4.8.

Answer.
  1. F′(60)≈1.56 degrees per minute.
  2. L(t)≈1.56(t−60)+324.5.
  3. F(63)≈L(63)≈=329.18 degrees F.
  4. Overestimate.

1.8.4.9.

Answer.
  1. s(9.34)≈L(9.34)=3.592.
  2. underestimate.
  3. The object is slowing down as it moves toward toward its starting position at t=4.

1.8.4.10.

Answer.
  1. x=1.
  2. On −0.37<x<1.37; f is concave up.
  3. f(1.88)≈−3.0022, and this estimate is larger than the true value of f(1.88).

2 Computing Derivatives
2.1 Elementary derivative rules
2.1.5 Exercises

2.1.5.8.

Answer.
  1. h(2)=27; h′(2)=−19/2.
  2. L(x)=27−192(x−2).
  3. p is increasing at x=2.
  4. p(2.03)≈−11.44.

2.1.5.9.

Answer.
  1. p is not differentiable at x=−1 and x=1; q is not differentiable at x=−1 and x=1.
  2. r is not differentiable at x=−1 and x=1.
  3. r′(−2)=4; r′(0)=12.
  4. y=4.

2.1.5.10.

Answer.
  1. w′(t)=3tt(ln⁡(t)+1)+211−t2.
  2. L(t)=(32−2π3)+(32(ln⁡(12)+1)+43)(t−12).
  3. v is decreasing at t=12.

2.1.5.11.

Answer.
  1. f′(x)=limh→0ax+h−axh=limh→0ax⋅ah−axh=limh→0ax(ah−1)h,
  2. Since ax does not depend at all on h, we may treat ax as constant in the noted limit and thus write the value ax in front of the limit being taken.
  3. When a=2, L≈0.6931; when a=3, L≈1.0986.
  4. a≈2.71828 (for which L≈1.000)
  5. ddx[2x]=2x⋅ln⁡(2) and ddx[3x]=3x⋅ln⁡(3)
  6. ddx[ex]=ex.

2.2 The sine and cosine functions
2.2.3 Exercises

2.2.3.5.

Answer.
  1. V′(2)=24⋅1.072⋅ln⁡(1.07)+6cos⁡(2)≈−0.63778 thousands of dollars per year.
  2. V″(2)=24⋅1.072⋅ln⁡(1.07)2−6sin⁡(2)≈−5.33 thousands of dollars per year per year. At this moment, V′ is decreasing and we expect the derivative’s value to decrease by about 5.33 thousand dollars per year over the course of the next year.
  3. See the figure below. Adding the term 6sin⁡(t) to A to create the function V adds volatility to the value of the portfolio.

2.2.3.6.

Answer.
  1. f′(π4)=−5(22).
  2. L(x)=3+2(x−π).
  3. Decreasing.
  4. The tangent line to f lies above the curve at this point.

2.2.3.7.

Answer.
  1. Hint: in the numerator of the difference quotient, combine the first and last terms and remove a factor of sin⁡(x).
  2. Hint: divide each part of the numerator by h and consider the sum of two separate limits.
  3. limh→0(cos⁡(h)−1h)=0 and limh→0(sin⁡(h)h)=1.
  4. f′(x)=sin⁡(x)⋅0+cos⁡(x)⋅1.
  5. Hint: cos⁡(α+β) is cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β).

2.3 The product and quotient rules
2.3.5 Exercises

2.3.5.11.

Answer.
  1. h(2)=−15; h′(2)=23/2.
  2. L(x)=−15+23/2(x−2).
  3. Increasing.
  4. r(2.06)≈−0.5796.

2.3.5.12.

Answer.
  1. w′(t)=tt(ln⁡t+1)⋅(arccos⁡t)+tt⋅−11−t2.
  2. L(t)≈0.740−0.589(t−0.5).
  3. Increasing.

2.3.5.13.

Answer.
  1. r′(−2)=5 and r′(0)=1.
  2. At x=−1 and x=1.
  3. L(x)=2.
  4. z′(0)=−14 and z′(2)=−1.
  5. At x=−1, x=1, x=−1.5, and x=1.

2.3.5.14.

Answer.
  1. C(t)=A(t)Y(t) bushels in year t.
  2. 1190000 bushels of corn.
  3. C′(t)=A(t)Y′(t)+A′(t)Y(t).
  4. C′(0)=158000 bushels per year.
  5. C(1)≈1348000bushels.

2.3.5.15.

Answer.
  1. g(80)=20 kilometers per liter, and g′(80)=−0.16. kilometers per liter per kilometer per hour.
  2. h(80)=4 liters per hour and h′(80)=0.082 liters per hour per kilometer per hour.
  3. Think carefully about units and how each of the three pairs of values expresses fundamentally the same facts.

2.4 Derivatives of other trigonometric functions
2.4.3 Exercises

2.4.3.8.

Answer.
  1. h′(2)=−2sin⁡(2)−2cos⁡(2)ln⁡(1.2)1.22≈−1.1575 feet per second.
  2. h″(2)=cos⁡(2)(−2+2ln2(1.2))+4ln⁡(1.2)sin⁡(2))1.22≈1.0193 feet per second per second.
  3. The object is falling and slowing down.

2.4.3.9.

Answer.
  1. f′(x)=sin⁡(x)⋅(−csc2⁡(x))+cot⁡(x)⋅cos⁡(x).
  2. False.
  3. f′(x)=−sin2⁡(x)sin⁡(x)=−sin⁡(x) for x≠π2+kπ for some integer value of k.

2.4.3.10.

Answer.
  1. p′(z)=(z2sec⁡(z)+1)(zsec2⁡(z)+tan⁡(z))−ztan⁡(z)(z2sec⁡(z)tan⁡(z)+2zsec⁡(z))(z2sec⁡(z)+1)2+3ez
  2. y−4=3(x−0).
  3. Increasing.

2.5 The chain rule
2.5.5 Exercises

2.5.5.11.

Answer.
  1. h′(π4)=322.
  2. r′(0.25)=cos⁡(0.253)⋅3(0.25)2≈0.1875>h′(0.25)=3sin2⁡(0.25)⋅cos⁡(0.25)≈0.1779; r is changing more rapidly.
  3. h′(x) is periodic; r′(x) is not.

2.5.5.12.

Answer.
  1. p′(x)=eu(x)⋅u′(x).
  2. q′(x)=u′(ex)⋅ex.
  3. r′(x)=−csc2⁡(u(x))⋅u′(x).
  4. s′(x)=u′(cot⁡(x))⋅(−csc2⁡(x)).
  5. a′(x)=u′(x4)⋅4x3.
  6. b′(x)=4(u(x))3⋅u′(x).

2.5.5.13.

Answer.
  1. C′(0)=0 and C′(3)=−12.
  2. Consider C′(1). By the chain rule, we’d expect that C′(1)=p′(q(1))⋅q′(1), but we know that q′(1) does not exist since q has a corner point at x=1. This means that C′(1) does not exist either.
  3. Since Y(x)=q(q(x)), the chain rule implies that Y′(x)=q′(q(x))⋅q′(x), and thus Y′(−2)=q′(q(−2))⋅q′(−2)=q′(−1)⋅q′(−2). But q′(−1) does not exist, so Y′(−2) also fails to exist. Using Z(x)=q(p(x)) and the chain rule, we have Z′(x)=q′(p(x))⋅p′(x). Therefore Z′(0)=q′(p(0))⋅p′(0)=q′(−0.5)⋅p′(0)=0⋅0.5=0.

2.5.5.14.

Answer.
  1. dVdh|h=1=7π cubic feet per foot.
  2. h′(2)=πcos⁡(2π)=π feet per hour.
  3. dVdt|t=2=7π2 cubic feet per hour.
  4. In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time.

2.6 Derivatives of Inverse Functions
2.6.6 Exercises

2.6.6.7.

Answer.
  1. f′(x)=12arctan⁡(x)+3arcsin⁡(x)+5⋅(21+x2+31−x2).
  2. r′(z)=11+(ln⁡(arcsin⁡(z)))2⋅(1arcsin⁡(z))⋅11−z2.
  3. q′(t)=arctan2⁡(3t)⋅[4arcsin3⁡(7t)(71−(7t)2)]+arcsin4⁡(7t)⋅[2arctan⁡(3t)(31+(3t)2)].
  4. g′(v)=1arctan⁡(v)arcsin⁡(v)+v2⋅(arcsin⁡(v)+v2)⋅11+v2−arctan⁡(v)⋅(11−v2+2v)(arcsin⁡(v)+v2)2

2.6.6.8.

Answer.
  1. f′(1)≈2.
  3. (f−1)′(−1)≈1/2.

2.6.6.9.

Answer.
  1. f passes the horizontal line test.
  2. f−1(x)=g(x)=4x−163.
  3. f′(x)=34x2; f′(2)=3. g′(x)=13(4x−16)−2/3⋅4; g′(6)=13. These two derivative values are reciprocals.

2.6.6.10.

Answer.
  1. h passes the horizontal line test.
  2. The equation y=x+sin⁡(x) can’t be solved for x in terms of y.
  3. (h−1)′(π2+1)=1.

2.7 Derivatives of Functions Given Implicitly
2.7.3 Exercises

2.7.3.12.

Answer.
Horizontal tangent lines: (0,−1), (0,−0.618), (0,1.618), (1,−1), (1,−0.618), (1,1.618), (0.5,−1.0493), (0.5,0.2104), (0.5,1.6139). Vertical tangent lines: (−0.1756,−0.379), (0.2912,−0.379), (0.7088,−0.379), (1.1756,−0.379), (−0.8437,1.235), and (1.8437,1.235).

2.7.3.13.

Answer.
y=π2−(x−π2).

2.7.3.14.

Answer.
  1. x=ln⁡(y)ln⁡(a).
  2. 1=1ln⁡(a)⋅1ydydx.
  3. ddx[ax]=axln⁡(a).

2.8 Using Derivatives to Evaluate Limits
2.8.4 Exercises

2.8.4.11.

Answer.
limx→3h(x)=−2.

2.8.4.12.

Answer.
Horizontal asymptote: y=35; vertical asymptote: x=c; hole: (a,3(a−b)5(a−c)). R is not continuous at x=a and x=c.

2.8.4.13.

Answer.
  1. ln⁡(x2x)=2x⋅ln⁡(x).
  2. x=11x.
  3. limx→0+h(x)=0.
  4. limx→0+g(x)=limx→0+x2x=1.

2.8.4.14.

Answer.
  1. Show that limx→∞ln⁡(x)x=0.
  2. Show that limx→∞ln⁡(x)xn=0.
  3. Consider limx→∞p(x)ex By repeated application of LHR, the numerator will eventually be simply a constant (after n applications of LHR), and thus with ex still in the denominator, the overall limit will be 0.
  4. Show that limx→∞ln⁡(x)xn=0
  5. For example, f(x)=3x2+1 and g(x)=−0.5x2+5x−2.

3 Using Derivatives
3.1 Using derivatives to identify extreme values
3.1.4 Exercises

3.1.4.9.

Answer.
  1. f′ is positive for −1<xlt1 and for x>1; f′ is negative for all x<−1. f has a local minimum at x=−1.
  2. A possible graph of y=f″(x) is shown at right in the figure.
  3. f″(x) is negative for −0.35<x<1; f″(x) is positive everywhere else; f has points of inflection at x≈−0.35 and x=1.
  4. A possible graph of y=f(x) is shown at left in the figure.

3.1.4.10.

Answer.
  1. Neither.
  2. g″(2)=0; g″ is negative for 1<x<2 and positive for 2<x<3.
  3. g has a point of inflection at x=2.

3.1.4.11.

Answer.
  1. h can have no, one, or two real zeros.
  2. One root is negative and the other positive.
  3. h will look like a line with slope 3.
  4. h is concave up everywhere; h is almost linear for large values of |x|.

3.1.4.12.

Answer.
  1. p″(x) is negative for −1<x<2 and positive for all other values of x; p has points of inflection at x=−1 and x=2.
  2. Local maximum.
  3. Neither.

3.2 Using derivatives to describe families of functions
3.2.3 Exercises

3.2.3.6.

Answer.
  2. x=0 and x=2a3.
  3. x=a3; p″(x) changes sign from negative to positive at x=a3.
  4. As we increase the value of a, both the location of the critical number and the inflection point move to the right along with a.

3.2.3.7.

Answer.
  1. x=c is a vertical asymptote because limx→c+e−xx−c=∞ and limx→c−e−xx−c=−∞.
  2. limx→∞e−xx−c=0; limx→−∞e−xx−c=−∞.
  3. The only critical number for q is x=c−1.
  4. When x<c−1, q′(x)>0; when x>c−1, q′(x)<0; q has a local maximum at x=c−1.

3.2.3.8.

Answer.
  1. x=m.
  2. E is increasing for x<m and decreasing for x>m, with a local maximum at x=m.
  3. x=m±s.
  4. limx→∞E(x)=limx→−∞E(x)=0.

3.3 Global Optimization
3.3.4 Exercises

3.3.4.7.

Answer.
  1. Not enough information is given.
  2. Global minimum at x=b.
  3. Global minimum at x=a; global maximum at x=b.
  4. Not enough information is provided.

3.3.4.8.

Answer.
  1. Absolute maximum p(0)=p(a)=0; absolute minimum p(a3)=−2a333.
  2. Absolute max r(1b)≈0.368ab; absolute min r(2b)≈0.270ab.
  3. Absolute minimum g(b)=a(1−e−b2); absolute maximum g(3b)=a(1−e−3b2).
  4. Absolute max s(π2k)=1; absolute min s(5π6k)=12.

3.3.4.9.

Answer.
  1. Global maximum at x=a; global minimum at x=b.
  2. Global maximum at x=c; global minimum at either x=a or x=b.
  3. Global minimum at x=a and x=b; global maximum somewhere in (a,b).
  4. Global minimum at x=c; global maximum value at x=a.

3.3.4.10.

Answer.
  1. Absolute max s(5π12)=8; absolute min s(11π12)=2.
  2. Absolute max s(5π12)=8; absolute min s(0)=5−332≈2.402.
  3. Absolute max s(5π12)=8; absolute min s(11π12)=2. (There are other points at which the function achieves these values on the given interval.)
  4. Absolute max s(5π12)=8; absolute min s(5π6)≈2.402.

3.4 Applied Optimization
3.4.3 Exercises

3.4.3.9.

Answer.
The absolute maximum volume is V(1518)=156(1518)−(1518)3≈1.52145 cubic feet.

3.4.3.10.

Answer.
The maximum possible area that each of the four pens can enclose is 351562.5 square feet.

3.4.3.11.

Answer.
172.047 feet of cable.

3.4.3.12.

Answer.
The minimum cost is $1165.70.

3.5 Related Rates
3.5.3 Exercises

3.5.3.9.

Answer.
The boat is approaching the dock at a rate of 136≈2.167 feet per second.

3.5.3.10.

Answer.
The depth of the water is increasing at
dhdt|h=5=1.28
feet per minute. The depth of the water is increasing at a decreasing rate.

3.5.3.11.

Answer.
dθdt|x=30=−0.24 radians per second.

3.5.3.12.

Answer.
dhdt|V=1000=10π(3000π3)2≈0.0328 feet per minute.

4 The Definite Integral
4.1 Determining distance traveled from velocity
4.1.5 Exercises

4.1.5.7.

Answer.
  1. At time t=1, 12 miles north of the lake.
  2. s(2)−s(0)=1 mile north of the lake.
  3. 40 miles.

4.1.5.8.

Answer.
  1. t=50032=1258=15.625 is when the rocket reaches its maximum height.
  2. A=3906.25, the vertical distance traveled on [0,15.625].
  3. s(t)=500t−16t2.
  4. s(15.625)−s(0)=3906.25 is the change of the rocket’s position on [0,15.625].
  5. s(5)−s(1)=1616; the rocket rose 1616 feet on [1,5].

4.1.5.9.

Answer.
  1. 12+14π≈1.285.
  2. s(5)−s(2)=−2 is the change in position of the object on [2,5].
  3. On the time interval [5,7].
  4. s is increasing on the intervals (0,2) and (5,7); the position function has a relative maximum at t=2.

4.1.5.10.

Answer.
  2. Think about the product of the units involved: ``units of pollution per day’’ times ``days’’. Connect this to the area of a thin vertical rectangle whose height is given by the curve.
  3. An underestimate is 336 units of pollution.

4.2 Riemann Sums
4.2.5 Exercises

4.2.5.8.

Answer.
  1. M4=43.5.
  2. A=872.
  3. The rectangles with heights that come from the midpoint have the same area as the trapezoids that are formed by the function values at the two endpoints of each subinterval.
    Mn will give the exact area for any value of n. Neither Ln nor Rn will be exact for any n.
  4. For any linear function g of the form g(x)=mx+b such that g(x)≥0 on the interval of interest.

4.2.5.9.

Answer.
  1. f(x)=x2+1 on the interval [1,3].
  2. If S is a left Riemann sum, f(x)=x2+1 on the interval [1.4,3.4]. If S is a middle Riemann sum, f(x)=x2+1 on the interval [1.2,3.2].
  3. The area under f(x)=x2+1 on [1,3].
  4. R10=∑i=110((1+0.2i)2+1)⋅0.2.

4.2.5.10.

Answer.
  2. M3=99.6 feet.
  3. L6=114,
    R6=84,
    and 12(L6+R6)=99.
  4. 114 feet.

4.2.5.11.

Answer.
  1. M4≈6.4.
  2. The total tonnage of pollution escaping the scrubbing process in the time interval [0,4] weeks.
  3. L5≈5.19620599.
  4. 6.4 tons.

4.3 The Definite Integral
4.3.5 Exercises

4.3.5.9.

Answer.
  1. The total change in position is P=∫04v(t)dt.
  2. P=−2.625 feet.
  3. D=3.375 feet.
  4. AV=−0.65625 feet per second.
  5. s(t)=−t2+t.

4.3.5.10.

Answer.
  1. The total change in position, P, is P=∫01v(t)dt+∫13v(t)dt+∫34v(t)dt=∫04v(t)dt.
  2. P=∫04v(t)dt≈2.665.
  3. The total distance traveled, D, is D=∫01v(t)dt−∫13v(t)dt+∫34v(t)dt.
  4. D≈8.00016.
  5. vAVG[0,4]≈0.66625
    feet per second.

4.3.5.11.

Answer.
  1. ∫01[f(x)+g(x)]dx=1−π4.
  2. ∫14[2f(x)−3g(x)]dx=−152−3π.
  3. hAVG[0,4]=58+3π16.
  4. c=−38+3π16.

4.3.5.12.

Answer.
  1. A1=∫−11(3−x2)dx.
  2. A2=∫−112x2dx.
  3. The exact area between the two curves is ∫−11(3−x2)dx−∫−112x2dx.
  4. Use the sum rule for definite integrals over the same interval.
  5. Think about subtracting the area under q from the area under p.

4.4 The Fundamental Theorem of Calculus
4.4.5 Exercises

4.4.5.11.

Answer.
  1. 20 meters.
  2. vAVG[12,24]=12.5 meters per minute.
  3. The object’s maximum acceleration is 3 meters per minute per minute at the instant t=2.
  4. c=5.

4.4.5.12.

Answer.
  1. −56.
  2. fAVG[0,5]=12.
  3. g(x)=f(x) for 0≤x<5 and g(x)=−54(x−5) on 5≤x≤7.

4.4.5.13.

Answer.
  1. h (feet) 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10,000
    c (ft/min) 925 875 830 780 730 685 635 585 535 490 440
    m (min/ft) 1925 1875 1830 1780 1730 1685 1635 1585 1535 1490 1440
  2. The antiderivative function tells the total number of minutes it takes for the plane to climb to an altitude of h feet.
  3. M=∫010000m(h)dh.
  4. It takes the plane aabout M5≈15.27 minutes.

4.4.5.14.

Answer.
Yes.

4.4.5.15.

Answer.
  1. G′(x)=1−x⋅(−1)=1x.
  2. Since for those values of x, G′(x)=1x.
  3. If x<0, then H(x)=ln⁡(|x|)=ln⁡(−x)=G(x); if x>0, then H(x)=ln⁡(|x|)=ln⁡(x)=F(x).
  4. For x<0, H′(x)=G′(x)=1x; for x>0, H′(x)=F′(x)=1x for all x≠0.

5 Evaluating Integrals
5.1 Constructing Accurate Graphs of Antiderivatives
5.1.5 Exercises

5.1.5.5.

Answer.
  1. s(1)=53, s(3)=−1, s(5)=−113, s(6)=−52.
  2. s is increasing on 0<t<1 and 5<t<6; decreasing for 1<t<5.
  3. s is concave down for t<3; concave up for t>3.
  5. s(t)=−2t+16(t−3)3+5.

5.1.5.6.

Answer.
  1. C measures the total number of calories burned in the workout since t=0.
  2. C(5)=12.5, C(10)=50, C(15)=125, C(20)=187.5, C(25)=237.5, C(30)=262.5.
  4. C(t)=12.5+7.5(t−5) on this interval.

5.1.5.7.

Answer.
  1. B(−1)=−1, B(0)=0, B(1)=12, B(2)=0, B(3)=−1, B(4)=−32, B(5)=−1, B(6)=0. Also, A(x)=1+B(x) and C(x)=B(x)−12.
    x −1 0 1 2 3 4 5 6
    A(x) 0 1 1.5 1 0 −0.5 0 1
    B(x) −1 0 0.5 0 −1 −1.5 −1 0
    C(x) −1.5 −0.5 0 −0.5 −1.5 −2 −1.5 −0.5
  3. A, B, and C are vertical translations of each other.
  4. A′=f.

5.2 The Second Fundamental Theorem of Calculus
5.2.5 Exercises

5.2.5.5.

Answer.
F is increasing on x<−1, 0.5<x<4, and 5<x<6.5; decreasing on −1<x<0.5 and 4<x<5; concave up on approximately −0.4<x<2 and 4.5<x<6; concave down on approximately 2<x<4.5 and x>6; F(2)=0; F(0.5)=−6.06; F(−1)=−1.77; F(4)=6.69; F(5)=6.33; F(6.5)=8.12.

5.2.5.6.

Answer.
  1. The total sand removed on this time interval is
    ∫06[2+5sin⁡(4πt25)]dt.
  2. The total amount of sand on the beach at time x is given by
    Y(x)=∫0x[S(t)−R(t)]dt=∫0x[15t1+3t−(2+5sin⁡(4πt25))]dt.
  3. Y′(4)=S(4)−R(4)≈−1.90875 cubic yards per hour.
  4. Y has an absolute minimum on [0,6] of Y(5.118)≈2492.368.

5.2.5.7.

Answer.
  1. m(h)=1c(h).
    h (feet) 0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10,000
    c (ft/min) 925 875 830 780 730 685 635 585 535 490 440
    m (min/ft) 1925 1875 1830 1780 1730 1685 1635 1585 1535 1490 1440
  2. The antiderivative function tells us the total number of minutes that it takes for the plane to climb to an altitude of h feet.
  3. The number of minutes required for the airplane to ascend to 10,000 feet of altitude is given by the definite integral
    M=∫010000m(h)dh.
  4. The number of minutes required for the airplane to ascend to h feet of altitude is given by the definite integral
    M(h)=∫0hm(t)dt.
  5. Estimating the desired integral using 3 subintervals and midpoints,
    ∫06000m(h)dh≈7.77.
    Using 5 subintervals and midpoints,
    ∫010000m(h)dh≈15.27.

5.3 Integration by Substitution
5.3.5 Exercises

5.3.5.11.

Answer.
  1. ∫tan⁡(x)dx=ln⁡(|sec⁡(x)|)+C.
  2. ∫cot⁡(x)dx=−ln⁡(|csc⁡(x)|)+C.
  3. ∫sec2⁡(x)+sec⁡(x)tan⁡(x)sec⁡(x)+tan⁡(x)dx=ln⁡(|sec⁡(x)+tan⁡(x)|)+C.
  4. sec2⁡(x)+sec⁡(x)tan⁡(x)sec⁡(x)+tan⁡(x)=sec⁡(x).
  5. ∫sec⁡(x)dx=ln⁡(|sec⁡(x)+tan⁡(x)|)+C.
  6. ∫csc⁡(x)dx=−ln⁡(|csc⁡(x)+cot⁡(x)|)+C.

5.3.5.12.

Answer.
  1. ∫xx−1dx=∫(u+1)udu.
  2. ∫xx−1dx=25(x−1)5/2+23(x−1)3/2+C.
  3. ∫x2x−1dx=27(x−1)7/2+45(x−1)5/2+23(x−1)3/2+C.
    ∫xx2−1dx=13(x2−1)3/2+C.

5.3.5.13.

Answer.
  1. We don’t have a function-derivative pair.
  2. sin3⁡(x)=sin⁡(x)(1−cos2⁡(x)).
  3. u=cos⁡(x) and du=−sin⁡(x)dx.
  4. ∫sin3⁡(x)dx=13cos3⁡(x)−cos⁡(x)+C.
  5. ∫cos3⁡(x)dx=sin⁡(x)−13sin3⁡(x)+C.

5.3.5.14.

Answer.
  1. The model is reasonable because it appears to be periodic and the rate of consumption seems to peak at the times of day where people are most active in their homes.
  2. The total energy consumed in 24 hours, measured in megawatt-hours.
  3. ∫024r(t)dt≈95.7809 megawatt-hours of power used in 24 hours.
  4. rAVG[0,24]≈3.99087 megawatt-hours.

5.4 Integration by Parts
5.4.7 Exercises

5.4.7.13.

Answer.
  1. F′(x)=xe−2x.
  2. F(x)=−12xe−2x−14e−2x+14
  3. Increasing.

5.4.7.14.

Answer.
  1. ∫e2xcos⁡(ex)dx=∫zcos⁡(z)dz.
  2. ∫e2xcos⁡(ex)dx=exsin⁡(ex)+cos⁡(ex)+C.
  3. ∫e2xcos⁡(e2x)dx=12sin⁡(e2x)+C
    • ∫e2xsin⁡(ex)dx=sin⁡(ex)−zcos⁡(ex)+C.
    • ∫e3xsin⁡(e3x)dx=−13cos⁡(e3x)+C.
    • ∫xex2cos⁡(ex2)sin⁡(ex2)dx=14sin2⁡(ex2)+C.

5.4.7.15.

Answer.
  1. u-substitution; ∫x2cos⁡(x3) dx=13sin⁡(x3)+C.
  2. Both are needed; ∫x5cos⁡(x3) dx=13(x3sin⁡(x3)+cos⁡(x3))+C.
  3. Integration by parts; ∫xln(x2) dx=x22ln⁡(x2)−x22+C.
  4. Neither.
  5. u-substitution; ∫x3sin⁡(x4) dx=−14cos⁡(x4)+C.
  6. Both are needed; ∫x7sin⁡(x4) dx=−14x4cos⁡(x4)+14sin⁡(x4)+C.

5.5 Other Options for Finding Algebraic Antiderivatives
5.5.5 Exercises

5.5.5.6.

Answer.
  1. ∫x3+x+1x4−1dx=−12arctan⁡(x)+14ln⁡|x+1|+34ln⁡|x−1|+C
  2. ∫x5+x2+3x3−6x2+11x−6dx=x33+3x2+25x+2552ln⁡|x−3|−39ln⁡|x−2|+52ln⁡|x−1|+C.
  3. ∫x2−x−1(x−3)3dx=ln⁡|x−3|−5x−3−52(x−3)2+C.

5.5.5.7.

Answer.
  1. ∫1x9x2+25dx=−15ln⁡|5+9x2+523x|+C.
  2. ∫x1+x4dx=12(x22x4+1+12ln⁡|x2+x4+1|)+C
  3. ∫ex4+e2xdx=ex2e2x+4+2ln⁡|ex+e2x+4|+C
  4. ∫tan⁡(x)9−cos2⁡(x)dx=13ln⁡|3+9−cos2⁡(x)cos⁡(x)|+C.

5.5.5.8.

Answer.
  1. Try u=1+x2 or u=x+1+x2.
  2. Try u=x+1+x2 and dv=1xdx.
  3. No.
  4. It appears that the function x+1+x2x does not have an elementary antiderivative.

5.6 Numerical Integration
5.6.6 Exercises

5.6.6.5.

Answer.
  1. u-substitution fails since there’s not a composite function present; try showing that each of the choices of u=x and dv=tan⁡(x)dx, or u=tan⁡(x) and dv=xdx, fail to produce an integral that can be evaluated by parts.
    • L4=0.25892
    • R4=0.64827
    • M4=0.41550
    • T4=L4+R42=0.45360
    • S8=2M4+T43=0.42820
  3. L4 and M4 are underestimates; R4 and M4 are overestimates.

5.6.6.6.

Answer.
  1. Decreasing.
  2. Concave down.
  3. ∫36f(x)≈7.03.

5.6.6.7.

Answer.
  1. ∫060r(t)dt.
  2. ∫060r(t)dt>M3=204000.
  3. ∫060r(t)dt≈S6=6190003≈206333.33.
  4. 160S6≈3438.89; 2000+2100+2400+3000+3900+5100+65007=250007≈3571.43. each estimates the average rate at which water flows through the dam on [0,60], and the first is more accurate.

6 Using Definite Integrals
6.1 Using Definite Integrals to Find Area and Length
6.1.5 Exercises

6.1.5.10.

Answer.
  1. A=∫3−323+32−2y2+6y−3 dy=3.
  2. A=∫π/43π/4sin⁡(x)−cos⁡(x) dx=2.
  3. A=∫−15/2y+12−(y2−y−2) dy=34348.
  4. A=∫m−m2+42m+m2+42mx−(x2−1) dx=−13(m+m2+42)3−13(m−m2+42)3+m2(m+m2+42)2−m2(m−m2+42)2+(m+m2+42−m−m2+42).

6.1.5.11.

Answer.
a=12.

6.1.5.12.

Answer.
  1. r=43.
  3. A1=A2=4627.
  4. Yes.

6.2 Using Definite Integrals to Find Volume
6.2.5 Exercises

6.2.5.7.

Answer.
  1. L=∫01.842571+(−3sin⁡(x34)⋅34x2)2dx≈4.10521.
  2. A=∫01.845273cos⁡(x34)dx≈4.6623.
  3. V=∫01.84527π⋅9cos2⁡(x34)dx≈40.31965.
  4. V=∫03π(4arccos⁡(y3))2/3dy≈23.29194.

6.2.5.8.

Answer.
  2. A=∫0π4(cos⁡(x)−sin⁡(x))dx.
  3. V=∫0π4π(cos2⁡(x)−sin2⁡(x))dx.
  4. V=∫022πarcsin2⁡(y)dy+∫221πarccos2⁡(y)dy
  5. V=∫0π4π[(2−sin⁡(x))2−(2−cos⁡(x))2]dx.
  6. V=∫022π[(1+arcsin⁡(y))2−12]dy+∫221π[(1+arccos⁡(y))2−12]dy

6.2.5.9.

Answer.
  1. A=∫01.51+12(x−2)2−12x2 dx=2.25.
  2. V=∫01.5π[(2+12(x−2)2)2−(1+12x2)2] dx=31532π
  3. V=∫01.125π(2y)2 dy+∫1.1253π(2−2(y−1))2 dy≈7.06858347.
  4. P=3+∫01.51+(x−2)2+1+x2 dx≈7.387234642.

6.3 Density, Mass, and Center of Mass
6.3.5 Exercises

6.3.5.5.

Answer.
  1. a=−10ln⁡(0.7)≈3.567 cm.
  2. Left of the midpoint.
  3. x―≈50.333830≈1.687.
  4. q=−10ln⁡(0.85)≈1.625 cm.

6.3.5.6.

Answer.
  1. M1=arctan⁡(10)≈1.47113; M2=10−10e−1≈6.32121.
  2. x1―≈1.56857; x2―≈4.18023.
    1. M=∫010ρ(x)dx+∫010p(x)dx≈1.47113+6.32121=7.79234.
    2. ∫010x(ρ(x)+p(x)))dx=28.73167.
    3. False.

6.3.5.7.

Answer.
  2. V=∫030π(2xe−1.25x+(30−x)e−0.25(30−x))2dx≈52.0666 cubic inches.
  3. W≈0.6⋅52.0666=31.23996 ounces.
  4. At a given x-location, the amount of weight concentrated there is approximately the weight density (0.6 ounces per cubic inch) times the volume of the slice, which is Vtextslice≈πf(x)2.
  5. x―≈23.21415

6.4 Physics Applications: Work, Force, and Pressure
6.4.5 Exercises

6.4.5.6.

Answer.
  1. W≈1353.55 foot-pounds.
  2. W=∫0h3744xcos⁡(x34)dx.
  3. F≈462.637 pounds.

6.4.5.7.

Answer.
  1. W=1404(19π−8)≈305179.3 foot-pounds.
  2. F≈1123.2
    pounds.

6.5 Improper Integrals
6.5.5 Exercises

6.5.5.11.

Answer.
  1. Diverges.
  2. Diverges.
  3. Converges to 1.
  4. ∫e∞1x(ln⁡(x))pdx diverges if p≤1 and converges to 1p−1 if p>1.
  5. Diverges.
  6. Converges to −1.

6.5.5.12.

Answer.
  1. converges
  2. diverges
  3. diverges

7 Differential Equations
7.1 An Introduction to Differential Equations
7.1.5 Exercises

7.1.5.7.

Answer.
  1. dTdt|T=105=−2; when T=105, the coffee’s temperature is decreasing at an instantaneous rate of −2 degrees F per minute.
  2. T decreasing at t=0.
  3. T(1)≈103 degrees F.
  5. For T<75, T increases. For T>75, T decreases.
  6. Room temperature is 75 degrees F.
  7. Substitute T(t)=75+30e−t/15 in for T in the differential equation dTdt=−115T+5 and verify the equality holds; T(0)=75+30e0=75+30=105; T(t)=75+30e−t/15→75 as t→∞.

7.1.5.8.

Answer.
  1. 1<P<3.
  2. P<1 and 3<P<4.
  3. P will not change at all.
  4. The population will decrease toward P=0 with P always being positive.
  5. The population will increase toward P=3 with P always being between 1 and 3.
  6. The population will decrease toward P=3 with P always being above 3.
  7. There’s a maximum threshold of P=3.

7.1.5.9.

Answer.
    1. y(t)=t+1+2et is a solution to the DE.
    2. y(t)=t+1 is a solution to the DE.
    3. y(t)=t+2 is a not solution to the DE.
  2. k=9.

7.2 Qualitative behavior of solutions to DEs
7.2.4 Exercises

7.2.4.6.

Answer.
  2. Sketch curves through appropriate points in the slope field above.
  3. y(t)=t−1.
  4. t and y are equal.

7.2.4.7.

Answer.
  2. Any solution curve that starts with P(0)>3 will decrease to P(t)=3 as t→∞; any curve that starts with 1<P(0)<3 will increase to P(t)=3; any curve that starts with 0<P(0)<1 will decrease to P(t)=0.
  3. P=0, P=1, and P=3. P=1 is unstable; P=0 and P=3 are stable.
  4. The population will stabilize either at the value P=3 or at P=0.
  5. P(t)=1 is the threshold.

7.2.4.8.

Answer.
  1. A graph of f against P is given in blue in the figure below. The equilibrium solutions are P=0 (unstable) and P=6 (stable).
  2. dPdt=g(P)=P(6−P)−1 ; the equilibrium at P≈0.172 is unstable; the equilibrium at P≈5.83 is stable.
  3. If P<6−322, then the fish population will die out. If 6−322<P, then the fish population will approach 6+322 thousand fish.
  4. dPdt=g(P)=P(6−P)−h; equilibrium solutions P=6+36−4h2, 6−36−4h2.
  5. 9000 fish; harvesting at that rate will maintain the number of fish we start with, provided it’s at least 3000.

7.2.4.9.

Answer.
  1. dydt=20y.
  2. dydt=20y−Cy2+y
  3. For positive y near 0, M(y)=y2+y≈0; for large values of y, M(y)=y2+y≈1.
  4. The only equilibrium solution is y=0, which is unstable.
  5. The equilibrium solutions are y=0 (stable) and y=1 (unstable).
  6. At least 41 cats.

7.3 Euler’s method
7.3.4 Exercises

7.3.4.6.

Answer.
  1. Alice’s coffee: dTAdt|T=100=−0.5(30)=−15 degrees per minute; Bob’s coffee: dTBdt|T=100=−0.1(30)=−3 degrees per minute.
  2. Consider the insulation of the containers.
  3. Alice’s coffee:
    dTAdt=−0.5(TA−(70+10sin⁡t)),
    with the inital condition TA(0)=100.
    t TA(t)
    0.0 100
    0.1 98.5
    0.2 97.12492
    0.3 95.86801
    0.4 94.72237
    ⋮ ⋮
    49.6 65.56715
    49.7 65.48008
    49.8 65.43816
    49.9 65.44183
    50 65.49103
  4. t TA(t)
    0.0 100
    0.1 99.7
    0.2 99.41298
    0.3 99.13872
    0.4 98.87689
    ⋮ ⋮
    49.6 69.39515
    49.7 69.33946
    49.8 69.29248
    49.9 69.25467
    50 69.22638
  5. Compare the rate of initial decrease and amplitude of oscillation.

7.3.4.7.

Answer.
  1. K=1.054; y(1)=2.6991.
  2. K=1.272; y(1)=2.7169.
  3. K=0.122 and y(0.3)=0.0412.

7.3.4.8.

Answer.
  1. y(1)≈y5=2.7027.
  2. y(1)≈y10=2.7141.
  3. The square of Δt.

7.4 Separable differential equations
7.4.3 Exercises

7.4.3.8.

Answer.
  1. dMdt=kM.
  2. M(t)=M0ekt.
  3. M(t)=M0e−ln⁡(2)5730t≈M0e−0.000121t
  4. t=5730ln⁡(4)ln⁡(2)≈11460 years.
  5. t=−5730ln⁡(0.3)ln⁡(2)≈9952.8 years.

7.4.3.9.

Answer.
  1. y=64−t2.
  2. −8≤t≤8.
  3. y(8)=0.
  4. dydt=−ty is not defined when y=0.

7.4.3.10.

Answer.
  1. dhdt=kh.
  2. The tank with k=−10 has water leaving the tank much more rapidly.
  3. k=−2.
  4. h(t)=(10−t)2.
  5. 10 minutes.
  6. No.

7.4.3.11.

Answer.
  2. P=3 is stable.
  3. P(t)=3eln⁡(13)e−t.
  4. P(t)=3eln⁡(2)e−t.
  5. Yes.

7.5 Modeling with differential equations
7.5.3 Exercises

7.5.3.6.

Answer.
  1. dAdt=1+0.05A
  2. A(25)=49.80686 million dollars.
  3. A(25)=34.90343 million dollars.
  4. The first.
  5. t=20ln⁡(2)≈13.86 years.

7.5.3.7.

Answer.
  1. dvdt=9.8−kv
  2. v=9.8k is a stable equilibrium.
  3. v(t)=9.8−9.8e−ktk
  4. k=9.8/54≈0.181481.
  5. t=ln⁡(0.5)−0.181481≈3.1894 seconds.

7.5.3.8.

Answer.
  1. dwdt=kw.
  2. w(t)=17t+64; w(12)=268≈16.37 pounds.
  3. The model is unrealistic.

7.5.3.9.

Answer.
  1. The inflow and outflow are at the same rate.
  2. 60 grams per minute.
  3. S(t)100gramsgallon
  4. 3S(t)100gramsminute.
  5. dSdt=60−3100S.
  6. S=2000 is a stable equilibrium solution.
  7. S(t)=2000−2000e−3100t.
  8. S(t)→2000.

7.6 Population Growth and the Logistic Equation
7.6.4 Exercises

7.6.4.5.

Answer.
  1. p(t)→1 as t→∞ provided p(0)>0.
  2. p(t)=19e−0.2t+1.
  3. t=−5ln⁡(1/9)≈10.986 days.
  4. t=−5ln⁡(0.25/9)≈17.19 days.

7.6.4.6.

Answer.
  1. dbdt=13000b(15000−b)
  2. b=15000.
  3. When b=7500.
  4. t=−15ln⁡(1/70)≈0.8497 days.

7.6.4.7.

Answer.
  1. 10000 fish.
  2. dPdt=0.1P(10−P)−0.2P.
  3. 8000 fish.
  4. P(1)≈8.7899 thousand fish.
  5. t=−1.25ln⁡(5/11)≈0.986 years.

8 Taylor Polynomials and Taylor Series
8.1 Approximating f(x)=ex
8.1.5 Exercises

8.1.5.3.

Answer.
  1. Table 8.1.21. Formulas and values for f(x) and its first two derivatives, plus their values at a=0.
    f(x)= 13x3+14x2−2x−1
    f′(x)= x2+12x−2
    f″(x)= 2x+12
    f(0)= −1
    f′(0)= −2
    f″(0)= 12
  2. See the bottom half of the table in the previous item.
  3. T1(x)=−1−2x.
  4. Table 8.1.22. Formulas and values for f(x) and T2(x) and their first two derivatives.
    f(x)= 13x3+14x2−2x−1 T2(x)= c0+c1x+c2x2
    f′(x)= x2+12x−2 T2′(x)= c1+2c2x
    f″(x)= 2x+12 T2″(x)= 2c2
    f(0)= −1 T2(0)= c0
    f′(0)= −2 T2′(0)= c1
    f″(0)= 12 T2″(0)= 2c2
    T2(x)=14x2−2x−1
  5. T2(x) is a better approximation of f(x) than T1(x) and is better on a wider interval.

8.1.5.4.

Answer.
  1. Table 8.1.26. Formulas and values for f(x) and T3(x).
    f(x)= 13x3+14x2−2x−1 T3(x)= k0+k1x+k2x2+k3x3
    f′(x)= x2+12x−2 T3′(x)= k1+2k2x+3k3x2
    f″(x)= 2x+12 T3″(x)= 2k2+6k3x
    f‴(x)= 2 T3‴(x)= 6k3
    f(0)= −1 T3(0)= k0
    f′(0)= −2 T3′(0)= k1
    f″(0)= 12 T3″(0)= 2k2
    f‴(0)= 2 T3″(0)= 6k3
  2. T3(x)=13x3+14x2−2x−1.
  3. T3(x)=f(x) so T3(x) is identical to the original function.
  4. T4(x)=T3(x)=f(x).
  5. That the degree 6 approximation will be the original polynomial function itself.

8.2 Taylor Polynomials
8.2.4 Exercises

8.2.4.7.

Answer.
  1. T4(x)=2−3x−12!x2−34!x4.
  2. f(0.5)≈T4(0.5)=2−3⋅0.5−12!(0.5)2−34!(0.5)4=0.3671875.
  3. From T4(x), it follows
    T3(x)=(2−3x−12!x2+0x3T2(x)=(2−3x−12!x2T1(x)=(2−3x.

8.2.4.8.

Answer.
  1. T2(x)=3+0(x−2)+12!(x−2)2
  2. f(x)=T2(x).
  3. (2,3) is the vertex of the quadratic function f(x).
  4. L(x)=T1(x)=3, so f has a horizontal tangent line at x=2, which corresponds to its global minimum at x=2.

8.2.4.9.

Answer.
  1. T6(x)=1−12!(x−12)2+14!(x−12)4−16!(x−12)6
  2. T6(x) appears to be the same as P6(x−π2), the degree 6 Taylor polynomial of cos⁡(x) centered at a=0, shifted π2 units to the right.
  3. Since sin⁡(x)=cos⁡(x−π2), this tells us that the sine function is simply a shifted version of the cosine function (the cosine function shifted π2 units to the right).

8.2.4.10.

Answer.
  1. ln⁡(1.5)=f(0.5)≈T5(0.5)=1(0.5)−12(0.5)2+13(0.5)3−14(0.5)4+15(0.5)5=0.40729166―.
  2. ln⁡(1.5)=g(1.5)≈P5(1.5)=1(1.5−1)−12(1.5−1)2+13(1.5−1)3−14(1.5−1)4+15(1.5−1)5=0.40729166―.
  3. The same. This occurs because f(x)=g(1+x).

8.3 Geometric Sums
8.3.5 Exercises

8.3.5.7.

Answer.
  1. For Sn=1+1+⋯+1,
    1. S2=1+1=2, S3=1+1+1=3, S4=4, and Sn=n.
    2. The infinite sum S=1+1+⋯+1+⋯ diverges.
  2. For Sn=1−1+1−1+⋯+(−1)n−1,
    1. S2=1−1=0, S3=1−1+1=1, S4=1−1+1−1=0, S5=1−1+1−1+1=1, and Sn=0 when n is even, and Sn=1 when n is odd.
    2. The infinite sum S=1−1+1−1+⋯+1−1+⋯ diverges.
  3. For Sn=1+2+4+⋯+2n−1.
    1. S2=1−22−1=3, S3=1−23−1=7, and S4=1−24−1=15, so Sn=2n−1.
    2. The infinite geometric series 1+2+4+⋯+2n−1+⋯ diverges.

8.3.5.8.

Answer.
  1. 30⋅500=1500 dollars.
  2. Day Pay on this day Total amount paid to date
    1 $0.01 $0.01
    2 $0.02 $0.03
    3 $0.04 $0.07
    4 $0.08 $0.15
    5 $0.16 $0.31
    6 $0.32 $0.63
    7 $0.64 $1.27
    8 $1.28 $2.55
    9 $2.56 $5.11
    10 $5.12 $10.23
  3. $0.01(230−1)=$10,737,418.23.

8.3.5.9.

Answer.
  1. h1=(34)h.
  2. h2=(34)h1=(34)2h.
  3. h3=(34)h2=(34)3h.
  4. hn=(34)hn−1=(34)nh.
  5. The distance traveled by the ball is 7h, which is finite.

8.4 Taylor Series
8.4.4 Exercises

8.4.4.8.

Answer.
  1. T4(x)=0−1(x−π2)+02!(x−π2)2+13!(x−π2)3+04!(x−π2)4; T(x)=0−1(x−π2)+02!(x−π2)2+13!(x−π2)3+04!(x−π2)4+⋯+(−1)n+11(2n+1)!(x−π2)2n+1+⋯; think about the Taylor series centered at a=0 for g(x)=sin⁡(x) and note that f(x)=cos⁡(x)=−sin⁡(x−π2)=−g(x−π2).
  2. T4(x)=12−122(x−1)+123(x−1)2−124(x−1)3+125(x−1)4; T(x)=12−122(x−1)+123(x−1)2−124(x−1)3+125(x−1)4+⋯+12n+1(x−1)n+⋯.

8.4.4.9.

Answer.
  1. P4(x)=x2.
    1. g(x)=T(x2)=x2−13!x6+15!x10−⋯.
    2. All real numbers.

8.5 Finding and Using Taylor Series
8.5.5 Exercises

8.5.5.7.

Answer.
  1. C(x)=x−15⋅2!x5+19⋅4!x9−113⋅6!t13+⋯.
  2. For all real numbers x.
  3. C(0.5)≈12−15⋅2!⋅25=159320=0.496875, which is accurate to within 19⋅4!⋅29=1110592=0.00000904….
  4. S(x)=13x3−7⋅3!x7+111⋅5!x11−⋯.
  5. For all real numbers x.
  6. S(0.8)≈13(45)3−17⋅3!(45)7=2718081640625=0.16567…, which is accurate to within 111⋅5!(45)11=0.00006507….

8.5.5.8.

Answer.
  1. a0=1.
  2. Both series expanstions for ex have “1” as their constant term, and thus a1=a0=1.
  3. By equating the coefficients of the linear terms a1x and 2a2x, it follows a1=2a2.
  4. a3=13a2=13⋅2=13!, a4=14a3=14⋅3!=14!, and a5=15a4=15⋅4!=15!; so in general, ak=1k!.

8.5.5.9.

Answer.
  1. ∫0141+x2dx=4−43+45−47+⋯.
  2. It would take n=200 terms in the sum to get an approximation within 0.01.
  3. ∫0141+x2dx=4arctan⁡(x)|01=π.
  4. π=4−43+45−47+⋯.

8.5.5.10.

Answer.

9 Supplementary material
9.1 Review of Prerequsites for Calculus I

Exercises

9.2 Review of Prerequsites for Calculus II

Exercises

9.3 Integrating Rational Functions
9.3.2 Exercises

9.4 Integration with Trigonometric Functions
9.4.2 Exercises

9.5 Parametric Curves

Exercises

9.6 Calculus in Polar Coordinates

Exercises

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