Calculus Notes

Functions
Limits
Derivatives
Implicit Differentiation
Mean Value Theorem
Maxima and Minima
Related Rates Problems
Optimization Problems
Integration by Substitution
Average Values of Functions
Work
Surfaces of Revolution
Volumes from Cross-Sections
Volumes of Revolution
Approximate Integration
Derivatives of Integrals
Infinite Sequences and Series

PROBLEM

Find the numbers at which the function

f(x) = x + 2 if x < 0
f(x) = e^x if 0 < x < 1
f(x) = 2 – x if x > 1

is discontinuous. At which of these points is f continuous from the right, from the left, or neither? Sketch the graph of f.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 2.4.29]

SOLUTION

A graph of f shows that it is discontinuous at x = 0 and x = 1. However, since f(x) = e^x at x = 0 and x = 1, f is continuous from the right at x = 0 and continuous from the left at x = 1.

Suppose we have two functions f(x) and g(x) that are differentiable near x = a, and either

(1) lim(x => a) f(x) = 0 and lim(x => a) g(x) = 0

or

(2) lim(x => a) f(x) = ± ∞ and lim(x => a) g(x) = ± ∞

Then lim(x => a) [f(x) / g(x)] is an indeterminate limit of type 0/0 or ± ∞/∞, respectively. In either of these cases, L’Hospital’s rule states that if g’(x) ≠ 0 near a, except possibly at a, then

lim(x => a) [f(x) / g(x)] = lim(x => a) [f’(x) / g’(x)]

PROBLEM

Use continuity to evaluate the limit

lim(x => 4) {[5 + sqrt(x)] / sqrt(5 + x)}

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 2.4.25]

SOLUTION

The square root function (sqrt(x)) is continuous at x = 4 and x = 9. Thus,

lim(x => 4) {[5 + sqrt(x)] / sqrt(5 + x)}
= {5 + sqrt[lim (x => 4) x]} / sqrt[5 + lim (x => 4) x]
= [5 + sqrt(4)] / sqrt(5 + 4) = 7/3

PROBLEM

Use continuity to evaluate the limit

lim(x => 1) e^x2-x

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 2.4.27]

SOLUTION

The exponential function (exp(x) or e raised to the power x) is continuous at x = 1. Thus,

lim(x => 1) e^x2-x = lim(x => 1) exp(x² - x) = exp[lim(x => 1) (x² - x)] = exp(0) = 1

PROBLEM

Use continuity to evaluate the limit

lim(x => 2) arctan[(x² – 4) / (3x² – 6x)]

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 2.4.28]

SOLUTION

The inverse tangent function (arctan(x)) is continuous at x = 2. Thus,

lim(x => 2) arctan[(x² – 4) / (3x² – 6x)]
= lim(x => 2) arctan{(x – 2)(x + 2) / [(3x)(x – 2)]}
= lim(x => 2) arctan[(x + 2) / (3x)] = arctan {lim(x => 2) [(x + 2) / (3x)]}
= arctan(2/3)

PROBLEM

Evaluate lim(x => 0) {[- ln(1 - x) - sin x] / (1 - cos² x)}.

SOLUTION

The numerator and denominator both approach zero as x approaches zero. Use l’Hospital’s rule.

lim(x => 0) {[- ln(1 - x) - sin x] / (1 - cos² x)}
= lim(x => 0) {[1/(1-x) - cos x] / (2 cos x sin x)}
= (1/2) lim(x => 0) {[1/(1-x) - cos x] / (cos x sin x)}
= (1/2) lim(x => 0) {[1/(1-x)² + sin x] / (cos x cos x - sin x sin x)}
= 1/2

Note: Here we use l’Hospital’s rule twice because the limits of the first derivatives as x approaches zero are both zero.

PROBLEM

Evaluate lim(x => 0) {[ln(1 - x) - sin x] / (1 - cos² x)}.

[from the movie Mean Girls (2004)]

SOLUTION

The numerator and denominator both approach zero as x approaches zero. Use l’Hospital’s rule.

lim(x => 0) {[ln(1 - x) - sin x] / (1 - cos² x)}
= lim(x => 0) {[- 1/(1-x) - cos x] / (2 cos x sin x)}

After applying l’Hospital’s rule, we see that the new numerator approaches – 2 as x approaches zero, while the denominator approaches zero as x approaches zero. We need to consider separately the cases where x approaches zero from the positive side and from the negative side. We get

lim(x => 0+) {[ln(1 - x) - sin x] / (1 - cos² x)}
= lim(x => 0+) {[- 1/(1-x) - cos x] / (2 cos x sin x)} = - 2/0+ = - ∞

lim(x => 0-) {[ln(1 - x) - sin x] / (1 - cos² x)}
= lim(x => 0-) {[- 1/(1-x) - cos x] / (2 cos x sin x)} = - 2/0- = + ∞

Thus, {[ln(1 - x) - sin x] / (1 - cos² x)} approaches - ∞ as x approaches zero from the positive side and + ∞ as x approaches zero from the negative side.

PROBLEM

If an initial amount A₀ of money is invested at an interest rate i compounded n times a year, the value of the investment after t years is

A = A₀(1 + i/n)^nt

If we let n => ∞, we refer to the continuous compounding of interest. Use l’Hospital’s Rule to show that if interest is compounded continuously, then the amount after n years is

A = A₀e^it

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 2.4.53]

SOLUTION

lim (n => ∞) A = lim (n => ∞) e^{ln A} = lim (n => ∞) exp(ln A)
= exp[lim (n => ∞) ln A] = exp[lim (n => ∞) ln A₀(1 + i/n)^nt]
= exp{lim (n => ∞) [ln A₀ + nt ln(1 + i/n)]}
= exp[ln A₀ + lim (n => ∞) nt ln(1 + i/n)]
= exp[ln A₀ + t lim (n => ∞) n ln(1 + i/n)]

Now

lim (n => ∞) n ln(1 + i/n) = lim (n => ∞) [ln(1 + i/n)] / (1/n)

This is an indeterminate limit of the form 0/0, so we use l’Hospital’s Rule to get

lim (n => ∞) n ln(1 + i/n) = lim (n => ∞) {(- i/n²)[1 / (1 + i/n)] / (- 1/n²)}
= lim (n => ∞) [i / (1 + i/n)] = i

Thus,

lim (n => ∞) A = exp(ln A₀ + it) = exp(ln A₀) exp(it) = A₀e^it

PROBLEM

Calculate the derivative of log_a x with respect to x.

SOLUTION

Write log_a x in terms of natural logs. Define y = log_a x. Then

a^y = x => y ln a = ln x => y = (ln x) / (ln a) => dy/dx = 1 / (x ln a)

PROBLEM

Calculate the derivative of a^x with respect to x.

SOLUTION

Write a^x in terms of e. Define y = a^x. Then

y = (e^{ln a})^x = e^{x ln a} => dy/dx = (ln a) e^{x ln a} = a^x ln a

PROBLEM

Find an equation of the tangent line to the following curve at the given point: y² = x³(2 - x) at (1, 1). The curve is a piriform.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 3.6.15]

SOLUTION

Taking the derivative of both sides, we get

2y dy/dx = 3x²(2 - x) - x³ = x²[3(2 - x) - x] = 2x²(3 - 2x)
=> dy/dx = (x²/y)(3 - 2x) = (1²/1)[3 - 2(1)] = 1

The tangent at (1, 1) has slope dy/dx = 1. The equation of a straight line having slope m and passing through the point (x₀, y₀) is

y - y₀ = m(x - x₀)

Thus, the equation of the required tangent is

y - 1 = x - 1 => y = x

See Stewart3.6.15.060413.xls for a graph of the curve and the tangent at (1, 1).

According to Rolle's theorem, if a function f(x) is continuous on the interval a < x < b, if f(a) =
f(b) = 0, and if f '(x) exists everywhere on the interval except possibly at the endpoints, then f '(x) = 0 for at least one value of x = x₀ between a and b.

According to the mean value theorem, if f(x) is continuous on the interval a < x < b, and if f '(x) exists everywhere on the interval except possibly at the endpoints, then there is at least one value of x = x₀ between a and b such that

[f(b) - f(a)] / (b - a) = f '(x₀)

Solving for f(b), we get

f(b) = f(a) + f '(x₀)(b - a)

---

PROBLEM

Find a value for x₀ as prescribed by Rolle's theorem, given:
(a) f(x) = x² - 4x + 3, 1 < x < 3
(b) f(x) = sin x, 0 < x < p
(c) f(x) = cos x, p/2 < x < 3p/2

[from Ayres, Frank Jr., and Mendelson, Elliot 1990, Differential and Integral Calculus, Third Edition (New York: McGraw-Hill), problem 26.16]

SOLUTION

(a) Find a value of x in the interval 1 < x < 3 for which f '(x) = 0.

f '(x) = 2x - 4 = 0 => x = 2

(b) Find a value of x in the interval 0 < x < p for which f '(x) = 0.

f '(x) = cos x = 0 => x = p/2

(c) Find a value of x in the interval p/2 < x < 3p/2 for which f '(x) = 0.

f '(x) = - sin x = 0 => x = p

PROBLEM

Find a value for x₀, as prescribed by the mean value theorem, given:
(a) y = x³, 0 < x < 6
(b) y = ax² + bx + c, x₁ < x < x₂
(c) y = ln x, 1 < x < 2e

[from Ayres, Frank Jr., and Mendelson, Elliot 1990, Differential and Integral Calculus, Third Edition (New York: McGraw-Hill), problem 26.17]

SOLUTION

(a) Find a value of x for which

y'(x) = [y(6) - y(0)] / (6 - 0) = 36

The derivative of y with respect to x is

y'(x) = 3x² = 36 => x = sqrt(12) = 2 sqrt(3)

(b) Find a value of x for which

y'(x) = [y(x₂) - y(x₁)] / (x₂ - x₁) = (ax₂² + bx₂ + c - ax₁² - bx₁ - c) / (x₂ - x₁)
= [a(x₂ - x₁)(x₂ + x₁) + b(x₂ - x₁)] / (x₂ - x₁) = a(x₂ + x₁) + b

The derivative of y with respect to x is

y'(x) = 2ax + b

Equating the two above expressions for y'(x), we get

a(x₂ + x₁) + b = 2ax + b => x = (1/2)(x₁ + x₂)

(c) Find a value of x for which

y'(x) = [y(2e) - y(1)] / (2e - 1) = (ln 2 + 1) / (2e - 1)

The derivative of y with respect to x is

y'(x) = 1/x

Equating the two above expressions for y'(x), we get

(ln 2 + 1) / (2e - 1) = 1/x => x = (2e - 1) / (ln 2 + 1)

PROBLEM

Use the mean value theorem to approximate 65^1/6.

[from Ayres, Frank Jr., and Mendelson, Elliot 1990, Differential and Integral Calculus, Third Edition (New York: McGraw-Hill), problem 26.4]

SOLUTION

Let f(x) = x^1/6, a = 64, and b = 65. The derivative of f(x) is

f '(x) = (1/6)x^-5/6

From the mean value theorem, we have

f(b) = f(a) + f '(x₀)(b - a) ≈ f(a) + f '(a)(b - a) = 64^1/6 + (1/6)(64^-5/6)(65 - 64) = 2 + (1/6)(1/32) = 2 + 1/192 = 2.005208

The exact answer, rounded off to six decimal places, is 2.005175.

PROBLEM

Use the mean value theorem to approximate
(a) sqrt(15)
(b) (3.001)³
(c) 1/999

[from Ayres, Frank Jr., and Mendelson, Elliot 1990, Differential and Integral Calculus, Third Edition (New York: McGraw-Hill), problem 26.18]

SOLUTION

(a) Let f(x) = x^1/2, a = 16, and b = 15. The derivative of f(x) is

f '(x) = (1/2)x^-1/2

From the mean value theorem, we have

f(b) = f(a) + f '(x)(b - a) ≈ f(a) + f '(a)(b - a) = 4 + (1/8)(15 - 16) = 4 - 1/8 = 3.875

The exact answer, rounded off to six decimal places, is 3.872983.

(b) Let f(x) = x³, a = 3, and b = 3.001. The derivative of f(x) is

f '(x) = 3x²

From the mean value theorem, we have

f(b) = f(a) + f '(x)(b - a) ≈ f(a) + f '(a)(b - a) = 27 + (27)(3.001 - 3) = 27 + 0.027 = 27.027

The exact answer, rounded off to five decimal places, is 27.02701.

(c) Let f(x) = 1/x, a = 1000, and b = 999. The derivative of f(x) is

f '(x) = - 1/x²

From the mean value theorem, we have

f(b) = f(a) + f '(x)(b - a) ≈ f(a) + f '(a)(b - a) = 1/1000 + (- 1/10⁶)(999 - 1000) = 0.001 + 0.000001 = 0.001001

The exact answer, rounded off to six decimal places, is 0.001001.

PROBLEM

Consider the function y(x) = e^2x+1 / ln x.
(a) Find its relative maxima and minima.
(b) Find its absolute maximum and minimum.

SOLUTION

(a) y(x) = e^2x+1 / ln x

Find the derivative of y(x), set it equal to zero, and solve for x.

y'(x) = - (1/x) e^2x+1 / (ln x)² + 2e^2x+1 / (ln x) = - (1/x) e^2x+1 / (ln x)² + (2 ln x)e^2x+1 / (ln x)²
= (- 1/x + 2 ln x)e^2x+1 / (ln x)² = 0 => - 1/x + 2 ln x = 0 => x = e^1/2x

Solve this iteratively for x. For example, try x = 1, substitute this into the right hand side, and get x = e^1/2 = 1.648721. Substitute this into the right hand side and get x = 1.354274. After several iterations, the solution converges to x = 1.42153.

Find the second derivative of y(x) and evaluate it at x = 1.42153 to see if y(1.42153) = 132.6801 is a relative minimum or maximum.

y"(x) = (1/x² + 2/x)e^2x+1 / (ln x)² + (- 1/x + 2 ln x)[2e^2x+1 / (ln x)² - (2/x)e^2x+1 / (ln x)³]

y"(1.42153) = 717.39 > 0 => y(1.42153) = 132.6801 is a relative minimum.

For a sketch of the function y(x) = e^2x+1 / ln x, see the file Calculus060114.xls.

(b) The absolute minimum is -∞ at x = 1-. The absolute maximum is +∞ at x = 1+ and x = ∞.

PROBLEM

A helicopter, hovering at an elevation of 15840 ft, drops a package vertically downward. The vertical position of the package at time t is given by

y(t) = 15840 - 16t²

where y is in feet and t is in seconds. The path of the package lies 100 ft in front of a cliff of height 600 ft. An observer standing at the top of the cliff watches the package fall.
(a) Find the angular position of the package, q(t), as seen by the observer. q(t) is the angle between the horizontal and the line of sight to the package, measured positive in the upward direction.
(b) Find the angular velocity of the package, dq/dt.
(c) At what time is the angular velocity maximized?

SOLUTION

(a) Let h = 600 ft and d = 100 ft. Then the elevation of the package relative to the top of the cliff is y(t) - h.

The tangent of the angle q(t) is

tan q(t) = [y(t) - h] / d (I)

q(t) = tan^-1{[y(t) - h] / d} = tan^-1[(15840 - 16t² - 600) / 100] = tan^-1(152.40 - 0.16t²) (in radians)

To convert to degrees, multiply by 180/p.

(b) Take the time derivative of (I) and solve for dq/dt.

(d/dt) tan q(t) = y'(t)/d => [sec² q(t)] dq/dt = - 32t/d
=> dq/dt = - (32t/d) cos² q(t) = - (32t/d) d² / {d² + [y(t) - h]²} = - (32td) / {d² + [y(t) - h]²}
= - [32t(100)] / {100² + [15840 - 16t² - 600]²} = - 3200t / [10000 + (15240 - 16t²)²]
= - 32t / [100 + (1524 - 1.6t²)²] (in radians/s)

To convert to degrees/s, multiply by 180/p.

(c) dq/dt = - 32t / [100 + (1524 - 1.6t²)²]

Find the second derivative of q(t), which is the first derivative of dq/dt, set it equal to zero, and solve for t.

d²q/dt² = - 32 / [100 + (1524 - 1.6t²)²] + 32t(2)(1524 - 1.6t²)(- 3.2t) / [100 + (1524 - 1.6t²)²]²
= - 32[100 + (1524 - 1.6t²)²] / [100 + (1524 - 1.6t²)²]²
- 204.8t(1524 - 1.6t²) / [100 + (1524 - 1.6t²)²]²
= 0

=> - 32[100 + (1524 - 1.6t²)²] - 204.8t²(1524 - 1.6t²) = 0
=> - 32(100 + 2322576 - 4876.8t² + 2.56t⁴) - 312115.2t² + 327.68t⁴ = 0
=> - 32(2322676 - 4876.8t² + 2.56t⁴) - 312115.2t² + 327.68t⁴ = 0
=> - 74325632 + 156057.6t² - 81.92t⁴ - 312115.2t² + 327.68t⁴ = 0
=> - 74325632 - 156058t² + 245.76t⁴ = 0
=> 245.76t⁴ = 74325632 + 156058t²
=> t = [(74325632 + 156058t²) / 245.76]^1/4

Solve this iteratively for t. Start with t = 50, for example. Substitute it into the right hand side, solve for a new value of t, and repeat until the solution converges to t = 30.86278 s. At this value of t, dq/dt has its minimum value of - 9.876 rad/s and its maximum magnitude of 9.876 rad/s.

For graphs of y(t) vs. t, q(t) vs. t, and dq/dt vs. t, see the file Calculus060122.xls.

PROBLEM

If a snowball melts so that its surface area decreases at a rate of 1 cm²/min, find the rate at which the diameter decreases when the diameter is 10 cm.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 4.1.5]

SOLUTION

The surface area S of a sphere is related to its radius r by

S = 4pr²

Since the diameter

D = 2r => r = D/2

we have

S = 4p(D/2)² = pD²

Take the derivative with respect to time.

dS/dt = p(2D)dD/dt = 2pD dD/dt => dD/dt = (1/2pD) dS/dt = [1/(2p)(10 cm)](- 1 cm²/min)
= - [1/(20p)] cm/min

PROBLEM

A plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 4.1.7]

SOLUTION

Define

h = altitude of plane = 1 mi
v = speed of plane = 500 mi/h
r = distance from radar station to plane
x = horizontal coordinate of plane relative to horizontal coordinate of radar station (i.e., radar station is located at x = 0)

The distance from the radar station to the plane is

r = sqrt(x² + h²)

The derivative of r with respect to time is

dr/dt = (1/2)(2x dx/dt) / sqrt(x² + h²) = (x dx/dt) / sqrt(x² + h²) = xv/r

x can be expressed in terms of r and h by writing

x = sqrt(r² - h²)

So

dr/dt = (v/r) sqrt(r² - h²) = [(500 mi/h) / (2 mi)] sqrt[(2 mi)² - (1 mi)²] = 250 sqrt(3) mi/hr

PROBLEM

Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P located 12 ft above the ground. The point Q is on the floor directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 2 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q?

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 4.1.17]

SOLUTION

Let (x_A, x_B) = distance from (A, B) to Q, h = distance from P to Q = 12 ft, and L = length of rope = 39 ft. The length of the rope is equal to the sum of the distances from A to P and from B to P,

L = sqrt(x_A² + h²) + sqrt(x_B² + h²) (1)

Take the derivative of both sides with respect to time.

0 = (1/2)(2x_A)(dx_A/dt) / sqrt(x_A² + h²) + (1/2)(2x_B)(dx_B/dt) / sqrt(x_B² + h²)
= (x_A)(dx_A/dt) / sqrt(x_A² + h²) + (x_B)(dx_B/dt) / sqrt(x_B² + h²)
=> dx_B/dt = - (x_A/x_B)(dx_A/dt)[sqrt(x_B² + h²)] / [sqrt(x_A² + h²)] (2)

Solve (1) for x_B using x_A = 5 ft.

x_B = sqrt{[L - sqrt(x_A² + h²)]² - h²} = [{39 ft - sqrt[(5 ft)² + (12 ft)²]}² - (12 ft)²] = sqrt(532) ft
= 2 sqrt(133) ft

Substitute this result into (2) using dx_A/dt = 2 ft/s.

dx_B/dt = - (x_A/x_B)(dx_A/dt)[sqrt(x_B² + h²)] / [sqrt(x_A² + h²)]
= - {(5 ft) / [2 sqrt(133) ft]}(2 ft/s) sqrt{[2 sqrt(133) ft]² + (12 ft)²} / sqrt[(5 ft)² + (12 ft)²]
= - {10 / [sqrt(133)]} ft/s = 0.8671 ft/s

PROBLEM

Two sides of a triangle are a = 4 m and b = 5 m in length and the angle between them is increasing at a rate of dq/dt = 0.06 rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is q = p/3.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 4.1.23]

SOLUTION

The area of the triangle is

A = (1/2)ab sin q

The rate at which the area is changing is

dA/dt = (1/2)ab cos q dq/dt = (1/2)(4 m)(5 m)(cos p/3)(0.06 rad/s) = 0.3 m²/s

PROBLEM

For what values of the numbers a and b does the function

f(x) = ax exp(bx²)

have the maximum value f(2) = 1?

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 4.3.42]

SOLUTION

f '(x) = ax(2bx) exp(bx²) + a exp(bx²) = 0 => 2bx² + 1 = 0 => x² = - 1 / 2b => x = ± sqrt(- 1 / 2b)

Requiring x = 2 yields b = - 1/8. Then requiring f(2) = 1 yields

f(2) = 2a e^-1/2 = 2a / e^1/2 = 1 => a = (1/2) e^1/2

A graph of f(x) vs. x is shown in Stewart4.3.42.060413.xls.

PROBLEM

The wall of a building is to be braced by a beam that must pass over a parallel fence 5 feet high and 4 feet from the building. Find the length of the shortest beam that can be used.

SOLUTION

(0,H)
  +
  |\
  | \
  |  \
  |   \
  |    \
  |     \
  |H     \
  |       \
  |       |\
  |       | \
  |      h|  \
  |       |   \
  |   d   |    \
  +-------+-----+
(0,0)         (x,0)

Let the intersection of the wall and the ground be at (x, y) = (0, 0). Let the beam extend from the wall at (0, H) to the top of the fence at (4, 5) to the ground at (x, 0). The length of the beam is

L(x) = sqrt(x² + H²) (1)

The triangle formed by the wall, ground, and beam is similar to the triangle formed by the fence, ground, and beam. Thus,

H / x = h / (x - d) => H = hx / (x - d)

where h = 5 ft and d = 4 ft. If we take the derivative of (1) with respect to x and set it equal to zero, we get

L'(x) = (1/2)(2x + 2H dH/dx) / sqrt(x² + H²) = (x + H dH/dx) / sqrt(x² + H²) = 0
=> x + H dH/dx = 0

The derivative of H with respect to x is

dH/dx = h / (x - d) - hx / (x - d)² = - hd / (x - d)²

So

x + [hx / (x - d)][- hd / (x - d)²] = 0 => x - h²dx / (x - d)³ = 0 => x[1 - h²d / (x - d)³] = 0

From this we see that either x = 0 or 1 - h²d / (x - d)³ = 0. x = 0 is physically impossible because the beam would not go over the fence. So

1 - h²d / (x - d)³ = 0 => x = (h²d)^1/3 + d = (5²4)^1/3 + 4 = (100^1/3 + 4) ft = 8.642 ft
=> H = hx / (x - d) = (5)(8.642) / (8.642 - 4) = 9.309 ft
=> L = sqrt(x² + H²) = sqrt(8.642² + 9.309²) = 12.702 ft

The length of the shortest beam that can be used is 12.702 ft.

PROBLEM

Find two positive integers such that the sum of the first number and four times the second number is 1000 and the product of the numbers is as large as possible.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), review problem 4.37]

SOLUTION

Let N₁ and N₂ be the two numbers. We require that

N₁ + 4N₂ = 1000

and

P = N₁N₂ is maximized.

N₁ = 1000 - 4N₂ => P = (1000 - 4N₂)N₂ = 1000N₂ - 4N₂²

Maximize P with respect to N₂.

P'(N₂) = 1000 - 8N₂ = 0 => N₂ = 125

The second derivative of P with respect to N₂ is

P"(N₂) = - 8 < 0 => P is maximized when N₂ = 125.

N₁ = 1000 - 4N₂ = 1000 - (4)(125) = 500 => The two numbers are N₁ = 500 and N₂ = 125.

PROBLEM

A hockey team plays in an arena with a seating capacity of 15,000 spectators. With the ticket price set at $12, average attendance at a game has been 11,000. A market survey indicates that for each dollar the ticket price is lowered, average attendance will increase by 1000. How should the owners of the team set the ticket price to maximize their revenue from ticket sales?

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), review problem 4.45]

SOLUTION

Let R = revenue, P = price per ticket, and N_tickets = number of tickets sold.

R = PN_tickets

From the market survey, we know that

N_tickets = 11,000 + 1000(12 - P)

This obeys the requirements that (1) the number of tickets sold when P = 12 is 11,000 and (2) for every dollar that the price per ticket is lowered 1000 additional tickets are sold. Thus,

R(P) = P[11,000 + 1000(12 - P)]

We have to maximize R with respect to P. Take the derivative of R with respect to P and set it equal to zero.

R'(P) = P[(1000)(-1)] + [11,000 + 1000(12 - P)] = - 1000P + 11,000 + 12,000 - 1000P = - 2000P + 23,000 = 0 => P = 11.5

The second derivative of R with respect to P is

R"(P) = - 2000 < 0 => P = 11.5 maximizes R => The ticket price should be set at $11.50 to maximize revenue from ticket sales.

PROBLEM

Evaluate ∫ cos 3x dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.1]

SOLUTION

Define u = 3x => du = 3 dx

∫ cos 3x dx = (1/3) ∫ cos u du = (1/3) sin u + C = (1/3) sin 3x + C

PROBLEM

Evaluate ∫ x² sqrt(x³ + 1) dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.3]

SOLUTION

Define u = x³ + 1 => du = 3x² dx

∫ x² sqrt(x³ + 1) dx = (1/3) ∫ sqrt(u) du = (1/3)u^3/2 / (3/2) + C = (2/9)u^3/2 + C = (2/9)(x³ + 1)^3/2 + C

PROBLEM

Evaluate ∫ [4 / (1 + 2x)³] dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.5]

SOLUTION

Define u = 1 + 2x => du = 2 dx

∫ [4 / (1 + 2x)³] dx = ∫ 2 du / u³ = - 1 / u² + C = - 1 / (1 + 2x)² + C

PROBLEM

Evaluate ∫ 2x(x² + 3)⁴ dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.7]

SOLUTION

Define u = x² + 3 => du = 2x dx

∫ 2x(x² + 3)⁴ dx = ∫ u⁴ du = (1/5)u⁵ + C = (1/5)(x² + 3)⁵ + C

PROBLEM

Evaluate ∫ [(ln x)² / x] dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.9]

SOLUTION

Define u = ln x => du = dx / x

∫ [(ln x)² / x] dx = ∫ u² du = u³ / 3 + C = (ln x)³ / 3 + C

PROBLEM

Evaluate ∫ sqrt(x - 1) dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.11]

SOLUTION

Define u = x - 1 => du = dx

∫ sqrt(x - 1) dx = ∫ sqrt(u) du = u^3/2 / (3/2) + C = (2/3)u^3/2 + C = (2/3)(x - 1)^3/2 + C

PROBLEM

Evaluate ∫ [1 / (5 - 3x)] dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.13]

SOLUTION

Define u = 5 - 3x => du = - 3 dx

∫ [1 / (5 - 3x)] dx = - (1/3) ∫ du / u = - (1/3) ln u + C = - (1/3) ln(5 - 3x) + C

This is only valid if 5 - 3x > 0 because we can only take the logarithm of a positive number. If 5 - 3x < 0, we write

∫ [1 / (5 - 3x)] dx = - ∫ [1 / (3x - 5)] dx

Define u = 3x - 5 => du = 3 dx

∫ [1 / (5 - 3x)] dx = - (1/3) ∫ du / u = - (1/3) ln u + C = - (1/3) ln(3x - 5) + C

In general,

∫ [1 / (5 - 3x)] dx = - (1/3) ln |5 - 3x| + C

PROBLEM

Evaluate ∫ [(1 + 4x) / sqrt(1 + x + 2x²)] dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.15]

SOLUTION

Define u = 1 + x + 2x² => du = (1 + 4x) dx

∫ [(1 + 4x) / sqrt(1 + x + 2x²)] dx = ∫ du / sqrt(u) = sqrt(u) / (1/2) + C = 2(1 + x + 2x²)^1/2 + C

PROBLEM

Evaluate ∫ [2 / (t + 1)⁶] dt.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.17]

SOLUTION

Define u = t + 1 => du = dt

∫ [2 / (t + 1)⁶] dt = 2 ∫ du / u⁶ = 2u^-5 / (-5) + C = - (2/5) / (t + 1)⁵ + C

PROBLEM

Evaluate ∫ sin 3q dq.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.19]

SOLUTION

Define u = 3q => du = 3 dq

∫ sin 3q dq = (1/3) ∫ sin u du = - (1/3) cos u + C = - (1/3) cos 3q + C

PROBLEM

Evaluate ∫ e^x sqrt(1 + e^x) dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.21]

SOLUTION

Define u = 1 + e^x => du = e^x dx

∫ e^x sqrt(1 + e^x) dx = ∫ sqrt(u) du = u^3/2 / (3/2) + C = (2/3)(1 + e^x)^3/2 + C

PROBLEM

Evaluate ∫ cos⁴ x sin x dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.23]

SOLUTION

Define u = cos x => du = - sin x dx

∫ cos⁴ x sin x dx = - ∫ u⁴ du = - u⁵ / 5 + C = - (1/5) cos⁵ x + C

PROBLEM

Evaluate ∫ sqrt(cot x) csc² x dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.25]

SOLUTION

Define u = cot x => du = - csc² x dx

∫ sqrt(cot x) csc² x dx = - ∫ sqrt(u) du = - u^3/2 / (3/2) + C = - (2/3)u^3/2 + C = - (2/3) cot^3/2 x + C

PROBLEM

Evaluate ∫ [(e^x + 1) / e^x] dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.27]

SOLUTION

∫ [(e^x + 1) / e^x] dx = ∫ (1 + e^-x) dx = ∫ dx + ∫ e^-x dx

Define u = e^-x => du = - e^-x dx

∫ e^-x dx = - ∫ du = - u + C = - e^-x + C

∫ [(e^x + 1) / e^x] dx = x - e^-x + C

PROBLEM

Evaluate ∫ sec³ x tan x dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.29]

SOLUTION

∫ sec³ x tan x dx = ∫ [(sin x) / (cos⁴ x)] dx

Define u = cos x => du = - sin x dx

∫ sec³ x tan x dx = - ∫ du / u⁴ = - u^-3 / (-3) + C = (1/3) / u³ + C = (1/3) / cos³ x + C

PROBLEM

Evaluate ∫ [(1 + x) / (1 + x²)] dx.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 5.5.31]

SOLUTION

∫ [(1 + x) / (1 + x²)] dx = ∫ dx / (1 + x²) + ∫ [x / (1 + x²)] dx

Define u = 1 + x² => du = 2x dx

∫ [x / (1 + x²)] dx = (1/2) ∫ du / u = (1/2) ln u + C = (1/2) ln(1 + x²) + C

∫ [(1 + x) / (1 + x²)] dx = arctan x + (1/2) ln(1 + x²) + C

PROBLEM

Evaluate ∫ {(sec x) / [ln(sec x + tan x)]^1/2} dx.

SOLUTION

Define u = ln(sec x + tan x) => du = [1 / (sec x + tan x)](sec x tan x + sec² x) dx = sec x dx

∫ {(sec x) / [ln(sec x + tan x)]^1/2} dx = ∫ du / u^1/2 = 2u^1/2 + C = 2[ln(sec x + tan x)]^1/2 + C

PROBLEM

Evaluate ∫ [(x - 2) / (x - 1)]² dx.

SOLUTION

Define u = x - 1 => du = dx

∫ [(x - 2) / (x - 1)]² dx = ∫ [(u - 1) / u]² du = ∫ (1 - 1/u)² du = ∫ (1 - 2/u + 1/u²) du = u - 2 ln u - 1/u + C = (x - 1) - 2 ln(x - 1) - 1 / (x - 1) + C

PROBLEM

In a certain city the temperature (in degrees Fahrenheit) t hours after 9 AM was approximated by the function

T(t) = 30 + 15 sin(pt/12)

(a) Determine the temperature at 9 AM.
(b) Determine the temperature at 3 PM.
(c) Find the average temperature during the period from 9 AM to 9 PM.

SOLUTION

(a) 9 AM: t = 0 => T(0) = 30 °F

(b) 3 PM: t = 6 => T = 30 + 15 sin(p/2) = 30 °F + 15 °F = 45 °F

(c) 9 PM corresponds to t = 12. The average temperature between 9 AM and 9 PM is

T_avg = [∫₀¹² T(t) dt] / (∫₀¹² dt) = (1/12) ∫₀¹² [30 + 15 sin(pt/12)] dt
= (1/12)[30t - (15)(12/p) cos(pt/12)] |₀¹² = (1/12)[360 - (15)(12/p) cos p + (15)(12/p)]
=(1/12)(360 + 360/p) = 30(1 + 1/p) °F

PROBLEM

Find the average value of the function f(x) = |8 - x| over the interval 7 < x < 11.

SOLUTION

f_avg = (∫₇¹¹ |8 - x| dx) / (∫₇¹¹ dx) = (1/4) ∫₇¹¹ |8 - x| dx
= (1/4)[∫₇⁸ (8 - x) dx] + (1/4)[∫₈¹¹ (x - 8) dx]
= (1/4)(8x - x²/2) |₇⁸ + (1/4)(x²/2 - 8x) |₈¹¹
= (1/4)(64 - 32 - 56 + 49/2) + (1/4)(121/2 - 88 - 32 + 64) = 5/4

PROBLEM

A force, F(x) = 5x³ + 1, moves a particle from x = 0 to x = 4. Calculate the work done by the force.

SOLUTION

W = ∫₀⁴ F(x) dx = ∫₀⁴ (5x³ + 1) dx = (5x⁴/4 + x) |₀⁴ = 320 + 4 = 324

PROBLEM

The force required to stretch a spring by 0.5 ft is 4 lb. Calculate the work needed to stretch the spring by 1.1 ft.

SOLUTION

Hooke's law states that

F(x) = kx

The spring constant is

k = F(0.5 ft)/ (0.5 ft) = (4 lb) / (0.5 ft) = 8 lb/ft

The work needed to stretch the spring by 1.1 ft is

W = ∫₀^1.1 F(x) dx = ∫₀^1.1 kx dx = kx²/2 |₀^1.1 = (8 lb/ft)(1.1 ft)²/2 = 4.84 ft lb

PROBLEM

A tank shaped like a cone of radius R = 14 m and height H = 4 m contains V₀ = 5 m³ of hot chocolate of density r = 1520 kg/m³. Calculate the work needed to empty the tank by pumping all of the hot chocolate to the top of the tank.

SOLUTION

         <---R--->
+----------------+^ ^
 \       <--r-->/ | |
  \------------/^ | |
   \ chocolate/ | | H-y
    \        /  | H |
    y\======/   h | v
      \    /    | |
       \  /     | |
        \/      v v

Each infinitesimal layer of chocolate has to be lifted a distance H - y against the force of gravity. The work needed to pump each layer to the top of the tank is

dW = dm g(H-y)

where dm is the mass of the layer, g = 9.8 m/s² is the acceleration of gravity, and y is the y coordinate of the layer, with y = 0 at the bottom of the tank. The mass of the layer is

dm = r dV = r A(y) dy

where dV is the volume of the layer, A(y) is the area of the layer, and dy is the thickness of the layer. The area of the layer is

A(y) = A₀(y/H)² = pR²(y/H)² = p(Ry/H)²

where A₀ is the area of the top of the tank. Thus,

dW = rp(Ry/H)²g(H-y) dy

The total work required to pump the hot chocolate to the top of the tank is

W = ∫ dW = ∫₀^h rp(Ry/H)²g(H-y) dy = rp(R/H)²g ∫₀^h (H-y)y² dy
= rp(R/H)²g(Hy³/3 - y⁴/4) |₀^h = rp(R/H)²g(Hh³/3 - h⁴/4) = rp(R/H)²(g/12)(4Hh³ - 3h⁴)
= rp(R/H)²(g/12)(4Hh³)(1 - 3h/4H) = rp(R/H)²(g/3)(Hh³)(1 - 3h/4H)
= rp(R²h³/H)(g/3)(1 - 3h/4H)

The total volume of the hot chocolate is

V₀ = (1/3)pr²h

Since initially the tank and the hot chocolate have the same shapes,

r/h = R/H => r = (R/H)h => V₀ = (1/3)p(R/H)²h³
=> h = [(3V₀/p)(H/R)²]^1/3 = {[(3)(5 m³)/p)](4 m / 14 m)²]}^1/3 = (60/49p)^1/3 m

The total work is

W = (1520 kg/m³)[p(14 m)²(60/49p m³)/(4 m)][(9.8 m/s²)/3]{1 - (3/4)[(60/49p)^1/3 m]/(4 m)} = (1520)(196)[1 - (3/16)(60/49p)^1/3] J = (297920)[1 - (3/16)(60/49p)^1/3] J = 2.571 x 10⁵ J

PROBLEM

Consider the two functions y₁(x) = sin 2x and y₂(x) = cos 3x.
(a) Calculate the lowest positive value of x at which the two functions intersect. Call this x₀.
(b) For x = a < x₀, consider the line segment connecting the points (x, y) = (a, sin 2a) and (a, cos 3a). Calculate the result of rotating the line segment about the following lines: (i) y = 0, (ii) x = 0, (iii) y = 1, (iv) x = -2, (v) x = p, (vi) y = -2, (vii) y = p, (viii) y = -p. In each case, describe the geometric shape that results and calculate the surface area.

SOLUTION

(a) y₁(x) = y₂(x) => sin 2x = cos 3x

We make use of the fact that sine of an angle is equal to the cosine of its complement. Thus, 2x and 3x add up to p/2 radians or 90°.

2x + 3x = p/2 => 5x = p/2 => x = p/10 = 18° = x₀

(b) (i) The result is a cylinder of radius a and height cos 3a - sin 2a.

A = 2pa(cos 3a - sin 2a)

(ii) The result is an annulus with inner radius sin 2a and outer radius cos 3a.

A = p(cos² 3a - sin² 2a)

(iii) y = 1 lies above both y = sin 2a and y = cos 3a. The result is an annulus with inner radius 1 - cos 3a and outer radius 1 - sin 2a.

A = p[(1 - sin 2a)² - (1 - cos 3a)²]

(iv) The result is a cylinder of radius 2 + a and height cos 3a - sin 2a.

A = 2p(2 + a)(cos 3a - sin 2a)

(v) The result is a cylinder of radius p - a and height cos 3a - sin 2a.

A = 2p(p - a)(cos 3a - sin 2a)

(vi) The result is an annulus with inner radius 2 + sin 2a and outer radius 2 + cos 3a.

A = p[(2 + cos 3a)² - (2 + sin 2a)²)]

(vii) The result is an annulus with inner radius p - cos 3a and outer radius p - sin 2a.

A = p[(p - sin 2a)² - (p - cos 3a)²]

(viii) The result is an annulus with inner radius p + sin 2a and outer radius p + cos 3a.

A = p[(p + cos 3a)² - (p + sin 2a)²]

PROBLEM

Find the volume, in ounces, of a cup which has a bottom diameter of 2", a top diameter of 3 1/16", and a height of 3 3/4".

SOLUTION

Let the bottom radius be r₁ = 1", the top radius be r₂ = 1 17/32" = 1.53125", and the height be h = 3 3/4" = 3.75". Break up the cup into infinitesimal disks of radius r and thickness dz which are perpendicular to the axis of the cup. Then the volume of one of the infinitesimal disks is

dV = pr² dz

The total volume of the cup is

V = ∫ dV = ∫₀^h pr² dz = ∫₀^h p[r₁ + (r₂ - r₁)z/h]² dz
= p ∫₀^h [r₁ + (r₂ - r₁)z/h]² dz = [ph / (r₂ - r₁)] ∫₀^h [h / (r₂ - r₁)][r₁ + (r₂ - r₁)z/h]² dz
= [ph / (r₂ - r₁)][r₁ + (r₂ - r₁)z/h]³ / 3 |₀^h = [ph / (r₂ - r₁)](r₂³ - r₁³) / 3
=(19.15 in³)[(1 gal) / (231 in³)](128 oz/gal) = 10.61 oz

PROBLEM

The base of a certain solid is the area bounded above by the graph of y = f(x) = 4 and below by the graph of y = g(x) = 36x². Cross-sections perpendicular to the x axis are squares. Use the formula

V = ∫_a^b A(x) dx

to find the volume of the solid.

SOLUTION

The cross-sectional area of the solid is a function of x,

A(x) = [f(x) - g(x)]² = (4 - 36x²)²

The volume of the solid is

V = ∫_a^b (4 - 36x²)² dx

The limits of integration are the values of x where f(x) = g(x).

f(x) = g(x) => 4 = 36x² => x = ±1/3 => a = -1/3, b = 1/3

Thus,

V = ∫_-1/3^1/3 (4 - 36x²)² dx = ∫_-1/3^1/3 4²(1 - 9x²)² dx = 16 ∫_-1/3^1/3 (1 - 18x² + 81x⁴) dx
= 16(x - 18x³/3 + 81x⁵/5) |_-1/3^1/3 = 16(x - 6x³ + 81x⁵/5) |_-1/3^1/3
= 16(1/3 - 6/27 + 81/1215) - 16(-1/3 + 6/27 - 81/1215) = 32(1/3 - 6/27 + 81/1215)
= 32(1/3 - 2/9 + 1/15) = 32(30/90 - 20/90 + 6/90) = 256/45

PROBLEM

Find the volume of a pyramid with height 27 and rectangular base with dimensions 4 and 9.

SOLUTION

The cross-sectional area of a pyramid of height h a distance y above its base is

A(y) = A₀(1 - y/h)²

where A₀ is the area of its base. The volume of the pyramid is

V = ∫ dV = ∫₀^h A(y) dy = ∫₀^h A₀(1 - y/h)² dy = A₀ ∫₀^h (1 - y/h)² dy = - A₀h ∫₀^h - (1/h)(1 - y/h)² dy
= - A₀h(1 - y/h)³/3 |₀^h = A₀h/3 = (4)(9)(27)/3 = 324

PROBLEM

Find the volume of the solid obtained by revolving the region bounded by y = 4x², x = 1, and y = 0 about the x axis.

SOLUTION

y^
 |        x y = 4x² => x = (1/2)y^1/2
 |       x
 |      x
 |     x|
 |    x |
 |   x  | x = 1
 |  x===|
 | x    |
 +x-------------->
                 x

Use cylindrical shells. Revolve infinitesimal horizontal strips about the x axis. The volume contributed by each of these strips is

dV = 2prh dr

where r = y, h = 1 - (1/2)y^1/2, and dr = dy. Integrate from y = 0 to y = 4, which is where y = 4x² and x = 1 intersect.

V = ∫ dV = ∫₀⁴ 2py[1 - (1/2)y^1/2] dy = 2p[y²/2 - (1/2)y^5/2/(5/2)] |₀⁴ = 2p(8 - 32/5) = 16p/5

PROBLEM

Find the volume of the solid obtained by rotating the region bounded by y = x², x = 1, and y = 0 about the y axis.

SOLUTION

y^
 |        x y = x²
 |       x
 |      x
 |     x|
 |    x |
 |   x| | x = 1
 |  x|| |
 | x || |
 +x-------------->
                 x

Use cylindrical shells.

dV = 2prh dr = 2pxy dx = 2px³ dx

V = ∫ dV = ∫₀¹ 2px³ dx = 2px⁴/4 |₀¹ = p/2

PROBLEM

Find the volume of the solid obtained by rotating the region bounded by y = x² and y = 1 about y = 5.

SOLUTION

         y^
 x       |       x y = x² => x = y^1/2
  x------|------x y = 5
   x     |     x
    x    |    x
     x   |   x
      x--|--x y = 1
       x | x
        x|x
---------x--------->
                 x

Use cylindrical shells.

dV = 2prh dr = 2p(5 - y)(2y^1/2) dy

V = ∫ dV = ∫₀¹ 2p(5 - y)(2y^1/2) dy = 4p ∫₀¹ (5y^1/2 - y^3/2) dy = 4p[5y^3/2/(3/2) - y^5/2/(5/2)] |₀¹
= 4p(10/3 - 2/5) = 4p(50/15 - 6/15) = 176p/15

PROBLEM

A ball of radius 15 has a round hole of radius 8 drilled through its center. Find the volume of the resulting solid.

SOLUTION

The volume of a sphere of radius R can be obtained by rotating the part of the circle x² + y² = R² above the x axis about the x axis. Use cylindrical shells.

dV = 2prh dr = 2py[2(R² - y²)^1/2] dy

V = ∫ dV = ∫₀^R 2py[2(R² - y²)^1/2] dy = 2p ∫₀^R 2y(R² - y²)^1/2 dy = - 2p(R² - y²)^3/2/(3/2) |₀^R
= - (4p/3)(R² - y²)^3/2 |₀^R = (4/3)pR³

The volume of the solid described is equal to the volume of the ball, without the hole, minus the volume of the hole.

V = (4/3)p(15³ - 8³) = (4/3)p(3375 - 512) = 11452p/3

PROBLEM

Let f and g be the functions given by

f(x) = 1/4 + sin(px)

and

g(x) = 4^-x

Let R be the region in the first quadrant enclosed by the y axis and the graphs of f and g, and let S be the region in the first quadrant enclosed by the graphs of f and g.
(a) Find the area of R.
(b) Find the area of S.
(c) Find the volume of the solid generated when S is revolved about the horizontal line y = - 1.

[from 2005 AP Calculus BC Free-Response Questions, problem 1]

SOLUTION

See Calculus060422.xls for graphs of f(x) and g(x). Find the values of x at which f(x) = g(x).

1/4 + sin(px) = 4^-x => x = (1/p) sin^-1(4^-x - 1/4) (A) or x = - {ln [1/4 + sin(px)]} / (ln 4) (B)

From the graph we see that the two values of x at which f(x) = g(x) are approximately 0.2 and 1. If we use (A) to solve iteratively for x, with 0.2 as the starting value, we get x = 0.178218. Neither (A) nor (B) converges if we take x = 1 as the starting value and iterate. By trial and error, we find that x = 1.000001 yields f(x) = g(x). Define

a = 0.178218
b = 1.000001

(a) R = ∫₀^a [g(x) - f(x)] dx = ∫₀^a [4^-x - 1/4 - sin(px)] dx = ∫₀^a [e^{-x ln 4} - 1/4 - sin(px)] dx
= (- 1 / ln 4) e^{-x ln 4} |₀^a - x/4 |₀^a + (1/p) cos(px) |₀^a
= (- 1 / ln 4)(e^{-a ln 4} - 1) - a/4 + (1/p)[cos(pa) - 1]
= (1 / ln 4)(1 - 4^-a) - a/4 + (1/p)[cos(pa) - 1] = 0.06475

(b) S = ∫_a^b [f(x) - g(x)] dx = ∫_a^b [1/4 + sin(px) - 4^-x] dx = ∫_a^b [1/4 + sin(px) - e^{-x ln 4}] dx
= x/4 |_a^b - (1/p) cos(px) |_a^b + (1 / ln 4) e^{-x ln 4} |_a^b
= (b - a)/4 + (1/p)[cos(pa) - cos(pb)] + (1 / ln 4)(4^-b - 4^-a) = 0.4104

(c) Use infinitesimal washers:

dV = p(r_max² - r_min²) dx = p{[f(x) + 1]² - [g(x) + 1]²} dx

V = ∫ dV = ∫_a^b p{[f(x) + 1]² - [g(x) + 1]²} dx = ∫_a^b p{[1/4 + sin(px) + 1]² - (4^-x + 1)²} dx
= ∫_a^b p{[5/4 + sin(px)]² - (4^-x + 1)²} dx
= ∫_a^b p[25/16 + (5/2) sin(px) + sin²(px) - 4^-2x - 2 · 4^-x - 1] dx
= p ∫_a^b [9/16 + (5/2) sin(px) + sin²(px) - e^{-(4 ln 2)x} - 2e^{-(2 ln 2)x}] dx
= (9p/16)(b - a) - (5/2)[cos(pb) - cos(pa)] + p ∫_a^b sin²(px) dx - p ∫_a^b e^{-(4 ln 2)x} dx
- 2p ∫_a^b e^{-(2 ln 2)x} dx
= (9p/16)(b - a) - (5/2)[cos(pb) - cos(pa)] + p ∫_a^b sin²(px) dx + [p/(4 ln 2)][e^{-(4 ln 2)b} - e^{-(4 ln 2)a}]
+ [2p/(2 ln 2)][e^{-(2 ln 2)b} - e^{-(2 ln 2)a}]

Using the trigonometric identities

1 = cos²A + sin²A
cos 2A = cos²A - sin²A

we get

sin²A = (1/2)(1 - cos 2A)

so

sin²(px) = (1/2)[1 - cos(2px)]

and

∫_a^b sin²(px) dx = (1/2) ∫_a^b [1 - cos(2px)] dx = (1/2)(b - a) - (1/4p)[sin(2pb) - sin(2pa)]

The volume of the solid is

V = (9p/16)(b - a) - (5/2)[cos(pb) - cos(pa)] + (p/2)(b - a) - (1/4)[sin(2pb) - sin(2pa)]
+ [p/(4 ln 2)][e^{-(4 ln 2)b} - e^{-(4 ln 2)a}] + [p/(ln 2)][e^{-(2 ln 2)b} - e^{-(2 ln 2)a}] = 4.5588

PROBLEM

(a) Show that the volume of a prolate spheroid is given by

V = 4pab²/3

where a is the semimajor axis and b is the semiminor axis.

(b) Show that the surface area of a prolate spheroid is given by

A = 2pb[b + (a/e) sin^-1 e]

where

e = sqrt[1 - (b/a)²]

(c) The American football has a long circumference between 27 3/4" and 28 1/2", a short circumference between 20 3/4" and 21 1/4" (Wikipedia), and a length between 11 1/8" and 11 3/8" (www.answerbag.com). Using the mean values for these dimensions, and assuming that a football is a prolate spheroid, use the formulas obtained in parts (a) and (b) to estimate the volume and surface area of a football.

SOLUTION

(a) An ellipsoid is a surface of the form

x²/a² + y²/b² + z²/c² = 1

where a, b, and c are constants. If two of a, b, and c are equal, the surface is a spheroid. If all three are equal, the surface is a sphere.

A prolate spheroid is a spheroid of the form

x²/a² + y²/b² + z²/b² = 1 (1)

where a > b. It is obtained by rotating an ellipse about its major axis.

Consider the ellipse

x²/a² + y²/b² = 1

which is the equation satisfied by the points of (1) which are in the xy plane (z = 0). If we solve for y in terms of x, we get

y = ± b sqrt[1 - (x/a)²]

The volume of a prolate spheroid is the volume obtained by rotating this curve around the x axis. Rotating the positive (y > 0) part of the curve around the x axis is sufficient. Consider the volume defined by rotating the part of the curve between x and x + dx around the x axis. This volume is

dV = pr² dx = py² dx

The total volume of a prolate spheroid is

V = ∫ dV = ∫_-a^a py² dx = pb² ∫_-a^a [1 - (x/a)²] dx = 2pb² ∫₀^a [1 - (x/a)²] dx = 2pb²(x - x³/3a²) |₀^a
= 2pb²(a - a/3) = 4pab²/3

(b) Consider once again the upper part of the ellipse,

y = b sqrt[1 - (x/a)²]

Consider an infinitesimal segment of this curve

ds = sqrt[(dx)² + (dy)²] = dx sqrt[1 + (dy/dx)²]

The surface area resulting from rotating this segment around the x axis is

dA = 2pr ds = 2py sqrt[1 + (dy/dx)²] dx

The total surface area resulting from rotating the entire curve around the x axis is

A = ∫ dA = 2p ∫_-a^a y sqrt[1 + (dy/dx)²] dx

The slope of the curve is

dy/dx = b(1/2)(- 2x/a²) / sqrt[1 - (x/a)²] = - (bx/a²) / sqrt[1 - (x/a)²]
=> (dy/dx)² = (b²x²/a⁴) / [1 - (x/a)²]

Substituting into the above equation, we get

A = 2p ∫_-a^a b sqrt[1 - (x/a)²] sqrt{1 + (b²x²/a⁴) / [1 - (x/a)²]} dx
= 2pb ∫_-a^a sqrt[1 - (x/a)² + b²x²/a⁴] dx = 2pb ∫_-a^a sqrt[1 - (1/a² - b²/a⁴)x²] dx
= 4pb ∫₀^a sqrt[1 - (1/a² - b²/a⁴)x²] dx = 4pb ∫₀^a sqrt(1 - ax²)

where

a = 1/a² - b²/a⁴ = (1/a²)[1 - (b/a)²] = (e/a)²

and

e = sqrt[1 - (b/a)²]

is the eccentricity. Define

sin² z = ax² => 2 sin z cos z dz = 2ax dx => sin z cos z dz = ax dx
=> dx = sin z cos z dz / ax = sin z cos z dz / [(sin z) sqrt(a)]

Then

∫₀^a sqrt(1 - ax²) dx = ∫₀^{asin[a sqrt(a)]} sqrt(1 - sin² z) sin z cos z dz / [(sin z) sqrt(a)]
= [1/sqrt(a)] ∫₀^{asin[a sqrt(a)]} cos² z dz

Next, we use the trigonometric identities

cos 2z = cos(z + z) = cos² z - sin² z

and

1 = cos² z + sin² z

Adding the two, we get

1 + cos 2z = 2 cos² z => cos² z = (1/2)(1 + cos 2z)

Thus,

∫ cos² z dz = (1/2) ∫ (1 + cos 2z) dz = (1/2)[z + (1/2) sin 2z] = z/2 + (1/4) sin 2z
=> ∫₀^{asin[a sqrt(a)]} cos² z dz = [z/2 + (1/4) sin 2z] |₀^{asin[a sqrt(a)]}
= (1/2) asin[a sqrt(a)] + (1/4) sin{2 asin[a sqrt(a)]}
= (1/2) asin[a sqrt(a)] + (1/2) sin{asin[a sqrt(a)]} cos{asin[a sqrt(a)]}
= (1/2) asin[a sqrt(a)] + (1/2)[a sqrt(a)] cos{asin[a sqrt(a)]}
= (1/2) asin[a sqrt(a)] + (1/2)[a sqrt(a)] cos f

where

f = asin[a sqrt(a)] => sin f = a sqrt(a) => cos f = sqrt(1 - sin² f) = sqrt(1 - aa²)
=> ∫₀^{asin[a sqrt(a)]} cos² z dz = (1/2) asin[a sqrt(a)] + (1/2)[a sqrt(a)] sqrt(1 - aa²)
= (1/2) asin e + (1/2)eb/a = (1/2)(eb/a + sin^-1 e)

Finally,

A = 4pb(a/e)(1/2)(eb/a + sin^-1 e) = 2pb[b + (a/e) sin^-1 e]

(c) The semimajor axis is

a = 11.25" / 2 = 5.625"

The semiminor axis is

b = 21" / 2p = 3.342"

The eccentricity is

e = sqrt[1 - (b/a)²] = sqrt[1 - (3.342" / 5.625")²] = 0.8043

The volume is

V = 4pab²/3 = 4p(5.625")(3.342")²/3 = 263.2 in³

The surface area is

A = 2pb[b + (a/e) sin^-1 e] = 2p(3.342")[3.342" + (5.625" / 0.8043) sin^-1 0.8043] = 207.4 in²

PROBLEM

A curve is specified by the function

y(x) = B sqrt[1 - (x/A)²] - C

where A = 5.535, B = 5.785, and C = 3.626.
(a) Find the volume enclosed by rotating the curve about the x axis between x = - 4.3 and x = 4.3 using (i) the midpoint rule, (ii) the trapezoidal rule, and (iii) Simpson's rule with a step size of Dx = 0.1.
(b) Find the area of the surface generated by rotating the curve about the x axis between x = - 4.3 and x = 4.3 using (i) the midpoint rule, (ii) the trapezoidal rule, and (iii) Simpson's rule with a step size of Dx = 0.1.

SOLUTION

(a) Consider rotating one infinitesimal segment of the curve around the x axis. The infinitesimal volume enclosed by this segment is

dV = pr² dx = py² dx

The volume enclosed by rotating the entire curve around the x axis between x = - 4.3 and x = 4.3 is

V = ∫ dV = ∫_-4.3^4.3 py² dx = ∫_-4.3^4.3 f(x) dx

where

f(x) = py² = p{B sqrt[1 - (x/A)²] - C}²

We divide the interval - 4.3 < x < 4.3 into 86 bins of width Dx = 0.1.

(i) By the midpoint rule, we get the value of f(x) at x = - 4.25, - 4.15,..., 4.25, which are at the centers of the 86 bins, multiply these values by Dx = 0.1, and sum the products to get the volume of revolution,

V = S_i=1^N f(x_i) Dx

where N = 86 and x_i is the x coordinate at the center of the ith bin. The result is 71.62578299.

(ii) By the trapezoidal rule, we get the value of f(x) at x = - 4.3, - 4.2,..., 4.3, which are at the edges of the 86 bins. For each bin, we multiply the average value of f(x) at the two edges by Dx = 0.1 and sum the products to get the volume of revolution,

V = S_i=1^N (0.5)[f(x_i) + f(x_i+1)] Dx
= (0.5)f(x₁) Dx + S_i=2^N f(x_i) Dx + (0.5)f(x_N+1) Dx

where N = 86 and x_i is the x coordinate at the left edge of the ith bin. The result is 71.62543808.

(iii) By Simpson's rule, the volume of revolution is

V = [Dx)/3][f(x₁) + 4f(x₂) + 2f(x₃) + 4f(x₄) + ... + 2f(x_N-1) + 4f(x_N) + f(x_N+1)]

where x_i is the x value at the left edge of the ith bin and x_i+1 is the x value at the right edge of the ith bin. The result is 71.62568695. See Integration060323.xls for the calculations that resulted in the answers to parts (i), (ii), and (iii).

(b) Consider rotating one infinitesimal segment of the curve around the x axis. The infinitesimal surface area swept out by the segment is

dA = 2pr ds = 2py ds

where ds is the length of the segment,

ds = sqrt[(dx)² + (dy)²] = dx sqrt[1 + (dy/dx)²]

Thus,

dA = 2py sqrt[1 + (dy/dx)²] dx

The surface swept out by rotating the entire curve around the x axis between x = - 4.3 and x = 4.3 is

A = ∫ dA = ∫_-4.3^4.3 2py sqrt[1 + (dy/dx)²] dx = ∫_-4.3^4.3 f(x) dx

where

f(x) = 2py sqrt[1 + (dy/dx)²]

Since

y = B sqrt[1 - (x/A)²] - C

we have

dy/dx = B(1/2)(- 2x/A²) / sqrt[1 - (x/A)²] = - (Bx/A²) / sqrt[1 - (x/A)²]

We divide the interval - 4.3 < x < 4.3 into 86 bins of width Dx = 0.1.

(i) By the midpoint rule,

A = S_i=1^N f(x_i) Dx

where N = 86 and x_i is the x coordinate at the center of the ith bin. The result is 88.78952454.

(ii) By the trapezoidal rule,

A = S_i=1^N (0.5)[f(x_i) + f(x_i+1)] Dx
= (0.5)f(x₁) Dx + S_i=2^N f(x_i) Dx + (0.5)f(x_N+1) Dx

where N = 86 and x_i is the x coordinate at the left edge of the ith bin. The result is 88.7566438.

(iii) By Simpson's rule, the surface swept out by rotating the curve around the x axis is

A = [Dx)/3][f(x₁) + 4f(x₂) + 2f(x₃) + 4f(x₄) + ... + 2f(x_N-1) + 4f(x_N) + f(x_N+1)]

where x_i is the x value at the left edge of the ith bin and x_i+1 is the x value at the right edge of the ith bin. The result is 88.77852806. See Integration060324.xls for the calculations that resulted in the answers to parts (i), (ii), and (iii).

PROBLEM

If f(x) = ∫_a^x g(t) dt, find f '(x).

SOLUTION

f '(x) = lim(e=>0) {[∫_a^x+e g(t) dt - ∫_a^x g(t) dt] / e} = lim(e=>0) {[∫_x^x+e g(t) dt] / e}
= lim(e=>0) {[g(x) ∫_x^x+e dx] / e} = g(x)

PROBLEM

If f(x) = ∫_a^h(x) g(t) dt, find f '(x).

SOLUTION

f '(x) = lim(e=>0) {[∫_a^h(x+e) g(t) dt - ∫_a^h(x) g(t) dt] / e} = lim(e=>0) {[∫_h(x)^h(x+e) g(t) dt] / e}
= g[h(x)] lim(e=>0) {[∫_h(x)^h(x+e) dt] / e} = g[h(x)] lim(e=>0) {[h(x+e) - h(x)] / e} = g[h(x)]h'(x)

PROBLEM

If f(x) = ∫₀^{cos x} ln y³ dy, find f '(p/4).

SOLUTION

From the definition of the derivative,

f '(x) = lim(e=>0) {[∫₀^cos(x+e) ln y³ dy - ∫₀^{cos x} ln y³ dy] / e}
= lim(e=>0) {[∫_{cos x}^cos(x+e) ln y³ dy] / e} = lim(e=>0) {[ln(cos³ x)][cos(x+e) - cos x] / e}
= [ln(cos³ x)] lim(e=>0) {[cos(x+e) - cos x] / e} = [ln(cos³ x)] (d/dx) cos x = - (sin x) ln(cos³ x)

Evaluating this at x = p/4,

f '(p/4) = - (sin p/4) ln(cos³ p/4) = - [sqrt(2) / 2] ln [sqrt(2) / 2]³

PROBLEM

Estimate the x coordinates of the points of intersection of the curves y = x⁴ and y = 3x - x². If D is the region bounded by these curves, estimate ∫∫_D x dA.

[from Stewart, James 1995, Multivariable Calculus, Third Edition (Pacific Grove, California: Brooks/Cole Publishing Company), exercise 13.3.29]

SOLUTION

If we sketch both curves (Stewart13.3.29.080309.xls), we see that one intersection point is (0, 0) and the other is at x ≈ 1.5. We can get a better estimate of the x coordinate of the second intersection point by doing the following. Set the two functions equal to each other,

x⁴ = 3x - x² => x³ = 3 - x => x = (3 - x)^1/3

Now, substitute the value x = 1.5 into the right hand side of the last equation, x = (3 - x)^1/3, and solve for a new value of x. Substitute the new value of x into the right hand side of the equation and solve for a new value of x again. If this is repeated, the x values converge to x = 1.213412.

Using this result,

∫∫_D x dA = ∫₀^1.213412 x(3x - x² - x⁴) dx = ∫₀^1.213412 (3x² - x³ - x⁵) dx
= (3x³/3 - x⁴/4 - x⁶/6) |₀^1.213412 = (1.213412)³ - (1.213412)⁴/4 - (1.213412)⁶/6 = 0.712639

Sir Isaac Newton expressed functions as sums of infinite series. Certain functions can thus be more easily integrated by integrating the individual terms of the infinite series.

PROBLEM

Determine whether the series a_n = (n^1/3 + n^1/4) / (n^1/2 + n^1/5) converges or diverges. If it converges, find the limit.

[from Stewart, James 1995, Multivariable Calculus, Third Edition (Pacific Grove, California: Brooks/Cole Publishing Company), exercise 10.1.18]

SOLUTION

lim(n=>∞) a_n = lim(n=>∞) (n^1/3 + n^1/4) / (n^1/2 + n^1/5)
= lim(n=>∞) (1 + n^1/4-1/3) / (n^1/2-1/3 + n^1/5-1/3) = lim(n=>∞) (1 + n^-1/12) / (n^1/6 + n^-2/15)
= lim(n=>∞) (1 + 1/n^1/12) / (n^1/6 + 1/n^2/15) = 0 (convergent)

PROBLEM

Determine whether the series a_n = 1/n² + 2/n² + ... + n/n² converges or diverges. If it converges, find the limit.

[from Stewart, James 1995, Multivariable Calculus, Third Edition (Pacific Grove, California: Brooks/Cole Publishing Company), exercise 10.1.37]

SOLUTION

lim(n=>∞) a_n = lim(n=>∞) (1/n² + 2/n² + ... + n/n²) = lim(n=>∞) (1/n²)(1 + 2 + ... + n)
= lim(n=>∞) (1/n²)(n + 1)n/2 = lim(n=>∞) (1/2n)(n + 1) = 1/2 (convergent)

PROBLEM

Determine if the following sequences are convergent or divergent. If convergent, find the limit.
(a) a_n = 1 · 3 · 5 · ... · (2n - 1) / (2n)ⁿ
(b) a_n = 1 · 3 · 5 · ... · (2n - 1) / n!

[from Stewart, James 1995, Multivariable Calculus, Third Edition (New York, New York: Brooks/Cole Publishing Company), exercises 10.1.47 and 10.1.48]

SOLUTION

(a) a_n = 1 · 3 · 5 · ... · (2n - 1) / (2n)ⁿ = (1/2n) · (3/2n) · (5/2n) · ... · [(2n - 1)/2n]

Each term in the sequence is the product of n factors, each of which is less than one. The smallest factor is the first one, 1/2n. As n approaches infinity, 1/2n approaches zero. Thus, a_n < 1/2n => 0 as n => ∞, and the sequence converges.

(b) a_n = 1 · 3 · 5 · ... · (2n - 1) / n! = 1 · (3/2) · (5/3) · ... · [(2n - 1)/n]

Each term in the sequence is the product of n factors, each of which is greater than or equal to one. The smallest factor is the first one, which is one. As n approaches infinity, (2n - 1)/n approaches two. As n increases, there are more and more factors which get closer and closer to two. Thus, the sequence diverges.

PROBLEM

For what values of r is the sequence {nrⁿ} convergent?

[from Stewart, James 1995, Multivariable Calculus, Third Edition (Pacific Grove, California: Brooks/Cole Publishing Company), exercise 10.1.49]

SOLUTION

The sequence {a_n} = {nrⁿ} consists of the numbers {a₁, a₂, a₃,..., a_k,...} = {r, 2r², 3r³,..., kr^k,...}, where k is an integer.

When r > 1, rⁿ increases as n increases. Thus, nrⁿ keeps increasing as n increases, so {nrⁿ} is divergent.

When r < - 1, the magnitude of rⁿ increases as n increases, although the sign alternates between positive and negative. Thus, the magnitude of nrⁿ keeps increasing as n increases, so {nrⁿ} is divergent.

When r = 1, rⁿ = 1. Thus, nrⁿ = n keeps increasing as n increases, so {nrⁿ} is divergent.

When r = - 1, rⁿ = (-1)ⁿ and always has a magnitude of one. Thus, nrⁿ = n(-1)ⁿ keeps increasing in magnitude as n increases, while alternating between positive and negative, so {nrⁿ} is divergent.

When |r| < 1, rⁿ decreases in magnitude as n increases. We have

a_n = nrⁿ (1)
a_n+1 = (n + 1)rⁿ⁺¹ (2)

Dividing (2) by (1), we get

a_n+1/a_n = [(n + 1) / n]r

If we take the limit of this ratio as n approaches infinity,

lim(n=>∞) a_n+1/a_n = lim(n=>∞) [(n + 1) / n]r = r

As n approaches infinity, the ratio between a given term and the preceding term approaches r. Successive terms get smaller and smaller in magnitude and approach zero.

From the above discussion, the only values of r for which {nrⁿ} is convergent are those for which |r| < 1 or - 1 < r < 1, and tbe limit is zero.

Additional Notes:

As n gets larger, the ratio (n + 1) / n gets smaller and approaches one. For n greater than some value, the product [(n + 1) / n]r is always less than one, so subsequent terms in the sequence get smaller and smaller. The critical value of n (the value for which [(n + 1) / n]r is equal to one) is determined by solving

[(n + 1) / n]r = 1

for n. Multiplying both sides of the equation by n, we get

(n + 1)r = n => nr + r = n => r = n - nr = n(1 - r) => n = r / (1 - r) (3)

If n, as calculated from this formula, turns out to be nonintegral, the critical value of n is the next larger integer. Equation (3) gives the value of n for which a_n+1/a_n = 1 and a_n+2/a_n+1 < 1; if n as calculated from (3) is nonintegral, then a_n+1/a_n < 1.

As an example, if r = 0.75 = 3/4, the critical value of n is calculated to be

n = r / (1 - r) = (3/4) / [1 - (3/4)] = (3/4) / (1/4) = 3

The first few terms in the sequence are

{3/4, (2)(3/4)², (3)(3/4)³, (4)(3/4)⁴, (5)(3/4)⁵, (6)(3/4)⁶,...}
= {3/4, 9/8, 81/64, 81/64, 1215/1024, 4374/4096,...}
= {0.75, 1.125, 1.265625, 1.265625, 1.1865, 1.0678711,...}

Thus, for n > 5, the terms start getting smaller.

PROBLEM

Find the limit of the sequence {sqrt(2), sqrt[2 sqrt(2)], sqrt(2 sqrt[2 sqrt(2)]),...}.

[from Stewart, James 1995, Multivariable Calculus, Third Edition (Pacific Grove, California: Brooks/Cole Publishing Company), exercise 10.1.59]

SOLUTION

Define

{a_n} = {sqrt(2), sqrt[2 sqrt(2)], sqrt(2 sqrt[2 sqrt(2)]),...} = {2^1/2, 2^1/22^1/4, 2^1/22^1/42^1/8,...}
={2^1/2, 2^1/2+1/4, 2^1/2+1/4+1/8,...} = {2^f(n)}

where

f(n) = S_k=1ⁿ (1/2)^k

The limit of the sequence is

lim(n=>∞) a_n = lim(n=>∞) 2^f(n)

which can be calculated by noting that

lim(n=>∞) f(n) = lim(n=>∞) S_k=1ⁿ (1/2)^k = 1/2 + 1/4 + 1/8 + ... = (1/2)(1 + 1/2 + 1/4 + ...)
= a(1 + r + r² + ...) ≡ s

where a = 1/2 and r = 1/2.

s is a geometric series, whose value is given by the formula

s = a / (1 - r) = (1/2) / (1 - 1/2) = 1

Thus,

lim(n=>∞) f(n) = 1

and

lim(n=>∞) a_n = 2

PROBLEM

Find the values of x for which the series S_n=1^∞ xⁿ/3ⁿ converges. Find the sum of the series for those values of x.

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 8.2.33]

SOLUTION

Write the series in the form of a geometric series.

S_n=1^∞ xⁿ/3ⁿ = S_n=1^∞ (x/3)ⁿ = S_n=1^∞ (x/3)(x/3)^n-1 = S_n=1^∞ ar^n-1

where a = x/3 and r = x/3. This is a geometric series, which converges if

|r| < 1 => |x/3| < 1 => |x| < 3 => -3 < x < 3

The sum is

S = a / (1 – r) = (x/3) / (1 – x/3) = x / (3 – x)

PROBLEM

What is the value of c if S_n=2^∞ (1 + c)^-n = 2?

[from Stewart, James 2001, Calculus: Concepts and Contexts, Second Edition (Pacific Grove, California: Brooks/Cole), exercise 8.2.43]

SOLUTION

S = S_n=2^∞ (1 + c)^-n = S_n=1^∞ (1 + c)^-(n+1) = S_n=1^∞ 1 / (1 + c)ⁿ⁺¹
= S_n=1^∞ [1 / (1 + c)]ⁿ⁺¹ = S_n=1^∞ [1 / (1 + c)]²[1 / (1 + c)]^n-1 = S_n=1^∞ ar^n-1

where a = [1 / (1 + c)]² and r = 1 / (1 + c). This is a geometric series, so the sum is

S = a / (1 – r) = [1 / (1 + c)]² / [1 – 1 / (1 + c)] = 1 / [(1 + c)c] = 2
=> 2c(1 + c) = 1 => 2c² + 2c – 1 = 0

Using the quadratic formula, we get

c = {- 2 ± sqrt[2² - (4)(2)(-1)]} / 4
= [- 2 ± sqrt(12)] / 4 = [- 2 ± 2 sqrt(3)] / 4 = [- 1 ± sqrt(3)] / 2

Solving the quadratic equation yields two values for c. If the lower value, [- 1 – sqrt(3)] / 2, is plugged into the original expression, the result is

S_n=2^∞ (1 + c)^-n = S_n=2^∞ {1 + [- 1 - sqrt(3)] / 2}^-n = S_n=2^∞ {[1 – sqrt(3)] / 2}^-n = S_n=2^∞ (- 0.3660)^-n

which diverges. The other value of c, [- 1 + sqrt(3)] / 2, is the value for which S = 2.