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

Find the numbers at which the function

f(x) = x + 2 if x < 0

f(x) = e

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,

SOLUTION

A graph of f shows that it is discontinuous at x = 0 and x = 1. However, since f(x) = e

(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)]

Use continuity to evaluate the limit

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

[from Stewart, James 2001,

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

Use continuity to evaluate the limit

lim(x => 1) e

[from Stewart, James 2001,

SOLUTION

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

lim(x => 1) e

Use continuity to evaluate the limit

lim(x => 2) arctan[(x

[from Stewart, James 2001,

SOLUTION

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

lim(x => 2) arctan[(x

= 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)

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

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 - cos

= 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)

= 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.

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

[from the movie

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 - cos

= 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 - cos

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

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

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

Thus, {[ln(1 - x) - sin x] / (1 - cos

If an initial amount A

A = A

If we let n => ∞, we refer to the

A = A

[from Stewart, James 2001,

SOLUTION

lim (n => ∞) A = lim (n => ∞) e

= exp[lim (n => ∞) ln A] = exp[lim (n => ∞) ln A

= exp{lim (n => ∞) [ln A

= exp[ln A

= exp[ln A

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/n

= lim (n => ∞) [i / (1 + i/n)] = i

Thus,

lim (n => ∞) A = exp(ln A

Calculate the derivative of log

SOLUTION

Write log

a

Calculate the derivative of a

SOLUTION

Write a

y = (e

Find an equation of the tangent line to the following curve at the given point: y

[from Stewart, James 2001,

SOLUTION

Taking the derivative of both sides, we get

2y dy/dx = 3x

=> dy/dx = (x

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

y - y

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).

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 = x

According to the mean value theorem, if f(x) is continuous on the interval a

[f(b) - f(a)] / (b - a) = f '(x

Solving for f(b), we get

f(b) = f(a) + f '(x

PROBLEM

Find a value for x

(a) f(x) = x

(b) f(x) = sin x, 0

(c) f(x) = cos x, p/2 < x < 3p/2

[from Ayres, Frank Jr., and Mendelson, Elliot 1990,

SOLUTION

(a) Find a value of x in the interval 1

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

(b) Find a value of x in the interval 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

Find a value for x

(a) y = x

(b) y = ax

(c) y = ln x, 1

[from Ayres, Frank Jr., and Mendelson, Elliot 1990,

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) = 3x

(b) Find a value of x for which

y'(x) = [y(x

= [a(x

The derivative of y with respect to x is

y'(x) = 2ax + b

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

a(x

(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)

Use the mean value theorem to approximate 65

[from Ayres, Frank Jr., and Mendelson, Elliot 1990,

SOLUTION

Let f(x) = x

f '(x) = (1/6)x

From the mean value theorem, we have

f(b) = f(a) + f '(x

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

Use the mean value theorem to approximate

(a) sqrt(15)

(b) (3.001)

(c) 1/999

[from Ayres, Frank Jr., and Mendelson, Elliot 1990,

SOLUTION

(a) Let f(x) = x

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

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) = x

f '(x) = 3x

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/x

From the mean value theorem, we have

f(b) = f(a) + f '(x)(b - a) ≈ f(a) + f '(a)(b - a) = 1/1000 + (- 1/10

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

Consider the function y(x) = e

(a) Find its relative maxima and minima.

(b) Find its absolute maximum and minimum.

SOLUTION

(a) y(x) = e

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

y'(x) = - (1/x) e

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

Solve this iteratively for x. For example, try x = 1, substitute this into the right hand side, and get x = e

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/x

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

For a sketch of the function y(x) = e

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

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 - 16t

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

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 => [sec

=> dq/dt = - (32t/d) cos

= - [32t(100)] / {100

= - 32t / [100 + (1524 - 1.6t

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

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

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

d

= - 32[100 + (1524 - 1.6t

- 204.8t(1524 - 1.6t

= 0

=> - 32[100 + (1524 - 1.6t

=> - 32(100 + 2322576 - 4876.8t

=> - 32(2322676 - 4876.8t

=> - 74325632 + 156057.6t

=> - 74325632 - 156058t

=> 245.76t

=> t = [(74325632 + 156058t

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.

If a snowball melts so that its surface area decreases at a rate of 1 cm

[from Stewart, James 2001,

SOLUTION

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

S = 4pr

Since the diameter

D = 2r => r = D/2

we have

S = 4p(D/2)

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 cm

= - [1/(20p)] cm/min

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,

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(x

The derivative of r with respect to time is

dr/dt = (1/2)(2x dx/dt) / sqrt(x

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

x = sqrt(r

So

dr/dt = (v/r) sqrt(r

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,

SOLUTION

Let (x

L = sqrt(x

Take the derivative of both sides with respect to time.

0 = (1/2)(2x

= (x

=> dx

Solve (1) for x

x

= 2 sqrt(133) ft

Substitute this result into (2) using dx

dx

= - {(5 ft) / [2 sqrt(133) ft]}(2 ft/s) sqrt{[2 sqrt(133) ft]

= - {10 / [sqrt(133)]} ft/s = 0.8671 ft/s

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,

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 m

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

f(x) = ax exp(bx

have the maximum value f(2) = 1?

[from Stewart, James 2001,

SOLUTION

f '(x) = ax(2bx) exp(bx

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

f(2) = 2a e

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

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(x

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(x

=> x + H dH/dx = 0

The derivative of H with respect to x is

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

So

x + [hx / (x - d)][- hd / (x - d)

From this we see that either x = 0 or 1 - h

1 - h

=> H = hx / (x - d) = (5)(8.642) / (8.642 - 4) = 9.309 ft

=> L = sqrt(x

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

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,

SOLUTION

Let N

N

and

P = N

N

Maximize P with respect to N

P'(N

The second derivative of P with respect to N

P"(N

N

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,

SOLUTION

Let R = revenue, P = price per ticket, and N

R = PN

From the market survey, we know that

N

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.

Evaluate ∫ cos 3x dx.

[from Stewart, James 2001,

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

Evaluate ∫ x

[from Stewart, James 2001,

SOLUTION

Define u = x

∫ x

Evaluate ∫ [4 / (1 + 2x)

[from Stewart, James 2001,

SOLUTION

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

∫ [4 / (1 + 2x)

Evaluate ∫ 2x(x

[from Stewart, James 2001,

SOLUTION

Define u = x

∫ 2x(x

Evaluate ∫ [(ln x)

[from Stewart, James 2001,

SOLUTION

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

∫ [(ln x)

Evaluate ∫ sqrt(x - 1) dx.

[from Stewart, James 2001,

SOLUTION

Define u = x - 1 => du = dx

∫ sqrt(x - 1) dx = ∫ sqrt(u) du = u

Evaluate ∫ [1 / (5 - 3x)] dx.

[from Stewart, James 2001,

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

Evaluate ∫ [(1 + 4x) / sqrt(1 + x + 2x

[from Stewart, James 2001,

SOLUTION

Define u = 1 + x + 2x

∫ [(1 + 4x) / sqrt(1 + x + 2x

Evaluate ∫ [2 / (t + 1)

[from Stewart, James 2001,

SOLUTION

Define u = t + 1 => du = dt

∫ [2 / (t + 1)

Evaluate ∫ sin 3q dq.

[from Stewart, James 2001,

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

Evaluate ∫ e

[from Stewart, James 2001,

SOLUTION

Define u = 1 + e

∫ e

Evaluate ∫ cos

[from Stewart, James 2001,

SOLUTION

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

∫ cos

Evaluate ∫ sqrt(cot x) csc

[from Stewart, James 2001,

SOLUTION

Define u = cot x => du = - csc

∫ sqrt(cot x) csc

Evaluate ∫ [(e

[from Stewart, James 2001,

SOLUTION

∫ [(e

Define u = e

∫ e

∫ [(e

Evaluate ∫ sec

[from Stewart, James 2001,

SOLUTION

∫ sec

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

∫ sec

Evaluate ∫ [(1 + x) / (1 + x

[from Stewart, James 2001,

SOLUTION

∫ [(1 + x) / (1 + x

Define u = 1 + x

∫ [x / (1 + x

∫ [(1 + x) / (1 + x

Evaluate ∫ {(sec x) / [ln(sec x + tan x)]

SOLUTION

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

∫ {(sec x) / [ln(sec x + tan x)]

Evaluate ∫ [(x - 2) / (x - 1)]

SOLUTION

Define u = x - 1 => du = dx

∫ [(x - 2) / (x - 1)]

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

= (1/12)[30t - (15)(12/p) cos(pt/12)] |

=(1/12)(360 + 360/p) = 30(1 + 1/p) °F

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

SOLUTION

f

= (1/4)[∫

= (1/4)(8x - x

= (1/4)(64 - 32 - 56 + 49/2) + (1/4)(121/2 - 88 - 32 + 64) = 5/4

A force, F(x) = 5x

SOLUTION

W = ∫

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 = ∫

A tank shaped like a cone of radius R = 14 m and height H = 4 m contains V

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/s

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) = A

where A

dW = rp(Ry/H)

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

W = ∫ dW = ∫

= rp(R/H)

= rp(R/H)

= rp(R

The total volume of the hot chocolate is

V

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

r/h = R/H => r = (R/H)h => V

=> h = [(3V

The total work is

W = (1520 kg/m

Consider the two functions y

(a) Calculate the lowest positive value of x at which the two functions intersect. Call this x

(b) For x = a < x

SOLUTION

(a) y

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° = x

(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(cos

(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)

(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)

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

A = p[(p - sin 2a)

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

A = p[(p + cos 3a)

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 r

dV = pr

The total volume of the cup is

V = ∫ dV = ∫

= p ∫

= [ph / (r

=(19.15 in

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) = 36x

V = ∫

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)]

The volume of the solid is

V = ∫

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

f(x) = g(x) => 4 = 36x

Thus,

V = ∫

= 16(x - 18x

= 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

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) = A

where A

V = ∫ dV = ∫

= - A

Find the volume of the solid obtained by revolving the region bounded by y = 4x

SOLUTION

y^

| x y = 4x

| 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

V = ∫ dV = ∫

Find the volume of the solid obtained by rotating the region bounded by y = x

SOLUTION

y^

| x y = x

| x

| x

| x|

| x |

| x| | x = 1

| x|| |

| x || |

+x-------------->

x

Use cylindrical shells.

dV = 2prh dr = 2pxy dx = 2px

V = ∫ dV = ∫

Find the volume of the solid obtained by rotating the region bounded by y = x

SOLUTION

y^

x | x y = x

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

V = ∫ dV = ∫

= 4p(10/3 - 2/5) = 4p(50/15 - 6/15) = 176p/15

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 x

dV = 2prh dr = 2py[2(R

V = ∫ dV = ∫

= - (4p/3)(R

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(15

Let f and g be the functions given by

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

and

g(x) = 4

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

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 = ∫

= (- 1 / ln 4) e

= (- 1 / ln 4)(e

= (1 / ln 4)(1 - 4

(b) S = ∫

= x/4 |

= (b - a)/4 + (1/p)[cos(pa) - cos(pb)] + (1 / ln 4)(4

(c) Use infinitesimal washers:

dV = p(r

V = ∫ dV = ∫

= ∫

= ∫

= p ∫

= (9p/16)(b - a) - (5/2)[cos(pb) - cos(pa)] + p ∫

- 2p ∫

= (9p/16)(b - a) - (5/2)[cos(pb) - cos(pa)] + p ∫

+ [2p/(2 ln 2)][e

Using the trigonometric identities

1 = cos

cos 2A = cos

we get

sin

so

sin

and

∫

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

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

V = 4pab

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

where

e = sqrt[1 - (b/a)

(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

x

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

x

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

Consider the ellipse

x

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)

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 = pr

The total volume of a prolate spheroid is

V = ∫ dV = ∫

= 2pb

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

y = b sqrt[1 - (x/a)

Consider an infinitesimal segment of this curve

ds = sqrt[(dx)

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

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

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

A = ∫ dA = 2p ∫

The slope of the curve is

dy/dx = b(1/2)(- 2x/a

=> (dy/dx)

Substituting into the above equation, we get

A = 2p ∫

= 2pb ∫

= 4pb ∫

where

a = 1/a

and

e = sqrt[1 - (b/a)

is the eccentricity. Define

sin

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

Then

∫

= [1/sqrt(a)] ∫

Next, we use the trigonometric identities

cos 2z = cos(z + z) = cos

and

1 = cos

Adding the two, we get

1 + cos 2z = 2 cos

Thus,

∫ cos

=> ∫

= (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 - sin

=> ∫

= (1/2) asin e + (1/2)eb/a = (1/2)(eb/a + sin

Finally,

A = 4pb(a/e)(1/2)(eb/a + sin

(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)

The volume is

V = 4pab

The surface area is

A = 2pb[b + (a/e) sin

A curve is specified by the function

y(x) = B sqrt[1 - (x/A)

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 = pr

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

V = ∫ dV = ∫

where

f(x) = py

We divide the interval - 4.3

(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

where N = 86 and x

(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

= (0.5)f(x

where N = 86 and x

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

V = [Dx)/3][f(x

where x

(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)

Thus,

dA = 2py sqrt[1 + (dy/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 = ∫

where

f(x) = 2py sqrt[1 + (dy/dx)

Since

y = B sqrt[1 - (x/A)

we have

dy/dx = B(1/2)(- 2x/A

We divide the interval - 4.3

(i) By the midpoint rule,

A = S

where N = 86 and x

(ii) By the trapezoidal rule,

A = S

= (0.5)f(x

where N = 86 and x

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

A = [Dx)/3][f(x

where x

If f(x) = ∫

SOLUTION

f '(x) = lim(e=>0) {[∫

= lim(e=>0) {[g(x) ∫

If f(x) = ∫

SOLUTION

f '(x) = lim(e=>0) {[∫

= g[h(x)] lim(e=>0) {[∫

If f(x) = ∫

SOLUTION

From the definition of the derivative,

f '(x) = lim(e=>0) {[∫

= lim(e=>0) {[∫

= [ln(cos

Evaluating this at x = p/4,

f '(p/4) = - (sin p/4) ln(cos

Estimate the x coordinates of the points of intersection of the curves y = x

[from Stewart, James 1995,

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,

x

Now, substitute the value x = 1.5 into the right hand side of the last equation, x = (3 - x)

Using this result,

∫∫

= (3x

Determine whether the series a

[from Stewart, James 1995,

SOLUTION

lim(n=>∞) a

= lim(n=>∞) (1 + n

= lim(n=>∞) (1 + 1/n

Determine whether the series a

[from Stewart, James 1995,

SOLUTION

lim(n=>∞) a

= lim(n=>∞) (1/n

Determine if the following sequences are convergent or divergent. If convergent, find the limit.

(a) a

(b) a

[from Stewart, James 1995,

SOLUTION

(a) a

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

(b) a

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.

For what values of r is the sequence {nr

[from Stewart, James 1995,

SOLUTION

The sequence {a

When r > 1, r

When r < - 1, the magnitude of r

When r = 1, r

When r = - 1, r

When |r| < 1, r

a

a

Dividing (2) by (1), we get

a

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

lim(n=>∞) a

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 {nr

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

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)

= {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

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

[from Stewart, James 1995,

SOLUTION

Define

{a

={2

where

f(n) = S

The limit of the sequence is

lim(n=>∞) a

which can be calculated by noting that

lim(n=>∞) f(n) = lim(n=>∞) S

= a(1 + r + r

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

Find the values of x for which the series S

[from Stewart, James 2001,

SOLUTION

Write the series in the form of a geometric series.

S

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)

What is the value of c if S

[from Stewart, James 2001,

SOLUTION

S = S

= S

where a = [1 / (1 + c)]

S = a / (1 – r) = [1 / (1 + c)]

=> 2c(1 + c) = 1 => 2c

Using the quadratic formula, we get

c = {- 2 ± sqrt[2

= [- 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

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