Acing AP Calculus AB and BC

Further Applications of Integration 241

f (1.5) ≈ y1 = y0 + h  dy 
 dx 
(x0 , y0 )

=−1+ (0.5)(2 − −1) =0.5 x0 = 1 , y0 = −1
1 x1 =x0 + h =1 + 0.5 =1.5 , y1 = 0.5
x2 = x1 + h = 1.5 + 0.5 = 2 , y2 = 4 3
f (2) ≈ y2 = y1 + h  dy  x3 = x2 + h = 2 + 0.5 = 2.5 , y3 = 2
 dx 
(x1, y1 )

=0.5 + (0.5)(2 − 0.5) =4
1.5 3

f (2.5) ≈ y3 = y2 + h  dy 
 dx 
(x2 , y2 )

=4 + (0.5)(2 − 4 3) =2
32

f (3) ≈ y4 = y3 + h  dy 
 dx 
(x3 , y3 )

=2 + (0.5)(2 − 2 ) =2.6
2.5

Example 2 □ Let y = f (x) be the solution to the differential equation dy = x − y + 2 with the
dx

initial condition f (0) = 2 . Use Euler’s method starting at x = 0 with a step size

of 0.5 to approximate f (2) .

Solution □ Given dy = x − y + 2 , x0 = 0, y0 = 2 , and h = 0.5 .
dx

f (0.5) ≈ y1 = y0 + h  dy 
 dx 
(x0 , y0 )

= 2 + (0.5)(0 − 2 + 2) = 2 x0 = 0 , y0 = 2

f (1) ≈ y2 = y1 + h  dy 
 dx 
(x1, y1 )

= 2 + (0.5)(0.5 − 2 + 2) = 2.25 x1 =0 + 0.5 =0.5 , y1 = 2

f (1.5) ≈ y3 = y2 + h  dy 
 dx 
(x2 , y2 )

= 2.25 + (0.5)(1− 2.25 + 2)= 2.625 x2 = 0.5 + 0.5 = 1 , y2 = 2.25

f (2) ≈ y4 = y3 + h  dy 
 dx 
( x3 , y3 )

= 2.625 + (0.5)(1.5 − 2.625 + 2=) 3.0625 x3 =1+ 0.5 =1.5 , y3 = 2.625

242 Chapter 7

Exercises - Euler’s Method BC
Multiple Choice Questions

1. Let y = f (x) be the solution to the differential equation dy =1+ 2x − y with the initial condition f (1) = 2 .
dx

What is the approximation for f (2) if Euler’s method is used, starting at x = 1 with a step size of 0.5?

(A) 2.5 (B) 2.75 (C) 3.25 (D) 3.75

2. Let y = f (x) be the solution to the differential equation dy= x − xy with the initial condition f (0.5) = 0 .
dx

What is the approximation for f (2) if Euler’s method is used, starting at x = 0.5 with a step size of 0.5?

(A) 0.825 (B) 0.906 (C) 1.064 (D) 1.178

3. Let y = f (x) be the solution to the differential equation dy = arctan(xy) with the initial condition
dx

f (0) = 1. What is the approximation for f (2) if Euler’s method is used, starting at x = 0 with a step
size of 1?

(A) π (B) 1+ π (C) 1+ π (D) π
2 4 2

x −1 −0.6 −0.2 0.2 0.6

f ′(x) 1 2 −0.5 −1.5 1.2

4. The table above gives selected values for the derivative of a function f on the interval −1 ≤ x ≤ 0.6 .
If f (−1) =1.5 and Euler’s method is used to approximate f (0.6) with step size of 0.8, what is the
resulting approximation?

(A) 1.9 (B) 2.1 (C) 2.3 (D) 2.5

Further Applications of Integration 243

x0 = 0 f (x0 ) = 1
x1 = 0.5 f (x1) ≈ 1.5
x2 = 1 f (x2 ) ≈ 3

5. Consider the differential equation dy = kx + y − 2x2 , where k is a constant. Let y = f (x) be the
dx

particular solution to the differential equation with the initial condition f (0) = 1. Euler’s method,
starting at x = 0 with step size of 0.5, is used to approximate f (1) . Steps from this approximation

are shown in the table above. What is the value of k ?

(A) 2.5 (B) 3 (C) 3.5 (D) 4

Free Response Questions

6. Consider the differential equation dy = 1 x − y − 1 .
dx 2 2

(a) Find d 2 y in terms of x and y .
dx2

(b) Let y = f (x) be the particular solution to the given differential equation whose graph passes
through the point (0, − 1) . Does the graph of f have relative minimum, a relative maximum,
2
or neither at the point (0, − 1) ? Justify your answer.
2

(c) Let y = g(x) be another solution to the given differential equation with the initial condition
g(0) = k , where k is a constant. Euler’s method, starting at x = 0 with a step size of 0.5,
gives the approximation g(1) ≈ 1 .Find the value of k .

244 Chapter 7

7. Consider the differential equation d=y 2x + y .
dx

(a) On the axis provided, sketch a slope field for the given differential equation at the twelve points
indicated, and sketch the solution curve that passes through the point (1,1) .

y

2

1

−1 O x
1

(b) Let f be the function that satisfies the given differential equation with the initial condition f (1) = 1 .
Use Euler’s method, starting at x = 1 with a step size of 0.1, to approximate f (1.2) . Show the work
that leads to your answer.

(c) Find the value of b for which y =−2x + b is a solution to the given differential equation. Show the
work that leads to your answer.

(d) Let g be the function that satisfies the given differential equation with the initial condition g(1) = −2 .
Does the graph of g have a local extremum at the point (1, −2) ? If so, is the point a local maximum
or a local minimum? Justify your answer.

Chapter8
Parametric Equations, Vectors, and

Polar Coordinates

8.1 Slopes and Tangents for the Parametric Curves

If x and y are both given as functions of a third variable t , then the equations
x = f (t) , y = g(t)

are called parametric equations, and t is called the parameter.

The set of points (x, y) = ( f (t), g(t)) defined by the parametric equations is called the
parametric curve.

When the points in a parametric curve are plotted in order of increasing values of t , the curve
is traced out in a specific direction. This is called the direction of path (or motion) of the
curve.

Parametric Formula for dy dx

If the equation x = f (t) , y = g(t) define y as a differentiable function of x and dx dt ≠ 0 ,

then

dy = dy dt , where dx ≠ 0 .
dx dx dt dt

Parametric Formula for d 2 y
dx 2

=d 2 y d=( dy ) d ( dy )
dx2 dx dx dt dx

dx

dt

Horizontal Tangent

If dy dt = 0 and dx dt ≠ 0 when t = t0 , the curve represented by x = f (t) and y = g(t) has
a horizontal tangent at ( f (t0 ), g(t0 )) .

Vertical Tangent

If dx dt = 0 and dy dt ≠ 0 when t = t0 , the curve represented by x = f (t) and y = g(t) has
a vertical tangent at ( f (t0 ), g(t0 )) .

246 Chapter 8

Example 1 □ A curve in the plane is defined parametrically by the equations x
x = t2 and y= t3 − 2t . 5

(a) Sketch the curve in the xy- plane for −2 ≤ t ≤ 2 . Indicate the
direction in which the curve is traced as t increases.

(b) For what values of t does the curve have a vertical tangent?
(c) For what values of t does the curve have a horizontal tangent?

(d) Find the equation of the tangent lines to the curve at t = ± 2 .

Solution □ (a) Use a graphing calculator to draw y
a parametric curve. 4
Set the graphing calculator as follows. −5
−4
○ Mode: Parametric mode

○ Window: Tmin = −2 , Tmax = 2 ,
X min = −5 , X max = 5
Ymin = −4 , Ymax = 4

○ Y = : X1T = T2 , Y1=T T3 − 2T

(b) dx = 2t , d=y 3t2 − 2
dt dt
dx = 0 ⇒ t = 0
dt

The curve has a vertical tangent when t = 0 .

(c) dy = 0 ⇒ 3t2 − 2 =0 ⇒ t =± 2 =± 6
dt 3 3
The curve has a horizontal tangent when t = ± 6
3

(d) If t = 2 ,=dy 3( 2)2=− 2 =4 2
dx 2( 2) 2 2

The equation of the tangent line is
y −=0 2(x − 2) .
If t = − 2 , dy = 3(− 2)2 − 2 = 4 = − 2

dx 2(− 2) −2 2
The equation of the tangent line is
y − 0 =− 2(x − 2) .

Parametric Equations, Vectors, and Polar Coordinates 247

Example 2 □ A particle moves in the xy- plane so that its position at any time t ,
0 ≤ t ≤ 4 , is given by the equations x=(t) cos t + t sin t and y=(t) sin t − t cos t .

(a) Sketch the curve in the xy- plane for 0 ≤ t ≤ 4 . Indicate the
direction in which the curve is traced as t increases.

(b) At what time t , 0 < t < 4 , does the line tangent to the path of
the particle have a slope of −1 ?

(c) At what time t , 0 < t < 4 , does x(t) attain its maximum value?
What is the position (x(t), y(t)) of the particle at this time?

(d) At what time t , 0 < t < 4 , is the particle on the y-axis?

Solution □ (a) Use a graphing calculator to draw y
4
the parametric curve. −5 x
−4 5
○ Mode: Parametric mode, Radian mode

○ Window: Tmin = 0 ,Tmax = 4 ,
X min = −5 , X max = 5
Ymin = −4 , Ymax = 4

○ Y = : =X1T cosT +T sinT ,
=Y1T sinT −T cosT

(b=) dy d=y dt cos t − (−t sin t + c=os t) t=sin t tan t
dx dx dt − sin t + (t cos t + sin t) t cos t
So dy = tan t = −1 ⇒ =t tan−1(−=1) 3π 4 .
dx

(c) x′(t) =− sin t + (sin t + t cos t) =t cos t
x′(t) = 0 ⇒ t = π 2 , for 0 < t < 4 .
x(t) attains its maximum value when t = π 2 .

x(π ) =cos π + π sin π =π
2 22 2 2

y(π ) =sin π − π cos π =1
2 22 2

The position when t = π 2 is (π ,1) .
2

(d) The particle is on the y- axis when x(t) = 0 .

x(t) =cos t + t sin t =0

Use a graphing calculator (in function mode) to find the value of t which
makes cos t + t sin t =0 .
For 0 < t < 4 , x(t) = 0 when t = 2.798 .

248 Chapter 8

Exercises - Slopes and Tangents to the Parametric Curves
Multiple Choice Questions

1. If x = tet and y = t + et , then dy at t = 0 is
dx

(A) 0 (B) 1 (C) 1 (D) 2
2

2. If x = tan t and y = sin t , then d2y at t = π is
dx2 6

(A) − 9 (B) − 27 (C) 13 (D) 7
11 32 16 8

3. A curve C is defined by the parametric equations x= t3 − 3 and y = 2t2 . Which of the following
is the equation for the line tangent to the graph of C at the point (5,8) ?

(A) =y 1 x + 8 (B) =y 2x − 8 (C) =y 2 x + 14 (D) =y 3x + 8
33 3 33

4. If x = cos t and y = 2sin2 t , then dy at t = 1 is
dx

(A) −2 cos1 (B) −4 cos1 (C) −2 tan1 (D) −2sin1

Parametric Equations, Vectors, and Polar Coordinates 249

5. For what value(s) of t does the curve defined by the parametric equations x = 1 5t and
+ t3

y = 2t 2 have a horizontal tangent?
1+ t3

(A) 0 only
(B) 3 2 only
(C) 0 and 4 only
(D) 0 and 3 2 only

6. A point (x, y) is moving along a curve y = f (x) . At the instant when the slope of the curve

is 3 , the x- coordinate of the point is decreasing at the rate of 2 units per second. The rate
45

of change, in units per second, of the y-coordinate of the point is

(A) − 15 (B) − 3 (C) − 3 (D) 3
8 5 10 10

7. An object moving along a curve in the xy-plane is in position (x(t), y(t)) at time t ≥ 0

with dx= 2 − sin(t2 ) . At time t = 3 , the object is at position (2, 7) . What is the x-coordinate
dt

of the position of the object at time t = 6 ?

(A) 8.135 (B) 9.762 (C) 10.375 (D) 11.308

250 Chapter 8

Free Response Questions

8. A particle moves in the xy- plane so that its position at any time t , 0 ≤ t ≤ 2π , is given
by x(t)= t − sin t and y(t) = t cos t .

(a) Sketch the path of the particle in the xy- plane below. Indicate the direction of motion
along the path.

y

2 x

O2
−2

(b) At what time t , 0 ≤ t ≤ 2π , does y(t) attain its minimum value? What is the position (x(t), y(t)) of
the particle at this time?

(c) Write an equation for the line tangent to the curve at time t = π .

9. A particle moving along the curve is defined by the equation y =x3 − 4x2 + 4 . The x-coordinate of
the particle, x(t) , satisfies the equation dx = t , for t ≥ 0 with initial condition x(0) = 1.
dt t2 + 9
(a) Find x(t) in terms of t .

(b) Find dy in terms of t .
dt

(c) Find the location of the particle at time t = 4 .
(d) Write an equation for the line tangent to the curve at time t = 4 .

Parametric Equations, Vectors, and Polar Coordinates 251

8.2 Arc Length (Distance Traveled Along a Curve) in Parametric Form

If a curve C is given by the parametric equations x = f (t) and y = g(t) such that C does

not intersect itself on the interval a ≤ t ≤ b then the arc length C , or the distance traveled
by a particle along the curve is given by

∫b  dx 2 +  dy 2 dt .
 dt   dt 
=L

a

The displacement of a particle is the distance between its initial and final positions.
The displacement of a particle between time t = a and t = b is given by

b x′(t )  2 b 2
 a
a
=Dispalcement ∫ ∫ dt +  y′(t ) dt 
=  


[ x(b) − x(a)]2 + [ y(b) − y(a)]2 .

Example 1 □ A particle moves in the xy-plane so that its position at any time t ,
for 0 ≤ t , is given by x(t) = et and y(t) = 2 cos(t) .

(a) Find the distance traveled by the particle from t = 0 to t = 2 .
(b) Find the magnitude of the displacement of the particle between

time t = 0 and t = 2 .

Solution □ (a) dx = et , dy = −2sin(t)
dt dt
 dx 2 +  dy 2 dt
∫Distance traveled by t=he particle b  dt   dt 
a

( )∫= 2 et 2 + (−2sin t))2 dt
0

≈ 7.035 Use a graphing calculator (in function mode)
to find the value of the definite integral.

 b x′(t)  2  b y′(t)  2
   
a a
∫ ∫(b) =Displacement dt + dt

∫ ∫=  2 et dt 2 +  2 (−2 sin t) dt 2
   0 
0

= (6.389)2 + (−2.832)2

≈ 48.8395 Use a graphing calculator.
≈ 6.988

252 Chapter 8

Exercises - Arc Length (Distance Traveled Along a Curve) in Parametric Form
Multiple Choice Questions

1. The position of a particle at any time t ≥ 0 is given by x= t − t2 and y = 4 t3 2 . What is the total
3

distance traveled by the particle from t = 1 to t = 3 ?

(A) 7.165 (B) 8.268 (C) 9.431 (D) 10.346

2. The position of particle at any time t ≥ 0 is given by=x(t) a(cos t + t sin t) and=y(t) a(sin t − t cos t) .
What is the total distance traveled by the particle from t = 0 to t = π ?

(A) 1 π a (B) π a2 (C) 1 π 2a (D) 1 π 2a2
2 2 2

3. The length of the path described by the parametric equations=x sin t + ln(cos t) and y = cos t ,
for π ≤ t ≤ π , is given by
63

∫(A) π 3 cos2 t + 2sin t + 2 dt
π6

∫(B) π 3 sin2 t + 2 cos t + 2 dt
π6

∫(C) π 3 cot2 t + 2 cos t dt
π6

∫(D) π 3 sec2 t − 2sin t dt
π6

Parametric Equations, Vectors, and Polar Coordinates 253

4. A particle moving in the xy- plane has velocity vector given by v(=t) et − t, t sin t for time t ≥ 0 .
What is the magnitude of the displacement of the particle between time t = 0 and t = 2 ?

(A) 4.722 (B) 4.757 (C) 4.933 (D) 5.109

Free Response Questions

5. A particle moving along a curve in the xy-plane is at position (x(t), y(t)) at any time t , where
dx = 2sin(t2 ) and dy = cos(t3 ) . At time t = 1, the object is at position (3, 2) .
dt dt

(a) Write an equation for the line tangent to the curve at (3, 2) .
(b) Find the total distance traveled by the particle from t = 1 to t = 3 .
(c) Find the position of the particle at time t = 3 .
(d) Find the magnitude of the displacement of the particle between t = 1 and t = 3 .

254 Chapter 8

8.3 Vector valued Functions

Vector Valued Function, Velocity Vector, Acceleration Vector, and Speed
If a particle moves in the xy-plane so that at time t > 0 its position vector is given by

r(t) =< x(t), y(t) > ,
then the velocity vector, acceleration vector, and speed at time t are as follows.

Velocity =v(t) =r′(t) =< x′(t), y′(t) >
Accleration =a(t) =v′(t) =< x′′(t), y′′(t) >

Sp=eed v=(t) [ x′(t)]2 + [ y′(t)]2

If x(t) is increasing dx is positive and if x(t) is decreasing dx is negative.
dt dt

If y(t) is increasing dx is positive and if y(t) is decreasing dx is negative.
dt dt

Example 1 □ A particle moving in the xy-plane is defined by the vector-valued function
f (t) =< t − sin t,1− cos t > , for 0 ≤ t ≤ π .

(a) Find the velocity vector for the particle at any time t .
(b) Find the speed of the particle when t = π .

3
(c) Find the acceleration vector for the particle at any time t .

(d) Find the average speed of the particle from time t = 0 to time t = π .

Solution □ (a) x(t)= t − sin t , y(t)= 1− cos t
x′(t)= 1− cos t , y′(t) = sin t
v(t) =< x′(t), y′(t) >=< 1− cos t,sin t >

(b) x′(π ) =1− cos π =1− 1 =1
3 3 22

y=′(π ) s=in π 3
3 32

Speed =v(π ) =  x′(π  2  y′(π  2  1 2  3 2
3  3   3   2   2 
) + ) = + =1

(c) x′′(t) = sin t , y′′(t) = cos t
a(t) =< x′′(t), y′′(t) >=< sin t, cos t >

∫(=d) Average speed 1 π [x′(t)]2 + [ y′(t)]2 dt

π −0 0

∫= 1 π (1− cos t)2 + (sin t)2 d=t 4
π0 π

Parametric Equations, Vectors, and Polar Coordinates 255

Exercises - Vector Valued Functions
Multiple Choice Questions

1. If a particle moves in the xy-plane so that at time t > 0 its position vector is (t3 −1, ln t2 +1) ,
then at time t = 1, its velocity vector is

1 1 1 1
(A) (0, ) (B) (1, ) (C) (3, ) (D) (3, )

2 2 2 4

2. A particle moves in the xy-plane so that at any time t its coordinates are x= t3 − t2 and y = t + ln t .
At time t = 2 , its acceleration vector is

(A) (4, 1) (B) (6, 1) (C) (8, 3) (D) (10, − 1)
2 4 4 4

3. A particle moves in the xy-plane so that its position at time t > 0 is given by x(t) = et cos t
and y(t) = et sin t . What is the speed of the particle when t = 2 ?

(A) 2e (B) 2e2 (C) 2e (D) 2e2

( )4. If f is a vector-valued function defin=ed by f (t) ln(sin t), t2 + e−t , then the acceleration vector is

(A) (− csc2 t, 2 + e−t )
(B) (sec2 t, 2 + e−t )
(C) (csc2 t, 2 − e−t )
(D) (− csc2 t ⋅ cot t, 2 + e−t )

256 Chapter 8

5. A particle moves on the curve y= x + x so that the x-component has velocity x′(t) = cos t
for t ≥ 0 . At time t = 0 , the particle is at the point (1, 0) . At time t = π , the particle is at the point
2

(A) (0, 0) (B) (1, 2) (C) π π + π (D) (2, 2 + 2)
(, )
22 2

6. In the xy- plane, a particle moves along the curve defined by the equation=y 2x4 − x with a constant
speed of 20 units per second. If dy > 0 , what is the value of dx when the particle is at the point (1, 1)
dt dt

(A) 2 (B) 2 (C) 2 2 (D) 4

Free Response Questions

7. An object moving along a curve in the xy-plane is at position (x(t), y(t)) at time t , where dx = 1+ cos(et ) .
dt

and dy = e(2−t2 ) for t ≥ 0 .
dt

(a) At what time t is the speed of the object 3 units per second?
(b) Find the acceleration vector at time t = 2 .
(c) Find the total distance traveled by the object over the time interval 1 ≤ t ≤ 4 .
(d) Find the magnitude of the displacement of the object over the time interval 1 ≤ t ≤ 4 .

Parametric Equations, Vectors, and Polar Coordinates 257

8. An object moving along a curve in the xy-plane has position (x(t), y(t)) at time t ≥ 0 , with

dx = t − sin(et ) . The derivative dy is not explicitly given. At time t = 1, the value of dy is 3
dt dt dt

and the object is at position (1, 4) .

(a) Find the x-coordinate of the position of the object at time t = 5 .

(b) Write an equation for the line tangent to the curve at the point (x(1), y(1)) .

(c) Find the speed of the object at time t = 1.

(d) Suppose the line tangent to the curve at (x(t), y(t)) has a slope of (t − 2) for t ≥ 0 . Find the
acceleration vector of the object at time t = 3 .

9. The position of a particle moving in the xy-plane is given by the parametric equations x(t)= t − sin(π t)
and y(t)= 1− cos(π t) for 0 ≤ t ≤ 2 .

(a) On the axis provided below, sketch the graph of the path of the particle from t = 0 to t = 2 .
Indicate the direction of the particle along its path.

y

1 x
O1

(b) Find the position of the particle when t = 1.

(c) Find the velocity vector for the particle at any time t .

(d) Write and evaluate an integral expression, in terms of sine and cosine, that gives the distance
traveled of the particle from t = 0 to t = 2 .

258 Chapter 8

initial
position

10 m x(t)
y(t)

Note: Figure not drawn to scale.

10. An object is thrown upward into the air 10 meters above the ground. The figure above shows the initial
position of the object and the position at a later time. At time t seconds after the object is thrown upward,
the horizontal distance from the initial position is given by x(t) meters, and the vertical distance from the
ground is given by y(t) meters, where dx = 1.4 and d=y 4.2 − 9.8t , for t ≥ 0 .
dt dt

(a) Find the time t when the object reaches its maximum height.

(b) Find the maximum vertical distance from the ground to the object.

(c) Find the time t when the object hit the ground.

(d) Find the total distance traveled by the object from time t = 0 until the object hit the ground.

(e) Find the magnitude of the displacement of the object from time t = 0 until the object hit the ground.

(f) Find the angle θ , 0 < θ < π , between the path of the object and the ground at the instance the object
2

hit the ground.

Parametric Equations, Vectors, and Polar Coordinates 259

y
2
1

x
O 123456
−1
−2

11. At time t , the position of particle moving in the xy- plane is given by the parametric functions

( x(t), y(t)) , where =dx e x − cos(x2 ) . The graph of y consisting of four line segments, is shown

dt
in the figure above. At time t = 0 , the particle is at position (2,1) .

(a) Find the position of the particle at t = 2 .
(b) Find the slope of the line tangent to the path of the particle at t = 2 .

(c) Find the magnitude of the velocity vector at t = 2 .

(d) Find the total distance traveled by the particle from t = 0 to t = 3 .

260 Chapter 8

8.4 Polar Coordinates and Slopes of Curves

Coordinate-System Conversion Formula

If a point P has rectangular coordinates (x, y) and polar coordinates (r,θ ) ,

then

x = r cosθ , y = r sinθ ,

r=2 x2 + y2 , ta=nθ y ( x ≠ 0) .
x

The slope of the tangent line to the graph of r = f (θ ) at point (r,θ ) is

=dy dy d (r sinθ ) dr sinθ + r cosθ , where dx ≠0 at (r,θ ) .
d=θ dθ = dθ

dx dx d (r cosθ ) dr cosθ − r sinθ dθ
dθ dθ dθ

If r > 0 and dr < 0 for α < θ < β , then r is decreasing on this interval, which means
dθ

the curve is getting closer to the origin.

If r > 0 and dr > 0 for α < θ < β , then r is increasing on this interval, which means
dθ

the curve is getting farther from the origin.

If dr = 0 , the curve is either closest to the origin or farthest from the origin.
dθ

Example 1 □ A curve is defined by the polar equation r = 4sin(2θ ) for 0 ≤ θ ≤ π .
2

(a) Graph the curve.
(b) Find the slope of the curve at the point where θ = π 4 .

(c) Find an equation in terms of x and y for the line tangent to the curve at
the point where θ = π .
4

(d) Find an interval where the curve is getting closer to the origin.

(e) Find the value of θ in the interval 0 ≤ θ ≤ π such that the point
2

on the curve has the greatest distance from the origin.

Solution □ (a)

yπ
(4, )
4

x

Parametric Equations, Vectors, and Polar Coordinates 261

=(b) x r=cosθ 4sin(2θ ) cosθ , dx =4(− sin(2θ ) sinθ + 2 cos(2θ ) cosθ )
dθ

dx =4(− sin π sin π + π π ) =−2 2
2 cos cos
dθ θ =π 24 24

4

=y r=sinθ 4sin(2=θ ) sinθ , dy 4(sin(2θ ) cosθ + 2 cos(2θ ) sinθ )
dθ

dy = 4(sin π π + 2 cos π sin π ) =2 2
cos
dθ θ =π 24 24

4

The slope of a tangent line to the polar curve is dy .
dx

dy = dy dθ = 2 2 = −1

dx θ =π dx dθ θ =π −2 2
4
4

=(c) x r=cosθ 4sin(2θ ) cosθ , x(π=) 4 π ) cos(π=) =4 2 2 .
sin(
4 2 42

=y r=sinθ 4sin(2θ ) sinθ , y(π=) 4sin(π ) sin=π =4 2 2 .
4 2 42

dy = −1
dx θ =π

4

The equation of tangent line is y − 2 2 =−1(x − 2 2) or

y =−x + 4 2 .

(d) The curve is getting closer to the origin when dr < 0 .
dθ

=dr 8cos(2θ ) < 0 ⇒ cos(2θ ) < 0 ⇒ π < 2θ < π
dθ 2
⇒ π <θ < π
42

(e) The point on the curve is farthest from the origin when dr = 0 .
dθ

dr =0 ⇒ cos(2θ ) = 0 ⇒ 2θ = π ⇒ θ =π
dθ 2 4

262 Chapter 8

Exercises - Polar Coordinates and Slopes of Curves
Multiple Choice Questions

1. If r= θ − 3sinθ then dr at (π ,π ) is
dθ

(A) 2 (B) π (C) 4 (D) 2π

2. If r = 2 then dr π
at (2, ) is
1− cosθ dθ 2

(A) −2 (B) −1 (C) 0 (D) 1

3. If r = 3sinθ then dy at the point where θ = π is
dx 3

(A) −2 (B) − 3 (C) −1 (D) 3
3

4. The equation of the polar curve is given by r = 8 . What is the angle θ that corresponds
1− cosθ

to the point on the curve with x-coordinate −3 ?

(A) 1.248 (B) 1.356 (C) 1.596 (D) 2.214

Parametric Equations, Vectors, and Polar Coordinates 263

Free Response Questions

y

=r θ + cos(2θ )

1

−2 −1 O x

1

5. The polar curve=r θ + cos(2θ ) , for 0 ≤ θ ≤ π , is drawn in the figure above.

(a) Find dr , the derivative of r with respect to θ .
dθ

(b) Find the angle θ that corresponds to the point on the curve with x-coordinate 0.5.

(c) For π < θ < 5π , dr is negative. What does this fact say about r ? What does this fact say
12 12 dθ

about the curve?

(d) Find the value of θ in the interval 0 ≤ θ ≤ π that correspond to the point on the curve in the
2

first quadrant with the least distance from the origin. Justify your answer.

264 Chapter 8

8.5 Areas in Polar Coordinates

Area in Polar Coordinates

The area of the region bounded by the graph of r = f (θ ) between the radial lines θ = α and

θ = β is given by

∫A = 1 β 2

2α [ f (θ )] dθ

∫= 1 β r 2dθ

2α

Example 1 □ Find the area of the region that lies θ =π 3
inside the circle r = 3cosθ and
r r = 3cosθ
outside the cardioid r = 1+ cosθ . r = 1 + cosθ

Solution □ Find the points of intersection of the
two curves.
O θ

3cosθ = 1+ cosθ

⇒ cosθ = 1 ⇒ θ = ± π
23

The desired area can be found by θ = −π 3
subtracting the area inside the cardioid
between θ = −π 3 and θ = π 3 from the

area inside the circle between θ = −π 3 and θ = π 3 .

∫A = 1 β r2dθ

2α

∫= 1 π3 (3cosθ )2 − (1+ cosθ )2  dθ
2 −π 3

∫= 2 ⋅ 1 π3 (3cosθ )2 − (1+ cosθ )2  dθ The region is symmetric about
0 the horizontal axis θ = 0 .
2
cos2 θ = 1 + cos 2θ
∫= π3 (8cos2θ − 2 cosθ −1) dθ 2
0

∫= π3 8( 1+cos2θ ) − 2 cosθ − 1 dθ
0 2

∫= π 3 (3+4cos2θ − 2 cosθ ) dθ
0

= [3θ +2 sin 2θ − 2 sin θ ] π 3
0

= π + 3 − 3= π

Parametric Equations, Vectors, and Polar Coordinates 265

Exercises - Areas in Polar Coordinates
Multiple Choice Questions

1. The area of the region enclosed by the polar curve r2 = 6sin(2θ ) is

(A) 2 (B) 4 (C) 6 (D) 12

y x
O

2. What is the area of the region enclosed by the loop of the graph of the polar curve r = 2 cos(2θ )
shown in the figure above?

π π (C) 3π (D) π
(A) (B) 4

4 2

y

r = sinθ

r = cosθ

x

O

3. The area of the shaded region that lies inside the polar curves r = sinθ and r = cosθ is

(A) 1 (π − 2) (B) 1 (π − 2) (C) 1 (π − 2) (D) 1 (π −1)
8 4 2 8

266 Chapter 8

4. The area of the region enclosed by the polar curve r= 2 + sinθ is

(A) 3π (B) 7π (C) 4π (D) 9π
2 2

y

r =θ

x

O

5. The area of the shaded region bounded by the polar curve r = θ and the x-axis is

π2 π3 π3 π3
(A) (B) (C) (D)

4 6 3 2

6. Which of the following gives the area of the region inside the polar curve r = 1+ cosθ and
outside the polar curve r = 2 cosθ ?

∫(A) 1 2π (1− cosθ )2 dθ

20

∫(B) 2π (1+2cosθ )2 dθ
0

∫(C) 2π (1+cosθ )2 dθ − π
0

∫(D) 1 2π (1+cosθ )2 dθ − π

20

Parametric Equations, Vectors, and Polar Coordinates 267

Free Response Questions

y

r=2 2

R2 P r= 2 + cos(2θ )
1

R3 R1

−3 −1 O 1 x

3

−1

 −2

7. The figure above shows the graphs of the polar curves r= 2 + cos(2θ ) and r = 2 . Let R1 be the
shaded region in the first quadrant bounded by the two curves and the x-axis, and R2 be the shaded
region in the first quadrant bounded by the two curves and the y-axis. The graphs intersect at point P
in the first quadrant.

(a) Find the polar coordinates of point P and write the polar equation for the line  .

(b) Set up, but do not integrate, an integral expression that represents the area of R1 .

(c) Set up, but do not integrate, an integral expression that represents the area of R2 .

(d) Let R3 be the shaded region in the second quadrant bounded by the two curves and
the coordinate axis. Find the area of R3 .

(e) The distance between the two curves changes for 0 < θ < π . Find the rate at which the distance
4

between the two curves is changing with respect to θ when θ = π .
6

268 Chapter 8

y

3

2

1

x

O 1234

8. The graph of the polar curve r= 2 + 2 cos(θ ) for 0 ≤ θ ≤ π is shown above.
(a) Write an integral expression for the area of the shaded region.
(b) Write expressions for dx and dy in terms of θ .
dθ dθ
(c) Write an equation in terms of x and y for the line tangent to the curve at the point where θ = π .
2

Chapter 9
Infinite Sequences and Series

9.1 Sequences and Series

A sequence is an ordered list of numbers
a1, a2 , a3,  , an , 

whereas a series is an infinite sum of a list of numbers
a1 + a2 + a3 + + an + .

Mathematically, a sequence is defined as a function whose domain is the set of positive
integers.

If a sequence {an} has the limit L , where L is a real number, it is written as

lim an = L,

n→∞

and we say the sequence converges to L . If the sequence does not have a limit we say

the sequence diverges.

Given a sequence of numbers of {an} , an expression of the form

∞

∑ an = a1 + a2 + a3 + + an +

n=1

is an infinite series.

∑For the infinite series an , the nth partial sum is given by

n

∑sn = ai = a1 + a2 + a3 + + an .
i =1

If the sequence of partial sums converges to a limit L , we say that the series converges,
and its sum is L . If the series does not converge, we say that the series diverges.

The geometric series a + ar + ar2 + + arn−1 + converges to the sum

∑∞ a , if r < 1.

ar n−1 =
n=1 1 − r

If r ≥ 1, the series diverges.

A series is a telescoping series if it is in the form
(a1 − a2 ) + (a2 − a3 ) + (a3 − a4 ) + (a4 − a5 ) + ⋅ ⋅ ⋅ .

Since the nth partial sum of the series is Sn= a1 − an+1 , a telescoping series converges

if and only if lim an = L , where L is a real number.

n→∞

The sum of the series is S= a1 − L .

270 Chapter 9

Limit of nth Term of a Convergent Series

∞
∑If the series
an converges, then lim an =0.

n=1 n→∞

The converse of this theorem is not true in general. But the contrapositive of this theorem
provides a useful test for divergent series.

nth Term Test for Divergence

∞
∑If
lim an does not exist or lim an ≠ 0 , then the series an diverges.

n→∞ n→∞ n=1

Example 1 □ Find the sum of the series.

∑(a) ∞ 2n+1 ∑∞ 1
n=1 3n
(b)
n=1 n(n +1)

Solution ∑ ∑ ∑□=(a) =n∞1=23nn+1 n∞=1 23⋅=n2n 2n∞1  2 n
 3



∑∞  2 n is an infinite geometric series with a=2 and r = 2.
3 3

n=1 3 

∑∞ 2n+1 =2 ⋅ 2 3 =4
3n 1−2 3
n=1

(b) an = 1=  1 − n 1 1  Partial fractions
n(n +1)  n +  Sum of the telescoping series

Sn =  1 − 1  +  1 − 1  +  +  1 − n 1 1  = 1− 1
 1 2   2 3   n +  n +1

∑∞ 1 = lim Sn = lim (1− 1) =1
n(n +1) n +1
n=1 n→∞ n→∞

Example 2 □ Determine whether the series is convergent or divergent.

∑∞ n ∞

(a) ∑(b) 2−n5n
n=1 2n + 3 n=1

Solution □ (a) lim an= lim n = 1 ≠0
n→∞ 2n + 3 2
n→∞

Therefore, the series diverges by the nth Term Test for Divergence.

2=−n5n ∞=5n ∞ n
=n 1=2n n 1 
∑ ∑ ∑∞  5 is an infinite geometric series with r = 5.
 2 2
(b)

=n 1

Since=r 5 2 ≥ 1 , the series diverges.

Infinite Sequences and Series 271

Exercises - Sequences and Series
Multiple Choice Questions

∑1. ∞ (3)n+1 = (B) 5 (C) 9 (D) The series diverges
n=1 5n 2 2
(A) 3
5

∞

2. If f (x) = ∑ (tan x)n , then f (1) =
n=1

(A) −2.794 (B) −0.61 (C) 0.177 (D) The series diverges

∑3. ∞ n 2 1 =
n=2 2−

(A) 0 (B) 1 (C) 1 (D) 3
2 2

4. The sum of the geometric series 2 + 4 + 8 + . . . is
21 63 189

(A) 5 (B) 2 (C) 4 (D) The series diverges
21 7 7

272 Chapter 9

5. If Sn =  3n−1   (7 + n)20  {Sn } converge?
 (4 + n)20   3n  , to what number does the sequence

(A) 1 (B) 7 (C)  7 20 (D) Diverges
3 4  4 

6. Which of the following sequences converge?

I. cos2 n  II.  en − 3 III. n
    
 (1.1)n   3n  9+ n 

(A) I only (B) II only (C) III only (D) I and II only

7. Which of the following series converge?

∑∞ n ∞ ∑∞ −6

I. II. ∑ arctan n III.
n=1 10(n +1) n=1 n=1 (−5)n

(A) I only (B) II only (C) III only (D) II and III only

Free Response Questions

∞  3 1 
 n(n + 7n 
Find the sum of the series  

n=1
∑8. 3) + .

Infinite Sequences and Series 273

9.2 The Integral Test and p-Series

The Integral Test

If f is positive, continuous, and decreasing on [1, ∞) and an = f (n) , then

∑ ∫∞ and ∞
an
f (x) dx
n=1
1

either both converge or both diverge. In other words:

∑ ∫∞ ∞ f (x) dx is convergent.
1. If an is convergent, then
1
n=1

∑ ∫∞ ∞
2. If an is divergent, then f (x) dx is divergent.

n=1 1

Example 1 □ Determine whether the series is convergent or divergent.

∑(a)∞1 (b) ∞ ln n
n=1 n2 + 1
∑
n=1 n

Solution □ (a) The function =f (x) 1 (x2 +1) is continuous, positive, and

decreasing on [1, ∞) so we can use the Integral Test:

∫ ∫=1∞ x21+1 dx bl=→im∞ 1b x21+1 dx lim  tan −1 x  b
1
b→∞

= lim  tan −1 b − tan −1 1 = π −π = π
2 4 4
b→∞

So the series converges.

Note : The fact that the integral converges to π 4 does not imply
that the infinite series converges to π 4 .

(b) The function f (x) = ln x x is continuous, positive, and

decreasing on [1, ∞) so we can use the Integral Test:

∫ ∫=∞ ln x dx l=im b ln x dx  ln2 x  b
b→∞ 1 x lim  
1x b→∞  2 1

= lim  ln2 b − ln 2 1 = ∞ − 0 = ∞
 2 
b→∞  2 

So the series diverges.

274 Chapter 9

p- Series and Harmonic Series

∑The p- series ∞ 1 = 1 + 1 + 1 + 1 + ⋅ ⋅ ⋅

n=1 n p 1p 2 p 3 p 4 p
is convergent if p > 1 and divergent if 0 < p ≤ 1 .

∑For p = 1 , the series ∞ 1 = 1+ 1 + 1 + 1 + ⋅ ⋅ ⋅ is called harmonic series.
n=1 n 234

A general harmonic series is of the form ∑ 1 b) .
(an +

Example 2 □ Determine whether the series is convergent or divergent.

(a) 1+ 1 + 1 + 1 + 1 +
3 4 3 9 3 16 3 25

∞

∑(b) n1−π
n=1

(a) 1+ 1 + 1 + 1 + 1 += ∞ =1 ∞ 1
3 4 3 9 3 16 3 25=n 1=3 n2 n 1 n2 3
Solution ∑ ∑□

The p- series is divergent since=p 2 3 < 1.

∑ ∑ ∑∞ ∞ ∞ 1
(b)=n1−π =n−(π −1) nπ −1

=n 1 =n 1 =n 1

The p- series is convergent since p = π −1 ≈ 2.14 > 1 .

Infinite Sequences and Series 275

Exercises - The Integral Test and p-Series
Multiple Choice Questions

∫1. If ∞ dx = π , then which of the following must be true?
1 x2 +1 4

∑∞ 1 diverges.

I.
n=1 n2 + 1

∑∞ 1

II. converges.
n=1 n2 + 1

∑∞ n 1 1 = π
2+ 4
III.

n=1

(A) none (B) I only (C) II only (D) II and III only

2. What are all values of p for which ∫∞ 1 converges?
1 3 xp

(A) P < −3
(B) P < −1
(C) P > 1
(D) P > 3

3. Which of the following series converge?

∑∞ n ∑II. ∞ ne−n2 ∞ 1
n=1
I. III. ∑
n=1 2n2 +1 n=2 x ln x

(A) I only (B) II only (C) III only (D) I and II only

276 Chapter 9

4. What are all values of p for which ∑∞ n converges?
n=1 n p + 1

(A) p > 0 (B) p > 1 (C) p > 1 (D) p > 3
2 2

5. What are all values of k for which the series 1+ ( 2)k + ( 3)k + ( 4)k + + ( n)k +
converges?

(A) k < −2 (B) k < −1 (C) k > 1 (D) k > 2

Free Response Questions

6. Determine whether the following series converge or diverge.

(a) 1+ 1 + 1 + 1 +
8 27 64

(b) 1+ (3 1 + 1 + (3 1 +
2)2 (3 3)2 4)2

Infinite Sequences and Series 277

9.3 The Comparison Test

Direct Comparison Test
Let 0 < an ≤ bn for all n.

∞∞

1. If ∑ bn converges, then ∑ an converges.
n=1 n=1
∞∞

2. If ∑ an diverges, then ∑ bn diverges.
n=1 n=1

Limit Comparison Test

If an >0, bn > 0 , and lim an = L , where L is finite and positive, then both series either
n→∞ bn

converge or both diverge.

Note: When choosing a series for comparison, you can disregard all but the highest powers
of n in both the numerator and denominator.

Example 1 □ Determine whether the series is convergent or divergent.

∑(a) ∞n ∑(b) ∞ sin2 n
n=2 n2 − 3
n=1 n3 + 1

Solution □ (a) n> n =1 for all n ≥ 2.
n2 − 3 n2 n

∞ 1 is divergent harmonic series.

∑
n=2 n

Therefore ∑∞ n is divergent by the Direct Comparison Test.
n=2 n2 − 3

(b) sin2 n = sin2 n ≤ 1 < 1 for n ≥ 1.
n3 +1 n3 2 +1 n3 2 +1 n3 2

∑∞ 1 is convergent because it is a p- series with=p 3 2 > 1 .

n=1 n3 2

∑Therefore ∞ sin2 n is convergent by the Direct Comparison Test.

n=1 n3 + 1

278 Chapter 9

Example 2 □ Determine whether the series is convergent or divergent.

∞ 1 ∑∞ 2n

(a) ∑ (b)
n=1 n2 + 4 n=3 3n + 1

Solution □ (a) Let an = 1 and bn = 1 .
n2 + 4 n

=lim an lim=1 n2 + 4 l=im n 1
n→∞ bn n→∞ 1 n n→∞ n2 + 4

Since ∑ bn = ∑1 n is a divergent (harmonic series), the given series

diverges by the Limit Comparison Test.

(b) Let an = 2n and bn = 2n
3n +1 3n

=lim an lim 2n=(3n +1) l=im 3n 1
n→∞ bn n→∞ 2n 3n n→∞ 3n + 1

Since ∑=bn ∑=2n ∑ ( 2)n is a convergent geometric series,

3n 3

the given series converges by the Limit Comparison Test.

Infinite Sequences and Series 279

Exercises - The Comparison Tests
Multiple Choice Questions

1. Which of the following series converge?

∑∞ 1 ∑II. ∞ cos2 n ∑III. ∞ 1+ 4n
n=1 n2 + 2 n=1 3n
I.
n=1 n2 + n + 3

(A) I only (B) II only (C) III only (D) I and II only

2. Which of the following series diverge?

∞ 1 ∞ 1 III. ∞ sin( 1)

I. ∑ II. ∑ ∑
n=1 n! n=1 n + 2 n=1 n

(A) I only (B) II only (C) II and III only (D) I, II, and III

3.Which of the following series converge?

∑∞ n3 2 ∞ 1 ∑III. ∞ n!

I. II. ∑ n=1 nn
n=1 3n3 + 7 n=1 3 n4 + 1

(A) I only (B) I and II only (C) I and III only (D) I, II, and III

280 Chapter 9

4. Which of the following series cannot be shown to converge using the limit comparison test

∑∞ 1?

with the series
n=1 2n

∑∞ 1

(A)
n=1 2n −1

∑(B) ∞ n
n=1 2n

∞ 2n

(C) ∑
n=1 2n+1 n2 + 1

∑∞ 2n2 − 3n

(D)
n=1 2n (n2 + n −100)

Free Response Questions
5. Determine whether the following series converge or diverge.

∑∞ cos(2n)

(a)
n=1 1+ (1.6)n

∑∞ 4n

(b)
n=1 2n + 3n

Infinite Sequences and Series 281

9.4 Alternating Series and Error Bound

Alternating Series Test

Let an > 0 . The alternating series

∞ ∞

∑ (−1)nan and ∑ (−1)n+1an

n=1 n=1

converge if the following two conditions are met.

1. lim an =0 2. an+1 ≤ an , for all n greater than some integer N .

n→∞

Alternating Series Estimation Theorem (Error Bound)

∞

∑If Sn is a partial sum and =S (−1)nan is the sum of a convergent alternating series that
n=1

satisfies the condition an+1 ≤ an , then the remainder Rn= S − Sn is smaller than an+1 , which
is the absolute value of the first neglected term.

Rn =S − Sn ≤ an+1

Definition of Absolute and Conditional Convergence

1. ∑ an is absolutely convergent if ∑ an converges.
2. ∑ an is conditionally convergent if ∑ an converge but ∑ an diverges.

Example 1 □ Determine whether the series is convergent or divergent.

∑(a) ∞ (−1)n ∑∞ n

n=1 n (b) (−1)n
n=1 2n −1

Solution □ (a) 1. =nl→im∞ an l=im 1 0 2. an+=1 1< 1= an
n→∞ n n +1 n

So the series is convergent by the Alternating Series Test.

(b) 1. lim an= lim n = 1 ≠0
n→∞ 2n −1 2
n→∞

So the series is divergent by the nth Term Test for Divergence.

Example 2 □ Determine whether the series is absolutely convergent, conditionally convergent,

or divergent.

∑(a) ∞ (−1)n n e ∞
n=1 n2
∑(b) (−1)n+1 n−2/3
n=1

282 Chapter 9

(a) Since 0 ≤ n e ≤ e =e( 1 ) and ∞1
n2 n2 n2 n=1 n2
Solution ∑□ is a convergent p- series ( p= 2 > 1) ,

∑ ∑∞ n e ∞ (−1)n n e is absolutely convergent.
n=1 n2
n=1 n2
converges, and so

(b) 1.=nl→im∞ an l=im 1 0
n→∞ n2/3

2. an+=1 1 < 1= an
(n +1)2/3 (n)2/3

∞

∑So the series (−1)n+1n−2/3 is convergent by the Alternating Series Test.
n=1

Now consider the series of absolute values.

∑ ∑∞ ∞ 2 < 1) .
3
(−1)n+1n−2/3 =n−2/3 is a divergent p- series ( p=
=n 1 =n 1

∞

∑Thus, (−1)n+1n−2/3 is conditionally convergent.
n=1

Example 3 □ Let f (x) = 1− x2 + x4 − x6 + + − (−1)n x2n + .
2! 4! 6! (2n)!

Use the alternating series error bound to show that 1− 1 + 1 approximates f (1)
2! 4!

with an error less than 1 .
500

Solution □ f (1) =1− 1 + 1 − 1 + + (−1)n +
2! 4! 6! (2n)!

Since series is alternating, with terms convergent to 0 and decreasing in absolute value,
the error is less than the first neglected term.

So, f (1) − (1− 1 + 1 ) ≤ 1= 1 < 1 .
2! 4! 6! 720 500

Infinite Sequences and Series 283

Exercises - Alternating Series Tests
Multiple Choice Questions

1. Which of the following series converge?

∑I. ∞ (−1)n n ∑II. ∞ (−1)n ∞
n=1 n n=1 ln n
III. ∑ cos(nπ )
n=1

(A) I only (B) II only (C) III only (D) I and II only

2. Which of the following series converge?

∞ (−1)n π ∞ sin( 2n −1)π ∑III. ∞ (−1)n+1 2n
∑I. cos(
) II. ∑ n=1 n2 + 1
n=1 n n=1 2

(A) I only (B) II only (C) III only (D) I and II only

(−1)kn n2
∑ ∑∞ ∞ n converge?

3. For what integer k , k > 1 , will both
n=1 n and n=1 nk + 1

(A) 3 (B) 4 (C) 5 (D) 6

∞ (−1)n+1 <1
n3 500
Let s =

n=1
∑4. and sn be the sum of the first n terms of the series. If s − sn what is

the smallest value of n ?

(A) 6 (B) 7 (C) 8 (D) 9

284 Chapter 9

5. Which of the following series converge?

∞ ∑∞ 3n+1 ∞

∑I. ( −1)n n 3 II. ∑III. ( tan−1(n +1) − tan−1(n))
n=2 n=1 π n n=1

(A) I only (B) II only (C) III only (D) II and III only

∑∞ (−1)n+1(n + 3) is true?

6. Which of the following statements about the series
n=1 n2

(A) The series converges conditionally.
(B) The series converges absolutely.
(C) The series converges but neither conditionally nor absolutely.
(D) The series diverges.

7. Which of the following series is absolutely convergent?

∑∞ (−1)n+1 n

(A)
n=1 n

∑∞ (−1)n+1n

(B)
n=1 n2 n

∑∞ (−1)n−1n

(C)
n=1 n2 − n

∑∞ (−1)n−1(n2 +1)

(D)
n=1 n3

Infinite Sequences and Series 285

∞ (−1)n−1
n2 + 3 . Let S3 be the sum of the first three terms of
An alternating series is given by S =

n=1
∑8.

the given alternating series. Of the following, which is the smallest number M for which the

alternating series error bound guarantees that S − S3 ≤ M ?

(A) 1 (B) 1 (C) 1 (D) 1
4 7 19 28

Free Response Questions

9. Let f (x) =1− 3x + 9x2 − 27x3 + + (−1)n (3x)n + .
2! 4! 6! (2n)!

Use the alternating series error bound to show that 1− 3 + 9 approximates f (1) with an
2! 4!

error less than 1 .
20

286 Chapter 9

9.5 The Ratio Test

Ratio Test

∑Let an be a series with nonzero terms.

∑1. an converges absolutely if lim an+1 <1.
n→∞ an

∑2. an diverges if lim an+1 > 1 or lim an+1 = ∞.
n→∞ an n→∞ an

3. The Ratio Test is inconclusive if lim an+1 = 1 .
n→∞ an

Example 1 □ Determine whether the series is convergent or divergent.

∑∞ 3n ∑(b) ∞ n3 ∑(c)∞ 3n
5n n=1 2n −1
(a) (−1)n
n=1 n!
n=1

Solution □ (a)=lim an+1 lim 3n=+1 (n +1)! lim 3n+1 ⋅ n!
n→∞ an n→∞ 3n n! n→∞ (n +1)! 3n

= lim 3 = 0 < 1
n→∞ n +1

Thus, by the Ratio Test, the series converges.

(b)=lim an+1 lim (−1)n+1(n=+1)3 5n+1 lim (n +1)3 ⋅ 5n
n→∞ an (−1)n n3 5n 5n+1 n3
n→∞ n→∞

= lim 1 ⋅  n + 1 3= 1 <1
5  n 5
n→∞

Thus, by the Ratio Test, the series converges.

(c)=lim an+1 lim 3n+1=/ (2n+1 −1) lim 3n+1 ⋅ 2n − 1
n→∞ an n→∞ 3n / (2n −1) 2n+1 − 3n
n→∞ 1

= lim 3(2n −1)= lim 3(1−1 / 2n )= 3 >1
2n+1 −1 n→∞ 2 −1/ 2n 2
n→∞

Thus, by the Ratio Test, the series diverges.

Infinite Sequences and Series 287

Example 2 □ Determine whether the series is conditionally convergent or absolute convergent.

∑∞ n ∑(b) ∞ (−1)n en
n=1 n!
(a) (−1)n
n=1 n + 3

Solution □ (a) =lim an+1 (−1)n+1 n +=1 (n + 4) lim n +1 ⋅ n + 3
n→∞ an lim n→∞ n + 4 n
n→∞ (−1)n n (n + 3)

= lim n +1 ⋅ n=+ 3 1
n→∞ n n + 4

The Ratio Test is in conclusive. Try a different test.

In this case, you can apply the Alternating Series Test.

1. =nl→im∞ an l=im n 0
n→∞ n + 3

2. an=+1 n +1 < =n an for n ≥ 3
(n +1) + 3 n + 3

So, the series is conditionally convergent.

(b)=lim an+1 lim en+=1 / (n +1)! lim en+1 ⋅ n!
n→∞ an en / n! (n +1)! en
n→∞ n→∞

= l=im e 0
n→∞ n +1

So, the series is absolutely convergent.

288 Chapter 9

Summary of Tests for Series

Test Series Conditions of Convergence or Divergence
n th-Term
Geometric Series ∞ The series is divergent if lim an ≠ 0. Test is inconclusive if lim an =0.
Telescoping Series
p-Series ∑ an x→∞ x→∞
Alternating Series
n=1 The series is convergent if r <1, divergent if r ≥ 1. S =a
Integral ∞ 1− r

∑ arn−1 The series is convergent if lim an= L. S= a1 − L

n=1 x→∞
∞
The series is convergent if p > 1, divergent if p ≤ 1.
∑ (an − an+1)
The series is convergent if lim a=n 0 and 0 < an+1 ≤ an .
n=1
x→∞
∑∞ 1
If f is positive, continuous, and decreasing for x ≥ 1 and an =f (n),
n=1 n p
∫∞
∞
then the series converges if f (x) dx converges, diverges
∑ ( −1)n−1an
1
n=1 ∞

∞ if ∫1 f (x) dx diverges.

∑ an

n=1

∞ The series is convergent if lim an+1 < 1, divergent if lim an+1 > 1,
n→∞ an n→∞ an
Ratio ∑ an
inconclusive if lim an+1 = 1.
n=1 n→∞ an

∞∞
∑ ∑∞ Let 0 < an ≤ bn for all n. If bn converges, then an converges.
∑ an =n 1 =n 1
Direct Comparison ∞∞

∑ ∑n=1 If an diverges, then bn diverges.
=n 1 =n 1

and lim ( an ) ∞
n→∞ bn
∑Suppose > an
∞ that an 0, bn > 0, =L. Then converges
n=1
Limit Comparison ∑ an ∞ ∞∞

∑ ∑ ∑n=1 if bn converges and an diverges if bn diverges.
=n 1 =n 1 =n 1

Infinite Sequences and Series 289

Exercises - The Ratio Tests
Multiple Choice Questions

1. Which of the following series converge?

∑∞ n! ∑∞ n ∑III. ∞ n  2 n
2n n=1  3
I. II. 
n=1 3n 
n=1

(A) I only (B) II only (C) II and III only (D) I, II, and III

2. Which of the following series converge?

∑∞ n! ∑∞ (n!)2 ∑∞ n9
nn
I. II. III.
n=1 (2n)! n=1 9n
n=1

(A) I only (B) II only (C) I and II only (D) I, II, and III

Free Response Questions

3. Determine whether the following series converge or diverge.

∑∞ n!

(a)
n=1 n 2n

∑∞ cosn x

(b)
n=0 2n

∑∞ 3k k !

(c)
k =1 (k + 3)!

290 Chapter 9

9.6 Convergence of Power Series

Power Series
A power series about x = 0 is an infinite series of the form

∞

∑ an xn = a0 + a1x + a2 x2 + a3 x3 + ⋅ ⋅ ⋅ + an xn + ⋅ ⋅ ⋅

n=0

where x is a variable and the an ’s are constants.
More generally, a series of the form

∞

∑ an ( x − c)n = a0 + a1( x − c) + a2 ( x − c)2 + ⋅ ⋅ ⋅ + an ( x − c)n + ⋅ ⋅ ⋅

n=0

is called a power series centered at c, or a power series about c.

Convergence of a Power Series
For a power series centered at c there are only three possibilities:
1. The series converges only at x = c .
2. The series converges for all x .
3. There exists a real number R > 0 such that the series converges for x − c < R , and

diverges for x − c > R .

The number R is called the radius of convergence of the power series.
In most cases, R can be found by using the Ratio Test. The Ratio Test
always fails when x is an endpoint of the interval of convergence, so
each endpoint must be tested separately for convergence or divergence.
If the series converges only at c, the radius of convergence is R = 0 .
If the series converges for all x, the radius of convergence is R = ∞ .
The set of all values of x for which the power series converges is the
interval of convergence of the power series.

Example 1 □ Find the radius of convergence and interval of convergence of

the series ∑∞ (−2)n xn .
n=0 n + 3

Solution □ Let an =(−2)n xn n + 3 .

lim =an+1 lim (−2)n+1 xn+1 ⋅ (−2n=)+n x3n lim (−2)x =n + 3 2x
n→∞ an n+4 n→∞ n + 4
n→∞

By the ratio test, the given series converges if 2 x < 1 or x < 1 .
2

Thus the radius of convergence is R = 1 .
2