Chapter 11
Power series
11.1 Approximating Functions with Polynomials
Power Series
∞
A power series is a series of the form
∑c x
n =0
∞
many terms.
∑c x
n =0
n
n
n
n
which represent a polynomial with infinitely
=c0 + c1x + c 2 x 2 + c3 x 3 + 
Depending on the values of x, a power series may converge or diverge.
∞
The series
∑ c (x − a)
n =0
n
n
is called a power series centered at x=a.
Linear and Quadratic Approximation
If f is a differentiable function at x=a then the first-degree
approximation of f at a is:
p1 (x)= f (a) + (x − a)f ′(a)
y = p1 (x) is the equation of the tangent line to f at x = a
hence, p1 (x) and f (x) share same value and same derivative at
x =a.
The quadratic approximation of f at a is
p 2 (x)= f (a) + f ′(a)(x − a) +
f ′′(a)
(x − a) 2
2
y = p 2 (x) is the equation of the parabola tangent to f at a. p 2 (x)
and f share the same first and second derivative at x = a .
https://www.youtube.com/watch?v=3d6DsjIBzJ4
120
 Ex (1) Find the linear approximation and quadratic approximation of f (x) = e − x at
a=0.
(i)
(ii)
Use both approximations to evaluate e
−
1
2
1
1
1
1
Find the absolute error in approximation f   − p1   and f   − p 2   to
2
2
2
2
three decimal places.
=
 Ex (2) Find the linear approximation and quadratic approximation of f (x)
a=0.
(i)
(ii)
1 + x at
Use both approximations to evaluate 1.08
Using scientific notation to find the absolute error in approximation
f ( 0.08 ) − p1 ( 0.08 ) and f ( 0.08 ) − p 2 ( 0.08 ) to two decimal places.
121
Taylor Polynomials
Suppose f is defined on [a − R,a + R] and f (n ) (x) exists on (a − R,a + R) and
Tn (x)
=
n
∑ c (x − a) is an n-th degree polynomial. If the n-th derivative of f and T (x)
i
i =0
i
n
Tn (x)
(n=0,1,2,…,n) share the same value at a , then =
n
∑ c (x − a) is called the n-th Taylor
i
i =0
i
polynomial of f.
n
Suppose Tn (x) = ∑ ci (x − a)i = c0 + c1 (x − a)1 + c 2 (x − a) 2 +  + c n (x − a) n . If the k-th
i =0
derivative (k ≤ n) of T and f share the same value at a, then:
=
Tn (a) f (a)
⇒
c0 = f (a)
n
Tn′ (x) = ∑ cii(x − a)i−1 = c1 + 2c 2 (x − a) +  + nc n (x − a ) n −1
Tn′ (a) = f ′(a ) ⇒ c1 = f ′(a)
i =1
n
Tn(2) (x) =∑ cii(i − 1)(x − a)i−2 =2c 2 +  + n(n − 1)c n (x − a) n −2 ; Tn(2) (a)= f (2) (a ) ⇒ c 2 =
i=2


T=
(x)
(k )
n



n

∑ c i(i − 1)(i − k + 1)(x − a)=
i=k
i−k
i

n!c n

;
f (2) (a)
2

T =
(a) f
(n )
n
(n )
f (n ) (a)
(a) ⇒ c n =
n!
Therefore, the n-th degree Taylor polynomial of f is:
f (k) (a)
=
Tn (x) ∑
(x − a) k
k!
k =0
n
It is a good idea to show the n-th degree Taylor polynomial with Tn,a (x) , however for
simplicity we assume that we know the center and we just write Tn (x) .
=
T(x)
In general, if f has derivatives of all orders, the series
∞
f (k) (a)
(x − a) k is called
∑
k!
k =0
the Taylor series of f at x=a.
∞
f (k ) (0) k
x is called the Maclaurin series of f.
If a = 0 then ∑
k!
k =0
122
 Ex (3) Find first four nonzero terms of the Maclaurin series of the following functions.
(i)
f (x) = e x
(ii)
g(x) = sin x
 Ex (4) Find a first, second and third-degree Taylor polynomial centered at 9 to
approximate 10 then fill the following table using scientific notation to find the absolute
error in approximation with one decimal places.
n
Approximation Tn (10)
Absolute error
10 − Tn (10)
1
2
3
123
Reminder in Taylor’s Polynomials
If Tn (x) is the n-th degree Taylor polynomial of y = f (x) (centered at a), the remainder in
(x) f (x) − Tn (x) (Just like Tn,a (x) for simplicity the
approximating f with Tn is R n=
remainder is shown by R n (x) instead of R n,a (x) .
Taylor’s Theorem (Reminder Theorem)
If a function f is differentiable of order n+1 on an open interval containing a, then for each
n +1
f ( ) (c)
n +1
R n (x)
=
(x − a)
x in the interval, there exists a number c in (a,x) such that
( n + 1)!
(This is called the Lagrange form of the reminder)
Corollary
If f (n +1) (x) ≤ M on [a − R,a + R] , then R n ( x ) ≤
M
| x − a |n +1
( n + 1)!
 Ex (5) Consider the function f (x) = cos x .
Approximate f at x = 0.48 by a Maclaurin polynomial of degree 2 at x = 0 .
Then find an upper bound for the remainder of this approximation and using scientific
notation find the absolute error in approximation with two decimal points.
124
3
 Ex (6) Consider the function f (x) = x 5 .
Approximate f by a Taylor polynomial of degree 3 centered at x = 1.
Then find an upper bound for the remainder of this approximation for 0.9 ≤ x ≤ 1.1
125
 Ex (7) Find the maximum error in the following approximation on the given interval.
(i)
(ii)
x2
5 5
e ≈1+ x +
; [− , ]
2
7 7
2
x
ln(1 − x) ≈ − x −
; [−0.3,0.3]
2
x
126
11.2 Properties of Power Series
∞
The series
∑ c (x − a)
n
n =0
n
is called a power series centered at x=a.
∞
Theorem For a power series
∑ c (x − a)
n =0
n
n
, there are only three possibilities:
(I) Series is convergent only for x=a and divergent for all x’s that x ≠ a . (Radius of
convergence=0)
(II) Series is convergent for all real x’s. (Radius of convergence = ∞ )
(III) Series is convergent when | x − a |< R and diverges when | x − a |> R
To find the interval of convergence use Ratio or Root test to find L = lim
n →∞
c n +1 (x − a) n +1
c n (x − a) n
Now Series converges if L<1 and diverges when L>1.
The interval of convergence is the interval containing all values of x for which the series
converges. When R is a real number, the interval of convergence is not (−R,R) in all cases.
The interval of convergence may or may not include one or both of the endpoints of the
interval, therefore the interval of convergence could be one of the followings:
(a − R,a + R)
[a − R,a + R)
(a − R,a + R]
[a − R,a + R]
To find interval of convergence, a conclusive test must be applied.
 Ex (8) Find the radius of convergence of each series.
∞
(i)
∑x
n
n =0
∞
(ii)
∑ n!x
n
n =0
127
∞
(iii)
xn
∑
n =0 n!
 Ex (9) Find the radius of convergence and interval of convergence of the series.
2n (x − 2) n
∑
n
n =1
∞
(i)
128
(ii)
(x − 1) n
∑
n2
n =1
(iii)
n 4 (x + 3) n
∑
3n
n =0
∞
∞
129
(x − 5) n
∑
n
n =1 2 ⋅ n
∞
(iv)
∞
(v)
∑
n =1
n(x − 1)3n
8n
130
Representations of Functions as Power Series
Derivatives and Indefinite Integrals of Power Series Functions
=
f (x)
Suppose f is defined by power series
∞
∑ c (x − a)
n =0
n
n
with radius of convergence R.
The function f(x) is a differentiable and integrable on (a − R,a + R) and
∞
∞
d
d
n
f=
(x) ∑ c n (x −=
a) ∑ nc n (x − a) n −1
dx
dx
n 0=
n 1
=
∞
∫ f (x)dx= C + ∑ cn
n =0
(x − a) n +1
n +1
∞
Remainder: A geometric series
∑ ar
n
converges if |r|<1, therefore we have:
n =0
∞
1
=∑ x n =1 + x + x 2 + x 3 + 
1 − x n =0
| x |< 1
The above facts can be used to represent some specific functions as power series.
 Ex (10) Use the geometric series to express the following functions as a power series
and find the interval of convergence.
(i)
1
1 + x2
131
(ii)
6
6+x
(iii)
x2
3+ x
 Ex (11) Find the power series representation for g centered at 0 by differentiating or
integrating the power series for f. Give the interval of convergence for the resulting series.
g(x)
=
1
1
, f (x)
=
2
(1 − 5x)
1 − 5x
132
 Ex (12) Express the following functions as a power series.
a) tan −1 (x)
1+ x
)
 b) ln(
1− x
133
 Ex (13) Find the function represented by the following series and find the interval of
convergence of the series.
∞
∑ 2e
− nx
n =0
11.3 Taylor Series
Suppose f is defined on [a − R,a + R] and f is infinitely many times differentiable on the
interval (a − R,a + R) , the Taylor series of f centered at a is:
∞
f (k) (a)
=
T(x) ∑
(x − a) k
k!
k =0
If a = 0 then
∞
f (n ) (0) n
x is called the Maclaurin series of f.
∑
n!
n =0
If T(x) is the Taylor series of f at x=a then usually T(x)=f(x) on an interval containing a,
but it is possible that T(x) does not converge to f(x) for x ≠ a .
If I is an interval containing a, ( a ∈ I ) , and T(x) converges to f (x) on I, y = f (x) is called a
real analytic function at a on I.
If lim R n (x) = 0 then f is equal to its Taylor polynomial. ( f (x) = T(x) )
n →∞
Just like power series we can determine the radius and interval of convergence of Taylor
series using differnet tests such as ratio or root test.
134
 Ex (14) Find the Maclaurin series of each function. Use the ratio test to find the interval
of convergence of each of the series.
(i)
f (x) = e x
(ii)
g(x) = cos x
135
 Ex (15) Find the first four nonzero terms of the Taylor series for the given function
centered at x=a. Write the power series using summation notation and determine the
interval of convergence.
(i)
f (x)
= ln(1 + 4x) centered at a = 0
136
(ii)
f (x) =
1
centered at a = 3
x −1
137
Remark: The Taylor series T(x) of a function f could have an interval of convergence
( −∞, ∞ ) but still T(x) ≠ f (x) .
If T(x) is the Taylor series of f at x=a then usually on an interval containing a, T(x) = f (x),
however it is possible that T(x) does not converge to f(x) for x ≠ a . If I is an interval
containing a and T(x) converges to f (x) on I, y = f (x) is called a real analytic function at
a on I.
 Ex (16) Show that the following function is not real analytic at x=0, that is T(x) ≠ f (x)

e −1/x
f (x) = 

0
2
x≠0
x=0
Roughly speaking f is too smooth around x=0 therefore no polynomial can estimate the
function in some interval containing 0.
138
Reminder in Taylor’s Polynomials
If Tn (x) is the n-th degree Taylor polynomial of y = f (x) (centered at a), the remainder in
approximating f with Tn at is R n=
(x) f (x) − Tn (x)
Taylor’s Theorem (Reminder Theorem)
If a function f is differentiable of order n+1 on an open interval containing a, and for each
M
| x − a |n +1
x in the interval [a − R,a + R] we have f (n +1) (x) ≤ M then R n ( x ) ≤
( n + 1)!
Theorem
If R n (x) is the n-th reminder of the Taylor series and lim R n (x) = 0 then f is equal to its
Taylor polynomial ( f (x) = T(x) ) .
n →∞
 Ex (17) Use the reminder theorem to show that the following functions are equal to its
Maclaurin series.
(i)
f (x) = e x
(ii)
f (x) = cos x
139
A Short List of the Maclaurin Series
∞
1
1 x + x 2 + x 3 + x 4 +  ; −1 < x < 1
=∑ x n =+
1 − x n =0
∞
1
1 x + x 2 − x 3 + x 4 −  ; −1 < x < 1
=∑ (−1) n x n =−
1 + x n =0
∞
xn
x x 2 x3 x 4
e =
1
=+
+ + + +  ; −∞ < x < ∞
∑
1! 2! 3! 4!
n =0 n!
x
x 2n +1
x3 x5 x7
sin x =∑ (−1)
=x − + − +  ; −∞ < x < ∞
(2n
1)!
3! 5! 7!
+
n =0
∞
n
∞
x 2n
x2 x4 x6
=−
+ − +  ; −∞ < x < ∞
cos x =
(−1)
1
∑
(2n)!
2! 4! 6!
n =0
n
x 2n +1
x3 x5 x7
=x − + − +  ; −1 ≤ x ≤ 1
tan x =∑ (−1)
2n + 1
3 5 7
n =0
−1
∞
n
(−1) n +1 x n x x 2 x 3 x 4
= − + − +  ; −1 < x ≤ 1
n
1 2 3 4
n =1
∞
ln(1 + x) =∑
∞
x n x x 2 x3 x 4
− ln(1 − x) =∑ = + + + +  ; −1 ≤ x < 1
1 2 3 4
n =1 n
x 2n +1
x3 x5 x7
sinh x =∑
=x + + + +  ; −∞ < x < ∞
3! 5! 7!
n =0 (2n + 1)!
∞
∞
x 2n
x2 x4 x6
=+
+ + +  ; −∞ < x < ∞
cosh x =
1
∑
2! 4! 6!
n =0 (2n)!
140
 Ex (18) show that
x 2n +1
x3 x5 x7
tan x =∑ (−1)
=x − + − +  ; −1 ≤ x ≤ 1
2n
+
1
3 5 7
n =0
−1
∞
n
141
 Ex (19) Use the list of Maclaurin series to obtain a power series for the following
functions.
(i)
x 2e − x
(ii)
 x − sin x

 x3
f (x) = 
1

6
2
x≠0
x=0
 Ex (20) Evaluate the integral as a power series.
ex − 1
∫ x dx
142
Binomial Series
n
n!
Combination   =
 k  (n − k)!k!
(n choose k, or k combination of n elements)
n  n 
Find the simpler for of a combination   = 

k n − k
n
How to evaluate   ⇒
k
Beginning from n, move k steps down



 n  n × (n − 1) ×  (n − k + 1)
k =
k × (k − 1) ×  × 1
 
Some examples:
6  6  6 6×5
= 15
=
  6 −=
 =

4
4
2
2
×
1
  
  
1
2
 
3
n
 0 =1
 
n
1 = n
 
1  1  3
 ×− ×−  1
2  2  2
=
3 × 2 ×1
16
 1  3  5
 1  − ×− ×− 
− 2 =  2  2  2 = − 5


3 × 2 ×1
16
 3 
If k ∈  the Maclaurin series of f (x)= (1 + x) k is the binomial expansion of (1 + x) k ,
k
k n
which is a finite polynomial. (1 + x) k =
∑
nx .
n =0  
If k ∉   {0} then the Binomial series is:
k n
k(k − 1) 2 k(k − 1)(k − 2) 3
(1 + x) =
1 kx +
x +
x +
∑
 n  x =+
2!
3!
n =0  
∞
k
143
The radius of convergence of the binomial series is 1, however the endpoints of the interval
of convergence may or may not be ±1 .
The interval of convergence is [-1,1] if k > 0
The interval of convergence is (-1,1] if −1 < k < 0
The interval of convergence is (-1,1) if k ≤ −1
 Ex (21) Find the Binomial series of the following functions then write the first five
terms. (Evaluate a 4th degree polynomial)
(i)
f (x)
=
(ii)
g(x) =
1+ x
1
(3 + x)3
144
11.4 Working with Taylor Series
 Ex (22) Find each limit using Taylor series.
(i)
x − sin x
x →0
x3
lim
x3
tan x − sin x +
6
lim
5
x →0
x
−1
(ii)
145
 Ex (23) Use Taylor series to find the solution of the differential equation
′− y 1 ; =
y=
y(0) 1
146
 Ex (24) Evaluate each sum.
(i)
π3
π5
π7
π− 2
+
−
+
6 ⋅ 3! 64 ⋅ 5! 66 ⋅ 7!
(ii)
1+
2 3 4
+ + +
e e 2 e3
If I is the interval of convergence of the power series=
f (x)
∞
∑ c (x − x )
n =0
n
0
n
and ( a,b ) ⊆ I
then
b
(x − x 0 ) n +1
c n ∫ (x − x 0 ) dx= ∑ c n
∫a f (x)dx= ∑
a
n +1 a
=
n 0=
n 0
b
∞
b
∞
n
147
 Ex (25) Use power series to find
 Ex (26) Use power series to find
∫
0.54
∫
0.8
0
0
ln(1 + x 2 )dx correct to within 10−4 .
x
dx correct to within 10−6 .
x +1
6
148
 Ex (27) Identify the function represented by the following power series.
∞
(i)
x 3n
(−1) n
∑
4
n =0
(ii)
x n +1
∑
n
n =1 2 ⋅ n
(iii)
nx n +1
(−1)
∑
2n
n =1
n
∞
∞
n
149
tan a + tan b
1 − tan a tan b
 Ex (28) Sum formula for tangent: tan(a + b) =
π
4
1
2
1
3
(i)
arctan + arctan
Using the sum formula of tangent show=
that
(ii)
By using the reminder theorem in alternating series evaluate π with error less
than 10−4 .
Remark: To find more decimal digits of π one efficient way is to use the Machin’s
π
4
1
5
= 4arctan − arctan
formula.
1
.
239
This formula was first found by John Machin. (1680–1751)
150
Chapter 12
Parametric and Polar Curves
12.1 Parametric Equations
Suppose x and y are both defined in terms of a variable t, in this case t is called a parameter.
x = t + 1
. For a known domain of t, set of all the points (x,y) represent the
 y = 2t
An example is 
graph of the parametric curve.
x = t + 1
by making a table of values.
 y = 2t
 Ex (1) Graph the parametric curve 
 Ex (2) Use a graphing calculator to graph the parametric curve:
 x = sin(2t)
(0 ≤ t ≤ 7)

y
=
sin(3t)

151
Eliminate the Parameter in Parametric Equation
 Ex (3) Eliminate the parameter in the following equations and determine the type of the
graph.
(i)
 x = 3t

 y= 2 − t
(ii)
 x = 3cos t
( 0 ≤ t ≤ 2π )

y
3sin
t
=

 Ex (4) Find the parametric equation of the circle with center at (1,3) with radius R=2.
152
Remark: Parametric equations generate oriented graphs. Corresponding to generic
direction of the interval containing t, there is a direction on the parametric graph.
 Ex (5) Show the direction of the curves by graphing them:
(i)
 x = cos t
( 0 ≤ t ≤ 2π )

y
=
sin
t

(ii)
 x = sin t
( 0 ≤ t ≤ 2π )

y
=
cos
t

 Ex (6) Find the parametric equation of a line through (1,3) and (2,4) with the direction
from right to left.
153
 Ex (7) Find the parametric equation of the path of an object that moves on the
1 counterclockwise twice around, starting at the point
circumference of the circle x 2 + y 2 =
(0, -1)
x =
−2t 2 + 1
( −1 ≤ t ≤ 1)
 Ex (8) Consider the parametric equations. 
y
t
2
=
−

(i) Eliminate the parameter to obtain an equation in x and y.
(ii) Describe the curve and indicate the positive orientation.
 Ex (9) Using a graphing calculator graph the family of the following curves with a>0.
x a(t − sin t)
=
( t ≥ 0 ) This is called a cycloid and it models the path of the “valve

y a(1 − cos t)
=
stem” of a tire when the car moves along a straight line.
154
Derivative of Parametric Curves
 x = f (t)
such that both f and g are differentiable
 y = g(t)
dy dy dt g′(t)
dy
= =
; f ′(t) ≠ 0
on [a,b]. The derivative
is given by
dx dx dt f ′(t)
dx
Suppose C is a parametric curve given by 
The slope of the tangent line to Curve at t = t 0 is given by:
dy
g′(t 0 )
=
dx t = t 0 f ′(t 0 )
( f ′(t 0 ) ≠ 0 )
 x = 3sin t
; 0 ≤ t ≤ 2π at
=
y
2cos
t

 Ex (10) Find the equation of the tangent line to the ellipse 
3

 , 3.
2

155
 x = sin t
; 0 ≤ t ≤ 2π the tangent line is a
y
=
sin
2t

 Ex (11) At what point(s) of the curve 
horizontal tangent line.
 x = 1 + ln t
d2y
; 0.1 ≤ t ≤ 5 find 2 .
 Ex (12) For the curve C : 
y
t
ln
t
=
−
dx

What is the concavity of C at t=1?
156
Area Under the Curve
The area under the curve of y = "function of x" bounded by x=axis and two vertical lines
b
x=a and x=b is given by A = ∫ ydx . If the parametric equation of the curve bounded
a
 x = f (t)
b
≥
a
between x=a and x=b (
) is given by  y = g(t) ; α ≤ t ≤ β then the area under the

curve is:
=
A
β
g(t)d(f (t))
∫=
α
∫
β
α
g(t)(f ′(t))dt
α
Provided that a= f (α) and b= f (β) , otherwise A = ∫ g(t)(f ′(t))dt which guaranties that in
β
the process of integration the curve direction is from left to right.
 Ex (13) Find the area under each curve.
(i)
 x= t 2 + 1
; 0 ≤ t ≤1

3
 y =− t + 3t + 1
157
(ii)
2
 x =− t + 2
; 0 ≤ t ≤1

3
 y =− t + t + 2
(iii)
 x = 3t 2 − 3t − 1
; 0 ≤ t ≤1

y
t +1
=
158
2 + cos t , y =
1 + sin t 0 ≤ t ≤ 2π to
 Ex (14) Use the parametric equation of a circle x =
find the enclosed area. Find the equation of the tangent line at the point where t =
π
.
4
Arc length
 x = f (t)
; a ≤ t ≤ b where f ′ and g′ are continuous
y
=
g(t)

The length of the parametric curve C : 
on [a,b] and C is traversed only once when a ≤ t ≤ b is:
L
=
∫
b
a
(f ′(t)) 2 + (g′(t)) 2 dt
 Ex (15) Show that circumference of a circle with radius R is C= 2πR.
159

t2
x
−t
 =
2
; 0 ≤ t ≤1
 Ex (16) Find the length of the curve 
4
 y = t 3/2

3
12.2 Polar Coordinates
In polar coordinate a point is given by a pair ( r,θ ) where r is the distance from the origin
(Pole) and θ is the trigonometric angle made by the positive direction of the x=axis (Polar
Axis). Both θ and r can be negative. When r is negative then the point is still r unit away
from the origin but on the opposite side of the ray.
 Ex (17) Plot the polar points.
 π
 7π 
A=
1,
B
=
 
 2, 
 2
 6 
3π 

C=
4,
−


4

 −5π 
D=
 −2,

6 

160
Relation Between Polar and Rectangular Coordinate (Cartesian Coordinates)
x=
r cos θ
y=
r sin θ
,
r = ± x 2 + y2
,
θ = arctan
y
y
or θ = π + arctan
x
x
Choose the right sign by checking the quadrants.
 Ex (18) Following points are given in polar
coordinates. Find the rectangular coordinate of each one.
(i)
 3π 
 2, 
 4 
(ii)
π

−
2,


4

 Ex (19) Following points are given in rectangular coordinate. Find the polar coordinates
of each one.
(i)
(1,1)
(ii)
( −1, 3 )
161
 Ex (20) Convert the following equation to Cartesian coordinates.
r = − 4 cos θ
 Ex (21) Convert the following equation to polar coordinates.
3y − 2x + 1 =0
Graphs in Polar Coordinate
 Ex (22) Graph each curve in polar coordinate.
π
2
(i)
θ=
(ii)
r =1
162
(iii)
r= θ
(iv)
r =+
1 cos θ
r sin 2θ
(v) =
163
Symmetry in Polar Graphs
Symmetry with respect to the x-axis
In this case replacing ( r,θ ) with ( r,−θ ) or ( − r, π − θ ) produces the same equation.
Symmetry with respect to the y-axis
In this case replacing ( r,θ ) with ( −r, −θ ) or ( r,π − θ ) produces the same equation.
Symmetry with respect to the Pole
In this case replacing ( r,θ ) with ( −r, θ ) or ( r,π + θ ) produces the same equation.
r sin 3θ
 Ex (23) Identify all types of the symmetries of the curve:=
164
12.3 Calculus in Polar Coordinates
Derivative and Slope of the Tangent Lines
In polar coordinate by chain rule the derivative
dy
is:
dx
dr
sin θ + r cos θ
dy dy / dθ dθ
= =
dx dx / dθ dr cos θ − r sin θ
dθ
dr
cos θ − r sin θ ≠ 0
dθ
 Ex (24) Determine the slope of the tangent line to r =−
1 cos θ
π
6
at θ = . At what point(s) the curve has vertical or horizontal
tangent line?
165
Area Bounded by a Polar Curves
The area bounded between r= f (θ) and two rays
θ = α and θ = β=
is: A
1 β 2
r dθ
2 ∫α
 Ex (25) Find the area bounded by one leaf of the lose
=
r sin 2θ
166
Area bounded between two curves
The area bounded between two curves r= f (θ) and
r= g(θ) and two rays θ = α and θ = β is:
A
=
1 β 2
f (θ) − g 2 (θ) )dθ
(
∫
2 α
( g(θ) < f (θ) )
 Ex (26) Find the area that is
1 cos θ .
(a) Inside the circle r = 1 and outside the cardioid r =+
1 cos θ .
(b) Outside the circle r = 1 and inside the cardioid r =+
167
Points of Intersection of Polar Graphs
- First set the curves equal to each other and solve the
equation.
- Check the work on graphs to avoid missing other possible
points of intersection.
r
 Ex (27) Find the point(s) of intersection of=
3sin θ and
r =+
1 sin θ . Find the area that is enclosed by both curves.
θ
r = 3sin θ
r = 1 + sin θ
0
0
1
π
6
π
4
π
3
π
2
2π
3
3π
4
5π
6
π
3
2
3
2
5π
4
3π
2
7π
4
2π
3 2
2
3
3
3
3 2
2
3
2
0
−
−
2
2
3
1+
2
2
1+
3
2
2
1+
2
3
2
1
1+
3 2
2
−1
1−
3 2
2
0
1−
2
2
0
2
2
1
168
Arc length of Polar Curves
The arc length of the curve r= f (θ) (continuously differentiable curve) on [α, β] traversed
only once when α ≤ t ≤ β is: L =
∫
β
α
(r) 2 + (
dr 2
) dθ
dθ
 Ex (28) Find the arc length of the cardioid
r =+
1 cos θ on [0,2π]
 Ex (29) Find the arc length of the cardioid
r =+
1 sin θ on [0,2π]
169
 Ex (30) Find the arc length of the cardioid r=
1 2
θ on [0,2π]
4
170