Chapter 3 Homework Determine The Angle For Which First Derive

3.1 The tolerance, , of the solution in the bisection method is given by , where and

are the endpoints of the interval after the nth iteration. The number of iterations n that are required for

obtaining a solution with a tolerance that is equal to or smaller than a specified tolerance can be determined

before the solution is calculated. Show that n is given by:

where and are the endpoints of the starting interval and

is a user-

specified tolerance

.

Solution

Starting interval

tol

tol 1

2

—–bnan

–()=

an

bn

nba–()log tol()log–

2log

—————–———————————–

≥

a

b

tol

i1=

ab,[] a1b1

,[]=

i2=

i3=

1

3.2 Determine the root of by:

(a) Using the bisection method. Start with and , and carry out the first three iterations.

(b) Using the secant method. Start with the two points, and , and carry out the first three iter-

ations.

(c) Using Newton’s method. Start at

and

carry out the first three iterations

.

Solution

Using 5 significant digits.

(a)

,,, , ,

fx() x2ex–

–=

a0=

b1=

x10=

x21=

x11=

fx() x2ex–

–=

i1=

a0=

b1=

f0() 02e0–

–2–==

f1() 12e1–

– 0.26424==

i2=

a0.5=

i3=

a0.75=

i4=

i2=

i3=

i4=

i1=

x11=

2

1

3.3 The location of the centroid of an arc of a circle is given by:

Determine the angle for which

.

First, derive the equation that must be solved and then determine the root using

the following methods:

(a) Use the bisection method. Start with and , and carry

out the first four iterations.

(b) Use the secant method. Start with the two points and

, and carry out the first four iterations.

Solution

Using 5 significant digits.

(a)

x

α

C

α

r

x

xrαsin

α

————-–

=

α

x3r

4

—-–

=

a0.5=

b1.5=

α10.5=

α21.5=

fα() αsin

α

———-– 3

4

—–

–=

2

(b)

αi1+ αi

fαi

()α

i1– αi

–()

fαi1–

()fαi

()–

—————-———————–

–=

1

3.4 The lateral surface area, S, of a cone is given by:

where r is the radius of the base and h is the height. Determine the radius of a

cone that has a surface area of 1800 m2 and a height of 25 m. Solve by using

the fixed-point iteration method with as the iteration func-

tion. Start with

m and calculate the first four iterations.

Solution

Using 5 significant digits.

,m

2, m

h

r

Sπrr

2h2

+=

rSπr2h2

+()⁄=

r17=

ri1+

S

πri

2h2

+

——————-—-–

=

S1800=

h25=

1

3.5 Determine the fourth root of 200 by finding the numerical solution of the equation .

Use Newton’s method. Start at

and

carry out the first five iterations

.

Solution

Using 5 significant digits.

,,

x4200–0=

x8=

fx() x4200–=

f′x() 4x3

=

xi1+ xi

fx

i

()

f′xi

()

————–

–xi

xi

4200–

4xi

3

——————––

–==

1

3.6 Determine the positive root of the polynomial .

(a) Plot the polynomial and choose a point near the root for the first estimate of the solution. Using New-

ton’s method,

determine the approximate solution in the first four iterations.

(b) From the plot in part (a), choose two points near the root to start the solution process with the secant

method.

Determine the approximate solution in the first four iterations.

Solution

(a)

Plot with MATLAB:

fplot(‘x^3+0.6*x^2+5.6*x-4.8′,[0 2])

xlabel(‘x’); ylabel(‘y’);

x30.6x25.6x4.8–++

15

20

2

1

3.7 The equation has a root between and . Use these values

for the initial two points and calculate the next four estimates for the solution using the secant method.

Solution

,

,, ,

1.2x32x220x–10–+0=

x4–=

x5–=

fx() 1.2x32x220x–10–+=

xi1+ xi

fx

i

()xi1– xi

–()

fx

i1–

()fx

i

()–

————————————–

–=

x15–=

x24–=

fx

1

() 1.2 5–()

325–()

220 5–()–10–+10–==

1

3.8 Find the root of the equation using Newton’s method. Start at

and

carry out the first

five iterations.

Solution

,,

xx

2

+7=

x7=

fx() xx

27–+=

f′x() 1

2x

———–2x+=

xi1+ xi

fx

i

()

f′xi

()

————–

–xi

xixi

27–+

1

2xi

———-– 2xi

+

—————–———–

–==

1

3.9 The equation has a root between and .

(a) Write four different iteration functions for solving the equation using the fixed-point iteration method.

(b) Determine which from part (a) could be used according to the condition in Eq. (3.30).

(c) Carry out the first five iterations using the determined in part (b), starting with .

Solution

(a)

First: ,

x3x–ex

–2–0=

x2=

x3=

gx()

x2=

xx

3ex

–2–=

g1x() x3ex

–2–=

2

1

3.10 The equation has a root between and . To find the root by

using the fixed-point iteration method, the equation has to be written in the form . Derive two pos-

sible forms for — one by solving for x from the first term of the equation, and the next by solving for

x from the second term of the equation.

(a) Determine which form should be used according to the condition in Eq. (3.30).

(b) Carry out the first five iterations using both forms of to confirm your determination in part (a).

Solution

(a)

Solving for x from the first term of the equation:

,, ,

fx() x25x13⁄

–1+0==

x2=

x2.5=

xgx()=

gx()

x5x13⁄1–()

12⁄

=

gx() 5x13⁄1–()

12⁄

=

g′x() 5

3x23⁄5x13⁄1–()

12⁄

———————————————––=

2

3.11 The equation has a root between and . Find the root by

using the fixed-point iteration method. Determine the appropriate form of according to Eq. (3.30).

Start the iterations with

and carry out

the first five iterations.

Solution

fx() 2x34x2

–4x–20–0==

x3=

x4=

gx()

x2.5=

x4x24x20++

2

—————–—————–





13⁄

=

gx() 2x22x10++()

13⁄

=

g′x() 4x2+

32x22x10++()

23⁄

—————–——————————–

=

3.12 Determine the positive root of the equation by using the fixed-point iteration

method. Carry out the first five iterations.

Solution

Plot with MATLAB:

fplot(‘cos(x)-0.8*x^2’,[0 2])

xlabel(‘x’); ylabel(‘f(x)’);

xcos 0.8x2

–0=

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

f(x)

x1.25 xcos()

12⁄

=

gx() 1.25 xcos()

12⁄

=

3.13 Solve the following system of nonlinear equations:

(a) Use Newton’s method. Start at

, , and

carry out the first five iterations

.

(b) Use the fixed-point iteration method. Use the iteration functions and

. Start at

, , and

carry out the first five iterations

.

Solution

(a) Using Newton’s method.

The Jacobian (Eq. (3.47)) is:

2x3

–3y242++ 0=

5x23y369–+0=

x1=

y1=

y5x2

–69+

3

——————-—––





13⁄

=

x3y242+

2

——————––





13⁄

=

x1=

y1=

f12x3

–3y242++ 0==

f25x23y369–+0==

,

The calculations of the first 5 iterations are calculated with the following MATLAB program (script file).

%Solution of Chapter 3 HW 13a

clear all

F1 = @ (x,y) -2*x^3+3*y^2+42;

F2 = @ (x,y) 5*x^2+3*y^3-69;

F1x = @ (x) -6*x^2;

When the program is executed, the following (solution) is displayed in the Command Window:

x = 1.0000 y = 1.0000

i = 1 x = 7.6053 y = 0.4386

i = 2 x = 5.0714 y = -15.5856

i = 3 x = 5.2462 y = -10.4217

i = 4 x = 4.4193 y = -6.9736

x21 6.6053+ 7.6053==

y21 0.5614– 0.4386==

The iteration function are:

Use the iteration functions and

,

,,

,

yi

5xi

2

–69+

3

—————-——––





13⁄

=

xi

3yi

242+

2

—————-—-–





13⁄

=

i1=

x11=

y11=

x2

31

2

⋅42+

2

—————-——––





13⁄

2.8231==

y2

5–1

2

⋅69+

3

—————–———-–





13⁄

2.7734==

i2=

i3=

i4=

i5=