86 Higher Engineering Mathematics
f (1) = 5
3
− e
1.92
+ 5 cos
1
3
− 9 ≈ 114
f (2) = 6
3
− e
3.84
+ 5 cos
2
3
− 9 ≈ 164
f (3) = 7
3
− e
5.76
+ 5 cos 1 − 9 ≈ 19
f (4) = 8
3
− e
7.68
+ 5 cos
4
3
− 9 ≈ −1660
From these results, let a first approximation to the root
be r 1 = 3.
Newton’s formula states that a better approximation to
the root,
r 2 = r 1 −
f (r 1 )
f (r 1 )
f (r 1 ) = f (3) = 7
3
− e
5.76
+ 5 cos 1 − 9
= 19.35
f
(x) = 3(x + 4)
2
− 1.92e
1.92x
−
5
3
sin
x
3
f
(r 1 ) = f
(3) = 3(7)
2
− 1.92e
5.76
−
5
3
sin 1
= −463.7
Thus, r 2 = 3 −
19.35
−463.7
= 3 + 0.042
= 3.042 = 3.04,
correct to 3 significant figures.
Similarly, r 3 = 3.042 −
f (3.042)
f (3.042)
= 3.042 −
(−1.146)
(−513.1)
= 3.042 − 0.0022 = 3.0398 = 3.04,
correct to 3 significant figures.
Since r 2 and r 3 are the same when expressed to the
required degree of accuracy, then the required root is
3.04, correct to 3 significant figures.
Now try the following exercise
Exercise 37 Further problems on Newton’s
method
In Problems 1 to 7, use Newton’s method to solve
the equations given to the accuracy stated.
1. x 2 − 2x − 13 =0, correct to 3 decimal
places.
[−2.742, 4.742]
2. 3x 3 − 10x = 14, correct to 4 significant
figures.
[2.313]
3. x 4 − 3x 3 + 7x = 12, correct to 3 decimal
places.
[−1.721, 2.648]
4. 3x 4 − 4x 3 + 7x − 12 =0, correct to 3 decimal places.
[−1.386, 1.491]
5. 3 ln x + 4x = 5, correct to 3 decimal places.
[1.147]
6. x 3 = 5 cos 2x, correct to 3 significant figures.
[−1.693, −0.846, 0.744]
7. 300e −2θ +
θ
2
= 6, correct to 3 significant
figures.
[2.05]
8. Solve the equations in Problems 1 to 5,
Exercise 35, page 81 and Problems 1 to
4, Exercise 36, page 84 using Newton’s
method.
9. A Fourier analysis of the instantaneous value
of a waveform can be represented by:
y =
t +
π
4
+ sin t +
1
8
sin 3t
Use Newton’s method to determine the value
of t near to 0.04, correct to 4 decimal places,
when the amplitude, y, is 0.880.
[0.0399]
10. A damped oscillation of a system is given by
the equation:
y =−7.4e
0.5t sin 3t.
Determine the value of t near to 4.2, correct
to 3 significant figures, when the magnitude
y of the oscillation is zero.
[4.19]
11. The critical speeds of oscillation, λ, of a
loaded beam are given by the equation:
λ
3
− 3.250λ
2
+ λ − 0.063 = 0
Determine the value of λ which is approximately equal to 3.0 by Newton’s method,
correct to 4 decimal places.
[2.9143]
f (1) = 5
3
− e
1.92
+ 5 cos
1
3
− 9 ≈ 114
f (2) = 6
3
− e
3.84
+ 5 cos
2
3
− 9 ≈ 164
f (3) = 7
3
− e
5.76
+ 5 cos 1 − 9 ≈ 19
f (4) = 8
3
− e
7.68
+ 5 cos
4
3
− 9 ≈ −1660
From these results, let a first approximation to the root
be r 1 = 3.
Newton’s formula states that a better approximation to
the root,
r 2 = r 1 −
f (r 1 )
f (r 1 )
f (r 1 ) = f (3) = 7
3
− e
5.76
+ 5 cos 1 − 9
= 19.35
f
(x) = 3(x + 4)
2
− 1.92e
1.92x
−
5
3
sin
x
3
f
(r 1 ) = f
(3) = 3(7)
2
− 1.92e
5.76
−
5
3
sin 1
= −463.7
Thus, r 2 = 3 −
19.35
−463.7
= 3 + 0.042
= 3.042 = 3.04,
correct to 3 significant figures.
Similarly, r 3 = 3.042 −
f (3.042)
f (3.042)
= 3.042 −
(−1.146)
(−513.1)
= 3.042 − 0.0022 = 3.0398 = 3.04,
correct to 3 significant figures.
Since r 2 and r 3 are the same when expressed to the
required degree of accuracy, then the required root is
3.04, correct to 3 significant figures.
Now try the following exercise
Exercise 37 Further problems on Newton’s
method
In Problems 1 to 7, use Newton’s method to solve
the equations given to the accuracy stated.
1. x 2 − 2x − 13 =0, correct to 3 decimal
places.
[−2.742, 4.742]
2. 3x 3 − 10x = 14, correct to 4 significant
figures.
[2.313]
3. x 4 − 3x 3 + 7x = 12, correct to 3 decimal
places.
[−1.721, 2.648]
4. 3x 4 − 4x 3 + 7x − 12 =0, correct to 3 decimal places.
[−1.386, 1.491]
5. 3 ln x + 4x = 5, correct to 3 decimal places.
[1.147]
6. x 3 = 5 cos 2x, correct to 3 significant figures.
[−1.693, −0.846, 0.744]
7. 300e −2θ +
θ
2
= 6, correct to 3 significant
figures.
[2.05]
8. Solve the equations in Problems 1 to 5,
Exercise 35, page 81 and Problems 1 to
4, Exercise 36, page 84 using Newton’s
method.
9. A Fourier analysis of the instantaneous value
of a waveform can be represented by:
y =
t +
π
4
+ sin t +
1
8
sin 3t
Use Newton’s method to determine the value
of t near to 0.04, correct to 4 decimal places,
when the amplitude, y, is 0.880.
[0.0399]
10. A damped oscillation of a system is given by
the equation:
y =−7.4e
0.5t sin 3t.
Determine the value of t near to 4.2, correct
to 3 significant figures, when the magnitude
y of the oscillation is zero.
[4.19]
11. The critical speeds of oscillation, λ, of a
loaded beam are given by the equation:
λ
3
− 3.250λ
2
+ λ − 0.063 = 0
Determine the value of λ which is approximately equal to 3.0 by Newton’s method,
correct to 4 decimal places.
[2.9143]