# The liberation point L 1 . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4.7, Problem 34E

(a).

To determine

## To calculate: The liberation point L1 .

Expert Solution

The location points are

x10.99x20.96682 .

### Explanation of Solution

Given information:

The x -coordinate of L1 is the unique root of the fifth degree equation

p(x)=x5(2+r)x4+(1+2r)x3(1r)x2

+2(1r)x+r1=0

where r3.04042×106

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

xn+1=xnf(xn)f'(xn) and so on.

Formula: s=rθ where s is arc length, r is the radius of the circle and θ is the central angle.

Law of cosine c2=a2+b22abcosθwhere a , b , c are sides of triangle .

Calculation:

Consider the equation ,

p(x)=x5(2+r)x4+(1+2r)x3(1r)x2

+2(1r)x+r1=0

p(x)=x5(2+(3.04042×106))x4+(1+2(3.04042×106))x3(13.04042×106)x2

+2(13.04042×106)x+3.04042×1061=0

p(x)=x52.00000304x4+1.000006081x30.999996959x

+1.999993919x0.999996959=0

p'(x)=5x48.00001216x3+3.000018243x21.99999391x

+1.999993919

Now, let initial approximation be x1=0.99

For n=1

x2=x1p(x1)p'(x1)

The second approximation is x2=0.96682 .

Following with the iterations after 14 iterations will get the approximate value of x .

Hence , the location for liberation points are :-

x10.99x20.96682

(b).

To determine

### To calculate: The liberation point L2 .

Expert Solution

The location points are

x12x21x30.5

### Explanation of Solution

Given information:

The x -coordinate of L2 is the unique root of the equation

p(x)2rx2=0

where r3.04042×106

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

xn+1=xnf(xn)f'(xn) and so on.

Formula: s=rθ where s is arc length, r is the radius of the circle and θ is the central angle.

Law of cosine c2=a2+b22abcosθwhere a , b , c are sides of triangle .

Calculation:

Consider the equation ,

p(x)2rx2=0p(x)=2rx2

p(x)=(2×3.04042×106)x2

p(x)=0.00000608084x2

p'(x)=0.00001216168x

Now, let initial approximation be x1=2

For n=1

x2=x1p(x1)p'(x1)

x2=x10.00000608084x120.00001216168x1

x2=20.00000608084×(2)20.00001216168×(2)x2=20.000024323360.00002432336x2=21x2=1

The second approximation is x2=1 .

x3=x2p(x2)p'(x2)

x3=x20.00000608084x220.00001216168x2

x3=10.00000608084×(1)20.00001216168×(1)x3=10.000006080840.00001216168x3=10.5x3=0.5

The third approximation is x3=0.5 .

Hence , the locations of the libration points are :

x12x21x30.5

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