BuyFind

Single Variable Calculus: Concepts...

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

Single Variable Calculus: Concepts...

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

Solutions

Chapter 4.7, Problem 32E
To determine

To calculate: The central angle θ in radians.

Expert Solution

Answer to Problem 32E

The angle’s are

  θ12.2

  θ22.2662

  θ32.2622

Explanation of Solution

Given information:

The length of the chord AB is 4 cm.

The length of the arc AB is 5cm.

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,

Length of the chord is 4cm.

Length of the arc is 5 cm.

Therefore, s=5

  =s=rθ=5=rθ=r=5θ (i)

Now , using law of cosine we have ,

  c=4a=rb=r .

   = (4) 2 = (r) 2 + (r) 2 2(r)(r)cosθ=16=r2+r22r2cosθ=16=2r22r2cosθ=16=2r2(1cosθ) (ii)

Putting (i) in (ii) we get,

  16=2[5θ]2(1cosθ)16=2(25θ2)(1cosθ)16θ2=50(1cosθ)16θ2=5050cosθ16θ250+50cosθ=0

Hence, we get

  g(θ)=16θ250+50cosθg'(θ)=32θ50sinθ

Now, let initial approximation be θ1=2.2

For n=1

  θ2=θ1g(θ1)g'(θ1)

  θ2=θ116θ2150+50cosθ132θ150sinθ1

  θ2=(2.2)16(2.2)250+50cos(2.2)32(2.2)50sin(2.2)θ2=(2.2)77.445029.4250558670.440.42482019

  θ2=2.2(1.98505586)29.97517981θ2=2.2+1.9850558629.97517981θ2=2.2+0.066223317θ2=2.266223318

The second approximation is θ2=2.2662 .

For n=2

  θ3=θ2g(θ2)g'(θ2)

  θ3=θ216θ2250+50cosθ232θ250sinθ2

  θ3=(2.2662)16(2.2662)250+50cos(2.2662)32(2.2662)50sin(2.2662)θ3=(2.2662)82.170599045032.034771572.518438.38975664

  θ3=2.2662(0.13582754)34.12864336θ3=2.26620.003979869301θ3=2.262220131

The third approximation is θ3=2.2622 .

Hence , the central angle in radians are :-

  θ12.2

  θ22.2662

  θ32.2622

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