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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 30E
To determine

To calculate: The roots of the curve,

  y=sinx .

Expert Solution

Answer to Problem 30E

The roots are

  x14.5

  x24.493613

  x34.493409 .

Explanation of Solution

Given information:

The infinitely many lines that are tangent to the curve y=sinx passes through the origin.

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.

Equation of tangent line : yy1=m(xx1)

Slope of the tangent line : derivative of the function.

Calculation:

Consider the curve ,

  y=sinx

Now,

  f(x)=sinxf'(x)=cosx

The line passes through the origin

  =yy1=m(xx1)=y0=m(x0)=y=mx (i)

While the tangent will touch the curve at some point. Let the x -coordinate be a.

Therefore , putting x-coordinate in equation y=sinx we get,

  y=sina .

The point at which tangent touches the curve is (a,sina) .

Put it in equation (i)

  =sina=ma

Slope of the tangent :-

  m=cosx .

At point (a,sina) the slope will be

  m=cosa

We get ,

  =sina=(cosa)a=sinacosa=a=sinacosa=a=tana=a=tanaa=0

Sketching the graph of function y=sinx and y=tanxx . We observe the graph is symmetric.

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.7, Problem 30E

  Figure 1.

Hence, we get the slope of the curve is y=tanxx . In other words g(x)=tanxx used to find out roots using Newton’s Method.

Therefore,

  g(x)=tanxx and

  g'(x)=sec2(x)1

Now, let initial approximation be x1=4.5

For n=1

  x2=x1g(x1)g'(x1)

  x2=x1tanx1x1sec2(x1)1

  x2=(4.5)tan(4.5)4.5[sec(4.5)]21x2=(4.5)4.6373320554.522.50484861

  x2=4.50.13733205521.5048486x2=4.50.006386097273x2=4.493613903

The second approximation is x2=4.493613 .

For n=2

  x3=x2g(x2)g'(x2)

  x3=x2tanx2x2sec2(x2)1

  x3=(4.493613)tan(4.493613)4.493613[sec(4.493613)]21x3=(4.493613)4.4977266124.49361321.229544651

  x3=4.4936130.00411361220.22954465x3=4.4936130.0002033467422x3=4.493409653

The third approximation is x3=4.493409 .

Hence , the roots of the curve are :-

  x14.5

  x24.493613

  x34.493409

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