   Chapter 4.8, Problem 37E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use Newton’s method to find the coordinates of the inflection point of the curve y = x2 sin x, 0 ≤ x ≤ π, correct to six decimal places.

To determine

To Find: The coordinate of the inflection point of the given curve correct to six decimal places by Newton’s method.

Explanation

Given:

f(x)=x2sinx,0xπ.

Choose initial approximation x1=1.5.

Result used: (Newton’s method).

xn+1=xnf(xn)f(xn) Where f(x) is function defined above.

Calculation:

The given equation is f(x)=x2sinx,0xπ.

Calculate the derivative of the above function f(x).

Let the derivative of f(x) is f(x)=df(x)dx.

f(x)=d(x2sin(x))dx=sin(x)d(x2)dx+x2d(sin(x))dx=2xsin(x)+x2cos(x)

Calculate the second derivative of the above function f(x).

Let f(x)=df(x)dx Where f(x) is mentioned above.

f(x)=d(2xsin(x)+x2cos(x))dx=2xd(sin(x))dx+sin(x)d(2x)dx+x2d(cos(x))dx+cos(x)d(x2)dx=2xcosx+2sin(x)x2sinx+2xcos(x)=x2sinx+4xcosx+2sin(x)

To find the coordinate of the inflection point at first obtain the second derivative of f(x).

For that take f(x)=0 and solve for critical point.

Calculate the third derivative of the f(x).

Let f(x)=df(x)dx, Where f(x) is function mentioned above.

f(x)=d(x2sin(x)+4xcos(x)+2sin(x))dx=x2d(sin(x))dxsin(x)d(x2)dx+4cos(x)dxdx+4xd(cos(x))dx+2d(sin(x))dx=x2cos(x)2xsin(x)+4cos(x)4xsin(x)+2cos(x)=x2cos(x)6xsin(x)+6cos(x)

The value of f(x) at the point x1=1.5 is,

f(1.5)=(1.5)2sin(1.5)+4(1.5)cos(1.5)+2sin(1.5)=2.25×0.997494+6×0.0707372+2×0.997494=2.244363+0.424423+1.994989=0.175049

Calculate the value of f(x1) at the point x1=1.5.

f(1.5)=(1.5)2cos(1.5)6(1.5)sin(1.5)+6cos(1.5)=2.25×0.0707379×0.997494+6×0.070737=0.1591588.977446+0.424422=8.712182

Calculate the value of x2 by using above result and set n=1 to obtain x2=x1f(x1)f(x1)

Substitute,x1=1

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