   Chapter 3.8, Problem 37E

Chapter
Section
Textbook Problem

# Use Newton’s method to find the coordinates, correct to six decimal places, of the point on the parabola y = ( x − 1 ) 2 that is closest to the origin.

To determine

To find:

Point on the given parabola which is closest to the origin

Explanation

1) Concept:

Working rule of Newton’s method is,

Start with initial approximation x1  which is obtained from graph. Evaluate fx1and f'(x1) and apply the Newton’s formula. After the first iteration we get x2. Repeat the process with x1  replaced by x2. Calculate the approximations x1, x2, x3,  until the successive approximations xn, xn+1  agree the eight decimal places.

2) Formula:

i) Newton’s formula for nth approximation

xn+1=xn-fxnf'xn for n=1,2,3,

ii) Power rule of differentiation:

ddxxn=nxn-1

3) Given:

y=(x-1)2

4) Calculations:

The problem states that, find the point (x, y) on the graph of parabola y=(x-1)2, that minimizes the distance, d between the curve and the point (0,0)

To find the distance between the point on curve and origin use distance formula,

d=(x2-x1)2+(y2-y1)2

d=(x2-0)2+(y2-0)2

d=(x2)2+(y2)2

For any point (x,y) on the parabola, given y coordinate as y=(x-1)2. Substituting y in distance formula,

d=(x)2+((x-1)2)2

d=(x)2+(x-1)4

Now, minimize the distance is equivalent to minimizing the square of distance, therefore

d2=(x)2+(x-1)4

dx=(x)2+(x-1)4

Using a graphing calculator, sketch the graph of d(x) to find the point at which minimum value of d occurs,

From the graph, the minimum value of d occurs approximately when x=0.5

So considered initial value as x1=0.5

Newton’s formula for nth approximation is

xn+1=xn-fxnf'xn

But, to minimize the function we need to find zeros of dx

So use d' and d'' in newton’s formula means,

xn+1=xn-d'(xn )d''xn

Now differentiating d(x) with respect to x,

d'x=2x+4(x-1)3

Again differentiating with respect to x,

d''x=2+12(x-1)2

For n=1,

To find value of x2,

x2=x1-d'(x1 )d''x1

x2=x1-d'x1 d''x1=x1-2x1+4(x1-1)32+12(x1-1)2

x2=0

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