How can I 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?

Answer 1

First, use the distance formula which is d = sqrt{ (x-x_o)² + (y-y_o)² }

But in this case x = x and y = (x-1)² and (x_o,y_o) = (0,0) the origin

Substituting, we have d = sqrt{ (x-0)² + ( (x-1)² - 0 )² } or d = sqrt{ x2 + x4 -4x3 +6x2 -4x + 1 }

Simplifying, we have d = sqrt(x4 -4x3 + 7x2 -4x +1)

This is the function to be minimized but you can minimize the square of d just same to get the same answer, so D = d² = x⁴ -4x³ +7x² -4x +1

In order to minimize D, we take the derivative

dD/dx = 4x³ - 12x² + 14x -4 = 0 to obtain the minimum x

To use Newton's formula, let dD/dx be the f(x) in Newton's formula

Then, f'(x) = 12x² -24x + 14

Newton's formula then for x is x = x_o - f(x_o)/f'(x_o)

where x_o is the first & then the successive guess using the answer x repeatedly.

When you do this you find that

x = .410245 and y = (x-1)² = .347811