Help with Calculus problem?

Question

Using Newton's method, find the closest point to the curve y=x^2 from the point (2,0).

Answer 1

The distance between y=x^2 and (2,0) is

Dsq = (x-2)^2 +(y-0)^2

Dsq = (x-2)^2+y^2

Dsq = (x-2)^2 +x^4

dDsq/dx = 2(x-2) +4x^3 =0

4x^3+2x-2 = 0

x=-1 is a root (using Newton's method or graphing the equation)

dDsq/dx^2 =12x^2+2

when x=-1, dDsq >0, indicating that d^2 is minimum.

D^2 is the square of the distance between y=x^2 and the point (2,0).

We want to minimize D. Minimizing D^2 is the same as minimizing D.

The minimum distance is

sqrt[Dsq] = sqrt[(x-2)^2 +x^4]

plug x=-1 into the equation.

sqrt[(-3)^2+1]=sqrt(10)

Answer 2

D = SQRT[(x - 2)^2 + y^2']

D = SQRT[(x - 2)^2 + x^4]

If D is a minimum then D^2 will also be a minimum and it is easier to work with. So use:

D = [(x - 2)^2 + x^4]

dD/dx = [2x - 4 + 4x^3] = f(x)

We want to solve this for f(x) = 0 which will give the minimum for the distance D. So:

df/dx = 2 + 12x^2

fnew = f(0) + (df/dx)(x - x0) = 0

x = x0 - f(0)/(df/dx)

The solution by inspection is between x = 0 and x = 1 so let us start with x0 = 1.

x0 = 1

f = 2

df/dx = 14

x = 0.857142857

x0 = 0.857142857

f = 0.233236152

df/dx = 10.81632653

x = 0.835579515

x0 = 0.835579515

f = 0.004742522

df/dx = 10.37831751

x = 0.83512255

x0 = 0.83512255

f = 2.09341E-06

df/dx = 10.36915609

x = 0.835122348

That is getting pretty close to 0 but you can continue for a couple more and get it very close.

The answer is:

x = 0.835122348

y = 0.697429336

Answer 3

Define the locus of the point (t, t^2).

Distance D = √[(t-2)^2 + t^4] ...(1)

D^2 = (t-2)^2 + t^4

We wish to solve dD/dt = 0

Differentiate implicitely

2D * dD/dt = 2*(t-2) + 4t^3 = 0

or f(t) = 2t^3 + t - 2 = 0 ... (2)

We can now use Newton's method to solve eqn (2)

f' (t) = 6t^2 + 1

t_new = t - f(t) / f'(t) = t - (2t^3 + t - 2) / (6t^2 + 1)

Let's start with an initial guess of t = 0. Successive iterations give the following results:

0

2

1.36

0.997042719

0.856423536

0.835550479

0.835122526

0.835122348

0.835122348

0.835122348

Sub t = 0.835122348 in (1) to get

D_min = 1.357699386

The closest point is (0.835122348, 0.835122348^2)

or (0.835122348, 0.697429337)

*EDIT*

It turns out that this is the same approach captain mephisto used. But since I already did the work, I'm not going to delete my answer. But since he got it first, pick his as the best one.