# Shortest distance?

1. Nov 27, 2007

### rhey

shortest distance??

1. The problem statement, all variables and given/known data

find the shortest distance of y=x^2 from (4,0)

2. Relevant equations

3. The attempt at a solution

if y=x^2, then
y'=2x

where x=4
the slope is 8

the solution is

y=8(x-4)
if that is the tangent line \, the normal line would be

y=(1/8)(x-4)

after that.. i'm stuck!!

2. Nov 27, 2007

### EnumaElish

One way is to write out the Euclidian distance from vector [x, x^2] to vector [4, 0], then minimize w/r/t x.

3. Nov 27, 2007

### HallsofIvy

Staff Emeritus
EnumaElish, what rhey is trying to do is, I believe, more fundamental and a "nicer" method.

The shortest distance from (4,0) to y= x2 will be along the line that is perpendicular to the graph. HOWEVER, rhey, x= 4 is for the point (4, 0), not any point on the graph! Let $(x_0, x_0^2)$ be the point on y= x2 closest to (4, 0). The derivative there, and the slope of the tangent line, is 2x0 so the slope of the normal line is $-1/(2x_0)$. The equation of the line through (4, 0), perpendicular to y= x2 at $(x_0,x_0^2)$ is $y= -1/(2x_0)(x- 4)$. If that is to pass through $(x_0, x_0^2)$, you must have y= $x_0^2= -1/(2x_0)(x_0- 4)$. Solve that equation for x0 and you are almost done!

Hmm, that leads to a rather difficult equation for x0! Darn it, EnumaElish may be right!

Last edited: Nov 27, 2007
4. Nov 27, 2007

### rhey

i really cant understand what are u trying to tell me.. honestly, i'm not that good in math! but i'm trying..

5. Nov 27, 2007

### colby2152

1) Setup an equation for distance between point (0,4) and the function, $$f(x) = x^2$$

$$d = \sqrt{(x_2 - x_1)^2 - (y_2 - y_1)^2}$$
$$d = \sqrt{(x_2 - 4)^2 - (y_2)^2}$$
$$d = \sqrt{(x_2 - 4)^2 - (x_2)^4}$$

2) Take the derivative of d and set it equal to zero

$$d' = 0 = \frac{4(x_2)^3 - 2(x_2 - 4)}{2\sqrt{(x_2 - 4)^2 - (x_2)^4}}$$
$$0 = 4(x_2)^3 - 2(x_2 - 4)$$
$$4(x_2)^3 = 2(x_2 - 4)$$
$${x_2}^3 = x_2 - 4$$

As others have said, the problem is still difficult to solve from here...

Last edited: Nov 27, 2007
6. Nov 28, 2007

### HallsofIvy

Staff Emeritus
And, as EnumaElish said, it's far easier to minimize the square of the distance, not the distance itself! Oh, and it should be y- 4, not x- 4.