Find Shortest Distance of y=x^2 from (4,0)

[rhey]

shortest distance??

Homework Statement

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

Homework Equations

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!

 

[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.

 

[HallsofIvy]

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!

 

[rhey]

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

 

[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...

 

[HallsofIvy]

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.