Calculus Help!!!

Question

Find the point on the parabola x^2 =2y that is closest to the point (4,1).

Answer 1

closest point is the intersection point of a line perpendicular to the slope of the parabola

use the distance formula, then take the derivative

(x, x^2/2) to (1,0)

d^2 = (x^2/2)^2 + (x-1)^2 = x^4/4+ x^2 -2x+1=0

3x^3/4 +2x -2 =0

3x^3 +8x -8 =0

graph and find the x intercept

x = about .8, y=about .64/2 = .32

the point (.8, .32)

Answer 2

You have to minimize the distance or equivalently the distance squared between a point (x,y) on the parabola and the point (4,1):

Distance squared: f = (x-4)² + (y-1)²

To make f a function of a single variable, solve the equation of the parabola for y: y = x²/2

Substitute that in f and then it will be f(x), a function of a single variable x.

Then, find the minimum of f(x) using calculus. That will give you the x value at which the minimum occurs. Find the corresponding y value by using y = x²/2