Problem 4: Find the point (𝑥, 𝑦) on the line 𝑦 = √𝑥 that is closest to the point (3,0).

Question

(Hint: Minimizing the square of the distance is the same as minimizing the distance itself.)

Answer 1

y = sqrx is the top half of a rightward opening parabola with vertex at the origin.

y^2 = x is the full parabola

the point (3,0) is on the parabola's axis of symmetry, and a distance of sqr3 = 1.732 directly below the parabola and a distance 3 to the right of the vertex, (0,0).

(x,y) on the parabola = (x, sqrx)

the distance from the parabola to (3,0) is, using the distance formula: d^2 = (x-3)^2 + (sqrx)^2 = x^2-6x+9+x

d^2 = x^2 -5x +9

take the derivative and set equal to zero

2x -5 = 0

2x = 5

x = 2.5

y= sqr2.5

the point on the parabola closest to (3,0) is (2.5,sqr2.5)

that distance is sqr(.25 + 2.5) = sqr2.75

the line from through those two points has slope = sqr2.5/-.5 = -2sqr2.5

and is y-0 = y = 2sqr2.5[x-3]

the slope of the parabola at (2.5, sqr2.5) is y' = (1/2)x^-1/2 = 1/2sqrx = 1/2sqr2.5 = which is the negative inverse of the line connecting the two points (2.5,sqr2.5) and (3,0). A perpendicular line to the parabola is the shortest distance. a perpendicular line has a negative inverse slope to the tangent line of the parabola at (2.5, sqr2.5)