Find the point P on the graph of the function y=sqrt{x} closest to the point (10,0)

Answer 1

the slope of the tangent is 1/(2√x), so the slope of the normal is -2√x
so you want the line with slope -2√x that goes through (10,0) and (x,y) on the curve.

Or, consider the distance z from (x,y) to (10,0).
z = √((10-x)^2+y^2) = √((10-x)^2 + x)
now find where dz/dx = 0

Answer 2

z = distance between^2, minimize z
z = (x-10)^2 + (y-0)^2 = x^2 - 20 x + 100 + y^2 but y^2 = x
z = x^2 - 19 x + 100
dz/dx = 2 x -19
= 0 at min
so
x = 19/2 = 9.5
y = sqrt (9.5)
check my arithmetic !