Question

Find the point on the curve y = x^3 which is nearest the point (4, 0).

Answer 1

the distance from (x,x^3) to (4,0) is

z = √((x-4)^2+(x^3-0)^2)
= √(x^6+x^2-8x+16)
so,
dz/dx = (6x^5+2x-8) / 2√(x^6+x^2-8x+16)

Now, dz/dx=0 when the numerator is zero. That is when

6x^5+2x-8=0
x=1

So, the point (1,1) is closes to (4,0)

Or, you could reason that the distance is minimum when the line normal to the curve passes through (4,0)
Since y'=3x^2, the line through (h,h^3) normal to the curve is

y-h^3 = -1/3h^2 (x-h)
We want h when that line passes through (4,0)
0-h^3 = -1/3h^2 (4-h)
3h^6 = 4-h
again, h=1 so the line through (1,1) and (4,0) is normal to the curve, making that the shortest distance available.