Choose any point (k,k^2) with k>0 on the parabola y=x^2. Draw the normal line to the parabola at that point. Then there is a closed region defined by the parabola and the line. Find the value of k so the area of the region is minimized.

Note: A normal line is a line perpendicular to a tangent at the point of tangency.

at point (k,k^2) the slope of the tangent is 2k so the slope of the normal is -1/(2k).  thus the equation of the normal line is
y=(-1/(2k))*(x-k)+k^2
to find the other point of intersection equate it with y=x^2
x^2=(-1/(2k))*(x-k)+k^2
2kx^2=-x+k+2k^3
2kx^2+x-k-2k^3=0
x=(-1+-sqrt(1+8k(k+2k^3))/(4k)
x=(-1+-sqrt(16k^4+8k^2+1))/(4k)
x=(-1+-sqrt((4k^2+1)^2))/(4k)
x=(-1+-(4k^2+1))/(4k)
x=k which we already knew
x=(-1-4k^2-1)/(4k)=-(1+2k^2)/(2k)
so we want to find the integral of
(-1/(2k))(x-k)+k^2-x^2 dx from x=-(1+2k^2)/(2k) to k
and this is easily found to be
(4/3)k^3+k+(1/4)k^-1+(1/48)*k^(-3)
and this is minimized when
4k^2+1-(1/4)*k^-2-(1/16)*k^-4=0    *16k^4
64k^6+16k^4-4k^2-1=0
(4k^2+1)^2(2k+1)(2k-1)=0
k=1/2 k=-1/2 k=i/2 k=-i/2
since we want positive real k then we are left with k=1/2
and this gives minimum area of 4/3

Edited on December 2, 2009, 10:02 pm

Posted by Daniel on 2009-12-02 12:13:35