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))*(xk)+k^2
to find the other point of intersection equate it with y=x^2
x^2=(1/(2k))*(xk)+k^2
2kx^2=x+k+2k^3
2kx^2+xk2k^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=(14k^21)/(4k)=(1+2k^2)/(2k)
so we want to find the integral of
(1/(2k))(xk)+k^2x^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^44k^21=0
(4k^2+1)^2(2k+1)(2k1)=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 20091202 12:13:35 