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A Normal and a Parabola (Posted on 2009-12-02) Difficulty: 3 of 5
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.

  Submitted by Brian Smith    
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Solution: (Hide)
k=1/2, area=4/3

The slope of the tangent at (k,k^2) is 2k, so the slope of the normal is -1/(2k). Then the equation of the normal is y = -x/(2k) + k^2 + 1/2.

The second point the normal intersects y=x^2 is
( -k + -1/(2k), k^2 + 1 + 1/(4k^2) ).

The area of the closed region is equal to the integral of the difference of the equations of the normal and parabola from x=-k-1/(2k) to k.
=Integ{-k-1/(2k), k} [(-x/(2k) + k^2 + 1/2) - x^2] dx

The easiest way to evaluate this integral is to use Simpson's rule which is exact for parabolas. To do that one more point is needed at
x=[(k)+(-k-1/(2k))]/2 = -1/(4k)

The three points to be used are:
(k, 0)
(-1/(4k), k^2 + 1/2 + 1/(16k^2)
(-k-1/(2k), 0)

Then the integral evaluates to
Area = ( 2k+1/(2k) )/6 * [1*0 + 4*(k^2 + 1/2 + 1/(16k^2)) + 1*0]

This simplifies to Area = 1/6 * [2k+1/(2k)]^3
2k+1/(2k) and its cube are at a minimum when k=1/2 (remember k>0). Then the area is 1/6 * 2^3 = 4/3

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(2): solution error foundDaniel2009-12-02 22:08:16
Hints/Tipsre: solution error foundAdy TZIDON2009-12-02 14:43:34
re: solutionCharlie2009-12-02 12:43:05
solutionDaniel2009-12-02 12:13:35
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