Brain Teaser - Area enclosed by a parabola and a line

[fhub]

(09-15-2015 03:29 AM)Gerson W. Barbosa Wrote:  Things start to get complicated for the biquadratic parabola, if I've done it right:

\(y=\frac{4u^{7}+u-x}{4u^{3}}\)

Hmmm, from where did you get this strange expression (u^7?)?

I'm also wondering about some very (unnecessarily) complicated solutions here, especially using 'numeric' solvers!?
This problem can be solved completely 'by hand' (or better 'by brain'), only for the definite integral it's maybe better to use a CAS just to avoid any mistakes and to get a simple (factored) result:

parabola: y=x^2
at special point x=x0 (>0) ==> y=x0^2 and slope k=2*x0
normal line: y=kn*x+d with normal slope kn=-1/k=-1/(2*x0)
entering the special point (x0,x0^2) and kn ==> d= x0^2+1/2
so the normal line is: y=-1/(2*x0)*x+x0^2+1/2

The 2 intersection points are simply the solutions of the quadratic equation
x^2=-1/(2*x0)*x+x0^2+1/2
with the 2 results
x1=-x0-1/(2*x0) and x2=x0 (of course).

Now the area between the 2 curves is
A(x0)=integral(-1/(2*x0)*x+x0^2+1/2-x^2,x,-x0-1/(2*x0),x0)
Although it's also easy to calculate this integral manually, it's indeed safer to use a CAS for it:
A(x0)=(4*x0^2+1)^3/(48*x0^3)

If you put now the x0^3 from the denominator under (...)^3 in the numerator, you get a much simpler result, for which it's very easy to calculate the minimum by hand:
A(x0)=1/48*(4*x0+1/x0)^3

A'(x0)=3/48*(4*x0+1/x0)^2*(4-1/x0^2) = 0 (for minimum)
In this expression only the last (...) can be 0, so you have immediately the answer:
x0=sqrt(1/4)=1/2 (x0=-1/2 excluded, because x0>0 was assumed)

Franz