Question

Okay, so this is a line l.
http://puu.sh/prQdV/8a46768079.jpg
It passes through points (a, a^2) (b, b^2) of the parabola y=x^2. b>a
I was asked to find S1, so I did, it's 1/6 (b-a)^3
Then I was asked to find a in terms of b if l is perpendicular to the tangent line at b. It's (-1/2b) -b.
It says when 2 holds true, fins the minimum area of S in terms of b, for what value of b is the area simplified?
I'm really sorry for asking, a friend of mine says it's minimum when b=a, but I don't know. Help?

Answer 1

yes technically the Area is 0 when b=a, thus at a minimum.
But i dont think thats what they are looking for.

To find min Area, take derivative of Area function in terms of "b" and set equal to 0.
\[A = \frac{1}{6} (2b + \frac{1}{2b})^3\]
\[\frac{dA}{db} = \frac{1}{2}(2 - \frac{1}{2b^2})(2b+\frac{1}{2b})^2 = 0\]
\[\rightarrow 2-\frac{1}{2b^2} = 0\]
\[b = \frac{1}{2}\]
Leading to a min Area of
\[A_{\min} = \frac{1}{6} (2 *\frac{1}{2} + \frac{1}{2*\frac{1}{2}})^3 = \frac{8}{6} = \frac{4}{3}\]