# The blackboards of Rushmore

In the movie Rushmore
some blackboards with math appear. Lets look at them: 1) The first blackboard deals with the derivation of the formula d/dx f(x) for the square root function f(x)= x ^{1/2}. We have seen that. Multiply both sides with f(x+h)+f(x) to get
f(x+h)-f(x) = (f(x+h)^{2} - f(x)^{2})/(f(x+h)+f(x) = h/(f(x+h)+f(x). Therefore
[f(x+h)-f(x)]/h is equal to 1/[f(x+h) + f(x)] which goes to 1/(2 f(x)) in the limit h to zero. 2) The side blackboard shows an Extra credit problem. Displayed is the ellipse x ^{2}/a^{2} + y^{2}/b^{2} = 1.
The problem is to show that A = int__{-a}^{a} (y_{+} - y_{-}) dx= pi a b.
[That this is the hardest problem in geometry is of course a joke. The scene is a day dream in a comedy.
It can be done easiest with a change of variable reducing it to the circle. In multivariable calculus
it is done in polar coordinates or using Greens theorem. In single variable calculus it is a prototype
of a problem for trig substitution and this is how it is done on the next blackboards. ] 3) Max Fischer first writes y = (b/a) (a ^{2} - x^{2})^{1/2} and then
integrates is from -a to a. He substitutes x=a sin(u) to get the integral of b a cos^{2}(t) from
0 to pi/2. He then uses the double angle formula cos^{2}(t) = [1+cos(2t)]/2 to get the area
4 a b int_{0}^{pi/2} (1/2+cos(2t)/2) dt = pi a b. |