calculus
posted by Anonymous .
consider the function f(x)=(36x^2)^(1/2) whose graph describes the upper semi circle with radius 6 centered at the origin.let a be any number in the open interval (6,6).for a not equal to 0 determine the slope of the line through the origin and the point (a,f(a))

The slope of f(x) at any point on the circle will be perpendicular to the radius (tangent to circle perpendicular to radius at intersection).
so the slope we are looking for = 1/f')x)
f' = (1/2)(36x^2)^(1/2) (2x)
= x/sqrt(36x^2)
the negative inverse of that is
(1/x)sqrt(36x^2)