The slope of the tangent to the curve y=f(x) is given by (1/9)(x^2)(y^2). The curve passes through the point (3,1). Find the value of y when x=3*cubedroot(3).
9 dy/dx = x^2 y^2
9 dy/y^2 = x^2 dx
-9 / y = (1/3) x^3 + c
when x = 3, y = 1
-9/1 = (1/3)(27) + c
-9 = 9 + c
c = -18
so
-9/y = (1/3) x^3 + c
if x = 3 * 3^(1/3) = 3^(4/3)
then x^3 = 3^4
so
-9/y = (1/3)* 3 * 3^3 - 18
-9/y = 27 - 18 = 9
-1/y = 1
y = -1