f(x,y) = x^3y^2
df = (df/dx) dx + (df/dy) dy
df/dt=(df/dx) dx/dt + (df/dy) dy/dt
but f is a constant = 2916
df/dx = 3x^2y^2 = 3*81*4= 972
df/dy = 2x^3y = 2*729*2= 2916
so 0 = 972 (dx/dt) + 2916 (2)
dx/dt = -6
In the problem, x and y are differentiable functions of t. Find dx/dt when x = 9, y = 2, and dy/dt= 2.
x^3y^2 = 2916
1 answer
1 answer