let f(x,y) = 3x^3(y) - 4x^2 (sqrt(y)) and let γ(t) = <x(t),y(t)> be a curve in the Oxy plane such that at some point t0, we have γ(t0) = (-1,4) and γ'(t0) = (-2,-3). Find the tangent vector r'(t0) to the curve r(t) = <x(t),y(t), f(x(t), y(t))> at the point t0.
The tangent vector r'(t0) to the curve r(t) = <x(t),y(t), f(x(t), y(t))> at the point t0 is given by:
r'(t0) = <-2, -3, 9(-1)^3(4) - 4(-1)^2(2)> = <-2, -3, -20>