At the point (2,-1,2) on the surface z = x(y^2), find the direction vector for the direction of greatest decrease of z. A) i - 2j B) i - 4j C) (i - 4j)/sqrt(17) D) -i + 4j E) i + j Do I need a function f(x,y,z)? If so then f(x,y,z) = z = xy^2. Then the gradient vector would be <Fx, Fy, Fz>. This would be give me. < -y^2, -2xy, 1> At the point (2,-1,2) we get < -1, 4, 1> But how do i get the direction of the vector of greatest decrease? From the choices though it seems they only find the gradient vector of z = F(x,y) = xy^2 In which case the gradient vector is < fx, fy> < y^2 , 2xy> . At the point (2, -1 ,2) we get < 1, -4> Again i am not sure how to find the vector of greatest decrease. It says in my book the direction of minimum increase is - norm( grad f(x,y,z)). But i am not getting any of these answers can some one help?