Consider z=f(x,y)=x^3 + 2xy + y
Find all points on the surface where the tangnent plane is parallel to the plane 5x+3y-z=0
So I took the gradient of f (x,y,z) and got (2x^2 i , 2x+1 j , -1 k) And this vector must be colinear to (5, 3, -1)
So (2x^2 , 2x+1 , -1 ) = (5, 3, -1) I get: (x, y, z) = (1, 3/2, -1)
My problem is I was supposed to get multiple points (from subsequent questions), not one, so I must have done something wrong but I can't find it.
The plane is the graph of the function $g(x, y) = 5x + 3y$. The gradient of this function is $(5, 3)$. The gradient of $f(x, y)$ is $(3x^2 + 2y, 2x + 1)$. Now set one equal to the other, and solve.