find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2, 0)

The surface is parameterized by

and the normal to the surface is given by the cross product of the partial derivatives of :

It looks like you're given

Then the normal vector is

Now, the point (2, 2, 0) corresponds to u and v such that

and solving gives , so the normal vector at the point we care about is

Then the equation of the tangent plane is

10 * 563.67 = 5636.7