Finding the tangent plane to of f(x, y) = x^2 − y^2 at the point (2, 1, 3)

[ld436]

Which of the following vectors is a normal vector to the tangent plane of f(x, y) = x^2 − y^2at the point (2, 1, 3)?

So the normal vector is (df/dx,df/dy,df/dz) which turns out to be (4,-2,0). However the correct answer is (4,-2,-1) and all the other answers are (4, -2, z something). Why is there a Z component to this vector?

 

[tkhunny]

If we are starting with z = f(x,y), do you believe that z' = 0 in all cases?

If you pick a value for z, you are restricting yourself to a specific plane and, of course, you will get \(\displaystyle f_{z} = 0\). This is no good. You picked the plane before you started.

A good look at the general form of a plane provides the answer... \(\displaystyle f_{x}(x_{0},y_{0},z_{0})(x-x_{0})+f_{y}(x_{0},y_{0},z_{0})(y-y_{0})+f_{z}(x_{0},y_{0},z_{0})(z-z_{0}) = 0\)

As can be seen, one needs z as part of the functional notation and that other side is zero. Typically, this is portrayed by simple subtraction. Define \(\displaystyle F(x,y,z) = f(x,y) - z\) and \(\displaystyle \nabla F = \langle F_{x},F_{y},F_{z} \rangle = \langle f_{x},f_{y},-1 \rangle\)

Give that some thought.

 

[ld436]

Thanks, your answer makes a ton of sense. I understand now.