equation of tangent plane using differentials
find the tangent plane to x^2 + y^2 - z^2 = 4 at the point (2,1,1).
I know how to do this using the gradient vector but I am trying to do this problem a different way using differentials. i define the function w = x^2 + y^2 - z^2 so x^2 + y^2 - z^2 = 4 is equivalent to saying the level curve w = 4. so i start with the definition of differentials which is that dw = 2x dx + 2y dy - 2z dz. then i can replace the differentials with deltas and plug in the point (2,1,1) to create an approximation formula which is Δw ≈ 4 Δx + 2 Δy - 2 Δz. The next step according to the lecture i watched would be since we want to stay on the level curve, Δw = 0 and it becomes 0 ≈ 4 Δx + 2 Δy - 2 Δz. then 4 Δx + 2 Δy - 2 Δz = 0 means that we are on the tangent plane to the level curve and it is that part that i am confused about. how do we know that setting 4 Δx + 2 Δy - 2 Δz = 0 means we are on the tangent plane. i don't see the intuition behind it.
what the approximation 0 ≈ 4 Δx + 2 Δy - 2 Δz tells me is that when the changes in x y and z are small enough, the expression 4 Δx + 2 Δy - 2 Δz will be very close to 0. if the changes are not small enough however, the expression will not be close to 0. but the equation 4 Δx + 2 Δy - 2 Δz = 0 tells me that no matter what changes in x y and z I input into the expression, it will always equal 0. i know that if the changes are too large they will not approximate Δw well and thus not be close to 0. but the equation 4 Δx + 2 Δy - 2 Δz = 0 tells me that the expression does equal to 0 no matter how large or small the changes are. this is where i am getting confused.