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Math Help - equation of tangent plane using differentials

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    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.
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    Quote Originally Posted by oblixps View Post
    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.
    You do understand, I hope, that \Delta x= x- 2, \Delta y= y- 1 and \Delta z= z- 1. 4\Deta x+ 2\Delta y- 2\Delta z= 4(x- 2)+ 2(y- 1)+ 2(z- 1)= 0 is precisely the equation of the tangent plane.

    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.
    If "the changes are too large", that is if we are far from the point (2, 1, 1), we cannot expect the tangent plane to be close to the surface.
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