yz = ln(x+z)(adsbygoogle = window.adsbygoogle || []).push({});

So I'm trying to find the tangent plane to the surface at a particular point (x0,y0,z0).

Here's the general formula:

Fx(x0,y0,z0)(x-x0) + Fy(x0,y0,z0)(y-y0) + Fz(x0,y0,z0)(z-z0) = 0,

where Fx, Fy, and Fz are the partial derivatives of the below F(x,y,z):

1. F(x,y,z) = ln(x+z) - yz

2. F(x,y,z) = yz - ln(x+z)

When I take the partial derivatives Fx,Fy,Fz for both of these functions, they turn out to be different:

For (1.), I get:

Fx = 1/(x+z)

Fy = -z

Fz = 1/(x+z) - y

For (2.), I get:

Fx = -1/(x+z)

Fy = z

Fz = y - 1/(x+z)

, which all have the opposite signs of (1.)

So my question is: how do I know which F(x,y,z) to use when trying to find the tangent plane through a particular point?

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# Homework Help: A contradiction? - Calculus III Question

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