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In just the XY plane, a derivative is the slope of the tangent line.

How does it work when you have a 3D shape? I recall 3D math, and the Z-axis. Was this Calc 3?

Let's say you have a "mound". Then a "tangent" to the mound at a point (x,y) will be a plane, right? Is the derivative of a 3d shape a 2d plane? How do you represent this plane?

Anyone have a basic link that will explain the general concepts I am broaching?

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Given a surface defined by z= f(x, y), the partial derivatives at, say, point $(x_0, y_0)$ are $f_x(x_0,y_0)$, $f_y(x_0,y_0)$ and the equation of the tangent plane to the surface at that point is $z= f_x(x_0,y_0)(x- x_0)+ f_y(x_0,y_0)(y- y_0)+ f(x_0,y_0)$.

For example, if the "mound" is given by $z= 4- x^2- y^2$ the partial derivatives are $f_x(x,y)= -2x$ and $f_y(x,y)= -2y$. At x= 1, y= 1, z= 2 and the partial derivatives are both -2. The equation of the tangent plane at (1, 1, 2) is z= -2(x- 1)- 2(y- 1)+ 2= -2x- 2y+ 6.

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  • $\begingroup$ Is there a simpler conceptual explanation, "in English", like I described in the question? Was I on the right path when saying the derivative of a 3D shape is a 2D plane? $\endgroup$ – JackOfAll Oct 19 '15 at 12:56

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