So I'm doing my assignment to calculate the equation of a tangent plane at some point. I stumble upon a page on the web that says that an equation of a tangent plane to the surface $z=f(x, y)$ at the point $(x_0, y_0, z_0)$ is $$z-z_0 = f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$ My professor also used this equation at some point. My question is, how was this equation derived? Can you explain to me the parts of this equation and why it works?



If the surface is described by $z =f(x,y)$, it can also be represented as the level surface of the function

$F(x, y, z) = f(x, y) - z \tag{1}$

corresponding to the value $F(x, y, z) = 0$. Since the tangent plane to a surface at a point $(x_0, y_0, z_0)$ consists of all vectors perpendicular to the surface normal at that point, we must have, for any tangent vector $\vec t$ at $(x_0, y_0, z_0)$,

$\vec t \cdot \nabla F(x_0, y_ 0, z_0) = 0. \tag{2}$

The tangent plane in question contains the point $(x_0, y_0, z_0)$, and if $(x, y, z)$ is any other point in that plane, the vector $\vec t = (x -x_0, y- y_0, z - z_0)$ satisfies (2). Writing (2) for this choice of $\vec t$ yields

$F_x(x - x_0) + F_y(y - y_0) + F_z(z - z_0) = 0. \tag {3}$

Now using $F = f - z$, we see that (3) becomes

$f_x(x - x_0) + f_y(y - y_0) - (z - z_0), \tag{4}$

which is basically the same as the required form of the equation. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!


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