My professor has introduced us to this equation when finding a tangent plane.
$$ z= f(x,y) + [\frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [\frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0) $$
I am asked to find the tangent plane of the function $ f(x,y,z)= \frac{xyz}{x^2+y^2+z^2}$. I get the answer $z=y$ when I solve it myself, but I saw other solutions which used the formula $$ 0 = [\frac{∂f}{∂x}(x_0,y_0,z_0)] (x-x_0) + [\frac{∂f}{∂y}(x_0,y_0,z_0)] (y-y_0)+[\frac{∂f}{∂z}(x_0,y_0,z_0)] (z-z_0) $$
I am not sure which one to use, because this solution gives me $ y=0 $. Can anybody help clear up the difference between these two equations?
Best Answer
You appear to be confusing implicit and explicit functions. The second example function that you’ve given is a function from $\mathbb R^3$ to $\mathbb R$, so its graph doesn’t have a tangent plane, but rather a tangent hyperplane, which you can find by extending your first formula by one more dimension.
I suspect that what you are really asking about with that second function is a surface in $\mathbb R^3$ that’s given implicitly by the equation $f(x,y,z)=c$ for some constant $c$. Such a surface is known as a level surface of $f$. That’s a different matter. If you can solve that equation for $z$ explicitly, i.e., describe the surface as the graph of a function of two variables, then your first formula applies. However, in many cases including this one that’s not possible, so you have to use a related, but different method.
There’s a theorem that says the gradient of a function at a point is always normal to its level surface at that point. The gradient is also orthogonal to the tangent plane (if it exists), so using the point-normal form of the equation of a plane, an equation for the tangent plane at a point $(x_0,y_0,z_0)$ on that surface is $$(x-x_0){\partial f\over\partial x}(x_0,y_0,z_0)+(y-y_0){\partial f\over\partial y}(x_0,y_0,z_0)+(z-z_0){\partial f\over\partial z}(x_0,y_0,z_0)=0.\tag{*}$$ The two formulas are related: If you have $z=f(x,y)$, then let $g(x,y,z)=f(x,y)-z$. It should be clear that the graph of $f$ is identical to the level surface $g(x,y,z)=0$. By the chain rule we then have for the gradient $\nabla g = \left({\partial f\over\partial x},{\partial f\over\partial y},-1\right)$ and if you plug this into (*) and rearrange you’ll end up with your first formula.