# What is the proper equation to use when finding the tangent plane?

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?

• Definitely the second one is the correct tangent plane. In the first case you have written f(x,y). What does that mean ? Because indeed f is a function on 3 variables. – Sota Antonino Jan 15 '19 at 0:24

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.
• So I have a question. From this explanation what I gather tis that there are two distinct ways we are characterizing our surface. 1) we can explicitly write it out as a level set: $F(x,y,z) = c$ and 2) if we cannot write it out as a level set we can write it as $z = f(x,y)$? Then taking the gradient at a specific point we can write out an equation of the tangent plane? I read it another question , but still didn't completely get it, what is the difference between the two characterizations in how we use them? – dc3rd Dec 6 '19 at 20:54
• @dc3rd It’s the other way around: you can always write $z=f(x,y)$ as a level set, namely $f(x,y)-z=0$. On the other hand, for the level set $F(x,y,z)=0$ of an arbitrary function $F(x,y,z)$, you generally can’t globally solve for one variable in terms of the others (the Implicit Function theorem tells you when you can do this locally). You might also want to have a look at math.stackexchange.com/q/1912660/265466, math.stackexchange.com/q/2459214/265466 and related questions for the geometric interpretations of $\nabla f$ and $\nabla F$. – amd Dec 6 '19 at 21:03
• Ah yes. So to try and extend what you're saying...IF I do have a characterization in the form $F(x,y,z) = c$, the gradient of this would help in expressing a tangent plane globally? – dc3rd Dec 6 '19 at 21:06
• @dc3rd I wouldn’t put it quite that way, but yes—you can use $\nabla F$ to find the tangent plane to a level surface of $F$ anywhere that $F$ is differentiable. – amd Dec 6 '19 at 21:09