Orthogonal family of curves to the level sets of $f(x_1, \cdots, x_n) = \prod_{j=1}^{n} x_j$

Question

I read this post and realized that if I can solve the related problem below that it might lead to an application. Their solution led to a hyperbola, so I suspect that my problem might lead to an $n$-dimensional hyperboloid or something similar. But I am unsure how to proceed from defining the level set and the derivative. Perhaps I need to consider all pairs of products of reciprocal derivatives $\frac{dx_i}{dx_j}\frac{dx_j}{dx_i} = -1$?

For the function $f(x_1, \cdots, x_n) = \prod_{j=1}^{n} x_j$ we can choose a level set:

$$\{(x_1, \cdots, x_n) \in \mathbb{R}^n : \prod_{j=1}^{n} x_j = K\phantom{space}\text{for some constant K} \}$$

The partial derivative with respect to variable $x_i$ is:

$$\frac{\partial f(x_1, \cdots, x_m)}{\partial x_i} = \sum_{k=1}^n \frac{\partial x_k}{\partial x_i} \prod_{j \neq k} x_j = 0$$

How do I complete the analsyis?

Answer 1

You want the family of curves orthogonal to these level hypersurfaces. The reasoning you use with orthogonal lines in the plane will not suffice. The best way to see these is to find curves that, at any given point $p$, are tangent to the gradient of $f$ (vector of partial derivatives) at $p$, as this is the normal vector to the hypersurface. In your case, the easiest way to proceed is the following.

I will write $\log$ (but if you're used to $\ln$, it's that). Note that $\log f(x) = \sum \log x_i$, so, by the chain rule, $$\frac 1{f(x)}\frac{\partial f}{\partial x_i} = \frac1{x_i}, \quad\text{so}\quad \frac{\partial f}{\partial x_i} = \frac{f(x)}{x_i}.$$ In particular, we want a curve whose tangent vector at each point $x$ is parallel to $f(x)\big(\frac1{x_1},\dots,\frac1{x_n}\big)$. Note how simple it is if we look for a curve $$x(t)=\big(x_1(t),\dots,x_n(t)\big) \quad\text{with}\quad \frac{dx_i}{dt} = \frac1{x_i(t)} \quad\text{for all }t.$$ You can solve this by separating variables, as in basic calculus. You get $$x_i(t)^2 = 2t + c_i, \quad i=1,\dots,n.$$ So, the orthogonal trajectories will be these quadratic curves $$x(t)=\big(x_1(t),\dots,x_n(t)\big) = \big({\pm}\sqrt{2t+c_1},\dots,\pm\sqrt{2t+c_n}\big)$$ for various values of $c_1,\dots,c_n$.