Triple product rule

An informal derivation follows. Suppose that f(x, y, z) = 0. Write z as a function of x and y. Thus the total differential dz is :dz = \left(\frac{\partial z}{\partial x}\right)dx + \left(\frac{\partial z}{\partial y}\right) dy Suppose that we move along a curve with dz = 0, where the curve is parameterized by x. Thus y can be written in terms of x, so on this curve :dy = \left(\frac{\partial y}{\partial x}\right) dx Therefore, the equation for dz = 0 becomes :0 = \left(\frac{\partial z}{\partial x}\right) \, dx + \left(\frac{\partial z}{\partial y}\right) \left(\frac{\partial y}{\partial x}\right) \, dx Since this must be true for all dx, rearranging terms gives :\left(\frac{\partial z}{\partial x}\right) = -\left(\frac{\partial z}{\partial y}\right) \left(\frac{\partial y}{\partial x}\right) Dividing by the derivatives on the right hand side gives the triple product rule :\left(\frac{\partial x}{\partial y}\right)\left(\frac{\partial y}{\partial z}\right) \left(\frac{\partial z}{\partial x}\right) = -1 Note that this proof makes many implicit assumptions regarding the existence of partial derivatives, the existence of the exact differential dz, the ability to construct a curve in some neighborhood with dz = 0, and the nonzero value of partial derivatives and their reciprocals. A formal proof based on mathematical analysis would eliminate these potential ambiguities. Alternative derivation Suppose a function , where , , and are functions of each other. Write the total differentials of the variables dx = \left(\frac{\partial x}{\partial y}\right) dy + \left(\frac{\partial x}{\partial z}\right) dz dy = \left(\frac{\partial y}{\partial x}\right) dx + \left(\frac{\partial y}{\partial z}\right) dz Substitute into dx = \left(\frac{\partial x}{\partial y}\right) \left[ \left(\frac{\partial y}{\partial x}\right) dx + \left(\frac{\partial y}{\partial z}\right) dz\right] + \left(\frac{\partial x}{\partial z}\right) dz By using the chain rule one can show the coefficient of on the right hand side is equal to one, thus the coefficient of must be zero \left(\frac{\partial x}{\partial y}\right) \left(\frac{\partial y}{\partial z}\right) + \left(\frac{\partial x}{\partial z}\right) = 0 Subtracting the second term and multiplying by its inverse gives the triple product rule \left(\frac{\partial x}{\partial y}\right) \left(\frac{\partial y}{\partial z}\right) \left(\frac{\partial z}{\partial x}\right) = -1. Short derivation This section is based on chapter 5 of Pippard. Suppose we are given four real variables (x, y, z, w), restricted to move on a 2-dimensional C^2 surface in \R^4. Then, if we know two of them, we can determine the other two uniquely (generically). In particular, we may take any two variables as the independent variables, and let the other two be the dependent variables, then we can take all these partial derivatives. Proposition: \left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y = -1 Proof. We can ignore w. Then locally the surface is just ax + by + cz + d = 0. Then \left(\frac{\partial x}{\partial y}\right)_z = -\frac{b}{a}, etc. Now multiply them. ==Applications==

Example: Ideal Gas Law The ideal gas law relates the state variables of pressure (P), volume (V), and temperature (T) via :PV=nRT which can be written as :f(P,V,T) = PV-nRT = 0 so each state variable can be written as an implicit function of the other state variables: : \begin{align} P &= P(V,T) = \frac{nRT}{V} \\[1em] V &= V(P,T) = \frac{nRT}{P} \\[1em] T &= T(P,V) = \frac{PV}{nR} \end{align} From the above expressions, we have : \begin{align} -1 &= \left( \frac{\partial P}{\partial V} \right) \left( \frac{\partial V}{\partial T} \right) \left( \frac{\partial T}{\partial P} \right) \\[1em] &= \left( -\frac{nRT}{V^2} \right) \left( \frac{nR}{P} \right) \left( \frac{V}{nR} \right) \\[1em] &= \left( -\frac{nRT}{PV} \right) \\[1em] & = -\frac{P}{P} = -1 \end{align} Geometric Realization A geometric realization of the triple product rule can be found in its close ties to the velocity of a traveling wave :\phi(x,t) = A \cos (kx - \omega t) shown on the right at time t (solid blue line) and at a short time later t+Δt (dashed). The wave maintains its shape as it propagates, so that a point at position x at time t will correspond to a point at position x+Δx at time t+Δt, :A \cos (kx - \omega t) = A \cos (k (x + \Delta x) - \omega (t + \Delta t)). This equation can only be satisfied for all x and t if , resulting in the formula for the phase velocity : v = \frac{\Delta x}{\Delta t} = \frac{\omega}{k}. To elucidate the connection with the triple product rule, consider the point p1 at time t and its corresponding point (with the same height) p̄1 at t+Δt. Define p2 as the point at time t whose x-coordinate matches that of p̄1, and define p̄2 to be the corresponding point of p2 as shown in the figure on the right. The distance Δx between p1 and p̄1 is the same as the distance between p2 and p̄2 (green lines), and dividing this distance by Δt yields the speed of the wave. To compute Δx, consider the two partial derivatives computed at p2, : \left( \frac{\partial \phi}{\partial t} \right) \Delta t = \text{rise from }p_2\text{ to }\bar{p}_1\text{ in time }\Delta t\text{ (gold line)} : \left( \frac{\partial \phi}{\partial x} \right) = \text{slope of the wave (red line) at time }t. Dividing these two partial derivatives and using the definition of the slope (rise divided by run) gives us the desired formula for : \Delta x = - \frac{\left( \frac{\partial \phi}{\partial t} \right) \Delta t}{\left( \frac{\partial \phi}{\partial x} \right)}, where the negative sign accounts for the fact that p1 lies behind p2 relative to the wave's motion. Thus, the wave's velocity is given by : v = \frac{\Delta x}{\Delta t} = - \frac{\left( \frac{\partial \phi}{\partial t} \right)}{\left( \frac{\partial \phi}{\partial x} \right)}. For infinitesimal Δt, \frac{\Delta x}{\Delta t} = \left( \frac{\partial x}{\partial t} \right) and we recover the triple product rule : v = \frac{\Delta x}{\Delta t} = - \frac{\left( \frac{\partial \phi}{\partial t} \right)}{\left( \frac{\partial \phi}{\partial x} \right)}. ==See also==