# Substitution of variables

{{ safesubst:#invoke:Unsubst||$N=Merge to |date=__DATE__ |$B= Template:MboxTemplate:DMCTemplate:Merge partner }} {{ safesubst:#invoke:Unsubst||$N=Unreferenced |date=__DATE__ |$B= {{#invoke:Message box|ambox}} }} In mathematics, substitution of variables (also called variable substitution or coordinate transformation) refers to the substitution of certain variables with other variables. Though the study of how variable substitutions affect a certain problem can be interesting in itself, they are often used when solving mathematical or physical problems, as the correct substitution may greatly simplify a problem which is hard to solve in the original variables. Under certain conditions the solution to the original problem can be recovered by back-substitution (inverting the substitution).

## Simple Example

Consider the system of equations

$xy+x+y=71$ $x^{2}y+xy^{2}=880$ ## Common examples

### Cylindrical coordinates

Some systems can be more easily solved when switching to cylindrical coordinates. Consider for example the equation

$U(x,y,z):=(x^{2}+y^{2}){\sqrt {1-{\frac {x^{2}}{x^{2}+y^{2}}}}}=0.$ This may be a potential energy function for some physical problem. If one does not immediately see a solution, one might try the substitution

$\displaystyle (x,y,z)=\Phi (r,\theta ,z)$ given by $\displaystyle \Phi (r,\theta ,z)=(r\cos(\theta ),r\sin(\theta ),z)$ .
$V(r,\theta ,z)=r^{2}{\sqrt {1-{\frac {r^{2}\cos ^{2}\theta }{r^{2}}}}}=r^{2}{\sqrt {1-\cos ^{2}\theta }}=r^{2}\sin \theta$ .

Note that, had we allowed $r=0$ , the origin would also have been a solution, though it is not a solution to the original problem. Here the bijectivity of $\Phi$ is crucial.

### Integration

{{#invoke:main|main}} Under the proper variable substitution, calculating an integral may become considerably easier. Consult the main article for an example.

### Momentum vs. velocity

Consider a system of equations

$m{\dot {v}}=-{\frac {\partial H}{\partial x}}$ $m{\dot {x}}={\frac {\partial H}{\partial v}}$ ${\dot {p}}=-{\frac {\partial H}{\partial x}}$ ${\dot {x}}={\frac {\partial H}{\partial p}}$ ## Lagrangian mechanics

$m{\ddot {x}}=\phi (t,x,v)$ .

He found that the equations

${\frac {\partial {L}}{\partial y}}={\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial {L}}{\partial {w}}}$ are equivalent to Newton's equations for the function $L=T-V$ , where T is the kinetic, and V the potential energy.

In fact, when the substitution is chosen well (exploiting for example symmetries and constraints of the system) these equations are much easier to solve than Newton's equations in Cartesian coordinates.