# What exactly is a dependent variable?

1. Nov 28, 2007

### Cyrus

In my course notes (Dynamics) I have the following written down:

(a) write as a set of implicit equations $$0=h(\dot{\vec{z}},\vec{z},\vec{\mu})$$

(b) pick a reference flight condition (RFC) $$0=h(\dot{\vec{z*}},\vec{z*},\vec{\mu*})$$

(c) Apply Taylor to linearize (a) about (b); Trick: Consider $$\dot{\vec{z*}}, \vec{z*}, \vec{\mu*}$$ to be independent variables.

The function $$h(\dot{\vec{z*}},\vec{z*},\vec{\mu*})$$ is found by finding the equations of motion. An example of a function h given in an example in class was:

$$0=-\frac{g}{L}sin(z_1)-\frac{b(z_2)}{mL}+\frac{\mu}{mL^2}-\dot{z_2}$$​

Based on this (or any) equation of motion, selection of a value of $$z_1,z_2,\mu$$ constrains a value of $$\dot{z_2}$$. In this sense at least one value has to depend on the other three. In my mind, by calling these four variables independent it means I can pick any value I want for them and input them into my function, regardless of the selection of the other values; however, it is obvious that any choice of those 4 variables is not going to satisfy the fact that h must equal 0: h=0. So, I ask how is it that these values really are independent of eachother?

Upon searching for a taylor series expansion in multi-variables, [1], it is clear that the input arguments into the function h, are independent.

So I tried to do more searching on what exactly are independent and dependent variables. It seems that an dependent variable is a function of an independent variable. This means that for each value of the independent variable there can be one and only one value for the dependent variable. So, does that mean if there are two possible solutions that satisfy the equation, there are no 'dependent/independent' variables, just, for a lack of a proper word, 'variables'? An example is the equation of a circle, $$x^2+y^2=R^2$$. This is another implicit function, but I cannot pick any value of x and y and have the sum of their squares equal $$R^2$$. Does it even make sense to use the word independent/dependent variables?

Any thoughts would be helpful.

[1] - http://www.chem.mtu.edu/~tbco/cm416/taylor.html

2. Nov 28, 2007

### Math Is Hard

Staff Emeritus
I don't know if it helps at all, Cyrus, but in psych research, independent variables are what we manipulate and dependent variables are what we measure. For instance, if I wanted to study the effects of sleep deprivation on math problem solving, I could assign three groups of subjects to get either 2 or 4 or 8 hours of sleep (the IV, with 3 levels) and then measure their scores on a math test I give them the next day (the DV).

3. Nov 29, 2007

### Chris Hillman

Indepependent and dependent variables in theory of DEs

Hi, Cyrus,

I am not sure I understand what you wrote down, but it seems to involve differential equations, and if so, then, for example:

• In the second order linear PDE
$$u_t = \Delta u = u_{xx} + u_{yy} + u_{zz}$$
(the heat equation), u is the dependent variable and x,y,t are the independent variables,
• in the system of first order nonlinear ODEs (where dot denotes differentiation wrt t)
$$\begin{array}{rcl} \dot{p}_1 & = & -q_1 - 2 \, q_1 \, q_2 \\ \dot{p}_2 & = & -q_2 - \left( q_2^2 - q_2^2 \right) \\ \dot{q}_1 & = & p_1 \\ \dot{q}_2 & = & p_2 \end{array}$$
(equations of motion derived from the Henon-Heiles Hamiltonian), $q_1, \, q_2, \, p_1, \, p_2$ are the dependent variables at t is the independent variable.
Does that help?

4. Nov 29, 2007

### D H

Staff Emeritus
Hi Cyrus,

Engineering perspective: Suppose you have some machinery that lets you control your "independent" variables. If you can control each parameter independently (changing "x" doesn't change "y", or "z", or whatever), you have independent controls. To an engineer (the best is enemy of good enough), that's close enough to independent variables.

Simple mathematical perspective: $\partial x_i/\partial x_j = \delta_{ij}$

Better mathematical perspective: Suppose your desired function $h$ is a function of an n-dimensional space. If you have $m>n$ supposedly independent variables they obviously are not independent. The supposedly independent variables are independent if (a) $m\le n$ and (b) they form an orthogonal basis for some m-dimensional subspace. If the variables aren't truly orthogonal there are some dependencies among them, and this in turn might invalidate your linearization.

5. Dec 2, 2007

### mathwonk

you make an interesting point. in the example x^2 +Y^2 = 1, neither X nor Y is a priori the dependent or independent variable. rather when a value is chosen for one of them, then (at most two) values are determined for the other. hence either one can be considered the independent variable.

6. Dec 2, 2007

### symbolipoint

The dependant variable must be the value that the function tells it to be. It must do what it is told.

7. Dec 2, 2007

### Cyrus

At the very least, read and pay attention to the discussion before you post. :uhh:

Last edited: Dec 2, 2007
8. Dec 2, 2007

### symbolipoint

At least it related to the title; seems enough.

9. Dec 3, 2007

### rs1n

If you are given

$$h(\dot{\vec{z*}},\vec{z*},\vec{\mu*}) = -\frac{g}{L}sin(z_1)-\frac{b(z_2)}{mL}+\frac{\mu}{mL^2}-\dot{z_2}$$

then the independent variables are $$\dot{\vec{z*}},\vec{z*},\vec{\mu*}$$ and the dependent variable is $$h$$ as its value depends on the inputs for $$\dot{\vec{z*}},\vec{z*},\vec{\mu*}$$. Generally speaking, independent variables are parameters whose values are unaffected by the values of the other parameters.

However, because you have now restricted $$h$$ to 0, you have placed a constraint on the previously independent variables. What you are essentially doing is asking: for which values of these independent variables is the right hand side of the formula above equal to 0. They are seemingly dependent on each other, but that is because of the constraint $$h=0$$. They are still independent variables within the formula above (i.e. with respect to the function $$h$$). On the other hand, taking the equation

$$-\frac{g}{L}sin(z_1)-\frac{b(z_2)}{mL}+\frac{\mu}{mL^2}-\dot{z_2} = 0$$

at face value (i.e. without the context of the function $$h$$), the parameters ARE dependent as only certain values will satisfy the equation above. The notion of "independent" relies on the fact that there are no constraints whatsoever.

Last edited: Dec 3, 2007
10. Dec 3, 2007

### rbj

i've always thought that the terminology meant "x" in

$$y = f(x)$$

where y is the "dependent variable" and x is the "independent variable".