Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My friends argue this $d_t( \partial_{\dot{x}} g)=1+2\dot{\dot{x}} \not = \partial_t (\partial_{\dot{x} }g)$ where $g=t\dot x + x^2 + \dot{x}^2$. Why?

share|improve this question
    
What exactly do you mean by $d_t$ for a multivariable function? –  gnometorule Feb 6 '13 at 13:39

2 Answers 2

Because they are derivatives of different functions.

Example $g(x,t)=t x^2$

When $x$ is itself a function of $t$, we have two options:

  • hold $x$ constant and take the partial derivative of $g(x,t)=tx^2$ with respect to $t$. This gives $\partial_t g = x^2$
  • treat $x$ as a function of $t$. Then we really look at the single-variable function $G(t):=g(x(t),t)$ but usually people don't bother introducing notation for it. The derivative is $G'(t)=x^2+t2x\dot x$.

You can treat the second computation as a partial derivative too; it's just that instead of holding $x$ constant, we hold an expression of $x,t$ constant. For example, if $x=t^3$, then we differentiate $g$ with respect to $t$ holding $x-t^3$ constant. This consideration occurs in mechanics when the change of coordinates is introduced.

Example with coordinate change

If your coordinates are $x,y$ and you decided to introduce a new coordinate $\tilde y=x+y$, then you should also introduce $\tilde x=x$ because the partial derivatives $\partial_x g$ and $\partial_{\tilde x}g$ will be different (even though $x$ and $\tilde x$ are the same thing). This takes a while to get used to.


Added later. The process of taking a partial derivative involves the following steps:

  1. Restrict the function to a curve
  2. Choose a parameter for that curve
  3. Differentiate the restricted function with respect to the chosen parameter.

For example, what is $\dfrac{\partial f}{\partial y}(1,2,3)$?

Step 1 is commonly expressed by saying "hold other variables constant". Here we take the partial derivative at $(1,2,3)$ "holding $x$ and $z$ constant", which means that we restrict the function to the line $x=1$, $z=3$. Notice that this step is not about the variable in which we will take the derivative.

Step 2: we choose the parameter for our line, namely $y$.

Step 3 is now unambigious: we differentiate a function of one variable.

But it does not have to be so rectangular all the time. Instead we can restrict $f$ to the curve formed by the intersection $z=x^2+y$, $x+y+z=6$. This means differentiation while holding the variables $u=z-x^2$ and $v=x+y+z$ constant. And our parameter could be $\tilde y = y$, or maybe $\tilde y = e^y-xyz$. The possibilities are infinite.

The traditional notation $\dfrac{\partial f}{\partial y}(1,2,3)$ hides step 1, taking for granted that the choice of restriction is obvious. This is the case in multivariable calculus, but often not the case in physics.

share|improve this answer
    
Where can I find more examples about this? I find it misleading that sometimes I need to consider variables as constants and sometimes not, misleading. I cannot understand why this is allowed, possible to do some mistake when assuming variable as constant? –  hhh Feb 11 '13 at 11:16
    
@hhh I added more explanation. I remember such discussions in analytical mechanics textbooks, but can't give a reference now. –  user53153 Feb 11 '13 at 21:18
up vote 0 down vote accepted

$g=t\dot x+x^2+\dot x^2$

Total derivative

$\frac{dg}{dt}=\dot x+2x\frac{dx}{dt}+2\dot x \frac{d\dot x}{dt}$

Partial derivative where we consider other variables as constants

$\frac{\partial g}{\partial t}=\dot x$

Hence for all $t$ and for all $x$ $$\frac{dg}{dt}\not = \frac{\partial g}{\partial t}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.