Taking Derivatives in Physics

  • #1
278
0

Main Question or Discussion Point

When I take derivatives in math I think of it as the amount of infinitesimals that change in one variable with respect to another, when the latter changes by one infinitesimal. But in physics those variables have real life meanings, so when I take the derivative of position with respect to time I feel like I am asking myself how many infinitesimals of time change in distance. Which is kind of weird.

Is that because all infinitesimals are the same? Like: dx=dt=dy=dz? And even if that's not the explanation for this particular question; are they all the same? Are the infinitesimals of all variables basically 1/∞?

Anyway; or is it that some mathematical equation just happens to explain a physical occurrence? Like, we just assign real meanings to abstract variables and the derivatives are actually speaking of the variables, and only as a consequence about the physical meaning we assign to them?
 

Answers and Replies

  • #2
Khashishi
Science Advisor
2,815
493
Derivative in physics means the same thing as derivative in math. It's the rate of change of something with respect to something else. Are you familiar with the definition of a limit?
"d" doesn't mean anything by itself. There must be at least 2 "d"s in an equation for it to mean anything, because "d" is really a device for comparing how two variables are changing with respect to each other.
dx = 2dy means that anytime you change y by a small amount, x changes by twice that amount. Infinitesimals like dx and dy are not the same. It only makes sense to equate them with other infinitesimals (or expressions containing infinitesimals) because they aren't real numbers.
 
  • #3
5,601
39
Same for physics:

In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's instantaneous velocity. The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value.
http://en.wikipedia.org/wiki/Derivative

[see the discussion there on differentiation]

Is that because all infinitesimals are the same? Like: dx=dt=dy=dz?
no; dx = dt only when dx/dt = 1.....

we can assign real physical meaning to variables and their derivatives....if x is a distance
measure, then dx/dt is the velocity [change in distance with respect to x]....and also
dx/dt = dx/dy times dy/dt
 
  • #4
256bits
Gold Member
3,114
1,133
Is that because all infinitesimals are the same? Like: dx=dt=dy=dz? And even if that's not the explanation for this particular question; are they all the same? Are the infinitesimals of all variables basically 1/∞?
I suspect you are mixing up Delta x with dx since you speak of infinitesmals.

An expression such as Delta y / Delta x = 2, would be more applicable to your inquiry about infinitesmals. Here, Delta y or Delta x has a certain size, value and can be made smaller and smaller until infinesimally small when the limit is reached whereDelta y or Delta x approach zero. For this expression Delta y will always be twice as large as Delta x. As the limit appraoches zero, you will recall, the slope of a graph at that certain point with which you are interested in will be the slope of a line tangent to the curve at that point.

On the other hand when you are speaking of dy/dx you are really expressing the slope of the tangent line as it follows the curve of your initial expression of which you have taken the derivative. dy/dx is not the same as Delta y / Delta x, but rather the change in x versus the change in y , and that can be calculated for values of x for a function y = f(x) throughout the whole curve.

For example, for a simple expression y = 2x, with derivative dy = 2 dx, or dy/dx = 2 , the slope, dy/dx, along the curve is always 2. and dx or dy on there own do not have a real value as mentioned in a previous post.

Until you encounter more indepth calculus this brief explanation may suffice for your immediate purposes.
 
Last edited:

Related Threads on Taking Derivatives in Physics

  • Last Post
Replies
1
Views
2K
Replies
5
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
15
Views
25K
Replies
16
Views
4K
  • Last Post
Replies
1
Views
1K
Replies
5
Views
14K
Replies
84
Views
5K
Replies
13
Views
3K
Replies
4
Views
802
Top