Why is it possible to do this

  • Thread starter ronaldor9
  • Start date
  • #1
92
1

Main Question or Discussion Point

How is it possible that one can break up the derivative operator such as this:

[tex]\frac{dv}{dt}=t^2[/tex], then integrate like this,
[tex] \int^v_{v_{0}}dv = \int^t_0 t^2 dt [/tex], where[tex] v=v_{o} [/tex] when [tex]t=0[/tex]. Especially in light of what most calculus teachers tell you; that the derivative symbol is not a fraction and should not be interpreted as a faction?
 

Answers and Replies

  • #2
1,838
7
The derivative is a limit of a fraction and the integral is a limit of a summation of function values times a step length.
 
  • #3
1,013
65
dv/dt is a function v'(t), and since it is a derivative, it is integrable by the fundamental theorem of calculus to v(t) + C where C is an undetermined constant.
If f(t) = g(t) and f(t) is integrable, then [itex]\int f = (\int g) + C[/itex] where C is an undetermined constant. Your use of boundary conditions (v(0) = v0) allows you to determine the constant.
That's all that's being done here. The Leibnitz notation is just a neat way of writing it out, but don't take it too seriously until you learn the proper way to manipulate differential forms.
 
  • #4
92
1
wow thanks slider. I never have thought about it that way, but now that you have explained it, it is very interesting!
 
  • #5
This was discussed in another thread.

What you're really integrating on the right side is [tex] \int^v_{v_{0}} \frac {dv}{dt} dt [/tex]
 

Related Threads for: Why is it possible to do this

Replies
1
Views
3K
Replies
20
Views
4K
Replies
2
Views
2K
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
4
Views
634
  • Last Post
Replies
4
Views
1K
Top