How do you integrate a term with a differential?

  • #1

Main Question or Discussion Point

In my Halliday Resnick Walker book, they derive the equations of kinematics using integrals in a small section after the algebraic derivation. I'm extremely confused by the calculus, however. Sorry if my questions are unclear.


They say that [tex]\frac{dx}{dt}=v⇒dx=v\cdot dt[/tex]. This first step itself confuses me. Are [itex]dx[/itex] and [itex]dy[/itex] supposed to be infinitesmally small values? If so, are they constants? I thought that this is just a notation that shows the derivative of position with respect to time, but is this for a specific time [itex]t[/itex] or does it represent some property that holds for all times?

Second, they integrate each side of [itex]dx=v\cdot dt[/itex], and end up with [itex]\int dx=\int v \: dt[/itex], with [itex]dt[/itex] as the integral term of the RHS. That made no sense to me; how does [itex]dt[/itex] go from being a constant to an integral term simply representing what variable you're integrating with respect to?


Again, sorry if my question is unclear.

Thanks in advance!
 
Last edited by a moderator:

Answers and Replies

  • #2
33,072
4,773
In my Halliday Resnick Walker book, they derive the equations of kinematics using integrals in a small section after the algebraic derivation. I'm extremely confused by the calculus, however. Sorry if my questions are unclear.


They say that [tex]\frac{dx}{dt}=v⇒dx=v\cdot dt[/tex]. This first step itself confuses me. Are [itex]dx[/itex] and [itex]dy[/itex] supposed to be infinitesmally small values? If so, are they constants?
No and no. dx and dt are differentials.
Vardaan Bhat said:
I thought that this is just a notation that shows the derivative of position with respect to time, but is this for a specific time [itex]t[/itex] or does it represent some property that holds for all times?

Second, they integrate each side of [itex]dx=v\cdot dt[/itex], and end up with [itex]\int dx=\int v \: dt[/itex], with [itex]dt[/itex] as the integral term of the RHS. That made no sense to me; how does [itex]dt[/itex] go from being a constant to an integral term simply representing what variable you're integrating with respect to?
As already said, neither of these is a constant. Both dx and dt indicate what the variable of integration is.

On the left side, you get x + C1. On the right side, you get vt + C2 (assuming v is a constant).
It's customary to have only one arbitrary constant, so after integration, you would have x = vt + C.
Vardaan Bhat said:
Again, sorry if my question is unclear.

Thanks in advance!
 
  • #3
No and no. dx and dt are differentials.
As already said, neither of these is a constant. Both dx and dt indicate what the variable of integration is.

On the left side, you get x + C1. On the right side, you get vt + C2 (assuming v is a constant).
It's customary to have only one arbitrary constant, so after integration, you would have x = vt + C.
Sorry, I'm still a little confused. I thought a differential is an infinitesmally small value.
 
  • #4
Specifically,

Cauchy, following d'Alembert, inverted the logical order of Leibniz and his successors: the derivative itself became the fundamental object, defined as a limit of difference quotients, and the differentials were then defined in terms of it. That is, one was free to define the differential dy by an expression

a40ea9db3d3d4907df2d90d281c7944a.png

in which dy and dx are simply new variables taking finite real values, not fixed infinitesimals as they had been for Leibniz.

So how does dy change from being a finite real value to the integral term?
 
  • #5
33,072
4,773
Sorry, I'm still a little confused. I thought a differential is an infinitesmally small value.
In an integral such as ##\int f(t)dt##, the role dt plays is to indicate what the variable of integration is.

In another context, suppose y = f(t). Then dy = f '(t)dt, by definition. Here dt is an infinitesimal, and the differential of y (dy) is f '(t) times dt.
 
  • #6
Okay, but then how does the infinitesmal turn into the term of an integral (how does it turn from a constant into something else)?
 
  • #7
33,072
4,773
Okay, but then how does the infinitesmal turn into the term of an integral (how does it turn from a constant into something else)?
In neither case is the differential a constant.

When used in an integral, as in ##\int f(t) dt##, you don't actually multiply anything by dt. To evaluate this integral, you look for a function F such that F' = f. It serves no purpose at all to think of dt in the context of an integral as being a differential.
 
  • #8
In neither case is the differential a constant.

When used in an integral, as in ##\int f(t) dt##, you don't actually multiply anything by dt. To evaluate this integral, you look for a function F such that F' = f. It serves no purpose at all to think of dt in the context of an integral as being a differential.
But in the differential equation where you multiply by dt, isn't it a differential? How does it magically turn into the integral term?

Sorry, I'm still getting used to all of this weird notation.
 
  • #9
33,072
4,773
But in the differential equation where you multiply by dt, isn't it a differential?
Yes, but what I'm saying is that the context is different between the equation involving differentials and an integral expression.
You're fine with ##y = f(t) \Rightarrow \frac{dy}{dt} = f'(t)##, right?
When you're first learning about the derivative in calculus, many textbooks take pains to tell you not to treat ##\frac{dy}{dt}## as a fraction. However, there's no harm in doing so, as the equation above can be rewritten in a way that defines the differential of y as ##dy = f'(t)dt##.

Differentiation if more-or-less the inverse operation of antidifferentiation, so from ##dy = f'(t)dt##, it's an easy step to go to ##\int dy = \int f'(t)dt \Rightarrow y = f(t) + C##

I don't know what more I can say.
Vardaan Bhat said:
How does it magically turn into the integral term?

Sorry, I'm still getting used to all of this weird notation.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Sorry, I'm still a little confused. I thought a differential is an infinitesmally small value.
Do you know what an "infinitesimally small value" is? That is a rather advanced topic in the foundations of mathematics and, from what you say, you don't appear to be that advanced.
 
  • #11
18
1
They aren't values neither fractions but these notations are treated as fractions. Remember Chain Rule? Yup.
Others explained the more advanced stuff already. I'm only a High Schooler..
 
  • #12
Do you know what an "infinitesimally small value" is? That is a rather advanced topic in the foundations of mathematics and, from what you say, you don't appear to be that advanced.
Sort of. They're numbers so close to 0 that they're virtually the same thing. Essentially it's what happens as [itex]x_1-x_0 \rightarrow 0[/itex]
 
  • #13
It can also be expressed as [itex]\frac{1}{\infty}[/itex]
 
  • #14
HallsofIvy
Science Advisor
Homework Helper
41,770
911
No, there are no such things. In any number system other than those used in "non-standard analysis", [itex]\frac{1}{\infty}= 0[/itex] and there are no "infinitesmals". While it is "convenient" to think of the "dy" and "dx" in dy/dx as separate quantities, and call them "infinitesmals", to prove things with them requires knowing "non-standard analysis", as I said rather an advanced field and certainly not taught in High School. In "standard" Calculus, the derivative, dy/dx, is first defined as a limit, not a fraction, then define the functions of x and y, not numbers, and certainly not "infinitesmals", such that "dy= (dy/dx) dx". I am just saying that you should use the word "differentials", not "infinitesmals".

(When Newton and Leibniz first developed the Calculus, they did it in terms of "infinitesmals" which y were never able to properly define. The Bishop Berkeley famously satirized them as "ghosts of departed quantities". Much later Cauchy and others developed Calculus using the "limit" procedure commonly used today. Abraham Robinson published a paper in 1961 and a text book in 1966, both titled "non-standard analysis", setting "infinitesmals" on a rigorous logical basis. However, to do so he had to use deep results from symbolic logic, in particular the "compactness property", that "if every finite subset of a set of axioms has a model then so does the entire set." Unless you understand what that means I recommend avoiding that path.)
 
Last edited by a moderator:
  • #15
Well differentials are just extremely small quantities, no? That's what I've been taught. I've been taught differentiation and integration using the standard limit-based method...
 
  • #16
HallsofIvy
Science Advisor
Homework Helper
41,770
911
It looks to me like you are remembering what you were taught not quite correctly. In Thomas' "Calculus" (the edition revised by Weir et. al.. Thomas, of course, passed away a few years ago. I actually had Thomas as my instructor in my first Calculus class. Yes, I really am that old!) a "differential" is defined by "Let y= f(x) be a differentiable function. The differential dx is an independent variable. The differential dy is dy= f'(x)dx. Notice that neither dx nor dy is a number at all. They are variables.
 
  • #17
It looks to me like you are remembering what you were taught not quite correctly. In Thomas' "Calculus" (the edition revised by Weir et. al.. Thomas, of course, passed away a few years ago. I actually had Thomas as my instructor in my first Calculus class. Yes, I really am that old!) a "differential" is defined by "Let y= f(x) be a differentiable function. The differential dx is an independent variable. The differential dy is dy= f'(x)dx. Notice that neither dx nor dy is a number at all. They are variables.
If we define [tex]f'(x)=\lim_{h\to\ 0}\frac{f(x+h)-f(x)}{h}[/tex] is [itex]h[/itex] the same thing as [itex]dx[/itex] and [itex]f(x+h)-f(x)[/itex] the same thing as [itex]dy[/itex]?
 
  • #18
33,072
4,773
If we define [tex]f'(x)=\lim_{h\to\ 0}\frac{f(x+h)-f(x)}{h}[/tex] is [itex]h[/itex] the same thing as [itex]dx[/itex] and [itex]f(x+h)-f(x)[/itex] the same thing as [itex]dy[/itex]?
No to both. f(x + h) - f(x) is the same as ##\Delta y##, which is the change in y between the points (x, f(x)) and (x + h, f(x + h)).
 
  • #19
HallsofIvy
Science Advisor
Homework Helper
41,770
911
If we define [tex]f'(x)=\lim_{h\to\ 0}\frac{f(x+h)-f(x)}{h}[/tex] is [itex]h[/itex] the same thing as [itex]dx[/itex] and [itex]f(x+h)-f(x)[/itex] the same thing as [itex]dy[/itex]?
No, if that were true you would be saying "[itex]f'(x)= \lim_{h\to 0}\frac{dy}{dx}[/itex]" and that I not true- f'(x) and [itex]\frac{dy}{dx}[/itex] are different ways of writing the same thing. Are you clear on what "[itex]\lim_{h\to 0}[/itex]" means?
 
  • #20
mathwonk
Science Advisor
Homework Helper
10,761
932
x and y are variables on the domain of the function, while dx and dy are variables on tangent space to the domain. thus at a given point x, the value of dx at a tangent vector v is equal to the x component of v. i.e. dx measures how much the tangent line to the graph changes in the direction. but i am celebrating. maybe this is wrong.
 
  • #21
No, if that were true you would be saying "[itex]f'(x)= \lim_{h\to 0}\frac{dy}{dx}[/itex]" and that I not true- f'(x) and [itex]\frac{dy}{dx}[/itex] are different ways of writing the same thing. Are you clear on what "[itex]\lim_{h\to 0}[/itex]" means?
Well, [itex]\lim_{x\to k} f(x)=L[/itex] means that for every [itex]\epsilon[/itex] such that [itex]|f(x)-L|<\epsilon[/itex], there is some [itex]\delta[/itex] such that [itex]0<|x-k|<\delta[/itex]. [itex]\lim_{h\to 0} f(h)=L[/itex] means that as [itex]h[/itex] is getting closer and closer to [itex]0[/itex], [itex]f(h)[/itex] is getting closer and closer to [itex]L[/itex].
 
  • #22
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Very good! So you realize that saying that [itex]A= \lim_{h\to 0} B(h)[/itex] is NOT the same as "A= B".
 
Last edited by a moderator:
  • #23
x and y are variables on the domain of the function, while dx and dy are variables on tangent space to the domain. thus at a given point x, the value of dx at a tangent vector v is equal to the x component of v. i.e. dx measures how much the tangent line to the graph changes in the direction. but i am celebrating. maybe this is wrong.
This made absolutely no sense to me...
 
  • #24
Very good! So you realize that saying that [itex] \lim_{h\to 0} B(h)[/itex] is NOT the same as "A= B".
So what's the next step to understanding what a differential is (dx and dy)?
 
  • #25
Because I don't understand how they freely multiply by dt and treat dx/dt as a fraction.
 

Related Threads for: How do you integrate a term with a differential?

Replies
9
Views
33K
  • Last Post
Replies
1
Views
893
Replies
7
Views
2K
Replies
13
Views
764
Replies
7
Views
2K
Replies
7
Views
3K
  • Last Post
Replies
2
Views
1K
Top