# Dv/dt must be a RATIO and here is why

1. Nov 11, 2014

### christian0710

Why do math professors always say that dy/dt is NOT a RATIO?

I’ve sat down and really analyzed it and I can only conclude that dy/dt is a ratio between the function delta value y and the corresponding delta t at ANY point on a continuous function.

Here is why: (This is a bit of a mix of topics I’ve posted which now make sense)

Δv = change in v = v(t2) - v(t1)
Δt = change in time = t2-t1

dv = infinitesemally small change in v
dt = infinitesimalle small change in t

If Δv/ Δt = 2, then this is a ratio between a chunk of function value and a corresponding chunk of time interval, It tells us that “At this specific interval some function the ratio between function value and time is 2”,

Now the term.

dv/dt = infinitesmally small change in both v and t as deltat approaches zero, which gives us the tangent to some point of the graph of v

Therefore if
dv/dt =1

then the change in dv is equal to the change in dt

dv=dt

But if dv/dt = 2

dv=2*dt

Then at any point of the graph the small interval dv is 2 times as large as the corresponding interval dt.

if dv/dt= 6t^2 then it must be true that dv=6t^2*dt, therefore you can interprete it like this: For the function v(t) you can see that an infitesimally small change in v (on the y-axis) is equal to 6*t^2 times the infinitely small change change in t (on the x-axis), so graphically when you divide the function V(t) up into intervals, the delta t (on the t axis) will correspond to a delta v (on the y-axis) whose vale is about 6*t^2 times larger in value than the delta t value? And this is true for ANY value of t, so for the complete function, so as the function progresses (as t increases) the change in t becomes much smaller compared to the change in v. This means the slope of the tangent increases. So Actually dv/dt can somehow be interpreted like a ratio because if dv/dt = 2, then the change in dv is always twice the change in dt at any given interval on the function.

Just to give an example to make sure I understood it 100%:

2. Nov 11, 2014

### lavinia

When you say "infinitesmally small change",what do you mean? What number is dt?

3. Nov 11, 2014

### Dr.D

The problem is that, if you take the limits of the numerator and the denominator separately, you wind up with a 0/0 situation. That is not defined.

4. Nov 11, 2014

### PeroK

It's good to think these things through for yourself, but what you are missing is the concept of a "limit".

$\frac{dy}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\Delta y}{\Delta t}$

5. Nov 11, 2014

### gopher_p

To say that $\frac{\mathrm{d}y}{\mathrm{d}x}$ is a ratio is to imply that $\mathrm{d}y$ and $\mathrm{d}y$ are members of a kind of algebraic structure called a division ring. Now that could be $\mathbb{R}$; in fact when modern calculus texts use so-called differentials to write $\mathrm{d}y=f'(x)\mathrm{d}x$, the differentials are assumed to be real numbers. But there are no infinitesimals in $\mathbb{R}$, so in order to say that $\mathrm{d}y$ and $\mathrm{d}x$ are "infinitesimals", you would need to stipulate the structure (which would necessarily have a "copy" of $\mathbb{R}$ sititing inside) in which you are working.

Now the way in which calculus is most commonly taught does not use infinitesimals; it uses limits. When your teachers tell you that $\frac{\mathrm{d}y}{\mathrm{d}x}$ is not a ratio, they do not mean that there is no way that it could be understood to be a ratio. What they mean is that notation as they are using it does not represent a ratio. They are giving you information about what that collection of symbols means (or, rather, doesn't mean) in their classrooms. In most calculus classrooms, $\frac{\mathrm{d}y}{\mathrm{d}x}$ is simply an alternative notation for a derivative and nothing more.

Just to give you an idea of what I'm saying, the notation $\sin^nx$ is used as an alternate notation for $(\sin x)^n$ when $n$ is a positive integer; i.e. for $n=1,2,3,\ldots$. It is reasonable for a student to attempt to extend this to negative integers and say, for instance, that $\sin^{-1}x$ should mean the same thing as $(\sin x)^{-1}=\frac{1}{\sin x}$; and maybe it should. However most math teachers will tell you that $\sin^{-1}$ denotes the inverse function of $\sin$, and that is the standard interpretation of that batch of symbols. Now, you are free to disagree with the standard meaning of $\sin^{-1}x$, but you'll be "wrong" because you're talking about something different than what everyone else is.

There are times (e.g. in many differential equations classes) where $\frac{\mathrm{d}y}{\mathrm{d}x}$ is treated as though it were a ratio. I would say that's more of a case where an abuse of notation is permitted because (a) it leads to the correct answer and (b) the alternative is rather cumbersome. There are some situations where similar abuses of notation can get you into trouble, though. For instance, treating a partial derivatives such as $\frac{\partial f}{\partial x}$ like a fraction will often lead you to an incorrect conclusion.

To reiterate my main point; $\frac{\mathrm{d}y}{\mathrm{d}x}$ is just notation. What that notation means depends on context. What your teachers are telling you is that, in the context of a standard calculus course, $\frac{\mathrm{d}y}{\mathrm{d}x}$ is not intended to be treated like a fraction. If you treat it like a fraction, you are not talking about the same thing as your teachers.

6. Nov 11, 2014

### Fredrik

Staff Emeritus
You got some very good answers already. I just want to add that $\Delta x/\Delta t$ is by definition the average velocity for the part of the journey that took place between $t_1$ to $t_2$. If you want the velocity at $t_1$ instead of some average velocity, you have to take the limit $t_2\to t_1$ (or equivalently $\Delta t\to 0$), and then we are no longer talking about a ratio.

I wouldn't say that this is how you calculate the velocity at $t_1$. Rather, I'd say that this is the definition of what we mean by "velocity at $t_1$". The average velocity from $t_1$ to $t_2$ is defined as $\Delta x/\Delta t$. The velocity at $t_1$ is defined as $\lim_{\Delta t\to 0} \Delta x/\Delta t$.
$$v(t_1)=x'(t_1)=\lim_{t_2\to t_1}\frac{x(t_2)-x(t_1)}{t_2-t_1} =\lim_{\Delta t \to 0}\frac{x(t_1+\Delta t)-x(t_1)}{\Delta t}.$$ The notation $dx/dt$ is simply misleading. It's better to use $x'(t)$. The only way I know to make sense of dx and dt separately is to define dx as a function that takes two real numbers as input: $dx(t,h)=x'(t)h$. Now for all real numbers $dt$, we have $dx(t,dt)=x'(t)dt$. To write the left-hand side as $dx$ is an abuse of notation, but it's OK if you know what you're doing.

7. Nov 12, 2014

### christian0710

Gopher_ p

Hi Gopher_P, thank you for some good answers. I just want to make sure I understand what you write: So If dy and dx are memebers of the division ring, then they are real numbers, and therefore you can write them as dy=f’(x)dx? But if they are NOT you should NEVER separate dy/dx into fx dy=dx? Do you know why physics teachers do it? Is it just because it’s easier? I think I might not fully grasp the idea of a limit. I know the geometric interpretation: we take deltay /deltax and let the delta x go toward zero, so dy/dx is the LIMIT when delta x goes toward zero in deltay/deltax. So what is actually meant the limit? I never really understood what it meant.

8. Nov 12, 2014

### christian0710

Hi Frederik,
Thank you for your help. So it’s only okay to abuse the dy/dx notatation if we define dx as a function who takes two real numbers as input, so in the case of of velocity and acceleration a*dt=dv would you then say dv is a function of (dt,a)?
The reason I ask is because my Physics book writes it out like this, so I really want to understand why and when it’s okay to master the math. I added an two images of two ways in which my physics abuses the dv/dt notation, and I really wan’t to understand why they can do it. Could I ask you to take a quick look at the two images?
If the first image he arranges a=dv/dt as a*dt=dv, and i just don't know how to interprete what "dv" standing for itself means, but maybe as you wrote it's a function of (dt,a)? and why can you integrate dv as it stands for itself?

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9. Nov 12, 2014

### zoki85

Probably becouse infinitesimals are not numbers

10. Nov 12, 2014

### Fredrik

Staff Emeritus
The book says something like this in the first picture: Since $a=dv/dt$, we have $dv=a\,dt$ and therefore
$$\int_{v_1}^{v_2}dv =\int_{t_1}^{t_2} a\, dt.$$ The result is correct, but the argument is pretty naive. I will show you both the non-rigorous justification for this that you'd get from a physicist, and the proof that you'd get from a mathematician.

The non-rigorous argument goes like this. When $\Delta t$ is small but positive, $a$ is approximately equal to $\Delta v/\Delta t$. So we can write $\Delta v\approx a\,\Delta t$. The following approximate equality between Riemann sums is explained in a similar way:
$$\sum_{i=1}^n (\Delta v)_i \approx \sum_{i=1}^n a_i\, (\Delta t)_i.$$ If the $(\Delta v)_i$ are small, the left-hand side is approximately equal to $\int_{t_1}^{t_2} dv$ (by definition of the Riemann integral). If the $(\Delta t)_i$ are small, the right-hand side is approximately equal to $\int_{t_1}^{t_2}a\,dt$. When we let the $(\Delta t)_i$ go to zero, the errors in the approximations should go to zero, so we should end up with $\int_{v_1}^{v_2} dv =\int_{t_1}^{t_2} a\, dt.$

A mathematician would tell you that the final result follows from the chain rule and the fundamental theorem of calculus, if we define $v_1=v(t_1)$ and $v_2=v(t_2)$:
\begin{align}
&\int_{t_1}^{t_2} f'(v(t))v'(t)\mathrm dt =\int_{t_1}^{t_2} (f\circ v)'(t)\mathrm dt =(f\circ v)(t_2)-(f\circ v)(t_1) =f(v(t_2))-f(v(t_1))\\
&=f(v_2)-f(v_1) =\int_{v_1}^{v_2} f'(v)\mathrm dv.
\end{align} Your result is the special case that we get if we take $f$ to be the identity function, i.e. the function defined by $f(x)=x$ for all $x\in\mathbb R$, so that $f'(x)=1$ for all $x\in\mathbb R$. We get
$$\int_{t_1}^{t_2} v'(t)\mathrm dt =\int_{v_1}^{v_2} \mathrm dv.$$
Note that the proof doesn't involve dv and dt separately. It just involves functions, derivatives, integrals, and two of the most important theorems about them.

11. Nov 12, 2014

### HallsofIvy

It's not a good idea to try to learn the "fundamentals" of mathematics from physics teachers!
Yes, this is a "short cut" language. dy/dx, as defined in standard Calculus classes, is NOT a fraction- but it is defined as the limit of a fraction. It is always possible to go back "before" the limit, use the fraction properties of the "difference quotient" and then take the limit again to show the fraction "properties" can be used for the derivative.

Actually, it is possible, through what is called "non-standard analysis" to define "infinitesmals" and then define the derivative, df/dx, as the "infinitesimal", "df", divided by the infinitesimal "dx". But "non-standard analysis" requires a lot of very abstract "logic" before it can be rigorously defined.

Last edited by a moderator: Nov 12, 2014
12. Nov 12, 2014

### gopher_p

http://en.wikipedia.org/wiki/Division_ring

The set of real numbers $\mathbb{R}$ is a division ring. So is the set of rational numbers $\mathbb{Q}$. The set of integers $\mathbb{N}$ is a ring that is not a division ring.

The reason I brought it up is that you were talking about these things called "infinitesimals" and assuming several properties without saying what they were. In particular, you were assuming that you could divide one infinitesimal by another. What is an infinitesimal? How do infinitesimals "interact" with real numbers? Is the ratio of two infinitesimals always a real number? If not always, then when?

I'm asking you these rhetorical questions because questions like these led to mathematicians basically abandoning the idea of infinitesimals in favor of limits. More recently (i.e. in the middle of the 20th century) people began to figure out some of the answers to the questions about infinitesimals, however they still remain part of nonstandard mathematics.

I am not a physics teacher, and I don't really know any. So I can't speak to why some insist on using infinitesimals. I will say that everything that I've seen a physicist do that looks like infinitesimal work is really just lazy limits from my perspective. For instance, when a physicist says something like
it's really just a shorthand for the "standard" Riemann sum set up; i.e. $$\Delta W_k\approx F(x_k)\Delta x_k$$ which gives $$W=\sum_{k=1}^n\Delta W_k\approx\sum_{k=1}^nF(x_k)\Delta x_k$$ which leads to $$W=\lim_{n\rightarrow\infty}\sum_{k=1}^nF(x_k)\Delta x_k=\int_{x_0}^{x_f}F(x)\mathrm{d}x$$ There's nothing inherently wrong with cutting out the Riemann sum "middleman" (at least for real-world problems) as long as you realize that's all you're doing.

$\lim\limits_{x\rightarrow a}f(x)=L$ means that, no matter how you define near to, $f(x)$ is near to $L$ provided $x$ is close enough to $a$. The $\epsilon$-$\delta$ definition makes the notions of near to and close enough more rigorous, but the basic idea is the same. I reckon the actual meaning of the limit matches the intuition you have that you think you need infinitesimals to express. In fact, the intuition that people think they have about "infinitesimals" is actually based on the very real notion of finite and extremely small, which is precisely what limits are actually talking about.

13. Nov 12, 2014

### Stephen Tashi

There are ways of remembering mathematical facts that are nonsensical - for example chanting "soak-ah toe-ah" to remember how the trig functions relate to ratios of the sides of a triangle. There are ways of remembering mathematical facts that are less nonsensical and more systematic. Thinking of the operations in calculus as operations on "infinitesimal numbers" makes it easy to remember some facts about calculus although it may lead to confusion about others. Some people develop the skill of reasoning about problems in physics by conceptualizing infinitesimal amounts. Historically, this type of thinking preceeded the modern definition of limit.

14. Nov 22, 2014

### MayCaesar

On a side note, I've always disliked this notation ($\frac{dx}{dt}$) since, first, it looks like $d$ is also a variable and, second, it, well, looks like a ratio which technically it is not. $x\prime_t$ looks much more clear. I think Feynman also complained about it (as well as about $\sin$ which looks like $s$ times $i$ times $n$). It is especially bad when one denotes a variable as $d$ and it may look like $\frac{dN}{dd}$.
Math is full of such awkward things. Which makes it both more fun and more frustrating. :)