# Why can you write a= dv/dt as a*dt = dv? and what does it mean?

1. Nov 11, 2014

### christian0710

If you write
a=dv/dt
Then you can write it as

dv=a*dt

And take the integral on both sides

integral(dv)=integral(a*dt)

My question is: What does dv=a*dt mean?? How do I interprete it? What does the individual term dt mean? And what does dv for it self mean?
To me it seems like “dv” means, differentiate t with respect to NOTHING. And dv means “Differentiate time with respect to nothing”

Can someone help me understand why this trick works and perhaps how to interprete it?

Last edited: Nov 11, 2014
2. Nov 11, 2014

### SteamKing

Staff Emeritus
If a = dv/dt, then a dt = dv, not dt = a dv. Always check your algebra.

3. Nov 11, 2014

### christian0710

Thank you, I changed it in the first post, but i still don't understand what "dv" in itself means and what "dt" means, does it meas "differentiate t with respect to nothing?" Why can you separate them like that?

4. Nov 11, 2014

### Vagn

dv and dt are infinitesimally small changes in v and t respectively. Revisiting the (geometric) definition of differentiation may help you grasp the justification.

5. Nov 11, 2014

### christian0710

Now i understand it !! Please reconfirm that this is correct? I'd really apreciate it :)

If deltav/deltat = slope=m is the same as the slope of a secant, m, then we must be able to find the slope of the secant as Δt*m=Δv so because this is true, then if delta t and delta v become infinitesimally small then we must also be able to find the slop of the tangent as dt*m=dv. Am i right?

Last edited: Nov 11, 2014
6. Nov 11, 2014

### christian0710

So is this also true?

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

dv/dt = infinitesmally small change in both 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 then change in the function value v is twice as large as the change in the time (geometrically speaking?

Last edited: Nov 11, 2014
7. Nov 11, 2014

### christian0710

Just to give an example to make sure I understood it 100% and also to proove that my argument that dv/dt can be seen as ratio (explanation follows) is true.

If v(t) = 3t^3 then dv/dt = 6t^2

Is it the true that dv/dt can be interpreted in the following way? Again help is truly appreciated :)

1) You can see that 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 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 v is always twice the change in t even as the delta t approaches zero.

You see, dv/dt = 6t tells us that every time time increases by one, dv/dt will change by a magnitude of 6, so the tangent dv/dt of velocity is in fact a RATIO, so why do people say, you can't treat it as a ration? You can isolate it like this dv=6tdt, which tells us, that dv must be 6*t times as big as dt, so every t (on the x-axis changes by t) then dv changes by 6t*dt, and if we divide them dv/dt we get a ratio of increasing magnitude = 6t nd this ratio of increasing magnitude is a function of acceleration.

Last edited: Nov 11, 2014
8. Nov 11, 2014

### Fredrik

Staff Emeritus
You're using the term "infinitesimal" in your arguments without defining the term. An argument that relies on an undefined term doesn't prove anything.

I'm guessing that you're a physics student. Physics books abuse that term in a horrible way. It's some sort of code for "what we're about to say involves a series expansion; we're only including terms up to some specific order, and we refuse to use notation that gives you a hint that something is being neglected". For example, if they say that $\sin x=x$ for "infinitesimal" x, what they really mean is that there's a sequence $(a_k)_{k=0}^\infty$ such that $\sin x=\sum_{k=0}^\infty a_k x^k$ for all $x\in\mathbb R$, and $a_0=0$, $a_1=1$.

In your example, no matter how small you make your change in t, if that change is greater than 0, the corresponding change "on the y axis" (I would call it the v axis) isn't equal to $6t^2\Delta t$ (or $9t^2\Delta t$, which is what you would have gotten if you had done the derivative correctly). $9t^2\Delta t$ is just what you get from the first-order approximation $v(t+\Delta t)\approx v(t)+\Delta t\, v'(t)$.
$$v(t+\Delta t)-v(t)\approx (3t^3+\Delta t 9t^2)-3t^3 =\Delta t 9t^2.$$ If v is interpreted as velocity, this gives you an approximation of the average velocity during the time from $t$ to $t+\Delta t$.
$$\frac{v(t+\Delta t)-v(t)}{\Delta t}\approx \frac{(3t^3+\Delta t 9t^2)-3t^3}{\Delta t} =9t^2.$$ This approximate result happens to be exactly equal to the velocity at t, which is defined as
$$\lim_{\Delta t\to 0}\frac{v(t+\Delta t)-v(t)}{\Delta t}.$$ The reason is that all the terms that were neglected in the approximation contain a factor of the form $(\Delta t)^n$ with $n\geq 2$. So when we divide by $\Delta t$, the contribution from the neglected terms is a sum of terms that all include at least one factor of $\Delta t$, and because of this, they all go to zero in the limit $\Delta t\to 0$.
$$\lim_{\Delta t\to 0}\frac{v(t+\Delta t)-v(t)}{\Delta t} =\lim_{\Delta t\to 0}\left(9t^2+\Delta t\left(\text{something that doesn't grow as t gets smaller}\right)\right)=9t^2.$$

9. Nov 11, 2014

### mathwonk

it means that if t and v are both functions of anything else like s, then a dt/ds = dv/ds.

If you also want to know what dt and dv are alone, they are functions on tangent vectors. But you don't need to know what that mens to use this rule. Actually it just means what I said first.

10. Nov 11, 2014

### WWGD

Why don't you look at a as acceleration for an interpretation?

11. Nov 12, 2014

### Simon Bridge

Relations like dv = a.dt don't make a lot of strictly logical sense by themselves, but are used as a shorthand to record the relationship.
Think of it as slang.

Treating the leibnitz notation for the derivative as a fraction is often seen a an abuse of the notation - but it is a popular one.
Physicists like to do this as a kind of shorthand and the hand-wave to a progression like:
$\Delta x$ is a change in x
$\delta x$ is a small change in x - because of the "small delta", get it?
... notice that "small" is poorly defined, and relies on context? It's informal.

$\text{d}x$ is an (also informally) "infinitesimal" change in x - and if someone demands a definition of that term, refer them to a dictionary.

Mathematicians, OTOH, like to be more rigorous.
The rigorous mathematical definition relates to limits - like we see above :)