christian0710 said:
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.
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##.
$$\lim_{\Delta t\to 0}\frac{v(t+\Delta t)-v(t)}{\Delta t} =\lim_{\Delta t\to 0}\left(9t^2+\Delta t\left(\text{something that doesn't grow as $t$ gets smaller}\right)\right)=9t^2.$$
