Is the following derivation of how to find $v$ given $a=v'$ wrong?

Question

My physics professor did the following:

Let $a(t)=v'(t)$ be a given function. Suppose $v(0)$ is known, then $$ \int_{v(0)}^{v(t)} dv=\int_0^ta(t)dt \iff v(t)=v(0)+\int_0^ta(t)dt $$

I believe this is incorrect in many ways:

1. The limits of integration (I don't understand why he got those!).
2. Having $t$ both as variable of integration and limit of integration.
3. "Multiplying by $dt$ on both sides"(??).

Could someone explain how to correctly get $v(t)$ from $v(0),a(t)$?

Answer 1

Integrating both sides of $a(t)=v'(t)$ with respect to tthe same variable $t$ and for the same interval $(0,x)$ we have:

$$ \int_0^x a(t)dt=\int_0^xv'(t) dt $$ and, by the fundamental theorem for the RHS; $$ \int_0^x a(t)dt=v(x)-v(0) $$

Answer 2

Little rant: sterilize this from the awful "$dv$" notation and it is clear what is going on.

The first equality states

$$\int_{v(0)}^{v(t)}1=\int_0^t a.$$

This clearly implies

$$v(t)-v(0)=\int_0^t a.$$

Hence,

$$v(t)=v(0)+\int_0^t a.$$

The other implication is also clear.

Your teacher probably just used the notation (also awful) that normal texts use for the change of variables. It bogs down to a "lifehack" for quick integral computations, but for conceptual purposes is not a good choice (in my opinion, of course).

Note also that what is written is somewhat ambiguous. The "iff" part is always true, regardless of whatever relationship $a$ and $v$ might have. So, if we read this as

"(...) then ($A \iff B$)",

we do not need the hypothesis for the thesis. However, he probably meant: "(...) then $A$, which happens to be equivalent to $B$".

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Summing up, what the text is trying to communicate is:

Let $v$ be a real function, and $a=v'$. Then

$$v(t)=v(0)+ \int_0 ^t a. $$

Note: The previous equality is equivalent to

$$\int_{v(0)}^{v(t)} 1=\int_0^t a.$$

Answer 3

Your professor isn't wrong per se, but his notation is not terribly precise, especially when teaching students who are new to calculus. Properly, the steps are the following.

Suppose we have $a(t) = v'(t)$. Now, let us integrate both sides with respect to $t$ from $0$ to some time $\tau$. Here $\tau$ is our time variable, which I've renamed from $t$ (which I will use as a dummy variable) to keep the two separate. This gives us $$\int_0^\tau a(t)\ dt = \int_0^\tau v'(t)\ dt.$$ Now consider the formula for integration by substitution $$\int^{g(b)}_{g(a)} f(t)\ dt = \int_a^b f(g(t))g'(t)\ dt.$$ We will take $f(t) = 1$, the constant function, and $g(t) = v(t)$. Then, taking the integration limits $a=0$ and $b=\tau$, the formula gives us $$\int_0^\tau v'(t)\ dt = \int_{v(0)}^{v(\tau)}\ dt.$$ And this is precisely the result your professor got. A few things to mention here:

1. First, you must realize that the integration variable we use is completely insignificant. It is a dummy variable, and we are free to rename it anything we want. In this example, I've chosen to use $t$ consistently, while your professor chose to rename the dummy variable to $v$ in $$\int_{v(0)}^{v(\tau)}\ dt = \int_{v(0)}^{v(\tau)}\ dv.$$ The two are completely equivalent, and in my experience, your professor's convention is in fact the common convention. The only source of confusion here is the fact that he chose to use $v$ both as the dummy variable and as the integration limits, which brings us to our second point.

2. The $t$ in the integration limit and the $t$ in the dummy variable of integration have nothing to do with each other. As I've stressed in point 1, the integration variable can be freely renamed without changing the value of the integral. It is good mathematical practice to keep the two separate just so confusion doesn't arise here. That's why I've chosen to rename my time variable as $\tau$ instead. But while using $t$ both as an integration limit and as an integration variable is confusion, it is not technically wrong. There is an unambiguous value that we can extract from the integral, and that's all we really care about.

3. Lastly, you mentioned about "multiplying by $dt$". This is not truly allowed of course, but it is formalized by integration by substitution. We can rewrite the integration by substitution formula in Leibniz notation $$\int_{g(a)}^{g(b)} f(g)\ dg = \int_a^b f(g(t))\frac{dg(t)}{dt}\ dt,$$ where I've renamed the dummy variable on the left-hand side to $g$ as an abuse of notation. This is very suggestive notation, and it seems to suggest that we've cancelled the $dt$ in $\frac{dg}{dt}\,dt$ to get just $dg$, but all that's really going on is integration by substitution. This is the source of the "multiply by $dt$" that you will very often see. There is nothing dubious going on, and I think you will find that this way of thinking is much quicker than going through the formal substitutions each and every time. The notation is suggestive, but completely justifiable.