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I've noticed in my physics textbook (and in a lot of other popular sources), that the process of integrating non-constant acceleration to get to a velocity formula, the integrating bounds imposed on the velocity-part seem wrong.

enter image description here

In the above snippet, they're using $v_0$ and $v$ as bounds in the definite integral on the left-hand side of 3-34. Why though? Why not use the (IMO) more appropriate bounds of $0$ and $t$? If I'm not mistaken, according to the fundamental theorem of calculus, that definite integral would evaluate to:

$$\int_{v_0}^{v}\mathrm{d}v_x(t) = \bigl[v_x(t)\bigr]_{v_0}^{v} = v_x(v)-v_x(v_0) \tag{1}$$

Which doesn't make much sense to me. However, using $t$ and $0$ as the bounds, we get:

$$\int_{0}^{t}\mathrm{d}v_x(t) = \bigl[v_x(t)\bigr]_{0}^{t} = v_x(t)-v_x(0) = v_x(t) - v_0 \, \, \text{(by definition of $v_x(0)=v_0$)}$$

As people way smarter than me all seem to write it as in (1), I'm probably the one who's wrong. But why?

And furthermore, why are we even integrating between $v$ and $v_0$? Shouldn't we be integrating between $t$ and $0$ as velocity is, in this scenario, derived from acceleration (and thus we're really just looking at the total area underneath an $a$-$t$ graph)?

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    $\begingroup$ Its because you are integrating over the differentials of velocity, so the bounds should be velocities. What you are saying (using time in the boundaries) should be used if you are integrating with respect to time (which is note the case). Also, you should not evaluate the velocity in the time parameter. The integral (1) should be something like $\int dv_x=v(t)-v_0(t)$ $\endgroup$ – Grego_gc Feb 17 at 14:53
  • $\begingroup$ @Grego_gc Right, but then that would be equivalent to $\int_{x_0}^{x} dx = \bigl[x \bigr]_{x_0}^{x}$, which I've been taught not to do in calculus, and instead to integrate over a different variable ($s$ for example) so that I don't end up with a situation where a variable becomes an input for itself (having to substitute $x$ into $x$ as above). $\endgroup$ – Ius Klesar Feb 17 at 14:59
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    $\begingroup$ There is nothing wrong with $ \int_{x_0}^{x} dx = \bigl[x \bigr]_{x_0}^{x}$, if you understand that it means $\int_{x_0}^{x_t} dx = \bigl[x \bigr]_{x_0}^{x_t}$, where $x_t$ is some specific x (in this case, at time t, so, it is $x(t)$ if you let the parameter t become a variable). $\endgroup$ – Wolphram jonny Feb 17 at 15:14
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    $\begingroup$ What @Wolphramjonny said above. I'll add that it's something maths teachers will generally tell you not to do (because it's a bit ambiguous to use the same symbol for two different meanings), but physicists do all the time (because it's a much cleaner notation and "everyone knows" what it really means). $\endgroup$ – Dronir Feb 18 at 8:13
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    $\begingroup$ @Dronir - The problem with "everybody knows what it really means" is that not everybody knows what it really means. I very, very regularly run into students who completely confuse themselves over this. And I disagree that it is in any way a "cleaner" notation. $\endgroup$ – Paul Sinclair Feb 18 at 17:36
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The other answers are correct in showing why what is being proposed in your text is correct, but I want to show you where your error is and how to easily fix it directly.

Your issue is that you are plugging velocities in for $t$ instead of $v_x(t)$ i.e. you have done

$$\int_{v_0}^{v}\mathrm{d}v_x(t) = \bigl[v_x(t)\bigr]_{t=v_0}^{t=v} = v_x(v)-v_x(v_0)$$

when we should have we should actually have

$$\int_{v_0}^{v}\mathrm{d}v_x(t) = \bigl[v_x(t)\bigr]_{v_x(t)=v_0}^{v_x(t)=v} = v-v_0$$

Putting the "function of $t$" part on the differential is confusing, but if you are going to do it that way, don't get mixed up :) You have made "$v_x(t)$" the entire "dummy variable". i.e. at this point you aren't really thinking of it as an actual function of $t$, since you are integrating with respect to $v_x(t)$. Usually the "function of $t$" part is dropped to avoid this confusion you are facing now.

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The fundamental theorem of calculus states: "For a continuous function $f:[a,b] \mapsto \mathbb{R}$

$$\int_a^b f(t) \, dt = F(b) - F(a)$$ where $F:[a,b] \mapsto \mathbb{R}$, differentiable with $F'(x) = f(x)$". The problem lies in equation (1). The integrand of that equation is a differential, not a derivative of some other function. If you want to be really rigorous, you can derive the result in your textbook in the following way:

By definition, $$a(t) = \frac{dv}{dt} = v'(t)$$ Integrating both sides on the interval $[0,t]$, we obtain: $$\int_0^tv'(t') \, dt' = \int_0^t \frac{dv(t')}{dt'} \, dt' = \int_0^ta(t') \, dt'$$ By the fundamental theorem of calculus, we transform the first hand side to: $$v(t)-v(0) = \int_0^ta(t') \, dt' \Longrightarrow v(t) = v(0) + \int_0^ta(t') \, dt'$$ or $$v(t) = v_0 + \int_0^ta(t') \, dt'$$ Now, a discussion on the use of differentials. Many authors choose such approaches and I remember being confused by them as well. I generally advise you to first formulate an equation with pure derivatives and then integrating with respect to the independent variable, as you suggested. But if you want to integrate differentials like $dv$, you do it by treating $v$ as an independent variable in the interval of interest (in this case, that would be [$v_0$, $v$]).

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    $\begingroup$ Small nitpick: You need continuity of $f$ for the fundamental theorem. $\endgroup$ – Jannik Pitt Feb 18 at 16:58
  • $\begingroup$ Thanks. I'll edit. $\endgroup$ – Andreas Mastronikolis Feb 18 at 21:24
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Notice that at time $t=0$ the velocity is $v_0$, and since there is only one independent variable (time) really, the integration starts at the same time for both sides of the equation.

Graphically I can explain this with the following timeline

pic1

where shaded is the integration range, showing how $t$, $x$ and $v$ vary with time.

So the integrations $\int \limits_0^t {\rm d}t$ for time,$\int \limits_{x_0}^x {\rm d}x$ for distance, and $\int \limits_{v_0}^v {\rm d}v$ for speed, are all over the same interval in time, covering the same events as they happen along the timeline.

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  • $\begingroup$ The problem is that since $x$ and $v$ are dependent variables, they may not be unique in the $\langle0;t\rangle$ interval, so anything expressed as depending on them (the implied $1(x)$ and $1(v)$) may not even be functions and therefore you can't integrate them in the strict sense. The symbolic manipulation with the differential still works thogh. $\endgroup$ – Jan Hudec Feb 18 at 6:53
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I think you are getting confused by notation. A cleaner way of writing 3-34 would be $$ \int_{v_0}^{v} d\tilde{v} = \int_{t=0}^{t} a(\tilde{t}) d\tilde{t} $$ then the LHS is $ [\tilde{v}]_{v_0}^{v}$ which does become $v-v_0$ (the tilde is just a a technique to distinguish between a variable and a parameter - a dummy variable).

It might seem odd (to me it did) to solve separable differential equations by making the bounds of the integrals the initial conditions instead of solving it as an indefinite integral and tweaking the resulting constant to fit the initial values.

Regardless, here is a resource I found helpful

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