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Let $x(t)$ and $a$ be in unit meters, and $t$ unit time. If we look at the indefinite integral $$ \int\frac{\frac{dx}{dt}}{x(t) + a} dx $$ then we note that the units of the above expression are $\frac{meters}{seconds}$. If we evaluate the integral, we get $$ \frac{dx}{dt}\log(a + x(t)) $$ But the log has units meters in its argument, which, as I understand, transcendental functions' arguments should be dimensionless. Changing the indefinite integral into a definite one does not necessarily fix this, because, as I understand $$ \int_{x=x_0}^{x=x_1}\frac{\frac{dx}{dt}}{x(t) + a} dx = \frac{dx}{dt}\log(a + x(t))|^{x=x_1}_{x=x_0} = \frac{dx}{dt}(x=x_1)\log(a+x_1) - \frac{dx}{dt}(x=x_0)\log(a+x_0) $$ so what gives?

Striker
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  • Where did this equation came from? Maybe the issue is with the problem setup. But if $dx/dt$ is constant, than the difference of logarithms yield a logarithm of a dimensionless number (the ratio $\frac{a+x_1}{a+x_0}$). – Andrei Aug 25 '18 at 16:39
  • @Andrei It came up while trying to find the potential energy in a spring at a point on it. x(t) is the position of the end of the spring oscillating about equilibrium. So dx/dt varies with time, but if you're evaluating at a certain position of the spring, dx/dt varies as well. In short, I don't think dx/dt is constant. – Striker Aug 25 '18 at 16:47
  • @AccidentalFourierTransform It's not. That can be solved by taking a definite integral. This cannot. – Striker Aug 25 '18 at 16:47
  • Potential energy does not usually depend on the velocity. I think you might have done something wrong. – Andrei Aug 25 '18 at 16:48
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    $$\int\frac{\frac{dx}{dt}}{x(t) + a} dx\neq\frac{dx}{dt}\log(a + x(t))$$ – AccidentalFourierTransform Aug 25 '18 at 16:49
  • @AccidentalFourierTransform Explain. If I take d/dx of the rhs then I get the integrand. – Striker Aug 25 '18 at 16:50
  • @Andrei Regardless, I think the above is a well-posed enough question to approach. – Striker Aug 25 '18 at 16:52
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    @Striker No, you don't. $$\frac{\mathrm d}{\mathrm dx}\left[\frac{dx}{dt}\log(a + x(t))\right]=\left[\frac{\mathrm d}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}\right]\log(a+x(t))+\frac{\mathrm dx}{\mathrm dt}\frac{1}{x(t)+a}$$ – AccidentalFourierTransform Aug 25 '18 at 16:53
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    What @AccidentalFourierTransform is saying is that either the velocity is a constant, then you get to my case, or is not, and then you cannot just ignore it when you do the integral – Andrei Aug 25 '18 at 16:55
  • @AccidentalFourierTransform The first term is 0 if x(t) is a continuous function... $\frac{d}{dx}\frac{dx}{dt} = \frac{d}{dt}\frac{dx}{dx} = 0$ – Striker Aug 25 '18 at 16:55
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    Nope, that's not how differentiation works. Consider e.g., $x(t)=\sin(t)$. Then $\frac{\mathrm d}{\mathrm dt}x(t)=\cos(t)=\sqrt{1-x(t)^2}$. Then $\frac{\mathrm d}{\mathrm dx}\frac{\mathrm dx}{\mathrm dt}=\frac{\mathrm d}{\mathrm dx}\sqrt{1-x^2}\neq0$. – AccidentalFourierTransform Aug 25 '18 at 16:59
  • If $\frac{d}{dx}\frac{dx}{dt}=0$, that means that the velocity does not depend on $x$ – Andrei Aug 25 '18 at 17:02
  • @AccidentalFourierTransform Ok, thanks. Was I missing some assumption in flipping the order of derivatives? I thought that continuity was sufficient. Ok, so I suppose I need to find how to evaluate the integral correctly to check if the problem remains. – Striker Aug 25 '18 at 17:04

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