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Apologies for the vague title I couldn't think of a way to be more explicit in this case. I am looking at a derivation for a physical quantity. The author states

$\frac{dE}{dt} = -\alpha \left(\frac{dx}{dt}\right)^{2}$

and goes on to say that

$\Delta E = - \int_{0}^{T} \frac{dE}{dt}dt$

Now I can't see how the author gets the above relation. Physically we are looking at the loss of energy in one period. I understand the equation in a physical way, but I don't know how to arrive at the $\Delta E$ expression mathematically.

The only thing I can think of is that the author is saying something like

$\frac{\Delta E}{\Delta t} = \frac{d E}{d t}$ and then rearranged and added the minus sign as we are losing energy...

If anyone can provide a more rigorous method to get to the second equation I would be grateful!

1 Answers 1

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Performing the integration, $-\int_0^T \frac{dE}{dt} dt = -\int_0^T dE = -[E(T)-E(0)] = E(0)-E(T) = \Delta E$ is the energy lost over one period.

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    No, I understand this, but how does one arrive at the integral in the first instance. My question is more how do you set that "problem" up in the first place? – 2017-02-27
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    Just read my explanation from right to left. The main thing here is to always associate a "final minus initial" situation with an integral. When you have an explicit function for E you can just use final minus initial, but if you only have it's derivative in time, you have to integrate to get E. – 2017-02-27
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    Do you mean that its just knowing the answer already and setting up the problem in reverse? – 2017-02-27
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    I wouldn't say it like that. It's more like you want to define this quantity of energy lost in one period $(0,T)$ called $\Delta E$. You can easily define it in terms of the function $E$ as $E(0)-E(T)$. If you already know the function $E$, then you can use this form right away. Since you don't have $E$, you need to find a new form of $\Delta E$ which is defined in terms of what you have, which is $\frac{dE}{dt}$. You could simply integrate this function to get $E$ and then use the original definition of $\Delta E$, or combine the two steps into one using the relationship in the original post. – 2017-02-28
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    Nice explanation thanks! – 2017-02-28