Is the Cancellation of Differentials Valid in Calculating Inductor Energy?

[BucketOfFish]

From http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html#c1, for calculating the energy stored in an inductor:

$$P=Li\frac{di}{dt}$$
$$E=\int_{0}^{t}Pdt=\int_0^ILidi=\frac{1}{2}LI^2$$

Is there a theorem that says it's okay to just cancel out the dt in that second equation, and then replace the limits of integration?

 

[LCKurtz]

From http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html#c1, for calculating the energy stored in an inductor:

$$P=Li\frac{di}{dt}$$
$$E=\int_{0}^{t}Pdt=\int_0^ILidi=\frac{1}{2}LI^2$$

Is there a theorem that says it's okay to just cancel out the dt in that second equation, and then replace the limits of integration?

Yes, it's just an ordinary "u substitution". To make it clear I will use $T$ instead of the dummy variable $t$, so you are doing the integral$$
\int_0^t L i \frac{di}{dT}dT$$Now let's change variables: $u = i(T),\ du=\frac{di}{dT}dT$. When $T=0,\ u=i(0)=0$ and when $T = t,\ u = i(t) = I$, so we get$$
\int_0^ILu du =\left . \frac{Lu^2}{2}\right|_0^I=\frac 1 2 LI^2$$

 

[BucketOfFish]

There must be some gaps in my basic knowledge of calculus (I learned this stuff in high school and it wasn't exactly rigorous), but why is it that you can say $du=\frac{di}{dT}dT$? Did you first have to say that $\frac{du}{dT}=\frac{di}{dT}$ and then move the $dT$ over to the right side? In that case, I guess the same question remains in a different format. Does it not cause problems to just move the derivative symbols around like that?

 

[Mute]

Consider that

$$\frac{1}{2}\frac{d}{dT}\left(i^2\right) = i \frac{di}{dT}$$
by the chain rule.

So, the integral is just

$$\int_0^t dT \frac{d}{dT}(i^2(T)) = i^2(t) - i^2(0)$$
by the fundamental theorem of calculus.

The 'cancellation' of differentials is in some sense just an abuse of notation that is allowable because it is really just a shorthand for changing variables, as LCKurtz showed, and the change of variables comes about through use of the chain rule.

 

[Whovian]

Consider that

$$\frac{1}{2}\frac{d}{dT}\left(i^2 \right) = i \frac{di}{dT}$$
by the chain rule.

So, the integral is just

$$\int_0^t dT \frac{d}{dT}(i^2(T)) = i^2(T) - i^2(0)$$
by the fundamental theorem of calculation.

The 'cancellation' of differentials is in some sense just an abuse of notation that it allowable because it is really just a shorthand for situations like the chain-rule trick above.

You forgot a \right. Fixed. EDIT: Looks like you beat me to it.

Anyway, it turns out in nonstandard analysis that this isn't just an abuse of notation, and that the "d"s are actual terms, though they are infinitesimals. But most teachers won't tell you that because you won't have the means to prove this in a long, long while.