# Question about the application of calculus in physics

The way I've been taught to apply calculus to physics problems is to consider a small element at a general position and write an equation for that element and then to integrate it.

For e.g

To find the moment of inertia of this rod as a function of y, I write the equation for the moment of inertia of this element at a distance of y

$dI = dm y^2$ here $dm = \rho dy$ where $\rho$ is the density Then I integrate this from $y = 0$ to $y = y$ to sum up the moment of inertia of these little elements up til the distance = $y$ to get the moment of inertia as a function of y

But I've never really understood this "procedure." Isn't $dI$ the infinitesimal $change$ in the moment of inertia function $I(y)$ for a change of $dy$ rather than the moment of inertia of this tiny element? Isn't what we should be trying to do is find the rate of change of $I(y)$ and integrate it w.r.t $y$ to the original function?

How can we just treat it as the moment of inertia of the tiny element and then sum these up using integration?

• You can think of dI(y) as the infinitesimal change in the moment of inertia as the distance y increases from y to y+dy. You can also think of dI(y) as being the moment of inertia of the tiny element of the beam between y and y+dy. The two interpretations are effectively equivalent. – Samuel Weir Feb 5 '17 at 5:20

This is a good question. You are misinterpreting the definition of the moment of inertia function $I(y)$. For a specific value of $y$, $I(y)$ does not mean "the contribution to the moment of inertia from the mass at distance $y$." It means "the total moment of inertia due to all the masses between distance $0$ and distance $y$." This is why, to find the moment of inertia of a rod of length $L$, you just plug in $y=L$ to get $I(L)$ instead of integrating the function. Since $I(y)$ is the total moment of inertia due to all the masses up to distance $y$, and $I(y + dy)$ is the total moment of inertia for a slightly longer rod, if you subtract the two, you get that $dI = I(y + dy) - I(y)$ is just the contribution to the moment of inertia from the length of rod at distance $y$.

In general, for these types of problems you have two functions you care about: a function $F(x)$ that represents "the sum of all the contributions from $0$ to $x$," and a function $f(x)$ that represents "the contribution just from the stuff at point $x$," which is typically some kind of density. The key mathematical relation is that $f(x) = \frac{dF}{dx}$, or equivalently $F(x) = \int_0^x f(y)\ dy$. In this case, $I(y)$ corresponds to the "integrated" function $F(x)$, not the "density" function $f(x)$.

There aren't enough letters in the alphabet to give every letter just one meaning in mathematics.

$d$ has a standard meaning in calculus. It is also sometimes used for the diameter of an object, or (as you said) to represent a small (but finite sized) change in some other quantity.

If you want to talk about "small changes" and "calculus" in the same equation, it would be better to use a different notation for a small change in $x$ like $\delta x$ or $\Delta x$, and only use $dx$ with its standard calculus meaning.

In your example, really you are saying that the inertia of a small part of the structure with mass $\Delta m$ is (approximately) $\Delta I = y^2 \Delta m$.

Summing over all the small parts, the inertia of the whole structure is then $$I = \sum_{\text{volume}} \Delta I = \sum_{\text{mass}} y^2 \Delta M = \sum_{\text{length}} y^2 \rho\, \Delta y.$$ In the limit as the individual $\Delta M$'s become smaller and smaller, you can replace the sum with an integral and (if the rod has length $L$) write $$I = \int_0^L y^2 \rho\, dy.$$

Note: this didn't refer to anything like "the moment of inertia function $I(y)$". The M.I. of one specific rod isn't a "function" of anything - it's a constant! If you really wanted to write a function for the M.I. of any rod, it would be something like $I(\rho, L)$ not $I(y)$, because the M.I. depends on the density and the length of the rod. It does not depend on the arbitrary coordinate system (i.e. $y$) that you chose to do the calculations.

But I've never really understood this "procedure." Isn't dI the infinitesimal change in the moment of inertia function I(y) for a change of dy rather than the moment of inertia of this tiny element?

Italics mine.

The moment of inertia of a body about a given (as in your example) principal axis, is a single number by definition. It is for rotational motions what the mass is for linear motions, and it does not change depending on the y, only the choice of the rotational axis can change I

To get the mass of the sugar in a box of 100 pieces you only need to know the weight of one piece, which is your dI.

## protected by Qmechanic♦Feb 5 '17 at 13:33

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