# Why/how can the product of 2 differentials be neglected?

1. Sep 11, 2012

### d.arbitman

I am looking at the derivation of the capstan friction equation and there is a term in there which the derivation claims can be neglected; my question is: why can it be neglected?

dT*sin(dθ/2)

source: http://www.jrre.org/att_frict.pdf

2. Sep 11, 2012

### Staff: Mentor

Differentials are assumed to be close to zero, so sin(dθ/2) is close to zero, as is dT. The product of two very small numbers is even smaller, so can be neglected.

3. Sep 11, 2012

### d.arbitman

That makes perfect sense, but with that reasoning we can neglect a single differential as well since it is close to 0 and then we will have no use for calculus.

4. Sep 11, 2012

### SteveL27

Another way to look at this intuitively is that if a variable x goes to zero, the variable x2 goes to zero much faster. In other words

$\displaystyle \lim_{x \to 0} \frac{x^2}{x} = 0$

Even though x2 and x both go to zero, x2 goes to zero so much faster than x, that the limit of their quotient evaluates to zero.

You can see this numerically. When x is 1/2, x2 is 1/4. When x = 1/10, x2 is 1/100. As the magnitude of x gets smaller, the magnitude of x2 gets a lot smaller.

But your question's a good one. Newton invented calculus in 1680 or so, but the rigorous theory of limits wasn't developed till two hundred years later. So your question shows good insight.

Last edited: Sep 11, 2012
5. Sep 11, 2012

### d.arbitman

That's a great example. Thank you. I guess what I take from that is that the product of two differentials changes so much slower than a single differential that we can almost forget about the product as it will have a miniscule effect. Maybe mathematically it can have an effect on the outcome, but in the context of friction it will have little to no effect.

6. Sep 12, 2012

### HallsofIvy

Staff Emeritus
Note that when you are talking about "differentials" just saying that differentials are 'close to 0' is sweeping a lot under the rug. Technically, one should have a whole new arithmetic ("non-standard analysis"), including various 'levels' of "differentials" (and various levels of "infinities" in order to have multiplicative inverses) such that the product of two differentials of the same 'level' gives a differential of a higher 'level'. Essentially we are saying that, for most calculus, we can work with "standard" numbers and first level differentials, treating higher level differentials as 0.

7. Sep 12, 2012

### Studiot

You are asking this in the mathematics forums, and have been answered.

However the question of when it is permissible to make that assumption is a matter of experience.

There are many areas of physics and engineering where we make this sort of assumption in the interests of a simple formula.

However sometimes we need to take second order (products of two small quantities) or even higher orders into account for more acurate work.
An example would be in beam theory where we obtain simple formulae applicable to long narrow beams (most are like this) by considering differentials of first and second order and disregarding higher orders.

Please note that there is a difference between saying

${\left( {\frac{{dy}}{{dx}}} \right)^2}$ is small

and

$\frac{{{d^2}y}}{{d{x^2}}}$ is small.

for sharply curving function the latter may not be true at all.