# I How is this rigorous?

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1. Mar 8, 2017

### EddiePhys

In the solution of this problem(121), dW = (kmgcosΦ + mgsinΦ) ds, where ds is the differential element along the curve. Now they have done kmg dscosΦ + mg ds(sinΦ) = kmgdx +mgdy. Makes sense intuitively, but I want to know how this is rigorous. What I'm thinking is, the curve is broken into N elements of the same length along the curve over which a Riemann sum gives a line integral. But here cosΦ is varying, so wouldn't all the Δx's and Δy's not be of the same length? And if your explanation is that as Δx tends to zero the two become the same, then why can't we simply treat all differential elements as the same since they all tend to zero after all?

Sorry for the rambling at the end. But can someone just show me why such an operation is rigorous?

2. Mar 8, 2017

### Khashishi

The lengths of your segments don't have to be the same for you to sum them up. It's easier if you take equal sized horizontal steps, but let the vertical step vary.

3. Mar 8, 2017

### sophiecentaur

If they don't actually tell you the shape of the curve then it must be irrelevant (?). The answer must be independent of shape.
Without being too rigorous, I would suggest the work done would have to be mgh + kmgl . That would be PE gained + resultant work against friction, equivalent to dragging it along the ground first or last. Your infinitesimal approach is 'better behaved' and I think that, if you examine what cos and sin theta represent, in terms of δx and δy, I think that things should fall out when you work out the work along the hypotenuse of an infinitesimal sloping section. You will need to get your hands dirty with some trig and basic friction. You have to believe it will all fall out nicely and it will.