How Does Calculus Explain the Changes in Notation in Rocket Equation Derivation?

[Jimmy87]

Homework Statement

Hi, need some help trying to understand calculus derivation of the rocket equation. If someone who be so kind as to look at the attachment where it is all laid out.

Homework Equations

All equations are listed in the attachment

The Attempt at a Solution

I have had a good go at this and understand all the derivatives and integrations just not the subtle details. I don't get how you can swap all the deltas for d's towards the end of the derivation. From the research I have done on the web, delta m (mass) is not the same as dm so how can you suddenly change from deltas to d's? For example the mass starts off as delta m, then goes to dm and then disappears! Also what is the reason for suddenly integrating both sides of the equation right at the end? I understand why you take the derivative as this gives you the force (dp/dt) but why then integrate the velocity on the left then the mass on the right?

Thank you for any guidance given![/B]

 

[Simon Bridge]

I don't get how you can swap all the deltas for d's towards the end of the derivation

You can change deltas to dees when the "change in" becomes infinitesimally small ... i.e. in the limit that the size of the change approaches zero. It's one way of thinking about what the infinitesimal interval means - but it is also a form of sloppy notation.

Physics is full of this sort of shorthand, you get used to it.

Also what is the reason for suddenly integrating both sides of the equation right at the end?

... that is just to add up the contributions of each infinitesimal.
$$A_i = B_i\implies \sum A_i = \sum B_i$$
The integration sign is like the summation sign, but for the continuous case rather than the discrete case.

If you have seen Reimann sums, then you may prefer to do it in the other order - add up all the deltas and then take the limit.

Consider a simple example:

$$\Delta x=v\Delta t \implies dx = v(t)\;dt \implies \int dx = \int v\; dt$$ ... see what happened?
If the velocity between $t_i$ and $t_i+\Delta t$ is $v(t_i)$, then $v(t)$ is the velocity between $t$ and $t+dt$.
All I've done is to change notation from the discrete case to the continuous case.