Talk:Calculus/Product and Quotient Rules

Physics Example I: Rocket Acceleration
First of all, Newton's second law of motion is also true for model rockets. $$F = ma$$, whatever (nonrelativistic) object is considered.

To understand the error in this example one should consider the physical meaning of each equation.

$$p = mv$$: The impulse of an object is its mass times its velocity. Also, the impulse of a closed system is always conserved. But the impulse of a non-closed system or object, such as a rocket loosing mass is not conserved.

For example, if we set the effective force (thrust, gravity, etc.) to zero, where the "rocket" is just loosing mass without propulsion,the problem becomes obvious:

$$F=0= v\frac{dm}{dt} + m\frac{dv}{dt}$$

$$a = -\frac{v}{m}\frac{dm}{dt}$$

But there obviously cant be any acceleration since the net force on the rocket is zero, and its changing mass term does not mean that the momentum of the disappearing mass is transferred to the remaining mass of the rocket. --194.106.245.213 (discuss) 23:25, 10 October 2019 (UTC)
 * I think the equation you don't like is this:
 * $$\vec a=\dfrac{\vec F-\vec v\dfrac{dm}{dt}}{m}$$
 * But this it is $$\vec F=\frac{d\vec p}{dt}$$, rewritten. Now this is one of the most established equations in physics. But there obviously cant be any acceleration since the net force on the rocket is zero is obviously wrong, the rocket does experience acceleration, look at any rocket that you choose to study. Considering both the rocket and its exhaust, there is no overall change in momentum, but both the rocket and exhaust will be accelerated in opposite directions. Hence there is an acceleration of the rocket. -- Jules (Mrjulesd) 12:43, 3 November 2019 (UTC)
 * The proplem I have with this equation ist that it physically doesn't make sense. The Article assumes a rocket to be an object with two properties:
 * A rocket has a net force acting on it (from thrust, aerodynamics, etc.)
 * A rocket has a changing (sinking) mass
 * But this is not a sufficient description of the system. Because there is no such thing as a changing mass term on its own. The mass, which the rocket looses, doesnt just disappear. It is expelled, and keeps existing on its own with its own momentum.
 * To prove the Article wrong, I will derive the true equation, as it is physically true.
 * Consider a rocket with following properties:
 * It is operated in vacuum, expelling a small fuel packet with mass $$-dm$$ with a velocity $$v_{exhaust}$$(relative to the rocket) every $$dt$$. It therefore has a mass $$m(t) = m(0) + t \frac{dm}{dt}$$ and net force $$F=-v_{exhaust} \frac{dm}{dt}$$
 * Now we insert into the conservation of momentum: $$\sum_i p_i(t=0) = \sum_i p_i(t=dt) \quad \Rightarrow \quad m_1 v_1 = m_2 v_2 - dm (v_1-v_{exhaust}) = m_1 v_1 = (m_1-dm) v_2 + dm (v_1-v_{exhaust})$$
 * now, to get the acceleration, we insert in $$a = \frac{dv}{dt} = \frac{v_2-v_1}{dt}$$
 * Solving for $$v_2$$:
 * $$(m_1-dm) v_2 = m_1 v_1 - dm (v_1-v_{exhaust}) \qquad v_2 = \frac{m_1 v_1 - dm (v_1-v_{exhaust})}{m_1-dm}$$
 * Inserting above:
 * $$a = \frac{v_2-v_1}{dt} = \frac{\frac{m_1 v_1 - dm (v_1-v_{exhaust})}{m_1-dm}-v_1}{dt} = \frac{m_1 v_1 - dm (v_1-v_{exhaust})-v_1(m_1-dm)}{dt(m_1-dm)} = $$
 * $$=\frac{m_1 v_1-dm v_1 + dm v_{exhaust} -v_1 m_1 + v_1 dm}{dt(m_1-dm)} = \frac{dm v_{exhaust}}{dt(m_1-dm)}$$
 * Now in the limit (because $$dm \ll m_1$$) this reduces to my mentioned formula:
 * $$\lim_{dm \to 0} a = \lim_{dm \to 0} \frac{dm v_{exhaust}}{dt(m_1-dm)} = \frac{dm}{dt} \frac{v_{exhaust}}{m_1} = \frac{F}{m_1}$$
 * This can of course be applied to any moment in time by inserting $$m(t)$$ for $$m_1$$:
 * $$a(t) = v_{exhaust}(t)\frac{\dot{m}(t)}{m(t)} = \frac{F(t)}{m(t)}$$
 * This proof, of course, is more complicated than my first example. My first example is a valid special case for a rocket, but it still is a (although bad) rocket. (You could imagine a rocket with a leaking fuel trank in space for example, so that it looses mass (large $$\frac{dm}{dt}$$ but gains neglegible (about 0) thrust)
 * And just rewriting any equation in physics doesn't mean that it represents the system. In fact, I can cite a even more established equation than $$\vec F=\frac{d\vec p}{dt}$$, which also represents the system of the rocket:
 * $$\vec{F}=m\vec{a} \Rightarrow \vec{a} = \frac{\vec{F}}{m}$$
 * --Johnaeph (discuss • contribs) 14:26, 25 November 2019 (UTC)
 * $$\vec{F}=m\vec{a} \Rightarrow \vec{a} = \frac{\vec{F}}{m}$$
 * --Johnaeph (discuss • contribs) 14:26, 25 November 2019 (UTC)


 * Yes, the equation only considers the rocket. Why? Well if we consider both the rocket and the exhaust the equation will be:
 * $$\vec F{system}=\frac{d\vec p}{dt} = 0$$
 * The reason why is Newton's third law: action = reaction, the force on the rocket is the equal and opposite of that on the exhaust. But this is not a terribly helpful, as it tells us nothing about the forces and acceleration on the rocket as opposed to the exhaust. So it is useless in telling us about the rocket's motion. To do this we have to look at the forces of the two components: the forces on the rocket and the force on the exhaust. To describe those we have to use two equations;
 * $$\vec F_{rocket}=\frac{d\vec p}{dt} = +ve$$
 * $$\vec F_{exhaust}=\frac{d\vec p}{dt} = -ve$$
 * (this assumes the rocket is accelerating along the positive x-axis). Now we don't know at this point what "ve" is, but we know it's a non-zero value, and the value is the same for both equations. Now if we isolate the equation for the rocket we get $$\vec F_{rocket}=\frac{d\vec p}{dt}$$, and the acceleration equation is just this manipulated.
 * Please note that $$\vec{F}=m\vec{a}$$ is not correct in a general sense, only if $$\frac{dm}{dt} = 0$$, which is obviously not the case here; in fact with special relativity it is actually never non-zero, but that's only becomes relevant near the speed of light. See Newton's laws of motion for a discussion on this.
 * Look if you still disagree with me, the best thing would be to discuss this with a third party. Reading room/Assistance would be one place to bring it up, but I also feel that Wikipedia:Wikipedia:Reference desk/Science is excellent for this sort of discussion, and would be a good choice. -- Jules (Mrjulesd) 19:35, 25 November 2019 (UTC)
 * Actually I will remove this section. Looking at Variable-mass system suggests another equation altogether. I was not aware of this not be applicable, but there now exist considerable doubt that my interpretation was correct. -- Jules (Mrjulesd) 20:28, 25 November 2019 (UTC)
 * Thank you for doing this despite me taking so long to answer!
 * (Although i would like to point out that my equation is coherent with the variable-mass-system-equation, as i had $$F_{external} = 0$$ and ther $$F_{net}=F_{external}+F_{thrust}=0+F_{thrust}=v_{exhaust}\frac{dm}{dt}$$--Johnaeph (discuss • contribs) 01:43, 29 November 2019 (UTC)