How to find the rate of change of an average

Question

Say I have a container with some mass $m$ of particles each with the same size, $x$. New particles with size $x_1$ are entering the container at a mass flow rate of $\dot m_1$. I need an expression for the rate of change of the average particle size in the container, $\frac{d\bar x}{dt}$.

I started by writing the equation for an average of the two sizes:

$$ \bar x = \frac{mx+m_1(t)\cdot x_1}{m+m_1(t)} $$

then taking the derivative to get

$$ \frac{d\bar x}{dt} = \frac{\dot m_1 \cdot m \cdot (x_1-x)}{(m+m_1(t))^2} $$

however the derivative still has an $m_1(t)$ term in it, for which I don't have an expression or a value. I'm not even sure what the physical significance of $m_1$ would be in the context of an instantaneous rate of change ($\dot m_1$). I'd appreciate any suggestions or other approaches to this problem.

Answer 1

Okay so right now you have $m$ mass of particles with the same size of $x$. Let's say you have $\displaystyle \frac{m}{x}$ particles.

Now, you have $m_1$ mass flow rate with particles of size $x_1$, so $\displaystyle \frac{m_1}{x_1}$ particles per second.

Your average mass size is $\displaystyle \overline{x}(t)=\frac{m+m_1t}{\frac{m}{x}+\frac{m_1t}{x_1}}$.

Now, the rate of change is $\displaystyle \frac{d\overline{x}(t)}{dt}$. $m,m_1x,x_1$ are all constants.

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If $m_1$ is a time variable, then we denote it as $m(t)$, and we have $\displaystyle \frac{m_1(t)}{x_1}$ particles per second.

Then, the average mass size is $\displaystyle \overline{x}(t)=\frac{m+\int_0^{t}m_1(a)\,da}{\frac{m}{x}+\frac{\int_0^{t}m_1(a)\,da}{x_1}}$.

Similarly, the rate of change is $\displaystyle \frac{d\overline{x}(t)}{dt}$. Little tougher, but doable using the Second Fundamental Theorem of Calculus.