Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Dear Professor Farnsworth,

We at D.O.O.P are trying to mathematically model a rocket ship fueled by your employee Leela's pet Nibbler's pooped Black matter. Obviously this rocket ship is fueled by black matter which along with the ship's combustion chamber has some special properties.

Black matter Properties We have discovered that there are two kinds of black matter the one naturally made (by Nibbler's poop) and the other which is synthetically made in our special rocket ship combustion engine. The Synthetically made Black matter exponentially decays at a rate of $r$, until all its mass eventually becomes nothing and is discarded.

Rocket Ship Properties When any Black matter is used as fuel in the combustion chamber. It produces black matter (synthetic kind) according to following equation. $$\dfrac {dP_a} {dt}\leq \dfrac {dP_c} {dt}=\left( 1-\dfrac {\alpha } {100}\right)B$$

Here $B$ represents the amount of Black matter currently in the combustion chamber. $\alpha$ is a percentage, a controlling mechanism in the ship to control production. $P_c$ is an upper bound of the new Black matter production (capacity), but we discover there is an inefficiency in the system such that the actual rate of Black matter production is in fact $P_{a}$.

Since we want to travel as fast as we can, as soon as any new Black matter $P_{a}$ is produced we add that to $B$ and we are able to do all of this in infinitesimally small amount of time(continuously).

We invite you to scientifically examine and model the processes of this rocket ship along with say $B_0$ amount of initial natural black matter. How can we model or represent this system with the least amount of equations while capturing the essence of the whole problem ?

From the desk of Zapp Brannigan
"And like all my plans, it's so simple an idiot could have devised it!"

Edit: Solution attempt Assuming $B_0$ to be the initial amount of black matter available. We start undertaking combustion with this initial amount $B_0$ we produce more black matter at the rate of $\dfrac {dP_a} {dt}\leq \dfrac {dP_c} {dt}=\left( 1-\dfrac {\alpha } {100}\right)B_{0}$. As $dt$ time period passes by we take the new $P_a$ amount produced and add it to $B_{0}$. We also observe that this newly created synthetic black matter $P_a$ is exponentially decaying. I am having trouble figuring out how to put these relations together so both of these processes can be carried out simultaneously.I'd be happy with if you wish, only consider the case when $P_a$ and $P_c$ are the same.

share|improve this question
1  
Have you got mass being destroyed in the decay and being created in the replication, or does it just change between martian cow dung and some neutral substance? –  Henry Mar 18 '12 at 20:44
    
No I was thinking of mass actually getting destroyed. –  Hardy Mar 18 '12 at 22:04
3  
In the events of Bender's Game Nibblonian poop (which is what dark matter really is) was rendered useless as fuel. –  Asaf Karagila Mar 18 '12 at 22:17
1  
I cannot make sense of your equation. –  Martin Argerami Mar 19 '12 at 0:38
1  
Nah, it's OK. I was just surprised that it changed so fast, with no one having the time to comment on it apparently. (Well, except me ;) –  Raskolnikov Mar 19 '12 at 8:39

1 Answer 1

up vote 1 down vote accepted

I think you have told us that $B$ is changing with respect to time for two reasons. One has positive influence: new dark matter is produced by your engine. The other has negative influence: the dark matter decays exponentially. I assume that it is this decay that powers the engine, because I don't see anything else explaining how the fuel is burned for the purposes of propulsion.

So $$\begin{align} \frac{dB}{dt} & = \frac{dP_a}{dt} - rB\\ \end{align} $$

Assuming $\frac{dP_a}{dt}=\frac{dP_c}{dt}$, $$\begin{align} \frac{dB}{dt} & =\left(1-\frac{\alpha}{100}\right)B - rB\\ &=\left(1-\frac{\alpha}{100}-r\right)B \end{align} $$ This equation has solution $B=B_0e^{\left(1-\frac{\alpha}{100}-r\right)t}$

With $\frac{dP_a}{dt}<\frac{dP_c}{dt}$, of course much more is possible. In the extreme, $\frac{dP_a}{dt}=0$, and $B=B_0e^{-rt}$.

So $$B_0e^{-rt}\leq B(t)\leq B_0e^{\left(1-\frac{\alpha}{100}-r\right)t}$$ Any curve that fits between these two and never has a steeper relative growth rate than the upper bound nor a less steep relative growth rate than the lower bound is a possible solution, depending on how exactly $\frac{dP_a}{dt}$ differs from $\frac{dP_c}{dt}$.

share|improve this answer
1  
Note that conservation laws would insist that $\frac{\alpha}{100}+r>1$, or a sufficiently efficient reactor would cause $B$ to grow over time, not decay. –  alex.jordan Mar 19 '12 at 8:20
    
Firstly thank you very much for your answer. I find it very clever though couple of questions. What if in reality B was allowed to grow over time and the synthetic Black matter was to decay even if it was sitting outside the combustion chamber. I know it does n't explain the propulsion but all of this is fictional. Would these two changes make any impact on your solution ? –  Hardy Mar 19 '12 at 8:29
1  
I'm not sure what you mean by "$B$ was allowed to grow over time". $B$ is allowed to grow if $\alpha/100+r<1$ and that changes nothing about the solution. For your second consideration, if the matter is not put back into the combustion chamber, then we have the same situation as the lower bound, where $\frac{dP_a}{dt}=0$. In this situation, there would be a second quantity to keep track of: the unused fuel that the chamber creates. –  alex.jordan Mar 19 '12 at 19:04
    
Buddy based on the different properties of black matter should n't your first equation be $$\begin{align} \frac{dB}{dt} & = \frac{dP_a}{dt} - r(B-B_0)\\ \end{align} $$ –  Hardy Mar 19 '12 at 19:15
1  
Is the original dark matter not subject to exponential decay? Then yes. In that case we have a nonhomogeneous first-order differential equation, which is almost as easy to solve. –  alex.jordan Mar 19 '12 at 19:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.