Question

Obtain a result for calculating the time required to consume a finite resource. dM/dt=-Pexp(at)+Fexp(-ut)

Answer 1

Whare are F, P, and M?

Answer 2

M is the current size of resources,Pexp(at) is rate of production of new resource and Fexp(-ut) is the rate new resource deposits are being discovered

Answer 3

Basically supposed to solve for the time t

Answer 4

You need to solve M(t) = 0.

Answer 5

So you'd have to integrate first.

Answer 6

this is where i'm stuck at.
M=(-1/a)Pexp(at) +P/a +F/a _(1/a)Fexp(-ut)

Answer 7

Okay, what is the anti derivative of \(\exp(at)\)

Answer 8

(1/a)exp(at)

Answer 9

Where is the constant of integration?

Answer 10

sorry. was integrated from t=0 to t=t

Answer 11

Alright so what are you getting \(M(t)\)?

Answer 12

M=(-1/a)Pexp(at) +P/a +F/a _(1/a)Fexp(-ut)

Answer 13

How are you getting that though?

Answer 14

the original equation was dM/dt=-Pexp(at)+Fexp(-ut)

Answer 15

Where is +P/a coming from though?

Answer 16

after the limits were put in

Answer 17

from 0 to t

Answer 18

I mean F/a

Answer 19

Sorry. That should be F/u

Answer 20

M=(-1/a)Pexp(at) +P/a +F/u -(1/u)Fexp(-ut)

Answer 21

Now you would find the roots.

Answer 22

Roots of which term?

Answer 23

roots of t

Answer 24

as in an equation of the exp(at) and exp(-ut) terms?

Answer 25

Basically M = 0, and you have to find t values where that expression is 0

Answer 26

The trivial case is t=0

Answer 27

\[
0=\frac Pa (1-e^{at}) + \frac Fu(1-e^{-ut})
\]

Answer 28

\[
\frac Pa (1-e^{at}) = - \frac Fu(1-e^{-ut})
\]

Answer 29

Hmmmm\[
-\frac{Pu}{Fa}(1-e^{at}) = 1-e^{-ut}
\]

Answer 30

This is the question. Obtain an analytic results that can be used for calculating the time required to consume a finite fuel resource. Currently the resource is of size, Mo, which is being produced at a rate ,P. The rate of consumption is growing exponentially at a relative rate ,Pexp(at), and new fuel resource deposits are being discovered at a rate, Fexp(-ut)

Answer 31

That changes things a bit.

Answer 32

So I formulated this equation dM/dt=-Pexp(at)+Fexp(-ut)

Answer 33

Well, if you integrate from 0 to t you will \(\Delta M\) rather than just \(M\).

Answer 34

\[
\Delta M = M-M_0
\]

Answer 35

I think to find the roots in this case, we'll have to use the natural log.

Answer 36

ln of -Pu/Fa=1-exp(-ut)/1-expt(at) ?

Answer 37

Yeah, but we'd have to factor in the \(M_0\) as well.

Answer 38

This problem is obnoxious.

Answer 39

\[
M(t) = M_0 +\int_0^t\frac{dM}{dt}dt
\]

Answer 40

Since we want \(M(t)=0\), we have: \[
-M_0 = \int_0^t\frac{dM}{dt}dt
\]

Answer 41

is that a possibility?

Answer 42

Yes, it's possible.