I have a problem calculating exponential decay, I’m not a mathematician, sorry when my notation is not correct. Imagine a radioactive substance is produced at a rate that increases linear with time:
N(t) = N0 + b*t
N0 is the production at t=0, b the slope of the increase of the production, t is time
I want to quantify the „cumulative“ amount of this substance, therefore I integrate N(t) within this time range.
My problem is that the radioactive substance also decays exponentially with a certain half life time.
The differential equation for this decay is (dN/dt) = -lamda*N
lama can be calculated using lama = ln(2)/ht
ht is the half life time
I have to combine these two equations, and I came to this:
N(t) = N0 + bt - lamda(N0 + b*t)
I think , when I integrate this equation for a certain time period I get the cumulative amount of the radioactive substance. However, it doesn’t seem logical for me: Every year a certain amount of N decays and is produced. But the decay happens not only for the currently produced amount of N, but also for the leftovers of N from the last time poimnt and so on. How can I solve this analytical? So far I used an iterative approach calculating the amount of N produced each year and for each of these cohorts the decay for each year. However, I want to use „real“ equations.
So if I am reading this correctly you want to calculate how much stuff you have ($n_{Total}$) in some time (t) by integrating the rate (net flux) of production and decay of the substance. We can write mathematically as: $$n_{Total}(t)= \int_{0}^{t} flux*dt=\int_{0}^{t} N_{Total}*dt$$ $$N_{Total}= N_{Made}+N_{Decay}$$ From the problem we know that the following are true: $$N_{Made}=N_0+b*t$$ $$\frac{d}{dt}N_{Decay}=-\lambda^2 N_{Decay}\rightarrow N_{Decay}= N_0e^{-\lambda^2t}$$ $$N_{Total}= N_0+b*t+N_0e^{-\lambda^2t}$$ exo facto: $$n_{Total}(t)= \int_{0}^{t} (N_0+b*t+N_0e^{-\lambda^2t})dt$$
From this some more math needs to be done, but that is left up to the reader to solve.