I am trying to create a decay factor that will decay a variable whenever the distance associated with the variable exceeds 5000 miles. And, I am trying to do so in two ways:
- Exponential. I am trying to decay a variable exponentially when the distance associated with the variable exceeds 5,000 miles so that the value of the variable is halved for ever 5,000 miles.
- Interval. I am trying to decay the variable by half for each interval of 5,000 miles (e.g., any variable associated with a distance between 5,001 to 10,000 miles would have a value of 1/2 the original).
In this problem, time is not a variable. All that matters is distance. Can I still use the half-life decay model to calculate the values for the variable? If so, does d just replace t in the model? In other words, is the cell function, =2(Value_At_Distance_Zero)(EXP((k)*(5000)))?
If the decay is continuously the you can use your model:
$y(d)=C\cdot e^{-k\cdot d}$, where $k$ is a constant. Then you have the condition $y(5000)=C\cdot e^{-k\cdot 5000}=\frac12\cdot C$
Dividing the equation by $C$
$e^{-k\cdot 5000}=\frac12$
Taking both sides as an argument of $\ln(\cdot )$
$-k\cdot 5000=\ln\left( \frac12 \right)$
$k=-\frac{\ln\left( \frac12 \right)}{5000}=\frac{\ln\left( 2 \right)}{5000}$
You need a initial condition to determine the constant $C$. For instance $y(0)=80$.
$$80=C\cdot e^{-\frac{\ln\left( 2 \right)}{5000}\cdot 0}\Rightarrow 80=C\cdot e^{0}\Rightarrow C=80$$
Thus you function is $Y(d)=80\cdot e^{-\frac{\ln\left( 2 \right)}{5000}\cdot d}$
On the graph below you can see that $y(0)=80$
After a distance of 5000 we the half of the initial value $Y(5000)=40$. And after additional distance of $5000$ the half of $40$ again: $Y(10000)=20$