Creating an Exponential decay equation for given parameters

[shakystew]

So, I am wanting to vary a parameter in an equation with respect to time.

• Vary mass flow [ m(t) ] for an exponential decay to half its original value in around 60 seconds.

I know the regular decay equation where:

m(t)=m₀*exp(-At) ​

but I want the value to approach a steady state at 60 seconds (i.e. I am decreasing my pump/mass-flow by one half over 60 seconds).

I need an equation which will allow this to occur.

 

[Mark44]

So, I am wanting to vary a parameter in an equation with respect to time.

• Vary mass flow [ m(t) ] for an exponential decay to half its original value in around 60 seconds.

I know the regular decay equation where:

m(t)=m₀*exp(-At) ​

but I want the value to approach a steady state at 60 seconds (i.e. I am decreasing my pump/mass-flow by one half over 60 seconds).

I need an equation which will allow this to occur.

Is this homework?

 

[shakystew]

It is not. It is for my current project for my research. I figured it out :)

 

[JJacquelin]

Why not m(t) = m₀ +B (exp(-A t)-1) with for example B=m₀/2 or other adjusted values of A and B for better fit to the given conditions.

 

[HallsofIvy]

So, I am wanting to vary a parameter in an equation with respect to time.

• Vary mass flow [ m(t) ] for an exponential decay to half its original value in around 60 seconds.

I know the regular decay equation where:

m(t)=m₀*exp(-At) ​

but I want the value to approach a steady state at 60 seconds (i.e. I am decreasing my pump/mass-flow by one half over 60 seconds).

I need an equation which will allow this to occur.

So you want m(60)= m₀ exp(-60A)= (1/2)m₀. Then you want exp(-60A)= 1/2 so -60A= ln(1/2), A= -ln(1/2)/60.
(Since 1/2< 1, ln(1/2)< 0 so A will be positive).
You have m(t)= m₀e^{(-tln(1/2)/60))}= m₀e^{tln((1/2)^(t/60))}= m₀(1/2)^(t/60).

In fact, we could have argued from the first that it must be of that form- since it decreases by 1/2 every 60 seconds (exponential decay always decreases by the same fraction over the same time interval) the original amount must be multiplied by 1/2 every 60 seconds. In time t seconds, there are t/60 "60 second time intervals" so the original amount is multiplied by 1/2 t/60 times: (1/2)^t/60.