how do i write exponential y=A(a-e^kt) in terms of t?

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bwind

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hi,

There are various formula for measuring blood alcohol absortion rates and synchronous metabolism rates.
Let's say blood alcohol level (y) at time t is given by:

y = A(1-ekt) - 0.015t

Where:
A is a constant that considers gender, weight and drink
K is a constant that considers the change in absorption rates if someone has eaten.

How can I write this in terms of t (time)? To querey some when questions? Like when is all alcohol metabolised so that y=0?

Many thanks.
Bwind
 
bwind said:
hi,

There are various formula for measuring blood alcohol absortion rates and synchronous metabolism rates.
Let's say blood alcohol level (y) at time t is given by:

y = A(1-ekt) - 0.015t

Where:
A is a constant that considers gender, weight and drink
K is a constant that considers the change in absorption rates if someone has eaten.

How can I write this in terms of t (time)? To querey some when questions? Like when is all alcohol metabolised so that y=0?

Many thanks.
Bwind
Click to expand...
For the given equation, you cannot "invert" and create a function for 't [e.g. t = f(y)]

Only possibility of estimating 't' would be through numerical approximation.
 
bwind said:
There are various formula for measuring blood alcohol absortion rates and synchronous metabolism rates.
Let's say blood alcohol level (y) at time t is given by:

y = A(1-ekt) - 0.015t

Where:
A is a constant that considers gender, weight and drink
K is a constant that considers the change in absorption rates if someone has eaten.

How can I write this in terms of t (time)? To querey some when questions? Like when is all alcohol metabolised so that y=0?
Click to expand...
You probably mean "solve for t (time)", making t the output. It is already written "in terms of t", which means that t is an input.

As has been said, this can't be done algebraically.
 
y is in terms of A, k and t. The only way to get y just in terms of t it would have to be that you can express A and K in terms of t which isn't true in this case.
 
0.015t is a very short time. I expect you could get results sufficiently accurate if you ignored it. If you were looking at bloodalcohol levels over a day the error would be only about 20 minutes.
I would suggest dropping the term 0.015t from the equation. You can then express t in terms of the other variables. Then try a few examples and see whether results are accurate enough.
 
Last edited:
JayJay said:
0.015t is a very short time. I expect you could get results sufficiently accurate if you ignored it. If you were looking at bloodalcohol levels over a day the error would be only about 20 minutes.
I would suggest dropping the term 0.015t from the equation. You can then express t in terms of the other variables. Then try a few examples and see whether results are accurate enough.
Click to expand...
ignore that, it's incorrect! Could you give the expected range of values for each unknown - there could be a way to get a good approximation.
 
You could linearize the exponential. The Maclaurin series for \(\displaystyle e^x\) is \(\displaystyle e^x= 1+ x+ \frac{1}{2}x^2+ \cdot\cdot\cdot+ \frac{1}{n!}x^n \cdot\cdot\cdot\). "Linearize" by dropping all higher power terms: \(\displaystyle e^x\) is approximately \(\displaystyle 1+ x\) for small x so \(\displaystyle e^{kt}\) is approximately \(\displaystyle 1+ kt\). Then \(\displaystyle y= A(1- e^{kt})- 0.015t\) is approximately \(\displaystyle y= A(1- 1- kt)- 0.015t= -Akt- 0.015t= (Ak- 0.015)t\).

From that it is easy to see that \(\displaystyle t= -\frac{y}{Ak+ 0.015}\), approximately.
 
Thanks for the replies. I will work my way through. Unhelpfully I have not written the initial equation correctly. It should be:

y = A(1-e-kt) - 0.015t [its -kt]

I have attached a basic graph of the application. The -0.015t represents the metabolising rate by decimal hour. The first half of the equation prior to this represents the alcohol absorption rate (the graph shows K=2, k=6 depending on food) and then straigt assumes that the alcohol is absorbed instantly.

I am trying to write a formula that will allow me to estimate how long in decimal hours it takes for blood alcohol % to get to a value (notably .05 , or zero). Thanks again.
 

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You could try the Newton-Raphson iterative method. Make a guess for \(t_0\) and this equation should give you a better value...

[math] t_{n+1} = t_n - \frac{y + 0.015 t_n - A \left(1-\mathrm{e}^{-k t_n}\right)}{0.015+ k A\mathrm{e}^{-k t_n} } [/math]
You can repeat for extra accuracy. Say A=1, k=2, y=0.05, and you make a good guess \(t_0= 50\). Plugging into the above gives \(t_1=63.3333333...\) which is VERY accurate in a single iteration.

If you start with a bad guess of t0=1, then t1=3.799254..., t2=59.572777..., t3=63.333333... three iterations gets an accurate answer.
 
Cubist said:
You could try the Newton-Raphson iterative method. Make a guess for \(t_0\) and this equation should give you a better value...

[math] t_{n+1} = t_n - \frac{y + 0.015 t_n - A \left(1-\mathrm{e}^{-k t_n}\right)}{0.015+ k A\mathrm{e}^{-k t_n} } [/math]
You can repeat for extra accuracy. Say A=1, k=2, y=0.05, and you make a good guess \(t_0= 50\). Plugging into the above gives \(t_1=63.3333333...\) which is VERY accurate in a single iteration.

If you start with a bad guess of t0=1, then t1=3.799254..., t2=59.572777..., t3=63.333333... three iterations gets an accurate answer.
Click to expand...

Thanks for this. It is beguiling. I can reproduce your example. The challange I have is that a is usually quite a small number (e.g. 0.06 is 3 standard drinks for a 70kg man). I would expect t to be a little over 1 (t is in decimal hours). With these numbers i dont get the same estimates by iterating.
 
Cubist said:
[math] t_{n+1} = t_n - \frac{y + 0.015 t_n - A \left(1-\mathrm{e}^{-k t_n}\right)}{0.015+ k A\mathrm{e}^{-k t_n} } [/math]
Click to expand...

I think I made a sign mistake in the denominator above, it ought to be...

[math] t_{n+1} = t_n - \frac{y + 0.015 t_n - A \left(1-\mathrm{e}^{-k t_n}\right)}{0.015- k A\mathrm{e}^{-k t_n} } [/math]
But this doesn't explain the convergence problem you observed in post11. The function's value never touches y=0.05 (shown by the green line below)...

graph.png

So the iterative process can't possibly find a solution! (blue curve is your function when A=0.06, k=2)

If you try again, this time searching for the t value where the function intersects y=0 instead of 0.05 then you'll get convergence.

While iterating, it's probably best if you check that the gradient remains -ve for the current t value, and start with a high initial guess of t. If the gradient becomes +ve then it's a fair bet that the function never touches the particular y value that you're searching for.

[math]\frac{dy}{dt} = kA\mathrm{e}^{-kt} - 0.015[/math]
 
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