How to calculate uncertainties of a natural exponential function?

(I apologize if this should be posted in mathematics, however I chose to post it here as it's technically about physics)

I conducted an experiment in which position of items were shifted on an object, either on the ends of wings of it, or on the base (I'd rather not get too much into what it's about), and the effect on its fall rate over a certain distance was measured. The result was a decay model of the form:

$T(N)=Ae^{-bN}+c$,

where $A=1.44,b=0.132,c=0.303$ and $T =$ Time,$N =$ Number of items added to wings.

However, for each of the times there is an uncertainty of between 0.08 and 0.09 seconds.

So, I asked my teacher for assistance and he explained the following: First you remove the 0.303, and then you can rearrange it as follows:

$T = 1.44*e^{-0.132N}$

$\ln{T} = \ln(1.44*e^{-0.132N})$

$\ln{T} = \ln{1.44} + \ln{e^{-0.132N}}$

$\ln{T} = 0.365 + -0.132N$

$\ln{T} = -0.132N + 0.365$

And thus you have a linear equation.

Then I calculated $\ln{T}$ and $-0.132N + 0.365$ for each value of N, and graphed it in a graphic software, and made error bars of $±((\ln(T+\delta T)-\ln{(T-\delta T))/2})$, and thereby can get a best-fit gradient, a maximum possible gradient, and a minimum possible gradient, all in terms of $(ln(T)/(-0.132N + 0.365))$ if I'm not mistaken.

But now for the questions:

1. Why could the (+0.303) simply be removed, and how can that be justified?
2. What do I do with my newly acquired values for the max. and min. gradients?

I'd truly appreciate any help on this!

• 1. Because it doesn't affect the gradient. 2. What do you mean "do with"? – Mike Dunlavey Jan 8 '13 at 20:00
• Okay, I now understand that first one fully, but I'm wondering how I should report the answer from these new ln(T) gradients. What are they useful for? @MikeDunlavey – DarkLightA Jan 8 '13 at 21:41
• Just report your results as graphs on semi-log paper, with T on the log axis. Then you've just got two straight lines for your fit, one low, and one high. – Mike Dunlavey Jan 8 '13 at 21:44

In general, the error in a function $$z$$; $$z=f(x_1,x_2,.. x_n)$$, in terms of the errors of its variables $$e_1,e_2,...e_n$$; is given by; $$(e_z)^2= (\frac{\partial f}{\partial x_1}e_1)^2+(\frac{\partial f}{\partial x_2}e_2)^2+.. +(\frac{\partial f}{\partial x_n}e_n)^2,$$assuming normal distribution of errors. If you've noticed, this formula is based on the well known total derivative of $$f$$.
For $$z=A \exp(-bx)+ c$$, where $$z=T$$,$$x=N$$, you get $$(e_z)^2 = (-Ab\exp(-bx)e_x)^2,$$where your $$c$$ disappears as it is a constant with zero derivative. Now substitute for $$A$$, $$b$$, and the error in $$x$$ to get the error in $$z$$.