I have position (S) vs time (t) data that is very well fit ($R^2$ > 0.9997) by the following equation:
$$S=a(b-\exp(-ct))$$
The instantaneous velocity ($v$) is just the derivative of this equation at a particular time, so:
$$v=ac\exp(-ct)$$
Now if I calculate $v^2$ from the second equation and graph it, I find that the following exponential yields a very good fit ($R^2$ > 0.98)
$$v^2 = f\exp(-gt)$$
The whole idea is to show a correlation of "g" with another different parameter, H. Plotting g vs H yields a pretty good fit.
But, it seems to me that it would be better to try to plot up c from the second equation vs H. So the second equation squared is,
$$v^2=a^2c^2\exp(-2ct)$$
Rather than "refit" this equation with a simpler exponential like the third equation, I wonder if I can somehow plot c or some mathematical expression of c vs H. It seems like I should be able to do something. For example, if I use the following:
$$c^2 = \exp(2\ln(c))$$
then I can get $c^2$ effectively into the exponential function. But, I couldn't get that to give me a useful expression (without $t$) that I could then plot against H.
$$v^2 = a^2[\exp(2\ln(c) - 2ct)]$$
Is there some way that I can make this work rather than do a "refit"?