It's been awhile since I've worked problems like these and I am a bit hazy on some of the rules. I was hoping someone could show me how these are solved so that I can make sure I'm on the right path:
Equations:
$$Y=\dfrac a x + \dfrac 1 b\\
Y=\dfrac a x + \dfrac x b\\
Y=axe^{-b/x^2}$$
I need to convert those equations into a form so that I can determine the values of '$a$' and '$b$' based on the slope and intercept. Any assistance you may provide would be much appreciated. Also, if you have a link to the rules you used so that I can use to help solve the rest of the problems, it would be much appreciated!
Best Answer
Not sure if this is what you are asking, but:
For the first equation, you can write it as $$Y = a\cdot\frac1x + \frac1b.$$ Think of plotting $Y$ against $\frac 1x$. If the plot turns out to be a straight line, then the slope of the line is $a$ and the intercept is $\frac1b$, so $b$ is determined as the reciprocal of the intercept.
For the second equation, multiply both sides by $x$ to obtain $$xY = a + \frac1b x^2.$$ So if plotting $xY$ on the vertical axis and $x^2$ on the horizontal axis produces a straight line, then the intercept of this line is $a$ and the slope is $\frac1b$.
For the third equation, you can divide both sides by $x$ to obtain $$\frac Yx = ae^{-b/x^2}$$ and then take logs of both sides, resulting in $$\ln\left(\frac Yx\right) = \ln a - \frac b{x^2}.$$ So if a plot of $\ln(Y/x)$ versus $\frac1{x^2}$ is a straight line, then the intercept is $\ln a$ and the slope is $-b$, from which you can determine $a$ and $b$.
As for hints on how to do this in general, the name of the game is to rearrange the equation and/or transform it until you can end up with something of the form $$y'=mx' + c$$ where I'm using $y'$ and $x'$ to represent new variables, and $m$ and $c$ are constants. It requires some trial and error, and knowledge about the effect of various transformations. (Note that it's not always possible to end up with an equation that isolates $m$ and $b$ in this way.)