0

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!

Frentos
  • 3,041
  • None of those equations seems to be a straight line, so what do you mean by "slope-intercept" form? Are you thinking of the equation of the tangent line to those functions at some point? – DonAntonio Feb 06 '16 at 15:54
  • x and y are measurable quantities and a and b are constants. I am trying to rearrange each of the equations to a linear relationship so I can indicate how 'a' and 'b' can be determined from the slope and intercept of the resulting straight line if I were to plot the data. –  Feb 06 '16 at 16:33

1 Answers1

0

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.)

grand_chat
  • 38,951
  • Thank you very much for your assistance. Your explanations make perfect sense. I have one question though: in the form of y=mx+c, can the 'y' actually be anything other than just 'y'? In one example you have the 'y' as 'y/x' and another you have 'ln(y/x)'. Would I not need to arrange it so only the Y exists on the left side? Any clarification you may provide would be much appreciated! –  Feb 07 '16 at 00:16
  • Yes, the $y'$ can be, and sometimes has to be something other than the original $Y$. If you insist on keeping only $Y$ on the left side, you may not be able to achieve the form $y'=mx'+c$. It was possible in the first equation to leave just $Y$ on the left side, but I don't see how to do that in the case of the second and third equations. – grand_chat Feb 07 '16 at 00:29
  • Also, in the first example you had Y=a*(1/x) + (1/b)

    would having (1/x) be ok? would it not need to take the format of y=mx?

     –  Feb 07 '16 at 00:31
  • I'm guessing that you are allowed to define new variables like $1/x$. If you define the variable $x' = 1/x$ and the constants $m=a$ and $c=1/b$, then your equation has the desired form $y = mx' + c$. In other words, the equation takes the form of a straight line provided you think in terms of this new variable; otherwise you won't get the desired form. – grand_chat Feb 07 '16 at 00:37
  • perfect! This line of thought simplifies everything. I would just need to keep any equation simply in terms of 'y' and 'x' and not have any instance where either 'y' or 'x' would be raised to a power higher than '1', correct? –  Feb 07 '16 at 20:31
  • I hope this doesn't make things more confusing, but even the equation $y^2 = ax + b$ has the desired form, provided you think of $y^2$ as the variable you are plotting on the vertical axis. In other words rather than plotting $y$ vs $x$, you square the value of $y$, obtaining $y^2$, which you plot vs $x$. As another example, the equation $y=ax^2 + b$ has the desired form, provided you think of $x'=x^2$ as the variable on the horizontal axis. What all these equations have in common is that they can be viewed as $(\mbox {some variable}) = m \times (\mbox{some other variable}) + c$. – grand_chat Feb 08 '16 at 01:39
  • Applying what you taught me, for the following equations I believe I have solved these two equations:$y=ae^\left(-bx\right)$ solved by: $ln(y)=-bx + ln(a)$ where $slope = -b$ and $intercept = ln(y)$. The other equation is: $y=ax^b$ solved by: $ln(y)=b*ln(x)+ln(a)$ where $slope = b$ and $intercept = ln(a)$... does this sound about right to you? (I apologize if my formatting is terrible as I am new to this site) –  Feb 08 '16 at 15:58
  • Yes, you have solved these correctly! (You have a typo in the first equation: the intercept is $\ln(a)$.) I think you're getting the hang of it – grand_chat Feb 08 '16 at 17:56