I have four data sets which comprise non-linear functions, and my teacher wants me to linearize them. Unfortunately, I am utterly confused.
The first data set is an inverse function. This is the data:
| V ($m^3$) | P (pa) |
|---|---|
| .1 | 40 |
| .5 | 8 |
| 1 | 4 |
| 2 | 2 |
| 3 | 1 |
| 4 | .8 |
| 5 | .5 |
| 8 | .4 |
This is the graph, which was made using Vernier Graphical Analysis:
The equation of this graph is $V=\frac4P$, or $V=4\frac1P$. Because I have no idea how to linearize the graphs, I tried finding the inverse of this:
$V=4\frac1{\frac1P}$, or $V=4P$, making it proportional and creating a straight line.
However, I don't know whether this is allowed--it seems to me that this isn't linearization, it's just outright changing the function. After linearizing, I'm also supposed to find the equation $V=m\frac1{\frac1P}+b$. I'm hoping that $b$ is $0$, and that the equation is just $V=4\frac1{\frac1P}$.
The second data set creates an exponential function:
| t (s) | x (m) |
|---|---|
| .1 | .03 |
| .2 | .12 |
| .5 | .75 |
| 1 | 3 |
| 2 | 12 |
| 3 | 27 |
| 4 | 48 |
| 5 | 75 |
The equation is $x=3t^2$. My teacher told me to plot $x=mt^2+b$. What does this mean? Would I ignore the exponent and just treat it as a linear equation?
The third data set seems to be linear, so I won't go over that.
The fourth data set is a square root function. The table is shown below:
| t (s) | v (m/s) |
|---|---|
| .3 | 10 |
| 1.2 | 20 |
| 2.7 | 30 |
| 4.8 | 40 |
| 7.5 | 50 |
| 10.8 | 60 |
| 14.7 | 70 |
| 19.2 | 80 |
The equation is $v=18.26\sqrt{t}$ or $v=18.26\,x^{1/2}$. My notes tell me to plot $v^2$ vs $t$, then find the linear expression $v^2=mt+b$.
How would I plot $v^2$?
Is there something that I'm misunderstanding?
Or am I supposed to change the scale of one of the axes to linearize the graph? If so, how?
I understand that these questions might not be clear, but please try to bear with me!
ad Question 1 (inversely proportional)
You already know or conjectured that $$v\sim p^{-1}$$ where $\sim$ means proportional. Two quantities $A$ and $B$ are proportional if they satisfy $A=\mathrm{const}\cdot B$, so in the case in question we have that $$p\cdot V=\mathrm{const}$$ In order to determine the constant $c$ of proportionality, you can take the arithmetic (or maybe geometric) mean of the $n$ values $p_k\cdot V_k$, i.e. $$c = \frac 1n\sum_{k=1}^n p_kV_k$$
For a graphical analysis, plot the values $p_k\cdot V_k$. You know the result should be a single point, but due to measurement error it's a cloud of points. So chose a point that's a goot match, like the center of mass of the points (which is their arithmetic mean).
ad Question 2 (quadratic) and Question 3 (square-root).
Compute$^1$ $v^2$ and plot that against $t$? The points should organize along an approximate line, and you can determine the parameters from the drawing or plot.
If you want to do it computationally: Suppose you have $n$ measurements $x_k$ and $y_k$ that are supposed to be in linear relation according to $$y_k = ax_k + b\tag 1$$ and you want to determine $a$ and $b$. Having $n$ equations usually means that (1) is over-determined, because you have $n$ equations that determine just 2 values: $a$ and $b$. So in general, there is no solution (just like the points won't lie on a perfect line in general, but you want a line anyway).
One can write the system (1) as linear system
$$ \underbrace{\begin{pmatrix}y_1\\y_2\\\vdots\\y_n\end{pmatrix}}_{\textstyle =:Y} = \underbrace{\begin{pmatrix}x_1 & 1\\x_2 & 1\\\vdots \\x_n& 1\end{pmatrix}}_{\textstyle =:X} \binom ab $$
This system $Y=X\binom ab$ is usually over-determined, but multiplying with $X^t$ from the left will generate a 2×2 matrix $X^tX$:
$$X^tY = X^tX \binom ab$$
That system will have a solution, and the values for $a$ and $b$ that you'll get are optimal in some sense. Solving that 2×2 system is almost trivial, and you get $a$ and $b$ as
$$\binom ab = (X^tX)^{-1} X^tY$$
where $$X^tY = \binom{\sum x_ky_k}{\sum y_k}$$ and
$$X^tX = \begin{pmatrix} \sum x_k^2 & \sum x_k \\ \sum x_k & n \end{pmatrix}$$
$^1$I cannot help you with Vernier Graphical Analysis or whatever you are using, maybe there's some shortcut and the software does the computation for you. But if not, you can always do the computation by hand.