1. The problem statement, all variables and given/known data How do you linearize a parabolic graph? the parabolic graph is obtained from an experiment relating to projectile motion. Angle (tan) Average Displacement 20.11 15.72 24.3 18.84 28.98 20.4 38.29 21.68 49.67 18.9 61.53 18.44 74.6 10.18 This is the data obtained from the experiment. The equation of the parabola is found to be: -0.01x^{2}+0.844x+3.723 3. The attempt at a solution I've tried squaring the x-axis datas... cos the data... and all sorts of things and still no linear graph... please help!
Funny, I was just helping a friend with the same thing. Try the mathematical method: L(x) = f(a) + f '(x)(x-a) Where x, is the point of interest and a is the point where the function is linearized.
There are two things people mean when they say "linearize" a graph or data. 1. As Topher925 said: approximate the graph as-is with a straight line. 2. What I think OhyesOhno means: transform the data so that the graph really is a straight line. For example, suppose we knew that y = mx^2 + b The graph is a parabola, requiring least-squares regression to find m and b. But if we let z=x^2, we get y = mz+b Plotting y vs. z gives a straight line. Simply using a ruler, we can get a pretty decent estimate of m and b. However, it's more complicated in general if we have y = ax^2 + bx + c (because of the "bx" term present) OhyesOhno, am I correct in your meaning of the word "linearize"?
Yeah, I agree with redbelly that this is probably a question designed to teach you how to manipulate data sets to get linear fits. So if you have a parabola, and you square one of your data columns (or take the square root of the other), you get a straight line.
yes, redbelly is correct. I'm supposed to transform the graph into a straight line... So is merryjman's method correct? or i should use Topher925's method? thanks again
oh and one more thing... I've squared each of them and and square rooted them as well... But still no straight line.
how exactly did you get this equation,-0.01x^{2}+0.844x+3.723? EDIT: is it y=-0.01x^{2}+0.844x+3.723? Where you can find the vertical displacement,y, given a horizontal displacement,x?
I'm guessing by "x" the meant the angle, while the displacement (whatever direction it's in) is the dependent variable. Use the least squares regression line. Are you allowed to use a graphing calculator? It's something like LinReg(???) L1 L2 with the angle in L1 and displacement in L2...
@rock.freak667 I found the equation using Microsoft Excel... I made a graph of the data and put a trendline in the shape of 'Polynomials' (this is Ms Excel 2007) with the power of 2. y is the horizontal displacement... x is the angle, just like what ephedyn said @ephedyn yes i'm allowed to use a graphing calculator (mine is TI-84 plus). What does it mean by least squares regression line?
The least square regression line is one where, given a set of data (which do not necessarily lie on a straight line as in this case) the {sum of [the squares of (the vertical distances from the data points to the line)]} is a minimum, where your vertical axis is your dependent variable (displacement), and your horizontal axis is your independent variable (angle). So, geometrically, we interpret that it basically gives you a line with a snug fit to your data points. If I am not wrong, go to - [2nd] [STAT] to bring up the editor of your lists. Put your values of x into column L1. Put your corresponding values of y into column L2. - then play around with [STAT] I think... find a function called LinReg(a+bx). Punch that in, then use [2nd] [1] and [2nd] [2] to get call out your L1/L2 variables. The syntax is: LinReg(a+bx) L2 L1. Notes: - The calculator will give you an equation in the form y = mx+c, where b is your gradient, a is your y-intercept. - In this case, LinReg(a+bx) L2 L1 because you are finding the regression line of y on x (L2 on L1). - There's also a LinReg(ax+b) if I recall. Don't get mixed up, but it works just as well but a and b and/or L1 and L2 have swapped their roles or something. Play around a little and you'll figure out how to use both. I can't describe further without a graphing calculator with me. Hope this helps.
I don't think just squaring or square-rooting will do it in this case. There could be some trig functions involved also. When I gave the parabola example earlier, it was just an example meant to illustrate the point. Often, knowing a general form for the equation that the data should follow is needed before figuring out how to linearize the graph. OhyesOhno, is there equation for range vs. angle already given in your textbook or class/lab notes?
well, I researched and there is an equation for range vs angle R= (u^2 sin2θ)/g R is horizontal range (or displacement), u is initial velocity, theta is angle and g is gravity (9.8 ms^-2) I asked my teacher about the relationship between angle and distance and he said R is proportional to theta...
Good, that's the equation we want. It suggests making a graph where y is R x is ___? p.s. That's weird what your teacher said. Well, you (and I also) have graphed the data and saw R is clearly not proportional to theta.
oh sorry redbelly, he meant was range is proportional to sin2theta... so I'm guessing.... x is sin2theta right??