Differentiation question(Involving LC Chemistry graph question)

JC 2K3

Was gonna post this in Physics & Chemistry, but this is really a maths question.

http://www.examinations.ie/archive/exampapers/2005/LC022ALPO00EV.pdf

Question 3 (e) (ii)
For those of you who don't want to bother opening that link it's basically draw a graph with the points: (0,0) ; (1,13.5) ; (2,23.4) ; (3,30.5) ; (4,35.4) ; (5,38.3) ; (6,39.6) ; (7,40) ; (8,40) and find the slope of the tangent at x=2.5.

The way you're supposed to do it is draw a graph, draw a tangent at x=2.5 and work out the slope based on tan of the angle between the tangent and the X axis. Pretty inaccurate way of doing it, they give a range of 6.0-8.0 for the answer.

Now I, preferring to be 100% accurate rather than relying on graphs, decided to try and use differentiation to solve it, so I took the points (1, 13.5), (2,23.4) and (3,30.5), put them into 3 y=ax^2+bx+c equations and got dy/dx when I got values for a and b. Here's how I got on:

HIGHLIGHTING MY MISTAKE IN RED, UGH, I FEEL SO STUPID
y=ax^2+b+c
13.5=a+b+c..........(1)
23.4=4a+2b+c.......(2)
30.5=9a+3b+c.......(3)

9.9=3a+b.....(2)-(1)
7.1=5a+b.....(3)-(2)

1.8=-2a.....((2)-(1))-((3)-(2))
a=-0.9
b=12.6

2.8=-2a.....((2)-(1))-((3)-(2))
a=-1.4
b=14.1

dy/dx=2ax+b
dy/dx(@x=2.5)=2(-1.4)(2.5)+14.1
dy/dx(@x=2.5)=7.1

EDIT: My error is corrected and it's within the range now, but is this the correct way to go about doing this?

Pherekydes

The question clearly states: (e) Use the graph to determine...

It doesn't say anything about using differential equations. Always answer the question you are asked, the way you are asked to answer it.

Additionally, there are just 9 marks for parts (i) and (ii). This is just 2.5% of the total for the paper, and wouldn't justify using differential equations.

Also, part of the skillset of a chemist or any scientist is estimating stuff from graphs rather than taking the pains to calculate a precise answer, which is what they are looking for in this case, I think.

Open to correction as I am no chemist/scientist.

JC 2K3

I didn't post this in the LC forum for a reason....

I won't be doing this on the day of the LC, I just went off on a bit of a tangent(I'm so witty) whilst doing my chemistry homework, and I just found it odd that the precise answer wouldn't be included in the range of accepted answers.

Forgetting that it originated as a chemistry question, is my maths right?

breadmonkey

How can you be sure that the graph is a quadratic?

seinstein

breadmonkey wrote:

How can you be sure that the graph is a quadratic?

I second this! I mean, how did you come up with this formula - this is like curve-fitting in statistics or something; what's the basis for using this formula?
Is it that the equation for a line is y=mx+c, and so a curve is y=ax^2+bx+c (actually y=ax^2+b+c, which you wrote, might be wrong!!)? But then, where did you come up with the idea to plug these values in?
Try fixing the formula with that value and then try the sum again.

Edit: actually I think you were right, but I think at the last part you have to calulate the slope m, with y=mx+c, and then plug it into the differential equation. I think the equation for the slope is m=(y2-y1)/(x2-x1) - might be wrong but just take the values from the table values - then that's your slope. No need for differential equations, I think that using it at this point wouldn't make sense as you have no reason to use one! Also the fact that the differential equation you resulted with (dy/dx=2ax+b) implies it is linear and describes only the slope at any particular point on a curve, but you calculated it incorrectly.

y=ax^2+bx+c
dy/dx=2ax+b

From (1),

30.5-9a-3b=c (FOUND c)

plugging into (2) gives

23.4=4a+2b+30.5-9a-3b=-5a-b+30.5, therefore

-b=5a-30.5+23.4
-b=5a-7.1
b=7.1-5a (FOUND b)

and plugging this and the above into (1) gives a:

13.5=a+7.1-5a+30.5-9a-21.3+15a=2a+16.3, or

13.5=2a+16.3
a=-1.4

Now we have a=-1.4, we find b=14.1 and c=85.4

From the equation y=ax^2+bx+c,

dy/dx=2ax+b
dy/dx=2(-1.4)x+14.1=-2.8x+14.1

At x=2.5, the slope is dy/dx=7.1.

You made a slight calculation blunder

A great way of 'testing' your results if you ask me!

breadmonkey

Given the context of the question, there's no way this can be modelled as a quadratic. I couldn't be arsed drawing the graph but I'd say you got an answer close to the correct one because the region you analysed further was quadraticish. Know what I mean?

JC 2K3

breadmonkey wrote:

Given the context of the question, there's no way this can be modelled as a quadratic. I couldn't be arsed drawing the graph but I'd say you got an answer close to the correct one because the region you analysed further was quadraticish. Know what I mean?

I've never attempted a question like this before, I presumed I could take 3 points from the graph and neglect the rest, and that would give me a definate curve in which I could find the slope of the tangent at x=2.5.

I'm reading your post now, seinstein.

breadmonkey

Think about it: if that were true then why did you choose to "fit" the points to the curve ax^2 + bx +c. Why not a parabola or something?

JC 2K3

Like I said, I've never done a question like this before, this isn't on the LC maths syllabus.

My reasoning behind what I did is that the slope of the tangent at x=2.5 would be the same as the slope of the tangent at x=2.5 in a quadratic function with the points (1, 13.5), (2,23.4) and (3,30.5).

Your knowledge is probably much better than mine, so I'll accept a "no, you just can't do that" if my reasoning is way off and it's too complex to explain.

breadmonkey

JC 2K3 wrote:

My reasoning behind what I did is that the slope of the tangent at x=2.5 would be the same as the slope of the tangent at x=2.5 in a quadratic function with the points (1, 13.5), (2,23.4) and (3,30.5).

Why? There are more curve functions than quadratics. I'm explaining this badly, I'll have another stab at it later maybe.......

JC 2K3

I've edited my original post, thanks seinstein, stupid mistake on my part.

I know there're more curve functions than quadratics, but with any 3 points you can define a quadratic curve right? And I can't see how the slope of the tangent wouldn't be the same...

I'm quite crap image making, but this graph I made in paint(attached) attempts to explain what I mean. The black curve is the original curve, the blue curve is a quadratic curve which fits the three enlarged black points. The light green point is the point I want the tangent for and the red line is the tangent I want the slope of.

(note that when the blue curve is increasing it wouldn't necessarily have to be that close to the black curve, just the way my example graph worked out)

Sev

breadmonkey wrote:

Think about it: if that were true then why did you choose to "fit" the points to the curve ax^2 + bx +c. Why not a parabola or something?

"ax^2 + bx +c" actually defines a parabola.

But anyway, I thought the method was pretty smart... A brief wikipedia has led me to believe what youre doing is called "polynomial interpolation".. it makes sense.. its one of the methods excel uses to fit points to a line.

http://en.wikipedia.org/wiki/Interpolation

Now.. of course at the beginning of all this comes the decision that a polynomial is the best choice of a prototype function for which the curve should fit too. I dont know much about this really.. but from common sense I would imagine that polynomial interpolation is fairly close approximation to exactly how the human brain might imagine connecting the dots with a smooth curve which is what you would be doing anyway if you sketched the graph with a pencil.

A criticism is, however, that you were given nine points. So you could have done this interpolation to 8th order, and gotten 9 equations each of the form

y = ax^8 + ax^7 + ax^6 + .. etc.

The way youve done it is taken a small window around 3 points.. and those 3 points could have been quite some distance from the rest of the other points, meaning it could have been a very pointy steep walled 8th order curve around that region, and not a parabola.. if that makes any sense.

I wouldnt take what Ive said as the absolute truth tho.. its my rash common sense talking, not experience or study.

breadmonkey

"ax^2 + bx +c" actually defines a parabola.

Sorry, I don't know what I was thinking.

OK, correct me if I'm wrong but I don't think anyone has got the equation right yet. This is what I've got:

y = ax^2 + bx + c

using point (0,0)
0 = a*0 + b*0 +c
c = 0

using point (1,13.5)
13.5 = a + b + c (c=0)
13.5 = a + b
a = 13.5 - b -->(1)

using point (2, 23.4)
23.4 = 4a + 2b +c
23.4 = 4a + 2b -->(2)

now using (1) and (2)
23.4 = 4(13.5 - b) + 2b
23.4 = 54 - 4b + 2b
2b = 30.6
b = 15.3

a = 13.5 - 15.3 =-1.8

so,
a = -1.8
b = 15.3
c = 0

This gives the equation
y = -1.8x^2 + 15.3x

I hope that's right because I'll feel like an idiot otherwise.

Anyway,

me wrote:

I couldn't be arsed drawing the graph but I'd say you got an answer close to the correct one because the region you analysed further was quadraticish.

Look at the graph in the attached Excel file.

I turns out that the quadratic actually matches the original data curve almost exactly up until around the point (3,27). You are analysing the graph at x = 2.5 so that's why it works.

I'm still not happy about it though since you can see from the graph that this won't work for x = 4 (for example). Also, as I sort of mentioned before, the entire curve could never be modelled as a quadratic since the volume of o2 plateaus and isn't going to start plummeting again.

JC 2K3

breadmonkey wrote:

Sorry, I don't know what I was thinking.

OK, correct me if I'm wrong but I don't think anyone has got the equation right yet. This is what I've got:

y = ax^2 + bx + c

using point (0,0)
0 = a*0 + b*0 +c
c = 0

using point (1,13.5)
13.5 = a + b + c (c=0)
13.5 = a + b
a = 13.5 - b -->(1)

using point (2, 23.4)
23.4 = 4a + 2b +c
23.4 = 4a + 2b -->(2)

now using (1) and (2)
23.4 = 4(13.5 - b) + 2b
23.4 = 54 - 4b + 2b
2b = 30.6
b = 15.3

a = 13.5 - 15.3 =-1.8

so,
a = -1.8
b = 15.3
c = 0

This gives the equation
y = -1.8x^2 + 15.3x

I hope that's right because I'll feel like an idiot otherwise.

Anyway,

Look at the graph in the attached Excel file.

I turns out that the quadratic actually matches the original data curve almost exactly up until around the point (3,27). You are analysing the graph at x = 2.5 so that's why it works.

I'm still not happy about it though since you can see from the graph that this won't work for x = 4 (for example). Also, as I sort of mentioned before, the entire curve could never be modelled as a quadratic since the volume of o2 plateaus and isn't going to start plummeting again.

We don't need the slope of the tangent at x=4 so it's irrelevant. You can't use the 3 points that you picked because x=2.5 isn't between x=0 and x=2. If I wanted the slope at x=4 I'd have used (3,30.5), (4,35.4), (5,38.3) and fit a curve to those points. Understand what I'm saying?

A criticism is, however, that you were given nine points. So you could have done this interpolation to 8th order, and gotten 9 equations each of the form

y = ax^8 + ax^7 + ax^6 + .. etc.

The way youve done it is taken a small window around 3 points.. and those 3 points could have been quite some distance from the rest of the other points, meaning it could have been a very pointy steep walled 8th order curve around that region, and not a parabola.. if that makes any sense.

But the slope is the same whether you take 3 points or 9 points, using 9 points is just making more work for yourself.....

JackieChan

JC 2K3 wrote:

We don't need the slope of the tangent at x=4 so it's irrelevant. You can't use the 3 points that you picked because x=2.5 isn't between x=0 and x=2. If I wanted the slope at x=4 I'd have used (3,30.5), (4,35.4), (5,38.3) and fit a curve to those points. Understand what I'm saying?

You would be kind of extrapolating here, inherently less acurate than interpolating

JC 2K3

The slope wouldn't be the same?

breadmonkey

Well, it certainly does seem to work, at least in this example. So, was this a special case or will it always work?

Sev

JC 2K3 wrote:

But the slope is the same whether you take 3 points or 9 points, using 9 points is just making more work for yourself.....

No.

Here's an example in which the slope is very different if you pick 6 points as opposed to only 3.

The longer curve represents 5th order interpolation. The shorter curve, interpolation to 2nd order,ie. using a quadratic expression, around the last 3 points.

Take a look at the slope in each case around 3.2.

The set of data points used is

0, 1
1, 2
2, 1
3, 4
4, 2
5, 1

JC 2K3

Hmm... But what if Y is always either increasing or staying the same? As in the question we're dealing with here.....

Is there an easy way to do an 8th order interpolation? Like using Excel or something?

breadmonkey

JC 2K3 wrote:

Is there an easy way to do an 8th order interpolation? Like using Excel or something?

Not in an exam;)

Sev

JC 2K3 wrote:

Hmm... But what if Y is always either increasing or staying the same? As in the question we're dealing with here.....

Is there an easy way to do an 8th order interpolation? Like using Excel or something?

Em yeah.. you make your "XY (Scatter)" graph.. then right click any one of the points on the graph and click "add trendline".

You choose polynomial, then choose to what order. I dont know exactly what excel does if you do the 2nd order interpolation with more than 3 points on your graph tho.. but if its 3 points.. it does exactly what you did in your question.. fits a quadratic around those 3 points.

Likewise if its 9 points in the graph.. it will fit an 8th order polynomial around those 9 points. If its like 11 points.. it does some kind of averaging thing and the line doesnt go exactly through the points anymore.

[EDIT] It turns out Excel only allows to 6th order [/EDIT]

Sev

JC 2K3 wrote:

Hmm... But what if Y is always either increasing or staying the same? As in the question we're dealing with here.....

Yes.. but if the question of whether or not to answer the question by fitting the points to a quadratic is based entirely on your visual judgement of the situation.. then using such a method to answer the question in an attempt to be clever or improve accuracy is useless. You might as well just estimate the slope from the rough graph which you would have had to draw anyway.

JC 2K3

lol, screw the exam for a second, does any sort of online interpolating Java applet exist?