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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: diam0nds on March 03, 2012, 06:03:18 pm

Title: Help? :)
Post by: diam0nds on March 03, 2012, 06:03:18 pm: Hey guys I have this analysis thing to do and it's due on Monday but I'm struggling with some of the questions -

A section of a roller coaster ride, as shown below, has the appearance of three quadratic functions joined together:

(http://i44.tinypic.com/2dmduts.png)
(sorry for the dodgy photo lol)

The points O, A, B and C indicate the start and finish of each of the three quadtratic functions used to model the track. By using a set of co-ordinate axes, the equations of each can be determined

* The first function starts at the origin and finishes at A and has a turning point at (4,4)
* The second function starts at A, and finishes at B with a turning point at (8,2)
* The final function starts at B and finishes at C. Note: the turning point is not at (15,12)

1. Find the equation of the curve joining the points BC
The 2 questions before this were the same but for OA and AB, and all I did for those was use three sets of coordinates, subbed them into the ax^2 + bx + c equation then solved for a, b and c on my calculator to get the equations, but because it says the turning point for this curve is not (15,12) I'm not sure how to get the third set of co-ordinates :S

2. Find the average rate of change of height between x=6 and x=12
I know you have to use differentiation and I need to look over my notes from last year, but what equation do you differentiate?

3. Model the track with two cubic equations. Discuss the inaccuracies of the cubic model in comparison to the quadratic model
Umm how do I approach this question? I'm not actually sure what it's asking lol

Sorry for the huge post, thanks so much if you can help :)
Title: Re: Help? :)
Post by: Stick on March 03, 2012, 07:40:38 pm: All I know is that for question 2, it does not involve differentiation. Because it's asking for the average rate of change, it's just like finding the gradient of a straight line.

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{7-3}{12-6}$

$m=\frac{4}{6}$

$m=\frac{2}{3}$

I don't know questions 1 and 3 though. :S
Title: Re: Help? :)
Post by: Phy124 on March 03, 2012, 07:49:01 pm: Do the three equations join smoothly (at A and B)? If so the gradient of the orange and green equations will be the same at B. So if you differentiate the orange equation at B you can get the answer of that and make it equal to the derivative of the green equation at this point.
Title: Re: Help? :)
Post by: diam0nds on March 03, 2012, 08:18:13 pm: Thanks guys @Stick ohhhh duh that makes sense haha thank you :)

@Phy124 yeah they do join smoothly I just highlighted the lines to look at the different curves, the equation I have for the orange one is $\frac{x^2}{4} - 4x +18$ , and I understand what you mean by differentiating it at B, but then how do I get the full equation for the green line?
Title: Re: Help? :)
Post by: Phy124 on March 03, 2012, 08:26:00 pm: Quote from: diam0nds on March 03, 2012, 08:18:13 pm
Thanks guys @Stick ohhhh duh that makes sense haha thank you :)

@Phy124 yeah they do join smoothly I just highlighted the lines to look at the different curves, the equation I have for the orange one is $\frac{x^2}{4} - 4x +18$ , and I understand what you mean by differentiating it at B, but then how do I get the full equation for the green line?
Ok so let orange line $= f(x) = \frac{1}{4}x^{2}-4x+18$

$f'(x)=\frac{1}{2}x-4$

$f'(14)=\frac{1}{2}(14)-4=3$

Now for the green line:

$f(x) = ax^{2}+bx+c$

$f'(x)= 2ax + b$

We can work out 3 simultaneous equations from the given information:

$11=196a+14b+c$

$3=324a+18b+c$

and finally since gradient of orange = gradient of green at B (x=14)

$3=28a+b$

Where 3 is the gradient and 28 is made up of 2 x 14 (the x-value)
Title: Re: Help? :)
Post by: diam0nds on March 03, 2012, 08:31:23 pm: Yay thank you so much that makes sense now :)
Title: Re: Help? :)
Post by: diam0nds on March 04, 2012, 02:50:17 pm: IS there any chance anyone knows how to do question 3? I have no idea how to model it with two cubic equations lol ???
Title: Re: Help? :)
Post by: Phy124 on March 04, 2012, 04:31:44 pm: Well you would have 2 equations of $ax^{3}+bx^{2}+cx+d$

To get values of a,b,c,d so that the graph would fit the shape you may put things such as:

It goes through the point (0,0)
It goes through the point (6,3)
It goes through the point (8,2)
It goes through the point (14,11)

Or you may wish to use the fact the gradient is 0 at x = 4, x= 8 instead of two of the coordinates stated above (putting them into $3ax^{2}+2bx+c$ )

You would also do the same for the second cubic.

The problem is, you have 4 unknowns so you're only going to be able to get 4 characteristics the same as the original. These would be determined whether you use coordinates which the graph should go through or turning points at certain coordinates etc.
Title: Re: Help? :)
Post by: diam0nds on March 04, 2012, 05:59:31 pm: Ahh you're a legend thanks! :D if I didn't attempt it my teacher would have eaten me