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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: ausyid on October 01, 2008, 11:29:00 am

Title: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ausyid on October 01, 2008, 11:29:00 am: Anyone worked out how to answer this ridiculous question?

I can scan it if people don't have it themselves.

It's the one about the graph of y versus x^3.

Thanks.
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: vce01 on October 01, 2008, 11:37:18 am: haven't got it, could you scan it?

and in an unrelated question, you don't go on bigfooty, do you?
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ausyid on October 01, 2008, 03:35:40 pm: I do go on BF actually :) What's your username on there?

Here is the question:

(http://img229.imageshack.us/img229/1733/furtherquestionpv2.jpg)

The answer is apparently C, but they give no explanation.
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: Collin Li on October 01, 2008, 03:38:01 pm: Let $z = x^3$ , so now you are looking at a $y\mbox{-}z$ graph.

Notice that it is a straight line:

$y = mz + c$

You have enough data to solve for $m$ and $c$ .

You'll get: $y = z - 15$

Now substitute back in $z = x^3$ when you're done.
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ausyid on October 01, 2008, 03:42:14 pm: You make it look so simple.

My teacher couldn't solve this question.

Thanks for the help.
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: Collin Li on October 01, 2008, 03:47:17 pm: No problem.

It's an important scientific technique for establishing empirical models. For example, if I believe that the relationship between something is an inverse relationship, i.e: $y = \frac{k}{x} + c$ , then I could do an experiment to get some data, plot the points, and try to fit it to such a curve.

A much better way to do this (especially without computers), is to let: $z = \frac{1}{x}$

Re-tabulate your data in terms of $y$ and $z$ instead (plotting $y$ and $\frac{1}{x}$ , essentially), and hopefully you should be able to draw a line of best fit to get a straight-line equation that confirms your model: $y = kz + c$ .
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ReVeL on October 01, 2008, 04:18:39 pm: Quote from: ausyid on October 01, 2008, 03:42:14 pm
You make it look so simple.

My teacher couldn't solve this question.

Thanks for the help.

Lol at that.
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ilovesuck on October 01, 2008, 04:26:43 pm: Quote from: coblin on October 01, 2008, 03:38:01 pm
Let $z = x^3$ , so now you are looking at a $y\mbox{-}z$ graph.

Notice that it is a straight line:

$y = mz + c$

You have enough data to solve for $m$ and $c$ .

You'll get: $y = z - 15$

Now substitute back in $z = x^3$ when you're done.

hm, i still dont get it :(

how did you get y = x^3 - 15 ?

i keep coming up with the answer: y = x^3 + 5 .. .. and thats not even one of the answers :(

this is really frustrating, since it SHOULD be easy sodafjwern
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ausyid on October 01, 2008, 04:31:00 pm: On closer inspection, I'm not quite sure you get y = z - 15 but rather y = z + 5. Or am I wrong?
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ausyid on October 01, 2008, 04:32:07 pm: Quote from: ilovesuck on October 01, 2008, 04:26:43 pm
Quote from: coblin on October 01, 2008, 03:38:01 pm
Let $z = x^3$ , so now you are looking at a $y\mbox{-}z$ graph.

Notice that it is a straight line:

$y = mz + c$

You have enough data to solve for $m$ and $c$ .

You'll get: $y = z - 15$

Now substitute back in $z = x^3$ when you're done.

hm, i still dont get it :(

how did you get y = x^3 - 15 ?

i keep coming up with the answer: y = x^3 + 5 .. .. and thats not even one of the answers :(

this is really frustrating, since it SHOULD be easy sodafjwern

You posted while I was posting!

I knew there was a reason why I couldn't get the answer. I kept getting y = x^3 + 5 as well.
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: Collin Li on October 01, 2008, 04:37:49 pm: Hmm... I agree. I didn't actually check the line's equation, to be honest. LOL

Yeah, I got that too!
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ReVeL on October 01, 2008, 04:41:47 pm: Same.

Must just be a shitty question.
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: Noblesse on October 01, 2008, 04:43:45 pm: I got that too :)

Anyone who does Accounting knows how many mistakes TSSM makes in those :P (a HEAP)
So it could be a mistake...
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: Collin Li on October 01, 2008, 04:46:40 pm: I didn't notice this was in the Further thread, and this was part of the graphs and relations module, otherwise I wouldn't have posted that mini-story about how scientists use this method to figure out equations... thought it was Methods. Oops!
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: Noblesse on October 01, 2008, 04:48:23 pm: What, are you saying us Further people can't cope? :P

...

*head explodes*
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ReVeL on October 01, 2008, 04:52:57 pm: Hahahaha.
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: Collin Li on October 01, 2008, 05:08:03 pm: No, I'm saying you guys have probably learnt about it already (preaching to the converted).
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: vce01 on October 01, 2008, 08:20:09 pm: Quote from: coblin on October 01, 2008, 03:38:01 pm
Let $z = x^3$ , so now you are looking at a $y\mbox{-}z$ graph.

Notice that it is a straight line:

$y = mz + c$

You have enough data to solve for $m$ and $c$ .

You'll get: $y = z - 15$

Now substitute back in $z = x^3$ when you're done.

ok im a bit confused,

when you get the

y = mz + c thing

you just work out the y-intercept and gradient in the normal way right?

so shouldnt it be

(69-45)/(64-0) = m = 3/8

and c = 45 = y-intercept

giving,

y = 3/8 z + 45. but you guys are getting something else by the looks of it...

im probably wrong and this should be easy but im really slow today :(
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: Collin Li on October 01, 2008, 08:22:22 pm: Hmm... we concluded this question was wrong, but note that the left point is actually (40, 45), so you should get $m=1$ , and $c=5$
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: vce01 on October 01, 2008, 08:40:20 pm: ohh right, shoot, i thought it was (0, 45) -_-

thankss for that lol
Title: Re: Question: TSSM 2006 Exam 1. Graphs and Relations Module. Q3.
Post by: ilovesuck on October 01, 2008, 10:05:08 pm: ^ ah yes, its quite poor they didnt put breaks in the axis...