ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: panicatthelunchbar on October 30, 2012, 12:44:05 pm
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Hey guys, need help with Question 2d, graphs/relations from the 2011 exam2....how do u do it?
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You pretty much just have to draw Katies line on the graph, then just look how long they are 3km's apart.
The answers was like 1.38 Hours which is ridiculous considering its from observation.
I did it another way which was way over complicated and was not taught in the FM course. If anyone has a similar more accurate way of doing it, confabulate :)
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Can you link me to the question? I've done it before, it's a little complex.
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http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011furmath2-w.pdf
pg 23
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My \LaTeX skills fail, so let me try this:
Her equation minus his equation has to be greater than or equal to 3 (for the first part of it): (-3x+16)-(5x) <= 3. Her's is first because it is on top and higher up on the graph.
Therefore, x >= 1.625 (13/8). So starting from x = 1.625 and going onwards is when they're going to have connectivity. So now we need a boundary on the right side.
His next equation is 10 (2<=x<=3) and her's is still -3x+16.
Therefore: (10)-(-3x+16) <= 3. His is on top now because it is higher up on the graph.
So x <= 3.
So our answer is our right boundary - our left boundary.
So 3 - 1.625 = 1.375.
Which to two decimal places will round to 1.38 hours.
I hope that helped! :)
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You pretty much just have to draw Katies line on the graph, then just look how long they are 3km's apart.
The answers was like 1.38 Hours which is ridiculous considering its from observation.
I did it another way which was way over complicated and was not taught in the FM course. If anyone has a similar more accurate way of doing it, confabulate :)
It's not just observation. Yes, you're able to obtain that there is at least one hour just by looking at the graph. Then, you set up an equation like this:
D(Katie)-D(Michael)=3.
Using that, you're able to find the time and subtract that from the highest time value which you found from observation. Voila, 1.375 hours. :)
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Oh no i did that. but my calculations were not from FM course it was from methods using hybrid notations and such, so i figured for a questions like that it was probably suppose to be something as straight forward as just looking at it.