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March 29, 2024, 10:24:00 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164419 times)  Share 

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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #810 on: October 16, 2012, 10:30:36 pm »
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Well you're right, they have the same asymptotes. =/
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martin1106

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Re: Specialist 3/4 Question Thread!
« Reply #811 on: October 17, 2012, 06:23:50 pm »
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Well you're right, they have the same asymptotes. =/

thanks!

generalkorn12

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Re: Specialist 3/4 Question Thread!
« Reply #812 on: October 19, 2012, 04:49:50 pm »
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For  dependence, how do we know which of these formulas to use?
i) ma+nb+oc=0
ii) a=mb+nc

I used (i) for the 2008 VCAA exam,  and got into a messy situation, and the solutions suggested using (ii), but i'm having trouble working out the difference on when switch between both .......
« Last Edit: October 19, 2012, 05:36:56 pm by generalkorn12 »

Lasercookie

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Re: Specialist 3/4 Question Thread!
« Reply #813 on: October 19, 2012, 05:03:57 pm »
+2
With ma+nb+oc=0, if m = n = o = 0 is the only solution, then the vectors are linearly independent.

If a=mb+nc is true, then the vectors are linearly dependent. This should make sense, because you have one of the vectors being expressed in terms of the other vectors (i.e. it "depends" on the other vectors). You can derive this expression from the first equation too.

soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #814 on: October 20, 2012, 10:15:57 am »
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VCAA 2011 MC 7 EX2
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf

Does anyone have any fast ways of doing this question?
I graphed each graph to see which intersects twice but it takes a long time to do it this way.
Also, i tried using the solve command, but when i solved the eqn with option c, how come it only gave one coordinate instead of two?

Thanks for any help :)
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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #815 on: October 20, 2012, 12:37:33 pm »
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I kind of just knew what the graphs looked like from experience, so they didn't take long to do quick sketches of. Do you know the perpendicular bisector thing? For lack of a better word =P
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soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #816 on: October 20, 2012, 12:45:30 pm »
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I think i came across it once, but i don't rmb much about it, could you perhaps explain to to me please?
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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #817 on: October 20, 2012, 01:36:34 pm »
+3
|z - u| = |z - v| (where u and v are complex numbers) represents the perpendicular bisector of the line joining u and v.

So for option B in that question they had |z + 3| = |z - 3i|, which is equivalent to |z - (-3)| = |z - 3i|. So it's the perpendicular bisector of the line joining -3 and 3i, which is the line Im(z) = -Re(z) (or alternatively, y = -x).

Option C is the perpendicular bisector of the line joining 3 and 3i, which is the line Im(z) = Re(z) (y = x). I think this was the correct answer.

Try and come up with a worded explanation of why that general expression gives the perpendicular bisector. If you can't get it then I'll tell you, but have a go.
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soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #818 on: October 20, 2012, 07:25:19 pm »
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Thanks a lot for your help :)

Okay, I'll try: Is it because u will represent the x coordinate and v will represent the y coordinate. So the point (u,v),now called (x,y), will form a right angled triangle when joining to the origin using a straight line? Hence, that general expression gives the perpendicular bisector?
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Jenny_2108

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Re: Specialist 3/4 Question Thread!
« Reply #819 on: October 20, 2012, 07:50:53 pm »
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Is it because u will represent the x coordinate and v will represent the y coordinate. So the point (u,v),now called (x,y), will form a right angled triangle when joining to the origin using a straight line? Hence, that general expression gives the perpendicular bisector?

what do you mean "the point (u,v)"?
Its the distance of z to the point u equal to the point v. Have a go again :P
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soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #820 on: October 20, 2012, 08:07:30 pm »
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Is it because u will represent the x coordinate and v will represent the y coordinate. So the point (u,v),now called (x,y), will form a right angled triangle when joining to the origin using a straight line? Hence, that general expression gives the perpendicular bisector?

what do you mean "the point (u,v)"?
Its the distance of z to the point u equal to the point v. Have a go again :P
Umm, so it forms a square? I'm confused now, i really don't know :S
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Re: Specialist 3/4 Question Thread!
« Reply #821 on: October 20, 2012, 09:26:17 pm »
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This might seem a little vague, but what does ' the acceleration is always directed towards the origin' exactly mean?
What context exactly? I guess it could refer to vertical motion (i.e., objects being thrown directly upwards) always accelerating downwards due to gravity.
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Re: Specialist 3/4 Question Thread!
« Reply #822 on: October 20, 2012, 09:34:39 pm »
+1
I'm guessing this is referring to the motion of an object moving in a circular path (about the origin). In which the displacement of the particle is the circular path, the velocity vector is tangential to the path, and the acceleration vector is always towards the origin. This is because this acceleration is needed to keep moving the object in the circular path, if no force were to act on it, it would just keep travelling in a straight line, a force is needed to keep 'pulling it around' to make it move in the circular path.
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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #823 on: October 21, 2012, 12:07:41 pm »
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Thanks a lot for your help :)

Okay, I'll try: Is it because u will represent the x coordinate and v will represent the y coordinate. So the point (u,v),now called (x,y), will form a right angled triangle when joining to the origin using a straight line? Hence, that general expression gives the perpendicular bisector?
u and v can be any complex numbers. So they won't necessarily only have one non-zero component.

Try this sort of thing:

Let , and let be the position vector of any point on the perpendicular bisector of the line joining the points and . What do and represent? What do their magnitudes represent?
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martin1106

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Re: Specialist 3/4 Question Thread!
« Reply #824 on: October 21, 2012, 05:21:32 pm »
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I am confused with the part c). The answer gives that frictional force acts in the same direction as the tension. The friction must be opposed to the tension, isn't it?

my answer is
                     
                     
« Last Edit: October 21, 2012, 05:28:04 pm by martin1106 »