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April 17, 2024, 06:53:27 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2170216 times)  Share 

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Mr. Study

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Re: Specialist 3/4 Question Thread!
« Reply #915 on: November 11, 2012, 12:22:18 pm »
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I can't remember if I have asked this already.

Are we allowed to use the third derivative to confirm a point of inflection?
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sodapop

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Re: Specialist 3/4 Question Thread!
« Reply #916 on: November 11, 2012, 12:36:05 pm »
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Hey guys, could someone explain this answer (the correct answer is in red)? I understand that the region defined by |z|= 2 is a circle with r=2. By why is |z| >or= |z-(-sqrt(3) + i ) | defined by that line? And why is the required region above the line?
Thanks!

Biceps

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Re: Specialist 3/4 Question Thread!
« Reply #917 on: November 11, 2012, 01:08:38 pm »
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Lasercookie

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Re: Specialist 3/4 Question Thread!
« Reply #918 on: November 11, 2012, 01:13:58 pm »
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Hey guys, could someone explain this answer (the correct answer is in red)? I understand that the region defined by |z|= 2 is a circle with r=2. By why is |z| >or= |z-(-sqrt(3) + i ) | defined by that line? And why is the required region above the line?
Thanks!



So we have bit. If we had That's just the perpendicular bisector of the line joining and the origin.

With perpendicular bisectors, each point on it is the same distance from the end points of the line that it's cutting. I think I've linked this before, but I like the picture here http://mathworld.wolfram.com/PerpendicularBisectorTheorem.html. That's what is saying isn't it? We have the line from the origin to - so those are our end points. So z being a set of complex numbers, |z| is the distance of some complex number from the origin, and it happens to be equal to the distance between some complex number and the point . Hence that's why we know it's that line. (I hope that reasoning is okay, hopefully someone else will be able to confirm).

That's what the parts of the question before this one was pushing you towards. If you couldn't recognise that straight away, the question before had you state that is equivalent to (which you could have shown by subbing in z=x+yi, but that'd just be going the long way around). 

Now note the greater than, that's why we're looking at the region that's above that line.

What about the bit? Well that's just a circle with radius 2.

Now the --> intersection. The region we want is the region when   and   overlap.
« Last Edit: November 11, 2012, 01:16:38 pm by laseredd »

sodapop

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Re: Specialist 3/4 Question Thread!
« Reply #919 on: November 11, 2012, 01:44:25 pm »
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Ah, thanks laseredd, that's a brilliant explanation, I completely get why the equation defines that particular line. But I'm still confused why the "greater than" part would specify that the region is above the line though.. (The equation is y >or= sqrt(3)x + 1) Thanks!
« Last Edit: November 11, 2012, 01:46:11 pm by sodapop »

Lasercookie

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Re: Specialist 3/4 Question Thread!
« Reply #920 on: November 11, 2012, 01:54:34 pm »
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Ah, thanks laseredd, that's a brilliant explanation, I completely get why the equation defines that particular line. But I'm still confused why the "greater than" part would specify that the region is above the line though.. (The equation is y >or= sqrt(3)x + 1) Thanks!


So we have something that looks like this right:


If the , we'll have something like this (pretend that the lines are of equal length):


But if we have , then we can have the |z| being longer than the |z-z_1| line:



In other words, the point z could be any point above that line.

edit: fixed error on diagram 3
« Last Edit: November 11, 2012, 02:07:40 pm by laseredd »

soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #921 on: November 11, 2012, 02:02:33 pm »
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http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specmaths2_assessrep_11.pdf

VCAA 2011 EXAM 2 Q5c)i)

To satisfy the differential eqn, on the LHS, why was (x/1+t) replaced with the expression for x(t) ? Are the two the same? How? Why?

Thank you for any assistance :)
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Biceps

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Re: Specialist 3/4 Question Thread!
« Reply #922 on: November 11, 2012, 02:11:40 pm »
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http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specmaths2_assessrep_11.pdf

VCAA 2011 EXAM 2 Q5c)i)

To satisfy the differential eqn, on the LHS, why was (x/1+t) replaced with the expression for x(t) ? Are the two the same? How? Why?

Thank you for any assistance :)
They are just asking you to differentiate. They purposely gave that formula for that reason.
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soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #923 on: November 11, 2012, 02:12:53 pm »
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http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011specmath2-w.pdf
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specmaths2_assessrep_11.pdf

VCAA 2011 EXAM 2 Q5c)i)

To satisfy the differential eqn, on the LHS, why was (x/1+t) replaced with the expression for x(t) ? Are the two the same? How? Why?

Thank you for any assistance :)
They are just asking you to differentiate. They purposely gave that formula for that reason.
Oh sorry i mean part c) ii) :)
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polar

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Re: Specialist 3/4 Question Thread!
« Reply #924 on: November 11, 2012, 02:19:59 pm »
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nah, they're not the same. when you solve the differential equation , x(t) is the solution
« Last Edit: November 11, 2012, 02:31:49 pm by polar »

sodapop

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Re: Specialist 3/4 Question Thread!
« Reply #925 on: November 11, 2012, 02:35:28 pm »
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Ah, thanks laseredd, that's a brilliant explanation, I completely get why the equation defines that particular line. But I'm still confused why the "greater than" part would specify that the region is above the line though.. (The equation is y >or= sqrt(3)x + 1) Thanks!


So we have something that looks like this right:
(Image removed from quote.)

If the , we'll have something like this (pretend that the lines are of equal length):
(Image removed from quote.)

But if we have , then we can have the |z| being longer than the |z-z_1| line:

(Image removed from quote.)

In other words, the point z could be any point above that line.

edit: fixed error on diagram 3

Ahh, okay! Get it now. Thanks laseredd, you're a genius. :D

soccerboi

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Re: Specialist 3/4 Question Thread!
« Reply #926 on: November 11, 2012, 02:36:20 pm »
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I'm still confused about it. I don't understand how to verify that x(t) satisfies the differential eqn. I don't see what they did on the report. Could someone perhaps provide a description of what they did for me?
Also, with satisfying the initial condition, we just sub t=0 and make it equal 0, and this would satisfy it because at t=0 the amount in the tank was 0?
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WhoTookMyUsername

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Re: Specialist 3/4 Question Thread!
« Reply #927 on: November 11, 2012, 02:37:57 pm »
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Anyway to convert a parametric Equation to a cartesian equation on cas?

paulsterio

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Re: Specialist 3/4 Question Thread!
« Reply #928 on: November 11, 2012, 02:40:00 pm »
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Anyway to convert a parametric Equation to a cartesian equation on cas?

I know of a way with the Classpad, but I don't think there's a way with the nSpire.

polar

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Re: Specialist 3/4 Question Thread!
« Reply #929 on: November 11, 2012, 02:44:30 pm »
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substitute the from 5(c)(i) and stated in the question, simplify and eventually, it simplifies to and thus, satisfies the differential equation.

yeah, that's what you do to show it satisfies the condition