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March 30, 2024, 01:24:09 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164692 times)  Share 

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dc302

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Re: Specialist 3/4 Question Thread!
« Reply #75 on: December 17, 2011, 03:24:34 pm »
0
Do you have to use vectors? There's another (probably less confusing) proof with just plain geometry.
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Re: Specialist 3/4 Question Thread!
« Reply #76 on: December 17, 2011, 03:34:13 pm »
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I guess not since the question never stated use vectors! but it is vector proofs so i dunno :p
but its made sense now, i just didnt think you'd need to go that deep into it haha, but ehhh its spesh!
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brightsky

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Re: Specialist 3/4 Question Thread!
« Reply #77 on: December 17, 2011, 05:21:02 pm »
+1
not sure if everything's been cleared up yet, but essentially what the guys have been trying to say is that in the formula: a.b = |a||b|cos(theta), the 'theta' is the angle made by the two vectors when you put their 'tails' together. so basically, the 'theta' in this case isn't angle B (since then you are taking the angle between the 'head' of vector a and the 'tail' of vector b), but the supplementary angle (picture moving vector a so that it's tail coincides with the tail of vector b. recall that you can 'move' normal vectors around and it will still remain the same vector). hence a.b = |a||b|cos(pi - theta) = -|a||b|cos(theta), with theta = angle B.

also, the geometric proof that dc302 alluded to is worth probing in order to consolidate your understanding of the cosine rule. (hint: try drawing the altitude of the triangle and do some trig off that). if i'm not mistaken, i think this particular proof is in your textbook.
« Last Edit: December 17, 2011, 05:23:23 pm by brightsky »
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Re: Specialist 3/4 Question Thread!
« Reply #78 on: December 18, 2011, 11:49:44 am »
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thanks, - yeah i figured out that I had to make them tail to tail for dot product to make sense :p
then i realized it was simply pi-theta as in... straight line- angle on one side.. therefore angle on other side = pi - theta.

which book ? this is off the heinneman
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Dominatorrr

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Re: Specialist 3/4 Question Thread!
« Reply #79 on: December 18, 2011, 04:12:42 pm »
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COMPLEX NUMBERS

If z=x+yi, determine the values of x and y such that z=sqrt(3+4i)

brightsky

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Re: Specialist 3/4 Question Thread!
« Reply #80 on: December 18, 2011, 04:25:22 pm »
+3
let sqrt(3+4i) = a + bi, for a,b E R
3+4i = a^2 + 2abi - b^2
3+4i = (a^2 - b^2) + (2ab) i

a^2 - b^2 = 3..(1)
2ab = 4 --> b = 2/a
sub into (1):
a^2 - (2/a)^2 = 3
a^2 - 4/a^2 = 3
a^4 - 4 = 3a^2
a^4 - 3a^2 - 4 = 0
(a^2 - 4)(a^2 + 1) = 0
since we are only looking for solutions in R,
a^2 - 4 = 0
(a-2)(a+2) = 0
a = 2 or a = -2
b = 1 or b = -1
so we have two solutions:
a = 2, b = 1
or
a = -2, b=-1
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Dominatorrr

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Re: Specialist 3/4 Question Thread!
« Reply #81 on: December 18, 2011, 04:44:05 pm »
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That does correspond with the answers in the book, thanks a lot.

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Re: Specialist 3/4 Question Thread!
« Reply #82 on: December 22, 2011, 10:59:45 am »
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COMPLEX NUMBERS

If z^n=(1+i)^n, determine the smallest value of n = element of N, so that z^n is equal to:

a) (sqrt(2))^n

abd123

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Re: Specialist 3/4 Question Thread!
« Reply #83 on: December 22, 2011, 11:30:09 am »
+5


let









a) if

and



Smallest is

hoped i helped im a bit rusty with complex numbers

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Re: Specialist 3/4 Question Thread!
« Reply #84 on: December 22, 2011, 11:36:49 am »
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Yes that helped, I was confused because I thought the answer was 0 but then I saw that the answer had to be a positive whole number so the arg(z) had to be 2pi and hence n=8. Thanks.

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Re: Specialist 3/4 Question Thread!
« Reply #85 on: December 22, 2011, 11:40:49 am »
0
This outflow/inflow question is troublesome for me :/.

Q9. A tank containing 600 litres of water has a sugar solution of concentration 0.1 kg/L pumped in at a rate of 10 L/min. if the mixture is kept uniform and is pumped out at a rate of 8 L/min, set up a differential equation for the amount of sugar, x kg, at any time t. (Do not attempt to solve).


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Re: Specialist 3/4 Question Thread!
« Reply #86 on: December 22, 2011, 11:49:46 am »
+1
It should be dx/dt=0.1*10-x*8/(600+2t)
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Re: Specialist 3/4 Question Thread!
« Reply #87 on: December 22, 2011, 11:50:24 am »
+4
For these questions where the flow in is not equal to the flow out, it is good to use this formula.











Hope that helps.

EDIT: Explaining the equation
The first part of the equation is giving how much of "x" is following in per second.
The second part of the equation is giving how much of "x" is following out per second.
You can see this by "rate"*"concentration"
In the second half of the equation, the concentration is  , think of it as c=mass/volume, mass is x and the volume is changing as t changes.
« Last Edit: December 22, 2011, 03:32:53 pm by b^3 »
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abd123

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Re: Specialist 3/4 Question Thread!
« Reply #88 on: December 22, 2011, 11:51:36 am »
0
It should be dx/dt=0.1*10-x*8/(600+2t)
For these questions where the flow in is not equal to the flow out, it is good to use this formula.









Hope that helps.

Thanks guys :).

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Re: Specialist 3/4 Question Thread!
« Reply #89 on: December 22, 2011, 11:55:58 am »
+1
I added a quick explanation of where the equation comes from (in an edit) if you wanted it.
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