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July 10, 2026, 10:16:19 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2806317 times)  Share 

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brightsky

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Re: Specialist 3/4 Question Thread!
« Reply #1770 on: June 09, 2013, 11:45:01 am »
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Thanks. I get it. Also, I just realised that B and C have the same sort of symmetry, but in B, the dot is lying on the Re(z) axis, so it is a real solution by itself. In C, the lone dot is on the Im(z) axis, so it has to be mirrored by another solution according to the conjugate root theorem, which it isn't.

One last question though, the coefficients in the polynomial are all z's, aren't they?, meaning they're not real coefficients.
Thanks.

z is the unknown in this instance. the coefficients are the numbers in front of the z. you have 1 in front of the z^5, 1 in front of z^2, -1 in front of z, and c, all of which are real.
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Re: Specialist 3/4 Question Thread!
« Reply #1771 on: June 09, 2013, 08:20:51 pm »
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I revised complex numbers today and tried doing the following problem by completing the square. It wasn't a very good method since I then had to go on and find the square root of the complex constant that forms at the end. The solutions suggested applying the factor theorem, but I would have had to have continued on to z=-5. Any feedback?
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availn

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Re: Specialist 3/4 Question Thread!
« Reply #1772 on: June 09, 2013, 08:34:46 pm »
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I revised complex numbers today and tried doing the following problem by completing the square. It wasn't a very good method since I then had to go on and find the square root of the complex constant that forms at the end. The solutions suggested applying the factor theorem, but I would have had to have continued on to z=-5. Any feedback?

If you split the equation into its real and complex parts before applying the factor theorem, instead of trial and error, then you would be faster in finding that z=-5 is a solution.
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #1773 on: June 09, 2013, 08:36:04 pm »
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If you split the equation into its real and complex parts before applying the factor theorem, instead of trial and error, then you would be faster in finding that z=-5 is a solution.

Yeah, the question says "find THE complex factor". Either this is a perfect square or there is a real factor as well. Split into components as availn said and from the imaginary component you'll find that z=-5.
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Alwin

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Re: Specialist 3/4 Question Thread!
« Reply #1774 on: June 09, 2013, 08:41:07 pm »
+1
I revised complex numbers today and tried doing the following problem by completing the square. It wasn't a very good method since I then had to go on and find the square root of the complex constant that forms at the end. The solutions suggested applying the factor theorem, but I would have had to have continued on to z=-5. Any feedback?



Since it's a quadratic, you could always use the quadratic formula. However, since "c" is a complex number, your determinant will be a complex number meaning you must square root a complex number (not much better than completing the square).

The "quick" way of choosing z=-5 is as follows:
Firstly break P(z) into its real and complex parts. Since I am finding real solution, z^2 is real





Now, for P(z) = 0, then z^2 + 2x - 15 = 0 and -z - 5 = 0
The only solution for both is


This only gives this real solution for P(z). This is because for the complex solution, z^2 is a complex number, not a real number.
After finding that P(-5)=0, you can then factorise and find the other solution.


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Re: Specialist 3/4 Question Thread!
« Reply #1775 on: June 09, 2013, 08:43:10 pm »
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Woah. I don't think I have come across such a question or such a solution before. I feel pretty rusty on these. :S
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #1776 on: June 09, 2013, 09:46:10 pm »
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I'll be fair and say that asking us to solve a quadratic equation with complex coefficients is pretty dog. I mean unless there's a shortcut like this one or the roots are conjugates, square-rooting a complex number will just take more time.
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Re: Specialist 3/4 Question Thread!
« Reply #1777 on: June 09, 2013, 10:06:46 pm »
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I can't get this question for some reason...

Heinemann 7.2 - Applications of the chain rule

8. A conical flask with a radius of and a height of is being filled with powder at a constant rate of . Find:

(a) The rate, in terms of height, at which the height of the powder is changing
(b) The rate at which the radius of the powder is increasing when the height is 8 cm


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SocialRhubarb

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Re: Specialist 3/4 Question Thread!
« Reply #1778 on: June 09, 2013, 10:49:13 pm »
+1




,   using similar triangles.

Substituting that into the original equation for volume:









At h = 8,



This is assuming the cone is inverted. For a non-inverted cone, substitute , I think, although I'm not too sure about that point.
« Last Edit: June 09, 2013, 10:55:51 pm by SocialRhubarb »
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lzxnl

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Re: Specialist 3/4 Question Thread!
« Reply #1779 on: June 09, 2013, 11:39:35 pm »
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Your last point would yield the similar triangles relation. However, the volume would then be volume of total cone - volume of small gap cone.
If h is the height of the frustum at any time
r/5 = (12-h)/12
r=(60-5h)/12

Volume of total cone = 1/3*pi*5^2*12=100 pi
Volume of empty part of cone = 1/3*pi*r^2*(12-h) = 1/3*pi*(60-5h)^2 /144*(12-h)=25pi/432*(12-h)^3
So volume of solid = 100pi minus the last expression.
That's your volume expression.
The awkward bit is, a conical flask isn't inverted...you do have to go through all of this.
Volume = 100pi - 25pi/432*(12-h)^3
dv/dh=25pi/144*(h-12)^2
You have dv/dt. You can use that, with dv/dh, to work out dh/dt. Note how the last two multiply to give dv/dt.
For the rate of change of the radius, sub in h=8 for dh/dt, then use dr/dh=-5/12 from the relationship between r and h established above. That will give you dr/dt.
I sure hope I haven't messed anything up.
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SocialRhubarb

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Re: Specialist 3/4 Question Thread!
« Reply #1780 on: June 09, 2013, 11:57:05 pm »
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You're right, I didn't notice it was looking for rate of change of radius, not rate of change of  height.

But since your expression for is essentially the same as substituting into my expression for , I think you can probably just substitute at the end, rather than at the start, which I think might be easier.
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Re: Specialist 3/4 Question Thread!
« Reply #1781 on: June 10, 2013, 11:15:27 am »
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Show that the function y(x)=(ln|x|)/x is a solution of the differential equation  x2y'+xy=1
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Re: Specialist 3/4 Question Thread!
« Reply #1782 on: June 10, 2013, 11:22:15 am »
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You could solve the differential equation! Oh wait VCE doesn't let us do that.
y'=(1-ln|x|)/x^2
x^2*y'=1-ln|x|
x*y=ln|x|
x^2*y+x*y=1
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Re: Specialist 3/4 Question Thread!
« Reply #1783 on: June 10, 2013, 12:25:10 pm »
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Haha yea i am going ENG1090 (Foundation Math, which is the equivalent of VCE Spesh)
Thanks for your help.
Still a bit confused though, in the second step x^2*y'=1-ln|x|, how did you remove the square from the x and cancel it down to x*y=ln|x|?

Also, how do I use differential equations to solve this? (see attachment below.)
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brightsky

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Re: Specialist 3/4 Question Thread!
« Reply #1784 on: June 10, 2013, 03:02:01 pm »
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dy/dx = y^1/2 / x^1/2
dy/dx (y^(-1/2)) = x^(-1/2)
int y^(-1/2)  dy/dx dx = int x^(-1/2) dx
int y^(-1/2) dy = int x^(-1/2) dx
2y^(1/2) = 2x^(1/2) + c
y^(1/2) = x^(1/2) + c
y = x + 2sqrt(x) + c
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