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April 19, 2024, 09:28:58 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2171418 times)  Share 

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rife168

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Re: Specialist 3/4 Question Thread!
« Reply #270 on: February 09, 2012, 08:25:51 pm »
+1
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.
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trinh

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Re: Specialist 3/4 Question Thread!
« Reply #271 on: February 09, 2012, 08:29:37 pm »
+2


, k∈Z
, k∈Z

now sub in values of k

Thus,

Hope this is right :s

EDIT: beaten as usual ><

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Re: Specialist 3/4 Question Thread!
« Reply #272 on: February 09, 2012, 08:38:29 pm »
0
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

It's just I had a lesson today with a new tutor and she had this totally different method to the book (with the 2kpi etc, letting k=1, k=2) and it confused me. She made sin(theta)=0 and cos(theta)=1, and then had like 0, 2pi, 4pi etc. Still got the right answers but it was weird and not sure if better.

edit:

to make it more clear, if z^3 = 8i was the thing, she'd make r=2 and then sin(theta)=1 and then cos(theta)=0. I kinda understand what she's doing with the expansion of cis(theta) and then equating real and imaginary sides, but I don't entirely get what was done from there.
« Last Edit: February 09, 2012, 08:45:37 pm by teewreck »
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rife168

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Re: Specialist 3/4 Question Thread!
« Reply #273 on: February 09, 2012, 08:49:53 pm »
0
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

It's just I had a lesson today with a new tutor and she had this totally different method to the book (with the 2kpi etc, letting k=1, k=2) and it confused me. She made sin(theta)=0 and cos(theta)=1, and then had like 0, 2pi, 4pi etc. Still got the right answers but it was weird and not sure if better.

edit:

to make it more clear, if z^3 = 8i was the thing, she'd make r=2 and then sin(theta)=1 and then cos(theta)=0. I kinda understand what she's doing with the expansion of cis(theta) and then equating real and imaginary sides, but I don't entirely get what was done from there.

I suggest you use the method that is taught in the book and make sure that method is clear to you, then have a go at your tutor's method. If you don't understand your tutor's method at all, don't let it concern you too much, you won't be at a disadvantage by only knowing one method, but you will be at an advantage by knowing 2 methods.
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Truck

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Re: Specialist 3/4 Question Thread!
« Reply #274 on: February 09, 2012, 08:53:18 pm »
0
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

It's just I had a lesson today with a new tutor and she had this totally different method to the book (with the 2kpi etc, letting k=1, k=2) and it confused me. She made sin(theta)=0 and cos(theta)=1, and then had like 0, 2pi, 4pi etc. Still got the right answers but it was weird and not sure if better.

edit:

to make it more clear, if z^3 = 8i was the thing, she'd make r=2 and then sin(theta)=1 and then cos(theta)=0. I kinda understand what she's doing with the expansion of cis(theta) and then equating real and imaginary sides, but I don't entirely get what was done from there.

I suggest you use the method that is taught in the book and make sure that method is clear to you, then have a go at your tutor's method. If you don't understand your tutor's method at all, don't let it concern you too much, you won't be at a disadvantage by only knowing one method, but you will be at an advantage by knowing 2 methods.

Yeah not my tutor's method I should have said, just a friend who did pretty decently in Spesh, but thank you very much! She's pretty bad at explaining things hence my hope someone here would know what to do haha.
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trinh

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Re: Specialist 3/4 Question Thread!
« Reply #275 on: February 09, 2012, 09:16:48 pm »
+2
Ok, I don't know if I'm helping here, but for the example you showed:

, where

=>




Equating coefficients,

AND , where , and thus

So solving ,
and solving ,

now we pick out the common solutions, which are of course

and then sub back into

If this indeed what you were talking about, I'd still stick to the classic method because it doesn't involve trig (which I found deadly sometimes)
« Last Edit: February 09, 2012, 09:18:31 pm by trinh »

Truck

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Re: Specialist 3/4 Question Thread!
« Reply #276 on: February 09, 2012, 09:20:15 pm »
0
Ok, I don't know if I'm helping here, but for the example you showed:

, where

=>




Equating coefficients,

AND , where , and thus

So solving ,
and solving ,

now we pick out the common solutions, which are of course

and then sub back into

If this indeed what you were talking about, I'd still stick to the classic method because it doesn't involve trig (which I found deadly sometimes)

That's it, yeh 2kpi seems way better, ty!
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Hutchoo

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Re: Specialist 3/4 Question Thread!
« Reply #277 on: February 09, 2012, 10:18:45 pm »
0
Have you guys done Vectors yet?
Some of the ER questions in essentials make me sad.

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Re: Specialist 3/4 Question Thread!
« Reply #278 on: February 09, 2012, 10:20:10 pm »
0
Have you guys done Vectors yet?
Some of the ER questions in essentials make me sad.

Yeh they were quite hard last year.
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TrueTears

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Re: Specialist 3/4 Question Thread!
« Reply #279 on: February 09, 2012, 10:22:31 pm »
0
Have you guys done Vectors yet?
Some of the ER questions in essentials make me sad.

although bora in your dp makes me quite happy hor..

the gif's from the mah boy mv right? lol

watch it now!@!@!@# http://www.youtube.com/watch?v=X6XXia5B2Wg
« Last Edit: February 09, 2012, 10:26:08 pm by TrueTears »
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Hutchoo

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Re: Specialist 3/4 Question Thread!
« Reply #280 on: February 09, 2012, 10:43:40 pm »
0
Have you guys done Vectors yet?
Some of the ER questions in essentials make me sad.

although bora in your dp makes me quite happy hor..

the gif's from the mah boy mv right? lol

watch it now!@!@!@# http://www.youtube.com/watch?v=X6XXia5B2Wg
I miss you, bebe. =(
How long did it take you past students to 'master' Vectors?

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Re: Specialist 3/4 Question Thread!
« Reply #281 on: February 09, 2012, 10:45:46 pm »
0
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

just wondering what the actual answer is ?!
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rife168

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Re: Specialist 3/4 Question Thread!
« Reply #282 on: February 09, 2012, 11:33:48 pm »
+1
Hi friends, I'm having a bit of trouble with this one application of de moivre's theorem...

Could someone please show me their method to solve z^3 + i = 0.

Thank you!

z3 + i = 0
z3 = -i          (Think of an Argand diagram. is represented by a point 1 unit vertically downwards)
so,
From there, it is in the standard form for these kinds of questions, so I'll let you work it out, if you have trouble working it out, reply again and myself or someone else will explain it to you.

just wondering what the actual answer is ?!




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Mr. Study

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Re: Specialist 3/4 Question Thread!
« Reply #283 on: February 10, 2012, 06:09:35 pm »
0
Okay.

This is to find the volume when rotated about the y-axis.



I get stuck at the last line. I can anti-differentiate but when I look at the answers I am wrong.

Could someone complete the last line and explain why it is?

Thanks!
« Last Edit: February 10, 2012, 06:24:12 pm by Mr. Study »
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Truck

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Re: Specialist 3/4 Question Thread!
« Reply #284 on: February 10, 2012, 06:13:44 pm »
+1
Not sure about the rest, but in the 2nd last line it should say (y^4 + 2y^2 + 1) instead of (y^4 + 2y + 1).
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