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May 01, 2026, 03:02:58 am

Author Topic: Hutchoo's spesh thread.  (Read 13069 times)  Share 

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Hutchoo

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Re: Hutchoo's spesh thread.
« Reply #15 on: October 01, 2012, 03:58:05 pm »
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Yeah, the mistake was with simplifying the integral.
Cause
(2+6x) * 1/ sqrt(4-x^2) = (2+6x)/ sqrt(4-x^2)
That was my logic, anyway.

So then I just did:
int (2x+6) dx * int (1/sqrt(4-x^2)) dx. Which didn't work.

Hutchoo

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Re: Hutchoo's spesh thread.
« Reply #16 on: October 01, 2012, 04:24:25 pm »
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Question 1 E : VCAA 2006 (SA)
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006specmaths2-w.pdf
How did they know that A = ?
I understand that's the area for a circle, but why do they assume a wine glass is a circle 0-o?


Also: wtf?

Question 3iii.
I found the correct time (31.8) second, so I continued to use straight line motion, which is what they do as well (they = VCAA).

I used s = ut + 1/2at^2, they used v^2 = u^2 + 2as.
How come they give  different values?!


How did they get the graph for 4d? :/
« Last Edit: October 01, 2012, 04:40:24 pm by Hutchoo »

Bhootnike

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Re: Hutchoo's spesh thread.
« Reply #17 on: October 01, 2012, 04:40:04 pm »
+1
Question 1 E : VCAA 2006 (SA)
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006specmaths2-w.pdf
How did they know that A = ?
I understand that's the area for a circle, but why do they assume a wine glass is a circle 0-o?

thats what i said..!
bullshit innit!


Question 3iii.
I found the correct time (31.8) second, so I continued to use straight line motion, which is what they do as well (they = VCAA).

I used s = ut + 1/2at^2, they used v^2 = u^2 + 2as.
How come they give  different values?!
they dont :p
i used the same forumula as you.
what values did you use?
i had t = 31.82...., a = 2.2, v = 70, u = 0
so x just = 0.5a^2
= 0.5 x 2.2 x 31.82^2
= 1113.636
=1114 m
2011: Biol - 42
2012: Spesh |Methods |Chemistry |English Language| Physics
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Hutchoo

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Re: Hutchoo's spesh thread.
« Reply #18 on: October 01, 2012, 04:46:22 pm »
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Ahhhhhh, my bad :P I meant iTute!

Wtf, I'm an idiot! I did a calculator error:
I typed t to be 31.8, not 31.82, which accounts for 1112 and not 1114 :P
@ that stupid circle thing:
:/ I know lol. That thing really confused me. Worth 3 marks, I was stumped tbh. It was a very easy question though, that's the worst part.


Also:
Question 5b. I don't really understand this question. Anyone care to explain? =)*looking at you bhootnike*
« Last Edit: October 01, 2012, 04:49:50 pm by Hutchoo »

Bhootnike

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Re: Hutchoo's spesh thread.
« Reply #19 on: October 01, 2012, 05:05:09 pm »
+1
Ahhhhhh, my bad :P I meant iTute!

Wtf, I'm an idiot! I did a calculator error:
I typed t to be 31.8, not 31.82, which accounts for 1112 and not 1114 :P
@ that stupid circle thing:
:/ I know lol. That thing really confused me. Worth 3 marks, I was stumped tbh. It was a very easy question though, that's the worst part.


Also:
Question 5b. I don't really understand this question. Anyone care to explain? =)*looking at you bhootnike*
lol its ok! i dot that!
and yea i know right, i still dont agree with that ..!

cos pi /8?
use double angle formula. cos 2 x pi/8 = cos pi/4 remember :p
2011: Biol - 42
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2014: Physiotherapy
khuda ne jab tujhe banaya hoga, ek suroor uske dil mein aaya hoga, socha hoga kya doonga tohfe mein tujhe.... tab ja ke usne mujhe banaya hoga

Hutchoo

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Re: Hutchoo's spesh thread.
« Reply #20 on: October 01, 2012, 05:07:00 pm »
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Oh, my bad, I meant a ii. xD

Bhootnike

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Re: Hutchoo's spesh thread.
« Reply #21 on: October 01, 2012, 05:10:36 pm »
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Oh, my bad, I meant a ii. xD
aii)
i didnt get it either lOL.
2011: Biol - 42
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Hutchoo

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Re: Hutchoo's spesh thread.
« Reply #22 on: October 01, 2012, 05:23:21 pm »
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pi

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Re: Hutchoo's spesh thread.
« Reply #23 on: October 01, 2012, 05:32:32 pm »
+1
Oh, my bad, I meant a ii. xD
aii)
i didnt get it either lOL.

/ping b^3!

I'm taking over this shit :D

Basically, for sake of me ceebsing latex, let z-line-on-the-top (complement) = y

Ok, so basically the question is asking for the line between points z and -z in terms of y.

So the equation is |z-y|=|-z-y|, which (and remember back to loci!) means that the distance between z and y is equivalent to the distance between -z and y.

(N.B. the equation can be further simplified to |z-y|=|z+y|)

Does that make sense? :/


edit: forgot a word in the sentence -_-
« Last Edit: October 01, 2012, 05:39:19 pm by ρнуѕικѕ ♥ »

Hutchoo

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Re: Hutchoo's spesh thread.
« Reply #24 on: October 01, 2012, 05:38:28 pm »
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OHHHHHHHHHHHHHHHHHH YEAH.
The distance thing!

Derrick Ha said the exact same thing in his lectures, can't believe I forgot that.
SO if you say that two points are equal distance, you're essentially showing that they form a line?

I need to understand this before I explode!

pi

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Re: Hutchoo's spesh thread.
« Reply #25 on: October 01, 2012, 05:49:46 pm »
+1
Yep :)

Hutchoo

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Re: Hutchoo's spesh thread.
« Reply #26 on: October 01, 2012, 05:50:24 pm »
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Yep :)

Thank you <3


Can someone explain this though?

Yeah, the mistake was with simplifying the integral.
Cause
(2+6x) * 1/ sqrt(4-x^2) = (2+6x)/ sqrt(4-x^2)
That was my logic, anyway.

So then I just did:
int (2x+6) dx * int (1/sqrt(4-x^2)) dx. Which didn't work.


pi

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Re: Hutchoo's spesh thread.
« Reply #27 on: October 01, 2012, 06:01:50 pm »
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So then I just did:
int (2x+6) dx * int (1/sqrt(4-x^2)) dx. Which didn't work.

Unsure on the context, but this is not a legitimate step.

For example, integrate x^2, answer is x^3/3. But if you did int(x)*int(x) you'd get (x^2/2)*(x^2/2)=x^4/4, which is incorrect. You can only split the integral for + or - .

Hutchoo

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Re: Hutchoo's spesh thread.
« Reply #28 on: October 01, 2012, 06:15:47 pm »
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Context: Go to page one.
And yeah, I understand that. But with fractions, you can simplify the numerator/denominator by taking it out of the integral.

I.e: int (2/x) = 2*int(1/x) = 2 loge (x)

That's why I don't understand why my method didn't work :/

pi

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Re: Hutchoo's spesh thread.
« Reply #29 on: October 01, 2012, 06:22:50 pm »
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Context: Go to page one.
And yeah, I understand that. But with fractions, you can simplify the numerator/denominator by taking it out of the integral.

I.e: int (2/x) = 2*int(1/x) = 2 loge (x)

That's why I don't understand why my method didn't work :/

Because removing a constant is different to removing (2x+6)


(I feel I'm still missing your point though :/ )