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August 23, 2025, 08:45:54 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2579493 times)  Share 

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Limista

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Re: Specialist 3/4 Question Thread!
« Reply #1635 on: May 18, 2013, 11:58:33 pm »
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Got another question.

r(t) = (4+3sin(t))i + (3+2sin(2t/3))j + (1.5 + cos(3t/4))k

s(t) = (6.923 + (t/2000)sin(t/12))i + 5j + (2.545 + (t/2000)cos(t/12))k

Find where the two bodies collide, given t > 0

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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #1636 on: May 19, 2013, 12:27:51 am »
+1
My calculator says they don't, are those rounded values?
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Limista

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Re: Specialist 3/4 Question Thread!
« Reply #1637 on: May 19, 2013, 07:50:55 am »
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That's what my calculator says as well.

No - those aren't rounded values.

So I'm safe to write that there is/was no collision?
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BubbleWrapMan

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Re: Specialist 3/4 Question Thread!
« Reply #1638 on: May 19, 2013, 03:32:17 pm »
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Yeah, but it's odd that they would ask it that way.
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LOLs99

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Re: Specialist 3/4 Question Thread!
« Reply #1639 on: May 19, 2013, 06:09:12 pm »
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Anti diff x /(x^4 + 9)  have a bit of trouble after factorising n breaking into partial fraction.
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Re: Specialist 3/4 Question Thread!
« Reply #1640 on: May 19, 2013, 06:12:36 pm »
+2
nah it's just arctan.

int x/(x^4+9) dx
let u = x^2
du = 2x dx
the integral becomes:
1/2*int 1/(u^2+9) du
= 1/6*int 3/(u^2+9) du
= 1/6 arctan(u/3) + c
= 1/6 arctan(x^2/3) + c
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Re: Specialist 3/4 Question Thread!
« Reply #1641 on: May 19, 2013, 06:15:33 pm »
+3
This is a little different, it will actually end up as an inverse tan.


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LOLs99

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Re: Specialist 3/4 Question Thread!
« Reply #1642 on: May 19, 2013, 08:50:08 pm »
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Oh yea. I was making it complicated. can u actually separate du/dx as du=dx? what if it the u= x^3 or higher power, does it work?
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Re: Specialist 3/4 Question Thread!
« Reply #1643 on: May 19, 2013, 08:58:06 pm »
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Should work for any function, i.e. if you can use . Though for the record I recommend leaving it as . :P
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Re: Specialist 3/4 Question Thread!
« Reply #1644 on: May 19, 2013, 09:40:46 pm »
+1
Remember the linear approximation change in y roughly equal to change in x times dy/dx?
If you treat dy and dx as infinitesimal changes in y and x, then dy/dx can be treated as a fraction.
Similarly, it allows for manipulations of integral substitutions.

For instance, if we integrate sqrt(tan x) dx, we let u^2 = tan x
2u du = sec^2 x dx = (tan^2 x+1) dx = (u^4+1) dx
So dx = 2u/(u^4+1) du
sqrt tan x dx = u dx = 2u^2/(u^4+1) du
Now we can integrate this by partial fractions (u^4+1=(u^2+1)^2-2u^2, use DOTS to factor).
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Re: Specialist 3/4 Question Thread!
« Reply #1645 on: May 19, 2013, 10:03:24 pm »
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Okay I guess it clears up my question :) Thanks
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Re: Specialist 3/4 Question Thread!
« Reply #1646 on: May 20, 2013, 08:01:39 pm »
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hey,
question 2 in the multi choice of the chapter 8 review in essentials, I know why D is correct, but would anyone be able to explain to me why A is incorrect?
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Re: Specialist 3/4 Question Thread!
« Reply #1647 on: May 20, 2013, 08:05:10 pm »
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if you expand the brackets, eg (a-b)^2 you get the -2ab term which is not needed
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Re: Specialist 3/4 Question Thread!
« Reply #1648 on: May 20, 2013, 08:11:05 pm »
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how do i find the volume generated when the region is rotated about the y - axis

ans:126pi/5
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Re: Specialist 3/4 Question Thread!
« Reply #1649 on: May 20, 2013, 08:26:21 pm »
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if you expand the brackets, eg (a-b)^2 you get the -2ab term which is not needed

yeah but doesnt minusing the 2 from the original function just translate it two down, making it easier to find the volume generated. Its a similar application to q16 in the volumes exercise isnt it ?
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