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March 28, 2024, 10:06:02 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164183 times)  Share 

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S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9315 on: November 04, 2018, 11:13:22 am »
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Is it worth doing any of the past VCAA exam papers from 2007-2012?

Yes, absolutely, almost all of the content is the same. The only significant thing that was removed was the conic sections (ellipses, hyperbolae) – although, arguably, you should still know these, since they come up in vector functions.

And, obviously, the exams from the previous study design won't have any probability on them. Although, all of the probability questions VCAA have shown us so far (in 2016 / 2017, and the NHT exams) have been quite straightforward.

Ramya63

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9316 on: November 05, 2018, 01:06:34 pm »
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Can anyone help with this question?
2017: Biology [46], Methods [43]
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S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9317 on: November 05, 2018, 02:51:48 pm »
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Can anyone help with this question?

Use the arc-length formula on the formula sheet.

Ramya63

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9318 on: November 05, 2018, 03:18:40 pm »
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Use the arc-length formula on the formula sheet.

yep, did that and it simplifies to the intergral of sqr(1-cos(t)). Any ideas on how to solve this in an exam1?
2017: Biology [46], Methods [43]
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S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9319 on: November 05, 2018, 03:25:34 pm »
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yep, did that and it simplifies to the intergral of sqr(1-cos(t)). Any ideas on how to solve this in an exam1?

Use cos(2x) = 1 – 2sin^2(x).

Unsplash

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9320 on: November 05, 2018, 03:39:58 pm »
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Use cos(2x) = 1 – 2sin^2(x).

Could you also multiply by

?

This will get you



Then you can use u substitution?

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9321 on: November 05, 2018, 03:54:26 pm »
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Could you also multiply by

?

This will get you



Then you can use u substitution?

I don't see how to make progress with that method, with either substitution (u = sin(x) or u = 1 + cos(x)). I don't think using the chain rule will give you an integral with respect to u.

Ramya63

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9322 on: November 05, 2018, 05:47:26 pm »
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Use cos(2x) = 1 – 2sin^2(x).

Thank you!!
2017: Biology [46], Methods [43]
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Unsplash

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9323 on: November 05, 2018, 10:49:26 pm »
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I don't see how to make progress with that method, with either substitution (u = sin(x) or u = 1 + cos(x)). I don't think using the chain rule will give you an integral with respect to u.

u = 1 + cos(x)
-du/dx=sin(x) (Numerator once the square root is simplified)

Or does this not work because you need to consider both + and - sin(x)


S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9324 on: November 06, 2018, 10:12:39 am »
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u = 1 + cos(x)
-du/dx=sin(x) (Numerator once the square root is simplified)

Or does this not work because you need to consider both + and - sin(x)

I played around with this a bit more, and yes, your method can be made to work – but you're right, you need to consider the fact that sin(x) changes from positive to negative in the interval [2pi/3, 2pi]. This is important because when we make the substitution u = 1 + cos(x) and end up with sqrt(sin^2(x)) / sqrt(1 + cos(x)), that integrand is always positive, so we can't just simplify sqrt(sin^2(x)) to sin(x).

So, using your method, you'd need to break it up into the sum of two definite integrals: one where you integrate sin(x) / sqrt(1 + cos(x)) over [2pi/3, pi] and another where you integrate –sin(x) / sqrt(1 + cos(x)) over [pi, 2pi].

fur2018

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9325 on: November 06, 2018, 12:16:36 pm »
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hi,
does anyone know how to do Q8c from 2018 NH VCAA Exam 1?

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9326 on: November 06, 2018, 12:39:18 pm »
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hi,
does anyone know how to do Q8c from 2018 NH VCAA Exam 1?

A ray with equation Arg(z) = a is a half-line with its endpoint at the origin.

A line passing through the origin that is perpendicular to the circle |z – 1 – i| = 2 is y = x. (We know this because the line must pass through the centre of the circle in order to be perpendicular, and the centre has cartesian coordinates (1, 1)).

What are the two half-lines, in the form Arg(z) = a, that have cartesian equation y = x?
« Last Edit: November 06, 2018, 12:40:52 pm by S_R_K »

fur2018

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9327 on: November 06, 2018, 01:33:56 pm »
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ah... that makes sense. i misread question believing it is asking for equation of all rays that are perpendicular to the circle

so it would be arg(z)= pie/4 and then -3pie/4 ?

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9328 on: November 06, 2018, 02:49:19 pm »
+1
ah... that makes sense. i misread question believing it is asking for equation of all rays that are perpendicular to the circle

That is exactly what it is asking for. The reason why there are only two answers is that a ray must have an endpoint at the origin.

Quote
so it would be arg(z)= pie/4 and then -3pie/4 ?

Yes.

Ramya63

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9329 on: November 07, 2018, 05:23:11 pm »
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A ray with equation Arg(z) = a is a half-line with its endpoint at the origin.

A line passing through the origin that is perpendicular to the circle |z – 1 – i| = 2 is y = x. (We know this because the line must pass through the centre of the circle in order to be perpendicular, and the centre has cartesian coordinates (1, 1)).

What are the two half-lines, in the form Arg(z) = a, that have cartesian equation y = x?

Weird question, but what does 'perpendicular to the circle' actually mean? I interpreted it as being tangent to the circle, but clearly that's wrong.
2017: Biology [46], Methods [43]
2018: Eng Lang [40] Spesh [40] Chem [45] Psych [47] ATAR: 99.40
2019: Med at Monash :)