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May 04, 2026, 04:43:12 am

Author Topic: VCE Methods Question Thread!  (Read 6073669 times)  Share 

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Brytz

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Re: VCE Methods Question Thread!
« Reply #3120 on: November 05, 2013, 01:29:42 pm »
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http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007mmCAS2.pdf

Q1.d How to answer this question? Apparently only 4% got marks for this question.
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Rectophobia

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Re: VCE Methods Question Thread!
« Reply #3121 on: November 05, 2013, 01:34:11 pm »
+1
This question was handled so badly because it was an end point maximum. Those who made the derivative equal to zero found the minimum and hence got 0 marks. When you graph the surface area as a function of V, this becomes more apparent. Try subbing in the two endpoints and finding which yields a larger surface area
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shadows

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Re: VCE Methods Question Thread!
« Reply #3122 on: November 05, 2013, 01:36:00 pm »
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Two rules to use here: and
Spoiler


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clıppy

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Re: VCE Methods Question Thread!
« Reply #3123 on: November 05, 2013, 01:38:02 pm »
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A question about algebra with powers
as an example. Is there any easy way to solve this without having to expand the 5 bit?
What are the general rules when trying to solve these kinds of questions?
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shadows

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Re: VCE Methods Question Thread!
« Reply #3124 on: November 05, 2013, 01:40:47 pm »
+1
A question about algebra with powers
as an example. Is there any easy way to solve this without having to expand the 5 bit?
What are the general rules when trying to solve these kinds of questions?

x= squareroot (5^5/2)

= 5^(5/4)

SocialRhubarb

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Re: VCE Methods Question Thread!
« Reply #3125 on: November 05, 2013, 01:52:24 pm »
+1
x= squareroot (5^5/2)

= 5^(5/4)

Basically. Remember the plus-minus sign though.

It's really just an application of power laws, like

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achre

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Re: VCE Methods Question Thread!
« Reply #3126 on: November 05, 2013, 02:13:42 pm »
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Can someone explain the highlighted justification for why the approximation of is less than the true value? The gradient isn't decreasing for [8, inf)?

Stevensmay

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Re: VCE Methods Question Thread!
« Reply #3127 on: November 05, 2013, 02:15:43 pm »
+6
Can someone explain the highlighted justification for why the approximation of is less than the true value? The gradient isn't decreasing for [8, inf)?

The gradient is getting smaller and smaller as it approaches infinity. This means that the gap between the gradient and the tangent line will be increasing, as the gradient curve moves away from it. Thus an overestimation.

Henreezy

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Re: VCE Methods Question Thread!
« Reply #3128 on: November 05, 2013, 02:42:40 pm »
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Can anyone explain to me how to do question 2d i) from the extended response of this paper?
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010mmcas2-w.pdf

I'm good with transformations with matrices and simultaneous equations, but I haven't a clue how to 'show' how to get this answer by hand. *panic mode*
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Re: VCE Methods Question Thread!
« Reply #3129 on: November 05, 2013, 02:59:00 pm »
+1
First find the probability of the third statue being superior given that the first one is. (ie P^2 +(1-p)(p-0.2)) and make this equal to 0.7
=> p^2+p-0.2-p^2+0.2p=0.7
=>1.2p=0.9
=>p=0.9/1.2 = 0.75
=>p = 0.75

*These values were obtained through a tree diagram
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #3130 on: November 05, 2013, 03:04:29 pm »
+2
[quote
Cosec = 1/cos
Sec = 1/sin


Sec and Cosec are the wrong way around. I think the double angle and compound angle formulas are useful (although I don;t know how often they pop up in methods) and possibly the altered versions of the Pythagorean identity (Although I think these are more so used in spesh)   


*Edit: I can't quote for shit

They are the wrong way around; cos=1/sec, sin=1/cosec
You'll just have to use sec as the (square root of the) derivative of tan x I think
There wasn't that much trig in Methods last year from memory.

Can someone explain the highlighted justification for why the approximation of is less than the true value? The gradient isn't decreasing for [8, inf)?

So, you're using linear approximation on x^1/3. Differentiating once yields 1/3*x^(-2/3). Differentiating again yields -2/9*x^(-5/3)
As you can see, for positive x, the gradient is decreasing for x>0 (as the gradient of the gradient, if you want, is decreasing). Linear approximation works by drawing a tangent line and extrapolating from that. As the gradient is decreasing, the tangent line becomes an overestimate.

Draw out your transition matrix. I don't know how to use latex to do that so bear with me.
From my rather limited knowledge of transition matrices, it would look like this:

Superior     | p      p-0.2  |
Regular      | 1-p   1.2-p  |

So what you'd do is that given you know the first statue is superior, your initial state vector looks like
|1|
|0|  from the way I've defined the matrix. If the transition state is T, the probabilities of the third state are given by T^2*initial state.
Read off the top row for the probability that the statue is superior. Set this equal to 0.7

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achre

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Re: VCE Methods Question Thread!
« Reply #3131 on: November 05, 2013, 03:16:14 pm »
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Random CAS question - is there a way to move back to the start of the line quickly without using the cursor? It feels like about ten seconds of wasted time if you make a typo/error on the start of a long-ish entry.

Damoz.G

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Re: VCE Methods Question Thread!
« Reply #3132 on: November 05, 2013, 03:19:49 pm »
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Random CAS question - is there a way to move back to the start of the line quickly without using the cursor? It feels like about ten seconds of wasted time if you make a typo/error on the start of a long-ish entry.

No, I don't think so. I've tried to find a way as well, but I can't.

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Re: VCE Methods Question Thread!
« Reply #3133 on: November 05, 2013, 03:40:57 pm »
+1
Random CAS question - is there a way to move back to the start of the line quickly without using the cursor? It feels like about ten seconds of wasted time if you make a typo/error on the start of a long-ish entry.
there sure is, press 2ND + <-
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achre

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Re: VCE Methods Question Thread!
« Reply #3134 on: November 05, 2013, 03:47:09 pm »
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Is that for nspire cx, I can't find the key