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September 22, 2019, 09:24:39 am

Author Topic: VCE Methods Question Thread!  (Read 2443515 times)  Share 

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kimokeeffe

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Re: VCE Methods Question Thread!
« Reply #18075 on: August 19, 2019, 05:50:40 pm »
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not really a specific question. just wondering if anyone has any good ways of remembering the circular functions exact values cause i feel like a bit of a idiot waving my fingers around during a test lol :)
thanks!!

Sine

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Re: VCE Methods Question Thread!
« Reply #18076 on: August 19, 2019, 05:55:57 pm »
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not really a specific question. just wondering if anyone has any good ways of remembering the circular functions exact values cause i feel like a bit of a idiot waving my fingers around during a test lol :)
thanks!!




for tan just find sin/cos.

This is really just for preliminary learning - ideally you won't have to think about it at the end of the year.

Source: https://atarnotes.com/forum/index.php?topic=161217.msg925985#msg925985

DrDusk

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Re: VCE Methods Question Thread!
« Reply #18077 on: August 20, 2019, 12:48:42 am »
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not really a specific question. just wondering if anyone has any good ways of remembering the circular functions exact values cause i feel like a bit of a idiot waving my fingers around during a test lol :)
thanks!!

There's a much easier way such that you won't even need to memorise it. All you need to do is firstly, type it into a calculator as you normally would, for e.g. cos(30). Then you will get a decimal number. You want to directly square this number. So for example if you square cos(30) your calculator will display 3/4. Now just by looking at the screen you can tell what cos(30) is... i.e. just it's square root sqrt(3)/2
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18078 on: August 20, 2019, 01:13:17 am »
+1
not really a specific question. just wondering if anyone has any good ways of remembering the circular functions exact values cause i feel like a bit of a idiot waving my fingers around during a test lol :)
thanks!!

Not a fan of either of the above methods. Personally, I like these triangles:



If you can memorise these, then the exact values come from the trig ratios. Eg,

sin(30)=O/H=1/2

And if you can't remember them, you can also logic them.

The triangle on the left is an isosceles triangle with two 45 degree angles, so the last one HAS to be 90. Then, just make the two same side lengths to be 1, and use pythag to find the last.

The triangle on the right is an equilateral triangle, and we've decided that all side lengths are 2. Since it's equilateral, all angles are the same - 60 degrees. If you cut it in, then the bottom side becomes length 1, the top angle becomes 30 degrees, and you can use pythag again to find the final side.

Just draw these triangles on your exam, use logic and pythag to find the side lengths, and you're good to go!
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #18079 on: August 20, 2019, 09:22:02 am »
+1
Bit late to the party but regarding improper integrals, I just wanted to add that the textbook is subtly wrong. Convergence of \(\displaystyle \int_{-\infty}^\infty f(x)\,\text{d}x\) actually must allow for the terminals to be sent to \(\pm\infty\) at different rates, not necessarily the same rate.

That is \[\int_{-\infty}^\infty f(x)\,\text{d}x=\lim_{L\to -\infty}\int_L^a f(x)\,\text{d}x+\lim_{U\to\infty}\int_a^U f(x)\,\text{d}x,\quad a\in\mathbb{R}.\] This avoids stupid cases which fixates \(\infty-\infty=0\) such as in the integral \(\displaystyle\int_{-\infty}^\infty \dfrac{1}{x}\,\text{d}x\), where if we send the terminals to \(\pm\infty\) at the same rate we obtain \(0\), which is not correct. The integral actually doesn't exist.
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blyatman

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Re: VCE Methods Question Thread!
« Reply #18080 on: August 20, 2019, 10:25:35 am »
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Bit late to the party but regarding improper integrals, I just wanted to add that the textbook is subtly wrong. Convergence of \(\displaystyle \int_{-\infty}^\infty f(x)\,\text{d}x\) actually must allow for the terminals to be sent to \(\pm\infty\) at different rates, not necessarily the same rate.

That is \[\int_{-\infty}^\infty f(x)\,\text{d}x=\lim_{L\to -\infty}\int_L^a f(x)\,\text{d}x+\lim_{U\to\infty}\int_a^U f(x)\,\text{d}x,\quad a\in\mathbb{R}.\] This avoids stupid cases which fixates \(\infty-\infty=0\) such as in the integral \(\displaystyle\int_{-\infty}^\infty \dfrac{1}{x}\,\text{d}x\), where if we send the terminals to \(\pm\infty\) at the same rate we obtain \(0\), which is not correct. The integral actually doesn't exist.

Yeh in the 1/x case, we cannot integrate across the discontinuity. However, if we define the integral using its Cauchy principle value, then the value between +-1 would just be equal to 0. Like, if there are singularities are infinity, then integrating between +- infinity is possible if and only if we take the Cauchy principle value of the integral (if there are no singularities then it doesn't matter, e.g. integrating 1/(1+x^2) between +- infinity). Although the questions don't often state this, it's usually implied (at least in the context of complex analysis) that we take the Cauchy principle value.
« Last Edit: August 20, 2019, 10:28:38 am by blyatman »
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Tau

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Re: VCE Methods Question Thread!
« Reply #18081 on: August 20, 2019, 05:52:15 pm »
+1
Quote
link=topic=128232.msg1129149#msg1129149 date=1566201040]
not really a specific question. just wondering if anyone has any good ways of remembering the circular functions exact values cause i feel like a bit of a idiot waving my fingers around during a test lol :)
thanks!!

Here's how I like to think about the exact values for trigonometric functions.

We know that \(\frac{\sqrt{3}}{2}>\frac{\sqrt{2}}{2}>\frac12.\), and that \(\cos\) is the 'x'-coordinate of a point on the unit circle, and \(\sin\) is the 'y'-coordinate of a point on the unit circle. Now, for example, to find \(\sin(\pi/3)\) we think about the 'height' of the unit circle at 60 (sin is the red arrow, the 'height', and cos is the green arrow, the 'width'.). Since \(\sin\) is going to be 'large' at that angle (not a maximum of 1, of course) we know it's going to be the 'biggest' of  \(\{\frac{\sqrt{3}}{2},\frac{\sqrt{2}}{2},\frac12\}\) which is \(\frac{\sqrt{3}}{2}\). That's my preferred way to visualise and remember the angles, there's a great list of techniques here: https://math.stackexchange.com/questions/1553990/easy-way-of-memorizing-values-of-sine-cosine-and-tangent/1556967

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S200

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Re: VCE Methods Question Thread!
« Reply #18082 on: August 20, 2019, 10:46:00 pm »
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not really a specific question. just wondering if anyone has any good ways of remembering the circular functions exact values cause i feel like a bit of a idiot waving my fingers around during a test lol :)
thanks!!
I personally like @Keltingmeith's answer...
...

But for me, it was just remembering that SIN was 1-2-3 and COS was 3-2-1.

I refer of course to the square root over two...
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persistent_insomniac

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Re: VCE Methods Question Thread!
« Reply #18083 on: August 22, 2019, 05:39:17 pm »
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I know you will probably be tired/annoyed hearing this question again but I recently got back my sac 2 results and did poorly (only 5% above the cohort average). My first sac was slightly better (in the 60's) and our school makes sacs insanely difficult. If I do well in the next sac and even better on the end-of-year exam, do I have a chance of getting 40+ (maybe even 45??). I feel so demotivated right now because methods is one of the subjects i want to do the best in.

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #18084 on: August 22, 2019, 06:23:45 pm »
+2
I know you will probably be tired/annoyed hearing this question again but I recently got back my sac 2 results and did poorly (only 5% above the cohort average). My first sac was slightly better (in the 60's) and our school makes sacs insanely difficult. If I do well in the next sac and even better on the end-of-year exam, do I have a chance of getting 40+ (maybe even 45??). I feel so demotivated right now because methods is one of the subjects i want to do the best in.

This question is almost impossible to answer without knowing actually how difficult the SACs are in relation to the exams, the relative distribution of scores in your cohort, and the overall strength of your cohort.

To get a good indication of how you're tracking, I suggest you do a VCAA past exam (both exam 1 and exam 2) from the current study design (2016-present) under exam conditions (since exams set the standard).

For 40 or above, you want at least 100 out of 120. For 45 or above, you want at least 110 out of 120.
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rr28

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Re: VCE Methods Question Thread!
« Reply #18085 on: August 22, 2019, 06:33:46 pm »
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To get a good indication of how you're tracking, I suggest you do a VCAA past exam (both exam 1 and exam 2) from the current study design (2016-present) under exam conditions (since exams set the standard).

For 40 or above, you want at least 100 out of 120. For 45 or above, you want at least 110 out of 120.

This is actually a good idea! I was feeling very demotivated after my last SAC, did pretty good on the first one, kinda flogged the last one. Feeling really anxious for exams now.......I really want something above 38 so this will help me!

Also, anyone done their probability SAC yet?

radiant roses

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Re: VCE Methods Question Thread!
« Reply #18086 on: August 23, 2019, 07:13:44 am »
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Hi  :D
I am having trouble with this question.
I am unsure why I am getting the same answer for part a and b.
Is it right?

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #18087 on: August 23, 2019, 09:49:25 am »
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Hi  :D
I am having trouble with this question.
I am unsure why I am getting the same answer for part a and b.
Is it right?

In part a you've incorrectly assumed that  \(\Pr(A\cap B')=\Pr(A)\times\Pr(B')\).  This is true if \(A\) and \(B\) are independent, but there is no mention of this in part a. \begin{align*}\Pr(A\cup B')&=\Pr(B')+\Pr(A\cap B)\\ &=1-\Pr(B)+0\quad \Big[\Pr(A\cap B)=0\ \text{since }A\ \text{and}\ B\ \text{are mutually exclusive}\Big]\\&=\dfrac{1}{2}\end{align*}

In part b you've plugged in the wrong numbers to multiply to obtain \(\Pr(A\cap B')\). Fix that and your solution becomes correct even though it's a bit cumbersome. Here's a better method: \begin{align*}\Pr(A\cup B')&=\Pr(B')+\Pr(A\cap B)\\ &=1-\Pr(B)+\Pr(A)\Pr(B)\quad \Big[\Pr(A\cap B)=\Pr(A)\Pr(B)\ \text{since }A\ \text{and}\ B\ \text{are independent}\Big]\\&=\dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}\\ &=\dfrac{2}{3}\end{align*}
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #18088 on: August 25, 2019, 08:35:02 pm »
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Do stationary points of inflections occur only if the derived equation factorised has a square or above? E.g 3x(x+3)^2 with the -3 being the x-value of the inflection? Will this work with ^3,^4 etc?
Sorry if that didn't really make sense.

Also, is there a simple way to factorise a cubic without using a calculator that you cannot group?
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^^^111^^^

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Re: VCE Methods Question Thread!
« Reply #18089 on: August 25, 2019, 09:42:20 pm »
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Do stationary points of inflections occur only if the derived equation factorised has a square or above? E.g 3x(x+3)^2 with the -3 being the x-value of the inflection? Will this work with ^3,^4 etc?
Sorry if that didn't really make sense.

Also, is there a simple way to factorise a cubic without using a calculator that you cannot group?
There are several methods to factorise a cubic without using a calculator:
1. The first is to use long division of polynomials. First of all we should become familiar with the factor theorem. This states that for a polynomial, if P(a) = 0, then (x-a) is a factor of P(x). So for e.g. for the polynomial  x3 + - 3x2 + 6x -4, P(1) [that is 13 - 3*12 + 6*1 - 4] is equal to 0. Thus (x-1) is a factor of the polynomial  x3 + - 3x2 + 6x -4. Anyway, using the factor theorem we can find one factor of the cubic polynomial, and then we can divide the cubic with the factor. For example,  (x3 + - 3x2 + 6x -4)/(x-1) is equal to x2 -2x +7 with a remainder of 3.  Take away the 3 from the result of the divison and then you can factorise the quadratic equation. :)

2. Now this is short divison. Using the same example as above, if we were to find one factor using the factor theorem we get (x-1). Actually, we know more than this. The resulting equation of the short division will have to be x^2 (since x^2 * x is x^3) and the constant will have to be 4, (since 4 *-1 is -4). So then we have (x^2 + ?x + 4). If we were to multiply the negative 1 of (x-1) with the ?x, we get -x. Then if we multiply x from (x-1) with the 4 from (x^2 + ?x + 4), we get 4x. (Basically I am expanding the equation (x-1)(x2 + [something]x + 4). The coefficient we receive from expanding, 3, is what "?" is equal to.

3.This is through a process called "finite differences" and requires a table of x and y values of the equation. But idk if it is still in the VCE study design, (I can show you the process if u wish though).

Sorry, it is kind of difficult to explain without Latex  :-\,

hope that helps...

Edit: Oh sorry, I didn't  see your question before that.. Anyway, I think what you mean is can there be stationary points of inflection with polynomials with degrees higher than that of cubics? Well if that is what you are asking, yes there can be inflection points with fourth degree polynomials (not too sure about higher than that though...)
« Last Edit: August 25, 2019, 09:58:13 pm by ^^^111^^^ »