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April 19, 2024, 06:40:34 pm

Author Topic: VCE Methods Question Thread!  (Read 4818375 times)  Share 

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knightrider

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Re: VCE Methods Question Thread!
« Reply #12285 on: October 03, 2015, 07:30:02 pm »
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For math methods, just learn this identity:
where is any antiderivative of .
You can verify this by differentiating the antiderivative - you will end up with the integrand.
How this is derived is covered in SM. IMO they should SO cover this in methods too.
It's called u-subsitution, and it goes like this:


Although this is not in the methods course, I would say it's beneficial for methods knowledge too :)

Thanks so much  zsteve  :)

StupidProdigy

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Re: VCE Methods Question Thread!
« Reply #12286 on: October 03, 2015, 07:57:15 pm »
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How do I from the top line to the next...? Thankyou! :)
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Escobar

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Re: VCE Methods Question Thread!
« Reply #12287 on: October 03, 2015, 08:48:02 pm »
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How do I from the top line to the next...? Thankyou! :)
first, factorise the 2 out, then use long division of polynomials (this is how i do it)


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popsicles

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Re: VCE Methods Question Thread!
« Reply #12288 on: October 03, 2015, 09:05:06 pm »
+1
For this question attached is my working out right?

Hey man i would solve df/dx <0, and then you get 5-2x>0, and then you get x <5/2. Then you write,"since the domain of f is a subset of x <5/2, the gradient of f is always negative over the given domain."

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Re: VCE Methods Question Thread!
« Reply #12289 on: October 04, 2015, 12:03:35 pm »
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When a question asks when something is strictly increasing or decreasing do include the turning point or exclude it?

knightrider

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Re: VCE Methods Question Thread!
« Reply #12290 on: October 04, 2015, 12:26:29 pm »
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Hey man i would solve df/dx <0, and then you get 5-2x>0, and then you get x <5/2. Then you write,"since the domain of f is a subset of x <5/2, the gradient of f is always negative over the given domain."

Thanks popsicles

babushka818

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Re: VCE Methods Question Thread!
« Reply #12291 on: October 04, 2015, 01:08:14 pm »
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When a question asks when something is strictly increasing or decreasing do include the turning point or exclude it?

I think for strictly increasing/decreasing it is included, but if just increasing or decreasing it is not included.
« Last Edit: October 04, 2015, 01:57:47 pm by babushka818 »
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Unblemished

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Re: VCE Methods Question Thread!
« Reply #12292 on: October 04, 2015, 01:39:17 pm »
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For math methods, just learn this identity:
where is any antiderivative of .
You can verify this by differentiating the antiderivative - you will end up with the integrand.
How this is derived is covered in SM. IMO they should SO cover this in methods too.
It's called u-subsitution, and it goes like this:


Although this is not in the methods course, I would say it's beneficial for methods knowledge too :)

According to my methods teacher who teaches both methods and spesh, you aren't allowed to use the U-substitution on the exam for meth.
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cosine

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Re: VCE Methods Question Thread!
« Reply #12293 on: October 04, 2015, 02:24:59 pm »
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Im pretty sure its the opposite, ie for strictly increasing the gradient should always be greater than 0, and for increasing the gradient should be equal or greater than 0.

This means exclude the stationary point for strictly increasing..

Incorrect, strictly increasing/decreasing you must include the turning points in the interval.
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #12294 on: October 04, 2015, 02:32:02 pm »
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Actually, any form of increasing/decreasing would include the turning point as an endpoint. Unless your function is actually constant at the turning point (what I mean is something like:
y = (x-2)^3 x>2
0 for |x|<=2
(x+2)^3 for x<-2; here, the function is actually constant for |x|<=2)

This is because the definition of increasing simply says:
If b>a, then f(b) > =f(a)
Even at one end of a turning point, the function still changes. Consider y = x^2. It is increasing on the interval [0, infinity) because on this interval, if you choose ANY b and a such that b>a, then f(b) >= f(a).

The difference between strictly increasing and increasing is that the set of increasing functions includes that weird example I gave above; that function is increasing on x >=-2, but strictly increasing on x>=2 only. Strictly increasing requires if b>a, f(b) > f(a), i.e. a strict inequality.
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pinklemonade

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Re: VCE Methods Question Thread!
« Reply #12295 on: October 04, 2015, 02:48:57 pm »
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Hey guys
So I need to solve the equation for

But I don't understand which quadrants the domain fits and how much times I would need to add 2 pi or subtract 2 pi

Could someone please explain?
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #12296 on: October 04, 2015, 03:50:07 pm »
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Hey guys
So I need to solve the equation for

But I don't understand which quadrants the domain fits and how much times I would need to add 2 pi or subtract 2 pi

Could someone please explain?

Let u = 2x + pi/2
Then your domain is transformed into u E [-7pi/2, 5pi/2]
Your equation also becomes cos u = -sqrt 3 / 2
Does that help somewhat?
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cosine

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Re: VCE Methods Question Thread!
« Reply #12297 on: October 04, 2015, 04:30:46 pm »
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Hey guys
So I need to solve the equation for

But I don't understand which quadrants the domain fits and how much times I would need to add 2 pi or subtract 2 pi

Could someone please explain?

Step 1: Rearrange the function to isolate the circular function:





Step 2: Rearrange the domain:



Thats what x is defined for, but clearly our new 'x' is actually

So what we have done is, multiplied all x values by 2, and then added on 90 degrees to each value, so do the same to our old domain:






 
Step 3: Solve, for the new domain:





Those are the positive solutions, ill let you do the negative ones xD
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cameotodd

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Re: VCE Methods Question Thread!
« Reply #12298 on: October 05, 2015, 07:33:56 pm »
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Anyone had a go at the Kilbaha methods exam 1 2015? How'd you find it?
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knightrider

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Re: VCE Methods Question Thread!
« Reply #12299 on: October 05, 2015, 08:27:26 pm »
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Is there a way to find the long term probability of something by hand for a transition matrix?