Login

Welcome, Guest. Please login or register.

March 29, 2024, 08:05:25 am

Author Topic: VCE Methods Question Thread!  (Read 4802776 times)  Share 

0 Members and 13 Guests are viewing this topic.

caqiu

  • Forum Regular
  • **
  • Posts: 84
  • Respect: +1
Re: VCE Methods Question Thread!
« Reply #18915 on: October 31, 2020, 04:51:57 pm »
0
Hey Guys,
Would someone be able to take a look at this question?
I subbed the values in and then made them the same, rather than doing what vcaa did.
Thank you!

james.358

  • MOTM: OCT 20
  • Forum Regular
  • **
  • Posts: 93
  • Respect: +110
Re: VCE Methods Question Thread!
« Reply #18916 on: October 31, 2020, 07:33:36 pm »
+4
Hey Caqiu!

When you're doing these types of "show that" problems, you should never assume that they're equal straight away by writing them as an equality. Even though each step of your working is technically correct, you would still lose a mark. You should always show LHS and RHS independently, and only then equate them.

Hope this helps!
James
« Last Edit: October 31, 2020, 09:02:23 pm by james.lhr »
VCE Class of 2022: 99.90 ATAR
Monash Medical School Class of 2026

RAW 50 Methods & Specialist High Yield Resources

keltingmeith

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 5493
  • he/him - they is also fine
  • Respect: +1292
Re: VCE Methods Question Thread!
« Reply #18917 on: October 31, 2020, 08:55:04 pm »
+4
Not sure if this is a misconception of yours or not, but I would first like to address this regardless: a circle is not a quadratic. Although there is a degree two x variable, there's also a degree two y variable in the general equation which prevents us from classifying a circle as a quadratic. You may then ask: what type of equation is a circle? Well, for the purposes of the syllabus, a circle is just a circle. That's all there is to it  :) (If you're interested, you could search up "conics" which can give you more insight into a circle, but that's beyond the Methods curriculum, so don't stress if you don't understand it all). Also apologies if you already know that a circle is not a quadratic.

What you did was find the midpoint of the two x-intercepts, (4,0) and (-4,0), of the circle, getting (0,0) and assuming that this is the center of the circle. The issue with doing this is that, using your x-intercepts, you can only deduce the x-coordinate of the center, which is 0, in this case. However, we cannot assume that the center has the same y-coordinate as these x-intercepts (i.e. we cannot assume the y-coordinate is zero). Using your method, we would actually have to find the midpoint of the y-intercepts, which are (0,8) and (0,-2), and take the y-coordinate (3) of the midpoint (which is (0,3) ) to get the y-coordinate of the circle's center. We would then put these pieces of information together to get the center of (0,3).

Really nice help, p0kem0n21! But I wanted to jump in on this line right here:

From this, you can see that the method is a bit lengthy and requires a bunch of substitution and quadratic solving.

There's actually a very easy thing you can do here once you know a point on the circle and the centre - after all, the radius is just the distance from the centre to a point on the circle. So, if I know the centre of the circle is (0,3), and that one of the x-intercepts is (4,0), then the distance from the centre of the circle to the x-intercept is:

\[
D=\sqrt{(4-0)^2+(0-3)^2}=\sqrt{16+9}=\sqrt{25}=5
\]

So, if the centre is (0,3), the radius is 5, and the equation of the circle is \(x^2+(y-3)^2=5^2\)

caqiu

  • Forum Regular
  • **
  • Posts: 84
  • Respect: +1
Re: VCE Methods Question Thread!
« Reply #18918 on: October 31, 2020, 09:05:32 pm »
0
Hey Caqiu!

When you're doing these types of "show that" problems, you should never assume that they're equal straight away by writing them as an equality. Even though each step of your working is technically correct, you would still lose a mark. You should always show LHS and RHS independently, and only then equate them.

Hope this helps!
James

Thanks for the help!!

caqiu

  • Forum Regular
  • **
  • Posts: 84
  • Respect: +1
Re: VCE Methods Question Thread!
« Reply #18919 on: October 31, 2020, 09:39:05 pm »
0
Hey guys,
Just another question do you write e.g Pr (x>10|x>8) = (pr (x>10 n x>8)) / pr (x>8) or do you write it out as (pr (x>10) n pr (x>8)) / pr (x>8)
Thank you  :)

keltingmeith

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 5493
  • he/him - they is also fine
  • Respect: +1292
Re: VCE Methods Question Thread!
« Reply #18920 on: October 31, 2020, 09:49:26 pm »
+5
Hey guys,
Just another question do you write e.g Pr (x>10|x>8) = (pr (x>10 n x>8)) / pr (x>8) or do you write it out as (pr (x>10) n pr (x>8)) / pr (x>8)
Thank you  :)


The first one is the correct one. You can't have \(\mathbb{P}(A)\cap\mathbb{P}(B)\), because the intersect symbol talks about two events happening at the same time. \(\mathbb{P}(A)\) is not an event, it's the probability of an event happening. If you were to write \(\mathbb{P}(A)\cap\mathbb{P}(B)\), it's equivalent to saying, "the probability of seeing a bird happening at the same time as the probability of seeing a duck" - it's just a statement that doesn't make sense. Probabilities are descriptions of things that can happen, they aren't things that can happen themselves. Kind of like how something can be green, but green itself is not an object, it's a colour

IThinkIFailed

  • Forum Regular
  • **
  • Posts: 97
  • Respect: +12
Re: VCE Methods Question Thread!
« Reply #18921 on: November 01, 2020, 09:58:50 am »
0
Hey guys, I have a quick question before I hop off atarnotes again to study :P

Are graphing circles and semi circles in the methods study design for unit 3 and 4? (I.e Relations I guess)
Also, are kinematics concepts in the methods 3/4 study design?
2019:
Biology [42]   Economics [46]

2020:
Methods[41 :( ]
English [42]
Chemistry [47]
Legal studies [44]

ATAR: 99.00

james.358

  • MOTM: OCT 20
  • Forum Regular
  • **
  • Posts: 93
  • Respect: +110
Re: VCE Methods Question Thread!
« Reply #18922 on: November 01, 2020, 10:46:51 am »
+8
Hey there!

Although the equations & graphing for circles and semi circles are not explicitly in the Methods 3/4 study design, they may still be assumed knowledge. So far no questions require you to know these equations, but being familiar with them is still useful.

For example in 2017 Exam 2 Q2e, they provided you the equation for a semi circle and asked you to differentiate it, but you didn't have to know what it was for.

Relations are definitely in the study design and I would recommend you to become familiar with them before the exam.

Hope this helps!
James
VCE Class of 2022: 99.90 ATAR
Monash Medical School Class of 2026

RAW 50 Methods & Specialist High Yield Resources

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Methods Question Thread!
« Reply #18923 on: November 02, 2020, 05:59:23 pm »
0
Not sure if this is a misconception of yours or not, but I would first like to address this regardless: a circle is not a quadratic. Although there is a degree two x variable, there's also a degree two y variable in the general equation which prevents us from classifying a circle as a quadratic. You may then ask: what type of equation is a circle? Well, for the purposes of the syllabus, a circle is just a circle. That's all there is to it  :) (If you're interested, you could search up "conics" which can give you more insight into a circle, but that's beyond the Methods curriculum, so don't stress if you don't understand it all). Also apologies if you already know that a circle is not a quadratic.


What you did was find the midpoint of the two x-intercepts, (4,0) and (-4,0), of the circle, getting (0,0) and assuming that this is the center of the circle. The issue with doing this is that, using your x-intercepts, you can only deduce the x-coordinate of the center, which is 0, in this case. However, we cannot assume that the center has the same y-coordinate as these x-intercepts (i.e. we cannot assume the y-coordinate is zero). Using your method, we would actually have to find the midpoint of the y-intercepts, which are (0,8) and (0,-2), and take the y-coordinate (3) of the midpoint (which is (0,3) ) to get the y-coordinate of the circle's center. We would then put these pieces of information together to get the center of (0,3).

From this, you can see that the method is a bit lengthy and requires a bunch of substitution and quadratic solving. Not to mention that calculating the radius could also be a disaster (not here, but in cases where neither the x-coordinate nor the y-coordinate of the center are 0). What's worse is that this method cannot be used with all circles, particularly those which have less than 2 x-intercepts and/or less than 2 y-intercepts. For example, try your method with a circle of this equation:

x2-4x+y2-4y+7=0

If you tried substituting x=0 or y=0 into the equation, you'll just end up getting a quadratic which cannot be solved. That's why the "complete the square" method is used with circles; you can complete the square with any quadratic you have, and deduce information from there. Ok, I just said that circles are not quadratics. However, you can treat the degree 2 x and y variables as quadratics. The thing is, you complete the square separately for your x and y variables. This process may be a bit confusing at times. In the above equation, you could write -4(x+y). Don't do that, that's keeping the x and y variables in the same brackets and will just lead to a load of confusion.

Hey pokemon, appreciate the help!
This cleared things up :) I thought that by there being no extra terms outside of the two binomial expressions, there wouldn't be any translation up in the graph and so the midpoint had to be (0,0) and radius had to be 4. I didn't realise I had this preconception until you corrected it.

Also, nice tip on using Pythagoras Kelting :) will remember that one.

a weaponized ikea chair

  • Trendsetter
  • **
  • Posts: 182
  • Respect: +12
Re: VCE Methods Question Thread!
« Reply #18924 on: November 02, 2020, 06:45:50 pm »
0
So I think I put this here a while back, but I'll include an image so I'm specific on what I'm asking.

For a bound reference, can you use tape and wrap it around two seperate books like in the image such that it is bound?

I have some notes in one BR and some in another, and I need them both.

AngelWings

  • Victorian Moderator
  • ATAR Notes Superstar
  • *****
  • Posts: 2456
  • "Angel wings, please guide me..."
  • Respect: +1425
Re: VCE Methods Question Thread!
« Reply #18925 on: November 02, 2020, 07:07:10 pm »
+3
So I think I put this here a while back, but I'll include an image so I'm specific on what I'm asking.

For a bound reference, can you use tape and wrap it around two seperate books like in the image such that it is bound?

I have some notes in one BR and some in another, and I need them both.
From memory, yes. As long as it has a single “spine” that holds it in place, which if you wrap duct tape or something to hold both books tightly together along their original spines, should do the trick. Your teachers/ invigilators might also use the “shake test” (basically shaking the bound reference to ensure that no additional flaps, pages can be  detached, etc.), so you might need to make it tough enough to endure that.
VCE: Psych | Eng Lang | LOTE | Methods | Further | Chem                 
Uni: Bachelor of Science (Hons) - genetics
Current: working (sporadically on AN)
VTAC Info Thread

dylan.kumar21

  • Trailblazer
  • *
  • Posts: 38
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #18926 on: November 03, 2020, 08:09:18 am »
0
Can someone help me with these?
 (Only part C for the first question)
« Last Edit: November 03, 2020, 09:04:28 am by dylan.kumar21 »

S_R_K

  • MOTM: Feb '21
  • Forum Obsessive
  • ***
  • Posts: 487
  • Respect: +58
Re: VCE Methods Question Thread!
« Reply #18927 on: November 03, 2020, 08:49:47 am »
+2
Can someone help me with these?
 (Part C for the first question)

\(\cos(\theta-\frac{\pi}{2})=\sin(\theta)\)

dylan.kumar21

  • Trailblazer
  • *
  • Posts: 38
  • Respect: 0
Re: VCE Methods Question Thread!
« Reply #18928 on: November 03, 2020, 09:03:22 am »
0
\(\cos(\theta-\frac{\pi}{2})=\sin(\theta)\)

i understand that property but how i would i go about showing it?

keltingmeith

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 5493
  • he/him - they is also fine
  • Respect: +1292
Re: VCE Methods Question Thread!
« Reply #18929 on: November 03, 2020, 09:10:35 am »
+1
i understand that property but how i would i go about showing it?

This isn't a result you have to prove - you can just use it directly. If you're still unsure, feel free to post an answer to that question, and we can let you know if you've shown enough working