VCE Methods Question Thread!

[silverpixeli]

and finally x=(pi*n)/2 +(-1)^n *pi/4
however the correct solution was x=n*pi + pi/4

The answers are basically equivalent, try n=0 and n=1 in your solution and you get:

x=(pi*0)/2 +(-1)^0 *pi/4 = pi/4

and

x=(pi*1)/2 +(-1)^1 *pi/4 = pi/2 - pi/4 = pi/4

and if you try n=0 in the correct solution you get:

x=0*pi + pi/4 = pi/4

if you then tried 2, 3 in your solution and 1 in their solution you'd get the same answers again


what's the reason? well your basic angle was pi/2, and normally you'd have to find the other angle that gives you the same sine value by subtracting pi/2 from pi, but in the special case where pi=1 (or in fact -1), that other angle is the same angle! there's only one angle that gives you sin(angle) = 1 in each cycle, whereas normally (for any other angle) there is a second in quadrant 2 (or two in quadrant 3 and 4 for negative values, with the exception that there's only one angle that gives you -1)

because of that, we can literally say 2x = pi/2 + n*2pi because we have our one angle and then we'll just add 2pi again and again, n times, to get solutions from other cycles
divide by 2 and you get x = pi/4 + npi which is the answer


it's always possible that the answer you get will appear to not be a solution to a multiple choice question, in that case you need to look at the solutions that are there and see if any are equivalent to your answer. (or look at your answer and see if it's the same as anything there)

[knightrider]

How would you do this question?

[silverpixeli]

we have the equation: perimeter = 11 = 2x + that arc at the top + 2x + y

we could solve for y if we knew the length of that arc at the top, which is half of the circumference of a circle with diameter y

once you realise that the arc's diameter is y you should be able to do it, here's my working:

Spoiler

C=piD=y*pi
but it's not a full circle, it's only half,
so 'that arc at the top' = 1/2 * y * pi

the equation becomes:

11 = 2x + 1/2 pi y + 2x + y
11 = 4x + (1/2 pi + 1)y
22 - 8x = (pi+2)y
(22-8x)/(pi+2) = y as required

hope that helps

[knightrider]

we have the equation: perimeter = 11 = 2x + that arc at the top + 2x + y

we could solve for y if we knew the length of that arc at the top, which is half of the circumference of a circle with diameter y

once you realise that the arc's diameter is y you should be able to do it, here's my working:

Spoiler

C=piD=y*pi
but it's not a full circle, it's only half,
so 'that arc at the top' = 1/2 * y * pi

the equation becomes:

11 = 2x + 1/2 pi y + 2x + y
11 = 4x + (1/2 pi + 1)y
22 - 8x = (pi+2)y
(22-8x)/(pi+2) = y as required

hope that helps

Thanks silverpixeli,

how would you do this question relating to the previous part
b) Hence, find an expression for the area, A, of the window in terms of x.

[silverpixeli]

Well in a similar manner to the above, we can get a formula for the area by considering all the pieces. In this case it will be the area of the rectangle plus the area of the semi-circle top:

A = rectangle + semicircle = rectangle + 1/2 * circle

So the rectangle's area is 2x * y

and 1/2 * circle is 1/2 * pi*r^2 where r is y/2
this simplifies to half circle area = (pi*y^2)/8

using our compound area thingy: A = 2x*y + (pi*y^2)/8

we're basically done, except for the fact that the question requires a formula in terms of x, not x and y

can you think of a way to express what we have in terms of x only?

here's my solution:

Spoiler

from part a, we know that  y = (22-8x)/(pi+2)

so wherever we see a y, we can replace it with (22-8x)/(pi+2) and this is just in terms of x,
so we just have to simplify and we've got it.

we know that this is okay information to use since we're told that the perimeter HAS to be 11, and we already worked out that if the perimeter is 11, y MUST be (22-8x)/(pi+2)

my working:

A = 2x*y + (pi*y^2)/8
A = 2x*(22-8x)/(pi+2) + (pi*[(22-8x)/(pi+2)]^2)/8
... simplifies to ...
A = (44x-16x^2)/(pi+2) + pi*[ (11-4x)/(pi+2) ]^2/2
which you can try to do by hand, or just use the CAS

(sorry for the lack of latex)

this structure is common for a question, where you need to first express one variable in terms of another, and then later express another value (like area) which is usually in terms of both, instead in terms of only 1 of the two

[knightrider]

Well in a similar manner to the above, we can get a formula for the area by considering all the pieces. In this case it will be the area of the rectangle plus the area of the semi-circle top:

A = rectangle + semicircle = rectangle + 1/2 * circle

So the rectangle's area is 2x * y

and 1/2 * circle is 1/2 * 2pi*r^2 where r is y/2
this simplifies to half circle area = (pi*y^2)/4

using our compound area thingy: A = 2x*y + (pi*y^2)/4

we're basically done, except for the fact that the question requires a formula in terms of x, not x and y

can you think of a way to express what we have in terms of x only?

here's my solution:

Spoiler

from part a, we know that  y = (22-8x)/(pi+2)

so wherever we see a y, we can replace it with (22-8x)/(pi+2) and this is just in terms of x,
so we just have to simplify and we've got it.

we know that this is okay information to use since we're told that the perimeter HAS to be 11, and we already worked out that if the perimeter is 11, y MUST be (22-8x)/(pi+2)

my working:

A = 2x*y + (pi*y^2)/4
A = 2x*(22-8x)/(pi+2) + (pi*[(22-8x)/(pi+2)]^2)/4
... simplifies to ...
A = (44x-16x^2)/(pi+2) + pi*[ (11-4x)/(pi+2) ]^2
which you can try to do by hand, or just use the CAS

(sorry for the lack of latex)

this structure is common for a question, where you need to first express one variable in terms of another, and then later express another value (like area) which is usually in terms of both, instead in terms of only 1 of the two

Thanks silverpixeli
But isnt the formula for the area of a circle              

[silverpixeli]

ooops, most definitely. i'll fix that right up

[knightrider]

ooops, most definitely. i'll fix that right up

Everyone makes mistakes even the best

Silverpixeli what are your tips on doing well in maths methods and how did you go about preparing for sacs and the exam and how many practice exams did you end up doing and when do you recommend doing the vcaa ones.

Also what are your thoughts on finishing  the methods course in the holidays and at what pace did you work through the course

[silverpixeli]

Silverpixeli what are your tips on doing well in maths methods

Spend your methods study time effectively by making sure there's nothing in the course that you don't get. Target one area at a time until you've mastered it and can handle any question. In this respect I include use of CAS as part of the course so you should work on that too if you're not great by the end of the year.

how did you go about preparing for sacs and the exam

Maths methods sacs can be prepared for best by keeping up/working ahead and working with plenty of practice questions, it's as simple as that. But just working through questions without understanding what you're doing and why you employed those formulas/strategies isn't helpful, at least for me, I need to understand why I'm doing what I'm doing because that means I'll know when I can and can't use that approach in an unfamiliar question environment, which is often what you get on SACs and certainly on the exam.

how many practice exams did you end up doing

I worked through lots of practice exams, over 40 exam 1's and over 20 exam 2's for methods last year (didn't keep an accurate count) but it's definitely more about what you get out of each paper than how many you do. I think it's worth spending time to really understand a small amount of questions rather than blazing through and rehearsing your strong points and leaving out the rest on a large number of papers.

when do you recommend doing the vcaa ones.

VCAA papers are good to leave till term4/swotvac so that you can get into the feel of the way the examiners like questions to be answered including working etc. It's not a bad idea to start with some as well as you'll have an idea what they're like compared to other exams that you do.

what are your thoughts on finishing  the methods course in the holidays and at what pace did you work through the course

I studied ahead and worked through 1/3 of the course prior to the start of term 1, then I finished the next part during term 1, then the probability section during term 1 holidays. You don't have to work ahead to do well but it can help you in terms of lightening the load in other subjects. I definitely don't think it's worth rushing, it's best to adopt an approach that has you learning wholesomely as you go because there will be less problems to go back and fix. Also, if you do work ahead, don't let yourself get complacent because I think that's a serious risk when working ahead in anything. SACs are pretty consistent and require constant preparation.

[knightrider]

Thanks so much silverpixeli

What book did you use for methods and do you think it is a good idea to get more than one book.

This is what idea i had do you think its good

I am going to finish the methods course in the summer holidays excluding the chapter reviews.

Then for sac preparation i will do all the chapter reviews from the relevant chapters, and also since i have other books as well i will do their chapter reviews as well.
This way the knowledge will stay in my head and i will be constantly preparing for sacs and the exams.
Do you think this is a good idea

[knightrider]

How would you do this question?

[silverpixeli]

Thanks so much silverpixeli

What book did you use for methods and do you think it is a good idea to get more than one book.

This is what idea i had do you think its good

I am going to finish the methods course in the summer holidays excluding the chapter reviews.

Then for sac preparation i will do all the chapter reviews from the relevant chapters, and also since i have other books as well i will do their chapter reviews as well.
This way the knowledge will stay in my head and i will be constantly preparing for sacs and the exams.
Do you think this is a good idea

sounds solid, but only if your working ahead is to exam standard and you don't leave anything out

as for books, one is fine, even maths quest, you get enough variety out of practice exams in term 3/4 anyway and there's no use burning out. dont hesistate to look for a better explanation in another book or online, though

[silverpixeli]

How would you do this question?

A function will only have an inverse that is a function if it is one-to-one. f(x) is a quadratic in his case, which is not one-to-one (some y values can be reached by more than one x value) so I'd say the answer is A. This is a theory question basically, make sure you're familiar with what an inverse function is and all of that terminology. don't let the a's and b's scare you, they might as well be a 3 and a 10 for all it matters.

[silverpixeli]

Thanks silverpixeli
i have another question that might be a bit off topic
How did you go about preparing for English and physics

That is off topic, I'll PM you, this thread's for methods 3/4 questions

[knightrider]

How are you expected to be able to graph this graph without a calculator.
How would you find the endpoints and xintercepts and yintercepts