VCE Methods Question Thread!

[james.358]

Hey Corey!

When solving simultaneous equations, there may be more variables than unique equations. In that case, you wouldn't be able solve it the traditional way.

This is where introducing a parameter comes in. You can kind of imagine it as assuming the value of one of the variables. In the screen-clipping you had, it defined y as the parameter, but you can choose any variable. Once you introduce the parameter, you can write the other solutions in terms of it. Just be careful to define the perimeter (e.g. λ ∈ R, k ∈ N, etc)

This concept might be a bit confusing at the start, but trust me you'll soon wrap your head around it!

Hope this helps,
James

[SS1314]

Hi can someone pls explain why this is the case. I don't understand how the boundaries of the integral change from m to m/k and from k to 1 when f(kx) become f(x)

[The Cat In The Hat]

Hi can someone pls explain why this is the case. I don't understand how the boundaries of the integral change from m to m/k and from k to 1 when f(kx) become f(x)

So f(kx) becomes f(x) (meaning kx -> x). This means that for all the information you already had about f(kx), you divide it by k to make it just x. Thus, m -> m/k, and k -> 1, and 1/k ->1.
I don't know how to explain it more, but if you're still confused I can try again (I had a question requiring something similar to this kind of thing in a practise exam I did recently, so it happens to be fresh in my memory)

[Corey King]

Hey Corey!

When solving simultaneous equations, there may be more variables than unique equations. In that case, you wouldn't be able solve it the traditional way.

This is where introducing a parameter comes in. You can kind of imagine it as assuming the value of one of the variables. In the screen-clipping you had, it defined y as the parameter, but you can choose any variable. Once you introduce the parameter, you can write the other solutions in terms of it. Just be careful to define the perimeter (e.g. λ ∈ R, k ∈ N, etc)

This concept might be a bit confusing at the start, but trust me you'll soon wrap your head around it!

Hope this helps,
James

I'll be honest I don't really understand. It just looks like they rearranged the equation to make x the subject, and then renamed y to .
Your answer was good I'm just not seeing it yet

[james.358]

Hey Corey,

I'll be honest I don't really understand.

All good. I'll try clarifying it further here.

It just looks like they rearranged the equation to make x the subject, and then renamed y to λ.

Yep you're spot on, that is exactly what it is. You might be asking "why on earth would anyone do this?!" After all, you already have a lot of equations to deal with, right? But by writing it this way you can establish the relationship between the variables, and how changing one of them (e.g. the parameter) can affect the value of the other variables. This also ensures that you will indeed get a solution (s)

Oh and if you didn't know the parameter is represented as c (for all real numbers) or n for natural numbers on the CAS, so don't get intimidated by it on a Tech Active test.

Hope this helps to clear it up!
James

[Corey King]

Thanks for helping out James.

But by writing it this way you can establish the relationship between the variables, and how changing one of them (e.g. the parameter) can affect the value of the other variables. This also ensures that you will indeed get a solution (s)

The definition of parameter here just sounds like a variable to me. What's the difference?

[james.358]

Yep! If it helps with your intuitive understanding, you can definitely think of it like that.

[Corey King]

Yep! If it helps with your intuitive understanding, you can definitely think of it like that.

If I get (6-a, a) as the parameter, is a the parameter, or is (6-a) the parameter?

When the textbook asks me to 'describe these solutions using a parameter' for two lines with infinite solutions, do I need to still include the two variables of the domain and codomain (x and y) in my answer?
I.e. x + y = 6

create a parameter:
let y = a
(6-a, a)
...
x = (6-a, a) where a is a real number.  ?

Many thanks,
Corey

[keltingmeith]

If I get (6-a, a) as the parameter, is a the parameter, or is (6-a) the parameter?

When the textbook asks me to 'describe these solutions using a parameter' for two lines with infinite solutions, do I need to still include the two variables of the domain and codomain (x and y) in my answer?
I.e. x + y = 6

create a parameter:
let y = a
(6-a, a)
...
x = (6-a, a) where a is a real number.  ?

Many thanks,
Corey

a is the parameter. It's like you said before - a parameter is basically a variable, so it makes sense from that, that a would be the parameter, not 6-a which is an expression involving the parameter - just like 6-x is an expression involving the variable, and the variable is x. And in the same ways, parameter equations are just like normal equations - so you wouldn't say, x=(6-a,a). You'd instead say, x=6-a, and y=a, listing the two equations separately as a solution.

This is going to be a massive tangent, but eh, YOLO. Parameterization is confusing when brought up this way, because you may honestly be thinking - "what's the point? Why not just say the solutions are on the line y=6-x?" And the answer for that is while you can do this for a straight line, you can't do this for other shapes.

For example, consider a particle moving around in a circle. It follows the path x^2+y^2=1. Cool. Where does the particle start? At x=0? Well, if it's x=0, then is it at y=1 or y=-1? What about y=0? Then is it x=1, or x=-1? Maybe it's when x=y - but is that at 1/sqrt(2), or -1/sqrt(2)? Here, x and y are positions, so we have no clue where the particle starts. Functions, like y=6-x, don't have this ambiguity - we know that everything starts at the lowest part of the domain, and finishes at the highest part. Well, what if I parameterise my equations like this:

x=cos(t)
y=sin(t)

Where t is time. Suddenly, we have two functions which tell you the x and y position separately based on what time it is. So, at t=0, the particle is at the position (cos(0),sin(0))=(1,0). At t=pi/2, the position is now (cos(pi/2),sin(pi/2))=(0,1). In fact, here's desmos graph for you to play around with, where you can change the value of t and look at what the particle's position is:

https://www.desmos.com/calculator/icx6fwpe1f

Notice what shape it makes as t increases? It's still a circle - but now, there's no ambiguity about where the particle starts, and where it might be at some other time. In fact, now we know something that x^2+y^2=1 could NEVER tell us - the particle is moving anti-clockwise around the circle as time goes forward. And using parameterisation like this, we can make even more, weirder, shapes, like this one:

https://www.desmos.com/calculator/cw1dafgcgv

This is why parameterising is useful, and why it makes sense to do it the way we are. The reason the textbook has introduced it this way to you is to teach it to you in a familiar context - as a way of drawing lines (and fun fact: in higher dimensions, the ONLY way to draw a line is to parameterise equations in the same way you are are learning). But, as you can see from these examples, the way of defining a parameterisation is exactly the same way you would do any other equation - by putting y=something, but also adding in x=something, and vice-versa.

[Corey King]

a is the parameter. It's like you said before - a parameter is basically a variable, so it makes sense from that, that a would be the parameter, not 6-a which is an expression involving the parameter - just like 6-x is an expression involving the variable, and the variable is x. And in the same ways, parameter equations are just like normal equations - so you wouldn't say, x=(6-a,a). You'd instead say, x=6-a, and y=a, listing the two equations separately as a solution.

This is going to be a massive tangent, but eh, YOLO. Parameterization is confusing when brought up this way, because you may honestly be thinking - "what's the point? Why not just say the solutions are on the line y=6-x?" And the answer for that is while you can do this for a straight line, you can't do this for other shapes.

For example, consider a particle moving around in a circle. It follows the path x^2+y^2=1. Cool. Where does the particle start? At x=0? Well, if it's x=0, then is it at y=1 or y=-1? What about y=0? Then is it x=1, or x=-1? Maybe it's when x=y - but is that at 1/sqrt(2), or -1/sqrt(2)? Here, x and y are positions, so we have no clue where the particle starts. Functions, like y=6-x, don't have this ambiguity - we know that everything starts at the lowest part of the domain, and finishes at the highest part. Well, what if I parameterise my equations like this:

x=cos(t)
y=sin(t)

Where t is time. Suddenly, we have two functions which tell you the x and y position separately based on what time it is. So, at t=0, the particle is at the position (cos(0),sin(0))=(1,0). At t=pi/2, the position is now (cos(pi/2),sin(pi/2))=(0,1). In fact, here's desmos graph for you to play around with, where you can change the value of t and look at what the particle's position is:

https://www.desmos.com/calculator/icx6fwpe1f

Notice what shape it makes as t increases? It's still a circle - but now, there's no ambiguity about where the particle starts, and where it might be at some other time. In fact, now we know something that x^2+y^2=1 could NEVER tell us - the particle is moving anti-clockwise around the circle as time goes forward. And using parameterisation like this, we can make even more, weirder, shapes, like this one:

https://www.desmos.com/calculator/cw1dafgcgv

This is why parameterising is useful, and why it makes sense to do it the way we are. The reason the textbook has introduced it this way to you is to teach it to you in a familiar context - as a way of drawing lines (and fun fact: in higher dimensions, the ONLY way to draw a line is to parameterise equations in the same way you are are learning). But, as you can see from these examples, the way of defining a parameterisation is exactly the same way you would do any other equation - by putting y=something, but also adding in x=something, and vice-versa.

Fantastic response as always.
I'll read it a few more times and play with those graphs until I absorb it.

Just to respond to the first paragraph, I tried to answer the problem that way as that is how it was posed in the textbook. It seems like I may have misunderstood the question?
Textbook Question: https://gyazo.com/bb44c00c5f7465bdb457b46ee09c2f52
Textbook Answer: https://gyazo.com/f68f194a1f1f4941e87f390a3868a1bc
I'm comfortable with what a parameter is now , but I'm not familiar with the notation (or the usefulness of a parameter) yet.

Many thanks Kelting,
Corey

[keltingmeith]

Fantastic response as always.
I'll read it a few more times and play with those graphs until I absorb it.

Just to respond to the first paragraph, I tried to answer the problem that way as that is how it was posed in the textbook. It seems like I may have misunderstood the question?
Textbook Question: https://gyazo.com/bb44c00c5f7465bdb457b46ee09c2f52
Textbook Answer: https://gyazo.com/f68f194a1f1f4941e87f390a3868a1bc
I'm comfortable with what a parameter is now , but I'm not familiar with the notation (or the usefulness of a parameter) yet.

Many thanks Kelting,
Corey

Yeah - so the notation they've used is identical to the notation that desmos is using. That is, instead of using equations, we say that all the solutions lie on the point (t,6-t). This means that the solutions for some values of t are:

t=0: (0, 6)
t=1: (1, 5)
t=2: (2, 4)

It's just like a normal ordered-pair - just like you'd normally describe points on the plane. If, however, you had written x=t, y=6-t, then your solutions for those same values of t are:

t=0: x=0, y=6
t=1: x=1, y=5
t=2: x=2, y=4

Notice the difference? Hopefully the answer is no, because there is NO difference. They both describe the exact same scenario, they're just using different ways of describing it. It's almost like walking up to something and you saying, "this is a very bright colour, it's very yellow", and someone else saying "I'd say it has the hex code of FFFF00, or 255 red, 255 green, and 0 blue" (plug this into paint to see what colour comes out if you don't believe that this is yellow) - sure, one's a little more odd than the other, and something you're not used to, but they both CORRECTLY identify the colour as yellow. Same thing with if you were to use the order pair notation as in the text book, or if you were to instead write the answer in terms of equations - both are correct, they're just describing the situation slightly differently. I answered this recently I think on this very thread? But you'll never be marked down for using a different notation in a VCAA exam, ONLY if you use that notation incorrectly

[Corey King]

Yeah - so the notation they've used is identical to the notation that desmos is using. That is, instead of using equations, we say that all the solutions lie on the point (t,6-t). This means that the solutions for some values of t are:

t=0: (0, 6)
t=1: (1, 5)
t=2: (2, 4)

It's just like a normal ordered-pair - just like you'd normally describe points on the plane. If, however, you had written x=t, y=6-t, then your solutions for those same values of t are:

t=0: x=0, y=6
t=1: x=1, y=5
t=2: x=2, y=4

Notice the difference? Hopefully the answer is no, because there is NO difference. They both describe the exact same scenario, they're just using different ways of describing it. It's almost like walking up to something and you saying, "this is a very bright colour, it's very yellow", and someone else saying "I'd say it has the hex code of FFFF00, or 255 red, 255 green, and 0 blue" (plug this into paint to see what colour comes out if you don't believe that this is yellow) - sure, one's a little more odd than the other, and something you're not used to, but they both CORRECTLY identify the colour as yellow. Same thing with if you were to use the order pair notation as in the text book, or if you were to instead write the answer in terms of equations - both are correct, they're just describing the situation slightly differently. I answered this recently I think on this very thread? But you'll never be marked down for using a different notation in a VCAA exam, ONLY if you use that notation incorrectly

Awesome, I can see exactly what they are doing now. I didnt even notice their answer was in coordinate form. I thought it was the rhs of the equation written out, and the little (t,) was just them displaying which pronumeral in the equation was the parameter or something.

I think this was largely what was confusing me.

[a weaponized ikea chair]

Question c.

I expanded the integrals but i do not know how to evaluate the integral of lower bound 2 and upper bound 4 of f(x), let alone the integral of lower bound 4 and upper bound 5 of f(x).

How can I know the value of those integrals?

Any help is appreciated.

[SS1314]

Question c.

I expanded the integrals but i do not know how to evaluate the integral of lower bound 2 and upper bound 4 of f(x), let alone the integral of lower bound 4 and upper bound 5 of f(x).

How can I know the value of those integrals?

Any help is appreciated.

You can combine two of the integrals with f(x), resulting in one integral with boundaries 5 and 2. You can try to visualise why this is true with a simple graph of f(x).

[a weaponized ikea chair]

You can combine two of the integrals with f(x), resulting in one integral with boundaries 5 and 2. You can try to visualise why this is true with a simple graph of f(x).

Ah, I get it now, thanks.