VCE Methods Question Thread!

[Frozone]

Sorry for my poor notation I'm on my mobile.


Solve for the following equation where the domain is between [ 0 , 2pi]

Root 3 sin x = cos 4x

[keltingmeith]

Do you know if this would be considered a tech-free or tech-able question? Because the tech-free method that comes to my mind would be beyond the scope of methods.

For methods, I'm pretty sure they'd just want you to chuck this into the calculator.

[EspoirTron]

Sorry for my poor notation I'm on my mobile.


Solve for the following equation where the domain is between [ 0 , 2pi]

Root 3 sin x = cos 4x

This question really made me cringe when I first looked at it. However, I am sure there are multiple methods, but if I remembered correctly use the fact that cos(4x)=cos^4(x)-6 cos^2(x) sin^2(x)+sin^4(x), then I am pretty sure you have a quartic solve from there. What you can do is just let a=sin(x) and solve through from there. Also, to make it a quartic in terms of sin use the fact that cos^2(x)=1-sin^2(x)

[EspoirTron]

Do you know if this would be considered a tech-free or tech-able question? Because the tech-free method that comes to my mind would be beyond the scope of methods.

For methods, I'm pretty sure they'd just want you to chuck this into the calculator.

I am pretty sure that question could be tech-free, highly unlikely though. The cos(4x) thing is just an expansion of periodicity and assumed knowledge, i.e trig identities you need to know for methods. If I do remember correctly I think you are expected to be able to solve anything up to a quartic by hand because you can just use long-division. Tedious, but possible I guess.

[keltingmeith]

I am pretty sure that question could be tech-free, highly unlikely though. The cos(4x) thing is just an expansion of periodicity and assumed knowledge, i.e trig identities you need to know for methods. If I do remember correctly I think you are expected to be able to solve anything up to a quartic by hand because you can just use long-division. Tedious, but possible I guess.

The method I was thinking of was along the lines of what you did - using compound angle formulas to get the angles to be the same. I didn't think that methods students were expected to know those formulas, though.

[EspoirTron]

The method I was thinking of was along the lines of what you did - using compound angle formulas to get the angles to be the same. I didn't think that methods students were expected to know those formulas, though.

Actually, you're quite right. Sorry, that was my mistake there. The only things students need to remember, iirm, is periodicity, the odd nature of sine and the even nature of the cosine function. I think the most you need to know in Methods is the identity: sin^x+cos^x=1.

[swagsxcboi]

Find the equation of the normal to  y= x sin(x) at x=pi/2

[e^1]

Find the equation of the normal to  y= x sin(x) at x=pi/2

Hint 1: As you know, the normal gradient is . This can become  at .
Use that to help find your normal.

Hint 2

Use formula.

[swagsxcboi]

got it! thanks, just made a silly error when diff-ing

[Billion]

Can someone algebraically solve for y with steps, thanks.
x = y² + 8y

[alchemy]

Can someone algebraically solve for y with steps, thanks.
x = y² + 8y

Complete the square first.
x=(y^2+8y+4^2)-4^2
x+16=(y+4)^2
y+4= plus or minus sqrt(x+16)
y= -4 plus or minus sqrt(x+16)

[Billion]

Complete the square first.
x=(y^2+8y+4^2)-4^2
x+16=(y+4)^2
y+4= plus or minus sqrt(x+16)
y= -4 plus or minus sqrt(x+16)

Ah, simple.
Thanks!

[jessss0407]

Hey Jess,

Limits, continuity and differentiability are all very intimately connected concepts. Here is a crash course:

Let's think a function and call it f(x). In this example you are free to think of f(x) as anything you would like. A limit is a way for me to say: as I take any value of x what value does f(x) approach? So basically I am saying as I take x arbitrarily small near a particular value what does f(x) approach? So basically if I had f(x)=x then the limit as x approaches 2 would just be 2. What that is basically saying is that if I take values really close to 2, but never equal to 2, f(x) will approach 2.

So continuity. A continuous function is just any function that is defined for all values within its given domain, or over all reals. For a function to be continuous at a point the partial limits must approach the same value for that number. So if again I have f(x) as I approach a particular value of x from the left-hand side I will get a value. As I approach from the right-hand side I will also get a value. For the function to be continuous at this point both these left-hand and right-hand limits must give me the same numerical value. A function is not continuous at jumps, endpoints or anywhere with hollow circles really.

Differentiability. I won't go too much into this, but basically when I am finding the derivative I am actually find the rate at which my function, f(x), is changing. This has a very intimate connection with limits and unfortunately because I am horrible with Latex I can't derive anything for you! However, understand your limits and everything in calculus will become a lot more easier!

On a side note: a partial limit does exist for an end point, but it isn't continuous there since the partial limits on both sides  don't equal each other.

Edit: I want to be more vigorous.

So the latter part of your question asked why a limit can exist if the function is discontinuous? Well let's think about Mr f(x) again. Let's say f(x) is defined over all reals excluding 3. Well intuitively f(3) doesn't exist. However, remember a limit is telling me what value f(x) approaches as I take x arbitrarily small close to 3. So if I ask you to find the limit as f(x) approaches 3, it exists - or, more precisely, one of the partial limits will assume a value the other partial limit does not. So I would be approaching a value when I approach 3 but f(x) would never equal that value. Think of it in this way: f(x) is a happy lad, but as he approaches 3 he gets scared because he sees the girl he likes so he tries going up to her but he gets so close to telling her but he just doesn't do it! So the poor guy never sees what is going on at x=3.

I hope my cheesy example shed some light on the topic

Oooh, so basically limits is sort of saying like what the value of f(x) is approaching if you take a number really close to x? Is if possible for a function to not have a limit?

Thanks so much, it cleared up a lot of things for me

[jessss0407]

Haha yeah depending on your textbook it might explain it or not (Maths Quest doesn't whilst Essentials does) but essentially Change of Base for logs is using the formula:

where 'b' is just any number that's not negative (so long as the top and bottom are the same then it'll work [personally I like to just use the natural log]). It's the same thing as what paper-back has written:
Prime example of this question being used was in the VCAA 2011 Exam 2 MCQ 22 where they asked a question where you needed to utilise the change of base rule.
If you use Essentials then Exercise 5F Questions 7-8 are quite similar to the VCAA question on Change of Base actually..

Examiner's report doesn't say much about the question but 45% only got that question right in that year so it's a good question to ask since it isn't really focussed on in Methods I guess (well not in class for me anyway)..

ahh okay, thanks! what do u mean by using natural logs?

[keltingmeith]

The natural logarithm is to the base e, where e is euler's number. It's also written as ln(x), and during differentiation of exponential and logarithmic functions, you'll learn that those two functions must be to the base e if you want to differentiate them.