VCE Specialist 3/4 Question Thread!

[RuiAce]

How do u do this question using volumes of revolution?

[RuiAce]

I have another question: I was wondering how they got the "n-1" part of the formula to get the sample variance. Or simply just what the formula means haha.


I thought it'd just be 'divide it by N' because mean X is equal to X/n... a bit confused with this tbh lmao.

Thanks in advance!

This is something I didn't expect to appear in the VCE so I'll let an actual former VCE student give you a response that's related to your subject. (Mostly because I don't think you guys need to deal with the student-t distribution?)

Technicalities are hard to describe, because we have to define what the bias is and what not to prove this is true.

[keltingmeith]

When finding confidence intervals, is it ok just to use the z-value up to two decimal places? I don't remember the question, but there was a question where using the calculated z-value (using inverse normal calculator) and 1.96 gave different 95% confidence intervals (to two decimal places)

Based on the study design, VCAA wants you to use 1.96 for the 95% confidence interval. What number did you use? It could simply be your method of calculating that z-value was wrong/you made a silly mistake somewhere, because you should've gotten 1.9600 (based on some random site's matlab calcs).

I have another question: I was wondering how they got the "n-1" part of the formula to get the sample variance. Or simply just what the formula means haha.


I thought it'd just be 'divide it by N' because mean X is equal to X/n... a bit confused with this tbh lmao.

Thanks in advance!

As a VCE student/stats major, see Rui's answer. He mentioned the t-distribution - you guys don't use it, you don't need to worry about it. As he said, the reason is somewhat complex. The easiest way to think of it first requires us defining the bias in a sample - and for that, we need to define an estimator.

Let's say our distribution has some parameter we're interested in (we'll call it lambda). Next, let's say we want to figure out what the value of lambda is. To do this, we might come up with a formula for it (as we do for the sample mean and sample variance). We call this an estimator of lambda, and normally notate it as lambda with a little arrow above it ("lambda hat"). Next, we can define the bias - that is, how much we expect this estimator to be different from the true value. We can calculate the bias in an estimator with the following equation:



Obviously, if the bias is 0, then the expected value of our estimator and the value of our parameter should be the same. It turns out that if you do this for the sample variance, but divided by n instead of n-1, then you'll find that the the variance is slightly underestimated, by a factor of (n-1)/n. A consequence of this is that if your sample size is big enough, you don't actually see the bias. A much better way to do this, however, is to divide the estimator by (n-1)/n to remove the bias in the first place - this results in the formula where it's divided by n-1.

All of this is, of course, as Rui mentioned, outside of VCE. But, there's the reasoning in case you were interested.

[StupidProdigy]

I have another question: I was wondering how they got the "n-1" part of the formula to get the sample variance. Or simply just what the formula means haha.


I thought it'd just be 'divide it by N' because mean X is equal to X/n... a bit confused with this tbh lmao.

Thanks in advance!

These videos should provide some basic understanding (without any formal details) about the choice of biased (/n) or unbiased (/(n-1)). For the Khan vid, the key point is at about 7:30. I find it useful to think of bias like a measure of the accuracy, for example a biased story is an inaccurate description of the story. As said above, don't get too caught up on this stuff
https://www.youtube.com/watch?v=ANsVodOu1Tg
https://www.youtube.com/watch?v=KkaU2ur3Ymw

[Gogo14]

havent been on AN in ages but got sum qss for everyone:
1. why can derivatives be treated like fractions (sometimes), and when does this principle not hold?
2. I don't really understand the arc length formula- what happens in vectors when you need to find the arc lenght of i,j,k components
3. why can some functions not be integrated? is there a special property/pattern of these functions? e.g. sinx/x
thanks

[Quantum44]

havent been on AN in ages but got sum qss for everyone:
1. why can derivatives be treated like fractions (sometimes), and when does this principle not hold?
2. I don't really understand the arc length formula- what happens in vectors when you need to find the arc lenght of i,j,k components
3. why can some functions not be integrated? is there a special property/pattern of these functions? e.g. sinx/x
thanks

dy/dx is simply not a fraction. We only treat it as a fraction when using the chain rule.

There is a formula for finding arc length of vectors using the derivatives of the parametric equations dx/dt and dy/dt (I think it is on the formula sheet)

There are plenty of functions which cannot be integrated using analytical methods used in spesh and even more functions that require numerical methods. There is no special property, it's just that in spesh we only look at a select few integration techniques.

Just a warning, I'm not an expert at maths so I'm sure there's a uni student who can explain these concepts better than me.

[Syndicate]

havent been on AN in ages but got sum qss for everyone:
1. why can derivatives be treated like fractions (sometimes), and when does this principle not hold?
2. I don't really understand the arc length formula- what happens in vectors when you need to find the arc lenght of i,j,k components
3. why can some functions not be integrated? is there a special property/pattern of these functions? e.g. sinx/x
thanks

1. Can you specify a little more? Any examples?
(Like Quantum said, you can only (in specialist) think of dy/dx as fractions when using the u-substitution method)
2. You use the parametric arc length formula (Remeber the i component represents the x-axis, j component represents the y-axis and the k component represents z-axis)
3. Usually if you can't use the u-Substitution method, or partial fractions method (or integrate by using trigonometric identies), you can't integrate an expression/ function in specialist. I think for this you can use integration by parts (??- but I am not entirely sure myself as I haven't covered it yet myself)

( Quantum44 beat me to it )

[RuiAce]

Having done a whole ton of integrals all my life, I'm going to jump in on Q3.

3. why can some functions not be integrated? is there a special property/pattern of these functions? e.g. sinx/x
thanks

No. There is no definitive property or rule governing the ability to find a primitive function of a composition of elementary functions.

The problem lies in the limitations of tools at our disposal. There is a saying that proofs of results involving derivatives is harder than those of integrals, however with computations it's usually the derivatives that are easier. And rightfully so - for single variable calculus, we have the product rule, chain rule and quotient rule filling pretty much every gap we need, when it comes to computing derivatives of elementary functions.

We don't have these tools for integration. Our techniques for integration are
- Substituting u = f(x)
- Substituting x = f(u) (includes trigonometric substitution)
- Integration by parts
- Decomposition partial fractions
- Common algebraic manipulation

Whereas the three derivative rules govern all kinds of differentiation problems we encounter, these integration techniques do not govern all possibilities with integration. There's no known algebraic methods to develop new integration techniques either.

Instead, to fill the gaps, we are forced to define non-elementary functions. These functions will only appear at university.

[lzxnl]

Having done a whole ton of integrals all my life, I'm going to jump in on Q3.No. There is no definitive property or rule governing the ability to find a primitive function of a composition of elementary functions.

The problem lies in the limitations of tools at our disposal. There is a saying that proofs of results involving derivatives is harder than those of integrals, however with computations it's usually the derivatives that are easier. And rightfully so - for single variable calculus, we have the product rule, chain rule and quotient rule filling pretty much every gap we need, when it comes to computing derivatives of elementary functions.

We don't have these tools for integration. Our techniques for integration are
- Substituting u = f(x)
- Substituting x = f(u) (includes trigonometric substitution)
- Integration by parts
- Decomposition partial fractions
- Common algebraic manipulation

Whereas the three derivative rules govern all kinds of differentiation problems we encounter, these integration techniques do not govern all possibilities with integration. There's no known algebraic methods to develop new integration techniques either.

Instead, to fill the gaps, we are forced to define non-elementary functions. These functions will only appear at university.

Side note: I think you'll find that measure theory is always taught after any formal course in analysis dealing with notions of differentiability

The sad bit is, to calculate antiderivatives, those techniques outlined are pretty much the only tools available. If you want definite integrals, then a whole host of techniques become available to use, like complex analysis, Fourier analysis, differentiation under the integral sign etc.

[RuiAce]

Side note: I think you'll find that measure theory is always taught after any formal course in analysis dealing with notions of differentiability

The sad bit is, to calculate antiderivatives, those techniques outlined are pretty much the only tools available. If you want definite integrals, then a whole host of techniques become available to use, like complex analysis, Fourier analysis, differentiation under the integral sign etc.

Oh yeah. Pretty keen for complex analysis next semester for stuff like that and they taught me Leibniz's rule last year. Even in high school there were definite integral tricks that are useless in computing antiderivatives.

[Gogo14]

Shouldnt the domain for this question be -sqrt(2pi)/2<x<sqrt(2pi)/2?
Answer says its 0<x<sqrt(2pi)/2 but I cant see why

[Syndicate]

Shouldnt the domain for this question be -sqrt(2pi)/2<x<sqrt(2pi)/2?
Answer says its 0<x<sqrt(2pi)/2 but I cant see why

The largest values y can hold is between -8 and 8. Using separation of variables, arcsin(y/8) = x^2 (also if x =0, y =0 then c = 0)

1) substitute y = 8:
arcsin(1) = x^2
x^2 = \( \frac{\pi}{2} \implies x \space = \space \pm \frac{\sqrt{2 \pi}}{2} \)
but when you differentiate in term of y (so y = 8sin(x^2)), you would get dy/dx = 2xcos(x^2).
After substituting in any negative number for x, you'll  get a negative value for dy/dx which cannot be possible by looking at the equation above. So x only equals to \( \frac{\sqrt{2 \pi}}{2} \) .

substitute y = -8
arcsin(-1) = x^2 (Remember that range of arcsin is between -pi/2 and pi/2)
x^2 =/= - \( \frac{\pi}{2} \)

This implies that y cannot be a negative number. So the smallest number y can be is 0.

substitute y = 0
arcsin(0) = x^2
x^2 = 0
therefore x = 0

Therefore the maximum domain = (0, \( \frac{\sqrt{2 \pi}}{2} \))

I found my explanation quite vague  If it doesn't make sense please do tell 

[peanut]

2016 Exam 2, MC Q18:
See images below.

Shouldn't the variance be:
(3^2)*(9^2) + (2^2)*(3^2), according to the formula Var(aX + bY) = a^2Var(X) + b^2Var(Y)

Thus, the sd shouldn't be 3root(85)?

[keltingmeith]

2016 Exam 2, MC Q18:
See images below.

Shouldn't the variance be:
(3^2)*(9^2) + (2^2)*(3^2), according to the formula Var(aX + bY) = a^2Var(X) + b^2Var(Y)

Thus, the sd shouldn't be 3root(85)?

You're thinking about this too deterministically.

What you've proposed would work if you took a single orange and tripled it's mass, which is what that formula is for - when you take a random number (in this case, the weight of the orange), and multiply it by some number. For example,

Billy decides to enter the lottery, with his chances of winning being 0.1. If his parents offer to double his winnings, what is the variance of his earnings?

In this case, you'd multiply by 4 (2^2, the parents are doubling his winnings) - because it's only one event, but it's now worth twice as much.


In the scenario on that exam (looking only at the oranges, but it's the exact same logic for the lemons), you don't have 1 orange that's with three times as much - you have 3 oranges that are each with one. This is why they've broken it up into var(O1) + var(O2) + var(O3).

Now, the next step is where the confusion arises, so I'm going to add an intermediate step. Since each of these oranges are only worth one, and they all have the same probability distribution as I. So, var(O1)=var(O), and the same for all the rest of them.

This means we have the following working out:

var(O1) + var(O2) + var(O3)
=var(O) + var(O) + var(O)
=3*var(O)

[Gogo14]

The largest values y can hold is between -8 and 8. Using separation of variables, arcsin(y/8) = x^2 (also if x =0, y =0 then c = 0)

1) substitute y = 8:
arcsin(1) = x^2
x^2 = \( \frac{\pi}{2} \implies x \space = \space \pm \frac{\sqrt{2 \pi}}{2} \)
but when you differentiate in term of y (so y = 8sin(x^2)), you would get dy/dx = 2xcos(x^2).
After substituting in any negative number for x, you'll  get a negative value for dy/dx which cannot be possible by looking at the equation above. So x only equals to \( \frac{\sqrt{2 \pi}}{2} \) .

substitute y = -8
arcsin(-1) = x^2 (Remember that range of arcsin is between -pi/2 and pi/2)
x^2 =/= - \( \frac{\pi}{2} \)

This implies that y cannot be a negative number. So the smallest number y can be is 0.

substitute y = 0
arcsin(0) = x^2
x^2 = 0
therefore x = 0

Therefore the maximum domain = (0, \( \frac{\sqrt{2 \pi}}{2} \))

I found my explanation quite vague  If it doesn't make sense please do tell 

thanks, but I dont undestand hte first part where you said that x cannot be negative to make a negative dy/dx. Why cant x be negative? If you sub it into both dy/dx equations, they should Still give a negative answer?