VCE Methods Question Thread!

[keltingmeith]

It would help if we had the question and not just the solution, but I'll do my best:

Firstly, these solutions are written weirdly and with a few typos... In i and j, the degrees symbol should be inside the brackets. For k and l, the final answer should not have a degrees symbol

Now then, for your questions I think you're actually referring to l and k?

For k,
For l, as I said - the degrees symbol should not be there, the answer is -1.

Finally, in terms of circle symmetry - because is one full revolution, so if you add that full revolution, you're just going to end up back where you started.

Just as an aside, you mentioned trying to find exact values by using the graphs - while this is okay, it always seemed as a strange way to find them. May I suggest instead using these triangles:



And also trying to understand the unit circle a bit more? (as this is where the graphs come from, anyway) At the end of the day, use whatever methods work for you, just trying to offer some alternatives.

[LiquidPaperz]

Thanks for the reply. But i actually needed help most with i and j, if you could show me how thats done, that would be great!

I also use those triangles

Finally, in terms of circle symmetry - because is one full revolution, so if you add that full revolution, you're just going to end up back where you started.

not sure what you meant here, like i know 2 pi is a full circle and adding to it, but how does it mean where ive started , because 375 degrees isnt same as 15 degrees? how would you do the questions applying what you have said?

[keltingmeith]

They're not the same? Are you sure? I suggest getting a protractor out and drawing them both.

For i, using circle symmetry:


For j, using circle symmetry:


EDIT: in terms of how to apply this to questions - consider the following:


You can quite quickly see that this isn't in any of the bands that you know (theta, pi - theta, pi + theta, 2pi - theta), and that it is pretty big. So, let's take a full revolution of the angle:

That's a little better - but it's still too big! So, let's take a full revolution off of it again:

Now we have something we can work with - and we can solve the initial problem:

[LiquidPaperz]

They're not the same? Are you sure? I suggest getting a protractor out and drawing them both.

For i, using circle symmetry:


For j, using circle symmetry:

Thanks, nah i know their the same, but their not, but they are haha.

Just a quick question you have sin(-180 + theta) = sin (180+ theta), how do you just get rid of the negative sign?

[keltingmeith]

This is going to sound weird, but I imagine the unit circle. If something is negative, it moves clockwise through the circle, if it's positive it moves anti-clockwise through the circle.

So first, I imagine where the -180 is - which is clockwise 180 degrees from the initial point. Then, I add theta, which is anti-clockwise from this new point - so, from my imaginings, I know we're in the third quadrant. From there, I just directly convert the angle to be of the form 0 + theta, pi - theta, pi + theta or 2pi - theta.

[Zealous]

Can someone show me how to do Q15, VCAA Exam 2 2013
(the y-average graph).

Sketch the line y=2 on each of the graphs and find the graph with an equal amount of area above and below the line. If there's an equal amount of area above and below, the areas will sort of "cancel" each other out, leaving an average value of y=2. Hopefully that makes sense.

[keltingmeith]

Slightly different answer to Zealous:

The longest, but easiest, method is to find an equation for each graph, and then use the formula for the average value of a function to find out which returned as 2. However, let's use some logic for this one:

A - this graph would have an average value of the median line (in this case, 2) if we had one full revolution. However, we have one and a half, so it can't be this graph.

B - there's less area above 2 than there is below 2, so the average value can't be 2 - rather, we'd expect it to be between 0 and 2. Feel free to double check using calculations.

C - This one looks like it could be 2 - 2 is certainly in the middle (as (6-2)/2=2), and since it's a linear graph, that would mean that 2 is the average value. So, let's just confirm the rest.

D - This function goes up to 8 from 0, so we can immediately say that the average value should be halfway between these - which is 4 (due to linear trends and such), so it's not this graph

E - Now, this would have you thinking it's two because of the whole linear thing again - but, you'll notice that from x=4 to x=6, there's a line on y=0, so this would bring the average value down. So, it can't be this graph.

So, by logical deduction, we have figured out that C is the graph with an average value of 2. You can also find the equation of the line, then chuck it into the average value function to double check.

[soNasty]

the monthly rainfall at the solaris resort follows a normal distribution with a standard deviation of 12mm. if the rainfaill is less than 25mm in 15% of the months, find the mean monthly rainfall.

[keltingmeith]

In cases like this, to find the mean or standard deviation, we can compare it to another normal distribution of known mean and standard deviation, and convert between the two to find the mean/sd of the first distribution - so, why not use the easiest distribution, the standard distribution:



Solving for mu (and putting the information we already have) gives:



Using the inverse normal function on our calculator, we can find the value z takes when we Pr(Z<z)=0.15, which is z=-1.03. Plugging this in, we have:



Finally, we double check this with our calculator, see that it's right, and our mean monthly rainfall is about 37.44mm

[soNasty]

thanks Euler!

Heres another im stuck on:
X~N(25,8). The shaded area of a bell curve with a region from a to b represents the middle 80% of observations. The values of a and b are best represented by what?

the answer is a=21.38 and b=28.62


****EDIT i got it sorry!
Is the invnorm function on my calc used for both standardised and normal values (n/z)?
because what i just did was realise Pr(X<a)=0.1 and put the info into invnorm and got the value of a
then Pr(X<b)=0.9, did the same thing and got the same answer

[keltingmeith]

Yep - that'll do it.

If you're interested, I've actually written a program for the TI-nspire that does this exact calculation for you in one step. Here's how to put it in yourself:

Push menu-->9-->1-->1, then:
Name: whatever (I called it midinvnorm)
Type: Function
Library Access: LibPub

Then, on the next screen, it should look like this:

Code: [Select]

Define LibPub midinvnorm()=
Func
[]
EndFunc

I want you to edit it so it looks like this:

Code: [Select]

Define LibPub midinvnorm(k,m,s)=
Func
Local z,x1,x2,x
z:=invNorm(0.5-k/2,0,1)
x1:=m+s*z
x2:=m-s*z
{x1.x2}-->x
Return x
EndFunc

Then, if you input midinvnorm(k,m,s) onto your calculator, where k is the probability of a middle range, m is the mean of the distribution and s is the standard deviation, you will get the two values of a and b where the probability k lies.

(note: there is a way to make it so you do it like how you normally do the inverse normal, but I forgot how to do the input boxes, so this works, hahah)

[soNasty]

ahhaah wow i did it, it works! thanks :DD

[Anchy]

Given g(x) = f(x^3) and that f(1) = f'(1) = 1. Use linear approximation on the function g to estimate g(1.1)

Thanks

[kinslayer]

Given g(x) = f(x^3) and that f(1) = f'(1) = 1. Use linear approximation on the function g to estimate g(1.1)

Thanks

Note that (chain rule)

Then

[LiquidPaperz]

does anyone have access to a online checkpoints for units 1/2 methods (circular functions)

Ive finished all the textbook questions on symmetry but i want to keep practicing and dont have any other resources.

Thanks