VCE Methods Question Thread!

[keltingmeith]

LOL but for exam 1s I was inconsistent as, I'd get like 40/40 on one then 36/40 on the next. I really should have taken them more seriously and actually done them all fully timed instead of finishing like 30 minutes in then marking it straight away.

Actually on that note: I'm pretty sure for study scores they don't look at your raw mark for each GA, but how you placed relative to everyone else (your Z-scores). This means that theoretically 1 mark dropped on exam 1 = 1 mark dropped on exam 2 due to the weighting, but in practice this isn't the case because exam 1s have higher averages and much more people full mark exam 1s, so if you want to score well you actually have to treat them seriously in your practice.

Yep - there's actually a bunch of statistical tools one can use to form a normal distribution (which specialist students will learn about next year~), and with a computer this process would be finished in like a day - probably less.

INSTEAD, VCAA go through an inverse normal on everybody, probably exponentially increasing the computation time... (basically, they see you scored better than 60% of students, so use an inverse normal for 0.6, then someone else beat 73% of students, so they inverse normal for .73) It's pretty stupid if you ask me, but whatever

[keltingmeith]

So i've sort of got my head screwed around composite functions, I can work them out, I can determine whether they're defined or not, but the thing i cannot understand is why is the domain on the inner function the domain of the entire function? Do we need to know 'why' this is, or do we just have to deal with trusting the formula, lol

Domain is all the numbers that you can plug in at the very start. Since the only numbers you can plug in at the start are those in the domain of the inner function, the inner function defines the domain of the composition.

[keltingmeith]

yeah but whats confusing me is that if we plug somehting into the 'start' function, we obtain the range of it, and then plug in the range of it into the second function.

DEFINITELY turn to the vTextbook video for composition of functions. Lisa explains it in a very nice manner using the machine analogy, and with pictures that will always beat the word of forum. If you still have trouble understanding after watching it, then come back and ask.

[Zues]

I havent read the earlier posts, if this is not applicable then dont worry. I am assuming you cant get your head around the domain of two functions, e.g. (f+g)(x) is equal to the intersection of the domain of f(x) and g(x) (the part they share).
Remember in probability when you have ( A intersect B ) = Pr (A) x Pr (B)

If that's not your question can you say your question again? ill try and help.

[Zues]

with trig symmetry, i know how pi - theta is Q2, 2pi - theta is Q4 etc.

can someone tell me each of these, i dont understand why there complementary etc.
a) sin (pi/2 - theta) = cos theta
b) cos (pi/2 - theta) = sin theta
c) sin (pi/2 + theta) = cos theta
d) cos (pi/2 + theta) = - sin theta
e) sin (3pi/2 - theta) = -cos theta
f) cos (3pi/2 - theta) = -sin theta
g) sin (3pi/2 + theta) = -cos theta
h) cos (3pi/2 + theta) = sin theta

and e.g. the last one, how does it become -cot(theta)

[Zues]

so with cos(-theta) is it cos theta cause its in q4? how can they assume it falls within and angle in the q4?

also cos(-240) is a clockwise rotation of 240 degrees which is in q2, thus -pi-60, therefore -[-cos(60)] = cos (60) i assume? but they say its q3 so im bit lost as to why its positive for all values of theta i.e. why is cos(-theta)=cos(theta)

[SE_JM]

Hello~
I don't understand this. I was wondering if you could help me?

The question is:
solve (13-4x)^1/2 ≥ x-2

First thing first though, is solving inequalities such as these (without actually graphing the graph) within the Methods Curriculum? I found the worksheet floating in the methods forum, but am unsure whether it's within the course or not

Secondly, (this is regarding the question itself) The answer says this:

"... Now we went to square, but we need that x ∈ [2,13/4]"
could someone tell me why?

Thank you


EDIT: I attached the worksheet. It's example 7
 Also what does it mean when it says: "Here we just intersected the sets" regarding the first solution S₁ = (-ve infinity, 2]?

[IntelxD]

so with cos(-theta) is it cos theta cause its in q4? how can they assume it falls within and angle in the q4?

also cos(-240) is a clockwise rotation of 240 degrees which is in q2, thus -pi-60, therefore -[-cos(60)] = cos (60) i assume? but they say its q3 so im bit lost as to why its positive for all values of theta i.e. why is cos(-theta)=cos(theta)

Studying the unit circle will enable you to understand why the relationship cos(-theta) = cos (theta) holds. In a unit circle, a line drawn from the origin to a point on the circle will produce a right-angled triangle. Within this triangle we know that the hypotenuse will equal to 1 (the radius of a unit circle) and the lengths of the opposite and adjacent sides will equal to the y and x coordinates of the original point respectively. By using trigonometric ratios we can establish rules for both cos(theta) and sin(theta). As cos (theta) = adjacent/hypotenuse, we can state that cos(theta) = x/1 = x. Similarly sin(theta) = y/1 = y. Now comes the easy part. When moving anti-clockwise along the unit circle (positive angles) the x coordinate is positive in Q1, negative in both Q2 and Q3, and finally positive again in Q4. When moving clockwise along the unit circle (negative angles), it is positive in Q1, negative in both Q2 and Q3, and positive again in Q4. This relationship proves that cos(-theta) will equal to cos(theta) as they are essentially the same thing, regardless of which direction you move along the unit circle. By applying this method you can also prove why sin(-theta) = -sin(theta).

As for your initial question, rotating 240 degrees clockwise is the same as rotating 120 degrees anti-clockwise. Therefore cos(-240) = cos(120) = -cos(60) = -0.5. Alternatively you can approach the question in the same manner as the worked example using cos(-theta) = cos(theta).

 I'm sorry if my answer was a little confusing, please let me know if you require any further clarification.

[SE_JM]

I believe I have a solution to it. I got 0.38m/s instead of 0.39m/s,  I'm not sure what happened there.

For the very last section where I divide by 108 metres squared, you can go into more algebraic proof as to why that works by setting up even more related rates for the rectangular prism. I just took a shortcut to save time.

(Image removed from quote.)

[EDIT] Explanation of last step, we want to find dh/dt for the swimming pool:







Sub dh/dv into dh/dt:



Substitude dV/dt=40.715



Hopefully that doesn't make it more confusing!

Hello, I found this question about related rates of change.

First I'll tell you the question
"A massive conical vat, 10 metres in height, is filled to its capacity of 120pi cubic metres of pure glacier water

A massive crane raises the conical vat above an empty swimming pool at an exclusive resort. (diagram of swimming pool provided) in attachment

The glacier water is siphoned into the swimming pool.

If the water level in the conical vat decreases at a constant rate of 1 meter per second, at what level does the water in the swimming pool rise at the point in time when the water level in the vat is 6 metres deep?"

Final answer: 0.39m/s

It's from a while back (like a year ago) but it was posted on this thread. I found it and made a go but don't understand the bit (very near the end) where it says:

Volume of swimming pool: length * width *height
                                           = 9*12*h

I'll attach the diagram of the swimming pool too, but if you have a look, it's not a rectangular shape (what to you call this in 3D? rectanguloid?) it's more like a triangular prism. I was wondering how the volume could be that.

Thanks in advance
Note: apparently, the cut off bit is 2m
You need to click on the link where it says (Image removed from quote.) otherwise, you won't see the whole post

[knightrider]

How would you do these questions?

[AirLandBus]

How would you do these questions?

Process of elimination dude, know how the transformations work.

[knightrider]

for equations of the form in my book it says  when b ≠ 0 they are 2 turning points. and it says when b=0 there is no turning points.How does this work?

[AirLandBus]

for equations of the form in my book it says  when b ≠ 0 they are 2 turning points. and it says when b=0 there is no turning points.How does this work?

did you work out your last post dude?

[IntelxD]

for equations of the form in my book it says  when b ≠ 0 they are 2 turning points. and it says when b=0 there is no turning points.How does this work?

When b=0, you have a point of inflection, hence no turning point. When b ≠ 0, you will have your conventional cubic graph with two turning points.

[knightrider]

When b=0, you have a point of inflection, hence no turning point. When b ≠ 0, you will have your conventional cubic graph with two turning points.

Thanks IntelxD 

When b=0, how do you get  a point of inflection