VCE Methods Question Thread!

[cosine]

Can anyone supply the answer to part b) and each sub section for this question. The answers in the book seem a bit suss and what to confirm whether or not there are right.
Taaaa

(i dont know the cube root rule here, so just imagine the sqrt being cube lol)
 
- Translation of 7 units in the positive direction parallel to the x-axis:

 

- 1 unit in the positive direction parallel to the y-axis:



Now you can do the rest, just know that translations parallel to the x-axis, or in the direction FROM the y-axis always mean you put the number after the x. Whereas if you have a translation parallel to the y-axis or FROM the x-axis, you place the number outside the function.

Say f(x) =

A translation of 3 units in the positive direction of the x-axis would be:

f(x-3) =

A translation of 3 units in the positive direction of the y-axis would be:

f(x) +3=

[Zues]

also having trouble with these two.

why cant the first one be y = -4cos(2x)?

[keltingmeith]

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]?

Interestingly enough, I can't actually find where in the study design it says one has to be able to solve inequalities at all... Of course, it's still useful to understand, so I'd definitely venture to get your head around this stuff (regardless of the graph/form of the equation!)

Also, I have no idea what to think of this example... According to Wolfram, it's wrong, and their logic is really weird and hard to follow. :\ Going through the question myself:



Using this, either x<=3 or x>=-3. By squaring, we would assume another solution was added, and so one of these must be false. When x=4, the original inequality breaks (because we'll get a negative number under the square root), so we ditch the x>=-3, and say that the above case holds for x<=3. Their solution implies that it doesn't hold for x=1, but:



Which is, of course, true - meaning that their solution interval is wrong.

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

This is just... Weird. Like, who the hell designs a swimming pool like that? Also, how is it 10m tall if the diagram shows it's 8m tall...?

Anyway, basically what Zealous has done here is considered the thing as a parrallelepiped (i.e, 3D parallelogram, with 3 sets of 2 parrallel planes instead of lines), in which the length downwards changes depending on where you are. For example, at the very front, the height is 2 - at the very back, the height is 8. This allowed him the ability to diff it and use some related rates to solve it.

Not gonna lie, it's definitely a weird question - I reckon I'd have had a bit of trouble with it if not for Zealous amazing help.

[keltingmeith]

also having trouble with these two.

why cant the first one be y = -4cos(2x)?

... Good question, because that graph can definitely be described by the rule y=-4cos(2x).

EDIT: This also marks my last post for until I return~ So, y'know, don't call for me specifically. You should all be fine with everyone else around here~ They are all just as smart (if not smarter) than me, so.

[AirLandBus]

(i dont know the cube root rule here, so just imagine the sqrt being cube lol)
 
- Translation of 7 units in the positive direction parallel to the x-axis:

 

- 1 unit in the positive direction parallel to the y-axis:



Now you can do the rest, just know that translations parallel to the x-axis, or in the direction FROM the y-axis always mean you put the number after the x. Whereas if you have a translation parallel to the y-axis or FROM the x-axis, you place the number outside the function.

Say f(x) =

A translation of 3 units in the positive direction of the x-axis would be:

f(x-3) =

A translation of 3 units in the positive direction of the y-axis would be:

f(x) +3=

Hmm
All good
Textbook must be wrong.
Its say for the cube root one (check the attachment)

[knightrider]

How would you do this question using the equate the gradient method?

Find the value(s) of a where a ∈ R, when the following equations below have a unique solution.

[keltingmeith]

How would you do this question using the equate the gradient method?

Find the value(s) of a where a ∈ R, when the following equations below have a unique solution.

Turns out I can't sleep... So, I'll be helping with one or two more questions until I leave in 4 hours.

First, we need to find the gradient of each. So, I transpose:



Now, we have one unique solution when these gradients aren't equal, so we consider when they are equal and ignore those situations:



So, there is a unique solution when , you could also write

Also, since cosine mentioned the cube root, you get the nth root of x in TeX by putting in \sqrt[n]{x}, eg:

EDIT: Urrrrp, apparently I read the question wrong. Look below for Steven's answer, using the right numbers~

[Stevensmay]

How would you do this question using the equate the gradient method?

Find the value(s) of a where a ∈ R, when the following equations below have a unique solution.

Just by inspection we can tell that these are both straight lines with no translations, so depending on a they will either have a unique solution or infinitely many.











Lines will have infinite solutions when their derivatives/gradients are equal (because no translation), so


Rearranging gives us a = 9, -3.
However these are the values for which we have infinite solutions, therefore the values for a unique solution is everything else.

a ∈ R\{-3, 9}

[Zues]

... Good question, because that graph can definitely be described by the rule y=-4cos(2x).

EDIT: This also marks my last post for until I return~ So, y'know, don't call for me specifically. You should all be fine with everyone else around here~ They are all just as smart (if not smarter) than me, so.

The equation of the graph given in the answer if I remember correctly is y= 4cos[2(x-pi/2)] since they wanted it in the translation form. How would you put it in the form y=acos[n(x-h)]

[Conic]

The equation of the graph given in the answer if I remember correctly is y= 4cos[2(x-pi/2)] since they wanted it in the translation form. How would you put it in the form y=acos[n(x-h)]

Both of those give the same graph. I would say y=-4cos(2x) is fine as an answer. The book doesn't say ε has to be non-zero, so your original answer is fine.

[knightrider]

Turns out I can't sleep... So, I'll be helping with one or two more questions until I leave in 4 hours.

First, we need to find the gradient of each. So, I transpose:



Now, we have one unique solution when these gradients aren't equal, so we consider when they are equal and ignore those situations:



So, there is a unique solution when , you could also write

Also, since cosine mentioned the cube root, you get the nth root of x in TeX by putting in \sqrt[n]{x}, eg:

EDIT: Urrrrp, apparently I read the question wrong. Look below for Steven's answer, using the right numbers~

Thanks eulerfan101 have a good rest and enjoy.

Just by inspection we can tell that these are both straight lines with no translations, so depending on a they will either have a unique solution or infinitely many.











Lines will have infinite solutions when their derivatives/gradients are equal (because no translation), so


Rearranging gives us a = 9, -3.
However these are the values for which we have infinite solutions, therefore the values for a unique solution is everything else.

a ∈ R\{-3, 9}

Thanks Stevensmay 

[knightrider]

i just wanted to actually know when were you meant to learn similar triangles.

Also are there any good explanation of it out there?

[Zues]

Both of those give the same graph. I would say y=-4cos(2x) is fine as an answer. The book doesn't say ε has to be non-zero, so your original answer is fine.

how did they get the y= 4cos[2(x-pi/2)] though?

[Conic]

Well, from the graph you can see that the amplitude is 4, and that the period is π, so you can write the equation in the form y=4cos(2(x-k)). Now, when x=0, y=-4, so



We know that cos(π) is -1, so π/2 is a possible choice for k. This is the answer given in the textbook.

[psyxwar]

i just wanted to actually know when were you meant to learn similar triangles.

Also are there any good explanation of it out there?

Probably like year 9, but I doubt anyone actually pays attention back then.

Check out this site: http://www.mathsisfun.com/geometry/triangles-similar.html

how did they get the y= 4cos[2(x-pi/2)] though?

Instead of treating it as a negative cosine graph they treated it as a positive one, and made the appropriate transformations to get to that? It's the same way that you got to your answer