VCE Methods Question Thread!

[Alwin]

Can I get some help trying to understand that p&q question geometrically?

I believe that what happens is (p,q) will always be the midpoint of the hypotenuse. I think in order for the area to be minimised, the line from (0,0) to (p,q) must split the area in half, which ends up . I kind of reverse-engineered this, so I don't really understand why. http://i.imgur.com/MfJT5ce.jpg is the way I've visualised it.

yes, that's exactly what I was getting at in my post Re: Methods [3/4] Question Thread! and the diagram


I did it by "intuition", in fact I even got my wording wrong the first time lol, don't really know how to explain it either. (Looking at your diagram) if increase the base of the green triangle then the area of the red on decreases. But, the area of the red on increases at a faster rate, so as the gradient of the line approaches horizontal the area goes to infinity. But, if we decrease the base of the green triangle, the height of the green area increases so as the line approaches vertical the area goes to infinity too...

It's kinda like a seesaw (well at least that's what I pictured) and that the point (p,q) is in the middle. Go too far one way boom to large and same for the other way. Just so happens when its the midpoint then it's a minimum value... and i've just gone on a wild tangent not answering your question sorry

[clıppy]

A bit lost again with inequalities. How would you find the maximal domain of

[abcdqd]

A bit lost again with inequalities. How would you find the maximal domain of

everything under the sq root has to be greater than or equal to 0. simplifying the function, we get . so . best way to solve this is to draw the graph and look at the appropriate x-values, and we end up with .

[stolenclay]

Can I get some help trying to understand that p&q question geometrically?

I believe that what happens is (p,q) will always be the midpoint of the hypotenuse. I think in order for the area to be minimised, the line from (0,0) to (p,q) must split the area in half, which ends up . I kind of reverse-engineered this, so I don't really understand why. http://i.imgur.com/MfJT5ce.jpg is the way I've visualised it.

I don't believe there is a straightforward geometric explanation as to why (p, q) ends up being the midpoint of the hypotenuse, but if you're interested, here's a proof without any calculus, using only geometry and algebra. It won't help you for Methods, but it is good to see that not all optimisation problems are limited to calculus methods, even though that's how you're taught to handle them in Methods.

The knowledge required should all be Methods-level, except maybe one formula for the area of a triangle (?).

Diagram

Proof

(Un)Necessary formalities

Assume is a straight line, where and are fixed points on the positive Y and X axes respectively, such that is the smallest right-angled triangle with the point (p,q) lyng on its hypotenuse and (0,0) being the vertex of the right angle.

Let and be arbitrary points on the positive Y and X axes respectively such that is a straight line.

Let be the point on the positive Y axis such that is parallel to the X axis.

Boring

Boring

In literally the same way:

When :


The result when we consider the case where , however, gives us (the process is pretty much identical, only with some flipping of inequality signs).

Together, and imply , leading to your observation through similar triangles.

[revcose]

Thanks anyway Alwin. Even though I'd thought of the same things, reading it helped me understand what I was looking for better last night.

Thank-you very much heaiyuo, that's amazing! I don't think is methods level, but this level of understanding is definitely above methods anyway.

[SocialRhubarb]

You're so amazing heaiyuo.

[darklight]

Can someone help me with VCAA 2002 Exam 2, 1cii and 4f

[teletubbies_95]

I was a bit confused about VCAA 2011  Exam 1 ( part b) :
It might be a stupid question , but why do we have to do ( x+d)(x+c) are greater than zero thing?
Instead of doing the rang g is a subset of dom f ?

Thanks !

[Stevensmay]

I was a bit confused about VCAA 2011  Exam 1 ( part b) :
It might be a stupid question , but why do we have to do ( x+d)(x+c) are greater than zero thing?
Instead of doing the rang g is a subset of dom f ?

In methods we can only take the square root of a positive number or zero, thus the expression (x+c)(x+d) must be greater than or equal to zero.

If we do it the way you suggest, we find the domain of f to be (-infinity,-3) union (3, infinity) and the range of g to be (-infinity, infinity), which also fails the subset test. As no domains/ranges are given in the question, we can simply find one that will allow the function to exist, which is where (x+c)(x+d) is greater than or equal to zero.

So for the new function to exist, its domain must be

[clıppy]

Consider the functional equation . Show that the function with rule satisfies this functional equation. Give reasons for your answer

I know it's got something to do with the symmetry of the unit circle and adding but I can't really show/explain it. Any help?

[Alwin]

Consider the functional equation . Show that the function with rule satisfies this functional equation. Give reasons for your answer

I know it's got something to do with the symmetry of the unit circle and adding but I can't really show/explain it. Any help?

You can state that    so that should be enough? Unless it's like a 5 mark explain question which I highly doubt

ADDED: Or, another way you could explain it is that the period of k(t) = cos(2t) is pi. Thus, when you translate it -pi units parallel to the x-axis you are superimposing the graph of k(t + pi) onto k(t) which gives the same graph. Just some food for thought

[lzxnl]

Thanks anyway Alwin. Even though I'd thought of the same things, reading it helped me understand what I was looking for better last night.

Thank-you very much heaiyuo, that's amazing! I don't think is methods level, but this level of understanding is definitely above methods anyway.

Wait, so you're telling me that maths taught in years nine and ten is beyond methods level? *shockhorror

[Alwin]

Wait, so you're telling me that maths taught in years nine and ten is beyond methods level? *shockhorror

No, at my school I had to wait till GMS in year 11 to get taught that *officially* D:

[darklight]

Can someone help me with VCAA 2002 Exam 2, 1cii and 4f

Anyone?

[Alwin]

Anyone?

erm 2002 exam 2? I couldn't find those questions lol... this one yeah?
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/02mmexam2.pdf