VCE Methods Question Thread!

[shadows]

Lol I realised I need to brush up on logarithmic manipulation etc....
This was a fairly simple question that I did in practise exam.  EPICFAIL

So just to confirm with you guys.

2log9(x-1) +log9(3) =1

what i did.

log9(x-1)^2 + log9(3) = log9(9)

(x-1)^2 + 3 = 9

then just solved for x...

so just to confirm you cannot cancel out until you the logs until you bring everything together? (so only a single expression)

cause my mentally when working through this question was

(3a + 4a)

= a(3+4)

[b^3]

When you equate the insides of the logs, you have to have them together as one long. When you have the sum of two logs of the same base, you can bring them together into one log of the product of what was inside the other two logs.
.
Then you can equate the inside. You can also think about this as 'undoing the logs on both sides'. You can't undo it if you have the sum of two logs, they need to be combined into a single log.
Think about it this way, if we have then by your working what would it give?
Lets say , then , so we have .
If , then , .
Does ?

(3a + 4a)

= a(3+4)

This doesn't work for logs, because they don't follow that expansion property.

So you should have

EDIT: Had product instead of sum in the first line. Fixed.

[Alwin]

I hate geometry
Can someone explain these questions please? Thanks

11. By intuition the min area is when the line is 45^o to the x-axis. But since we're all mathematicians on this thread , I'll stick in a short proof:

Proof

let a be a point on the x-axis such that the area of the triangle formed by the axes and the line connecting the point (a,0) and the point (p,q) is a maximum


The gradient of this line is given by:


Now, the line goes through the point (0,a) so:


So, the y-intercept is given by:


The area is given by:


To find the minimum area,




Now, subbing a=2p into the area function, we get:


Okay, so I lied a little, this proof ain't that short

Since we know that the angle between the line and the x-axis is 45^o, the point (p,q) must be the mid point of the hypotenuse. Thus, the base is 2p and the height is 2q




@SocialRhubarb: just got a chance to read your soln... area of a triangle is half base times height, no need to integrate lol

10 For this one I can't see an outright intuitive way like q11, so here's my solution (only one way of may possible ones)

Spoiler

diagram!

We can obtain two formulas for c and d immediately in terms of a and b respectively:


Also:

useless atm

But we also know that:

We want to maximise the rectangle area = 2x the trapezium area = A = 2 x (1/2 (a+b) (c+d) )
But, the rectangle area can also be written as: A = triangles with sides a, c  + triangles with sides b,d + rectangle in the middle =  1/2(ac + db) x 2 + 12




Using the second pair of equations we get:

Hm and I made it much more difficult than it needed to be. Put it in a spoiler since it's only half done. Need to have dinner, I'll bbl to finish it ^^

[shadows]

When you equate the insides of the logs, you have to have them together as one long. When you have the sum of two logs of the same base, you can bring them together into one log of the product of what was inside the other two logs.
.
Then you can equate the inside. You can also think about this as 'undoing the logs on both sides'. You can't undo it if you have the sum of two logs, they need to be combined into a single log.
Think about it this way, if we have then by your working what would it give?
Lets say , then , so we have .
If , then , .
Does ?
This doesn't work for logs, because they don't follow that expansion property.

So you should have

EDIT: Had product instead of sum in the first line. Fixed.

Thanks man

[revcose]

11. By intuition the max area is when the line is 45^o to the x-axis.

It's actually the minimum which was asked for, and taking SocialRhubarb's and then doing gives . q^2 will always be positive, m has to be negative, so is positive, and so it is a minimum.

I'm not sure where confusion arose for you, but have a think about it.

[Alwin]

Preamble

It's actually the minimum which was asked for,
I'm not sure where confusion arose for you, but have a think about it.

LOL typo man, sorry! I was reading the other q when I was doing it (you might have noticed we got the same answer ) There is no max answer (unless you count infinity) for that question.

Alright, my working for rhinwarr's second question is at http://i.imgur.com/5AjDN9N.jpg
I especially hope I didn't mess up after how many issues there have been with uploading this.   

Also, for Imgur images you can use the

Code: [Select]

[img][/img]It's right next to the (LaTex) button ^^

Fixed up version (just replaced the word "max" with "min") + Q10 here:


11. By intuition the min area is when the gradient of the line is -q/p. But since we're all mathematicians on this thread , I'll stick in a short proof:

Full Proof for those interested

let a be a point on the x-axis such that the area of the triangle formed by the axes and the line connecting the point (a,0) and the point (p,q) is a minimum


The gradient of this line is given by:


Now, the line goes through the point (0,a) so:


So, the y-intercept is given by:


The area is given by:


To find the minimum area,




Now, subbing a=2p into the area function, we get:


Okay, so I lied a little, this proof ain't that short

Since we know that min area is when the gradient of the line is -q/p, the point (p,q) must be the mid point of the hypotenuse. Thus, the base is 2p and the height is 2q




@SocialRhubarb: just got a chance to read your soln... area of a triangle is half base times height, no need to integrate lol
10 For this one I can't see an outright intuitive way like q11, so here's my solution (only one way of may possible ones)

Diagram!

Using similar triangles and Pythagoras theorem, we can proof that:

Full Proof

First up, all the triangles on the outside are similar. Thus the ratios of green to red triangles are:




Thus, the area of the external rectangle is:



Now,


Thus,

To maximise the area of the rectangle, maximise A_red which occurs when:
Working omitted, you can prove this yourself easy





EDIT: Found yet another typo I'm known for
EDIT 2: Thanks Rhubarb, found another mistake for me ty ^^

[SocialRhubarb]

What - you didn't use integration to do all the rectangle area problems in grade 3?

But yeah, haha, point taken. However, I don't think that the line will always be 45 degrees to the x-axis. What if p is really big and q is close to 0?

[Alwin]

What - you didn't use integration to do all the rectangle area problems in grade 3?

But yeah, haha, point taken. However, I don't think that the line will always be 45 degrees to the x-axis. What if p is really big and q is close to 0?

yeah, hmm. I think I'm gonna change that (it is pretty dodgy) now I look back on it And nah! was too busy using rotation integrals to find the volume of a cone

This was the relationship I was trying to explain:

[clıppy]

Stuck on this question:
For what values of k, where k is a real constant, does the equation have two distinct solutions
I've turned it into a quadratic by turning into y and finding the determinant but where do I go from there in proving it?

[rhinwarr]

For it to have two distinct solutions, the determinant must be bigger than 0.
So:

[b^3]

Now as we said before, has to be greater than zero, since we cannot power something and get zero or a negative number. So we need to check for this.

Now for both solutions to be positive, we need the part before the to be greater than the part after the , as if this occurs, when we take the second part away we will still get a positive solution (as long as our denominator is positive).


So the possible values of are

This is represented by the following graph, where the intercepts have to be greater than zero.
https://www.desmos.com/calculator/gxxmhbo2v6

EDIT: Added in the dropped explanation.

[clıppy]

The y > 0 bit clears up a lot but you've lost me at one point:

How did you go from that quadratic equation solve, to that?

[b^3]

The y > 0 bit clears up a lot but you've lost me at one point:How did you go from that quadratic equation solve, to that?

Well damn, I had an explanation there but where'd it go...
Anyways, since both solutions need to be positive, this will only occur if the part in front of the is greater than the part after the , as when we take the second part away we still want a positive number (provided the denominator is positive).
So we want (the first part) to be greater than the second part , hence .

I'll add it into the post above.

[clıppy]

Ah, it's for making sure it stays positive.
Cheers b^3

[revcose]

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.