VCE Methods Question Thread!

[Rod]

At a factory, the time T seconds spent in producing a certain size metal component is related to its weight, w kg, by T=k +2w^2, where k is a constant.

c) write down an expression for the average time A (in seconds per kilogram)

thanks

[Rod]

If the dimensions of a box is x by x by h, shouldn't the area be 2x^2 + 4xh? the textbook reckons its 3x^2+4xh

[keltingmeith]

Sorry Rod - for the first one, could I get the first two parts? I have a feeling I'll need those...

For the second one, assuming you mean TSA, you're certainly right, unless there's something special about this box that you/they forgot to mention?

[Rod]

Sorry Rod - for the first one, could I get the first two parts? I have a feeling I'll need those...

For the second one, assuming you mean TSA, you're certainly right, unless there's something special about this box that you/they forgot to mention?

thanks for the help euler

for the ox, the only thing special is that the double flaps of the box are open.

For the first question

a)if a 5kg component takes 75 seconds to produce, find k (sorry should have put this in)

[keltingmeith]

Then it would be 3x^2 + 4xh. You see - the top area doubles if the flaps are open, and so instead of 2x^2, we get 3x^2.

So, I get k = 25, I'm still not sure I have enough information to find the average time A... Y'see:




EDIT: nevermind, just read you want an expression, not a value... So the above should be fine.

[Rod]

Then it would be 3x^2 + 4xh. You see - the top area doubles if the flaps are open, and so instead of 2x^2, we get 3x^2.

So, I get k = 25, I'm still not sure I have enough information to find the average time A... Y'see:



EDIT: nevermind, just read you want an expression, not a value... So the above should be fine.

thanks euler

I got another question though sorry

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011mmcas2-w.pdf

just 3.a.ii

not sure how to explain and don't understand it

[nhmn0301]

thanks euler

I got another question though sorry

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011mmcas2-w.pdf

just 3.a.ii

not sure how to explain and don't understand it

You mean the question where f'(x)>=5 for all values of x right?
What they are asking you is basically can you prove that, the derivative of the function is always larger or equal to 5 for all values of x.
Since f'(x) = 12x^2 + 5
You know that    x^2 >= 0 for all x of R
                    => 12x^2 >=0
                    =>  12x^2 + 5  >= 5 for all x of R  (I added 5 to both sides)

[keltingmeith]

Similarly to nhmn, but slightly differently, you could also just do a quick sketch of the function, which also proves the point. Not as fulfilling, but certainly works if you get stuck, and I've done it before and been given the mark.

Otherwise, commenting that as a quadratic, we know that as a parabola, we expect it to have either an absolute maximum or minimum and branch off to negative or positive infinity (respectively) otherwise. By finding the turning point, all you have to do is show that it is an absolute minima, and that the quadratic will then branch off to positive infinity with limits on either side to prove that it is >= 5 for all x. (but honestly, this is a lot of work, and for 1 mark, definitely not worth it).

[Rod]

thanks guys but I still don't get how I would exactly prove it.

[keltingmeith]

Okay, so don't think of it as a proof. A proof is a formal mathematical method in which theorems come to life and the most important mathematics is shown to work.

Instead, you're just showing the examiners that f'(x) >= 5 for all x. You're not proving anything, you're just showing something.

So, you know that x^2 >= 0, and if you quickly sketch this, you'll see that that's true. You also know if you only perform linear operations, you won't break the inequality, so let's go with that:

x^2 >= 0
12x^2 >= 0
12x^2 + 5 > = 5
So, f'(x) > =5 for all x
This is the algebraic method.

If that doesn't work for you, draw up a cartesian plane. Draw the parabola 12x^2 +5, and then draw the line y = 5. Now, if 12x^2 + 5 is greater than 5 for all x, if should not dip below the line, yes? So, colour in all the area between those two lines. If you see any white space below the parabola, then you'll know that the quadratic must not be greater than or equal to 5 for all x.

Here's the two biggest things to get your head around:

• You're not trying to prove something here - so don't. You only need to show something is true, just do that.
• Whenever you DO have questions that seem like you need to prove something, you always sit there thinking, "wait - DID I prove that?". Don't stress about that - if you think you've proven it, assume you have and move on. You don't have time to sweat the details. If you think you have, you have.

[Rod]

thanks so much

[soNasty]

surely this cant be 3 marks?

its question 1 D of the 2012 methods exam 2.
http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2012/2012mmcas2-w.pdf

Wouldnt i just diff the equation theyve given with my calculator and express it in the correct form, or would i have to show working?

[keltingmeith]

1 mark for validation of differentiation technique (essentially, writing dV/dx), 1 mark for a, 1 mark for b. So, 3 marks. The moment they say "put it in the form ax^2 + bx + c", I'm pretty sure they give you a mark for each variable. Looking at the assessment report, using a calculator was not only fine, but generally encouraged. (they don't say that, but it's definitely implied)

[soNasty]

alright, thanks

[Rod]

Hey euler and others

can someone please help me

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009mmCAS2-w.pdf

question 4d ii, 4e i and 4e ii. Been stuck for ages