VCE Methods Question Thread!

[Phy124]

That's more like it!

Good thing I stopped tutoring because I don't remember these laws anymore

[Orb]

Hey guys

On VCAA 2010 Q2a (Exam 1)

It asks you find "an" antiderivative of an expression.

However, on the examiner's report a "+c" was included.

I'm incredibly confused as to whether you can/should put a "+c" if it asks you to find an antiderivative

[GeniDoi]

Hey guys

On VCAA 2010 Q2a (Exam 1)

It asks you find "an" antiderivative of an expression.

However, on the examiner's report a "+c" was included.

I'm incredibly confused as to whether you can/should put a "+c" if it asks you to find an antiderivative

If it says "an" anti derivative, you can choose any value of "c", eg 490, 29, c, Pi or 0 (which means having nothing there). Any of these will be valid.

This is supported by assessors report on the question being "where is a real constant (but not necessary)"

[Yacoubb]

Q5c from VCAA 2009 exam 1.

How is the answer 1/4? The question is:

Given that the sum of the numbers on the two balls is 5, what is the probability that the second ball drawn is numbered 1?

So, the probability that the sum of two balls=5 is 1/3 as calculated from the previous q (part b). Then, to calculate probability that the second ball that is drawn is 1, wouldn't that mean finding the probability that we get 2 then 1 + 3 then 1 + 4 then 1, giving a probability of 1/4?

So, 3/12 / 4/12 = 3/4...

I'm quite stunned, so help would be much appreciates guys!! Thank you!

[Yacoubb]

Never mind, just realised that we were required to use the restricted sample space!

[soNasty]

5a) (4,1) ---> 1/4*1/3 = 1/12
5b) (Sum of 5)
= (3+2) + (2,3) + (4,1) + (1,4)
= 1/12*4
= 1/3

5c) Second ball drawn is 1 = 1/12
and Pr(sum of 5) = 1/3

therefore 1/12  /  1/3
= 1/4

[Blondie21]

I don't understand what they're done.

If there are stationary points at (1,1) and (p,p), wouldn't f'(1) = 0 and f'(p)= 0 ?

This is 4f from exam 2 btw

[Valyria]

Hey,

For part ii of this question, I'm trying to get my head around why dy/dx must be less than or equal to 0 when x=10. I understand that a local minimum requires the derivative to be equal to 0 but why can the derivative be less than 0 at x=10? Is it because it's a decreasing function and so at x=10, it doesn't need to be a local minimum but rather, the absolute minimum. Thus, to ensure the function is decreasing at x=10, the derivative at that point is negative?

Clarification would be appreciated

[lzxnl]

Hey,

For part ii of this question, I'm trying to get my head around why dy/dx must be less than or equal to 0 when x=10. I understand that a local minimum requires the derivative to be equal to 0 but why can the derivative be less than 0 at x=10? Is it because it's a decreasing function and so at x=10, it doesn't need to be a local minimum but rather, the absolute minimum. Thus, to ensure the function is decreasing at x=10, the derivative at that point is negative?

Clarification would be appreciated

So yeah, if your function's derivative is less than zero, its value is dropping. Hence, if you increase x, your function value gets smaller and smaller. Therefore, a right endpoint is a minimum if the derivative is zero (local minimum) or negative.

[kawfee]

If a time/watch appears on my CAS, how do I 'escape' it and resume calculations? Yesterday I typed something in and the clock popped up [guessing because the calculation is too 'big' and CAS is busy]...and I left it to disappear, then this morning my CAS needed recharging. Thanks,

[keltingmeith]

If a time/watch appears on my CAS, how do I 'escape' it and resume calculations? Yesterday I typed something in and the clock popped up [guessing because the calculation is too 'big' and CAS is busy]...and I left it to disappear, then this morning my CAS needed recharging. Thanks,

On my old CAS, if this happened I gave it a "hard reboot" by taking the batteries out and putting them back in again - don't think you'll have time for this a rechargeable one, though. Depends on how easy it is to get the batteries out.

If your CAS DOES enter this mode, it's because it's doing a very hard a difficult calculation, and will not respond to anything else. Curiously, what were you trying to do that made it kersplat like that?

[soNasty]

If a time/watch appears on my CAS, how do I 'escape' it and resume calculations? Yesterday I typed something in and the clock popped up [guessing because the calculation is too 'big' and CAS is busy]...and I left it to disappear, then this morning my CAS needed recharging. Thanks,

Hold the 'on' button

[Blondie21]

Hold the 'on' button

I learnt this lesson the hard way during a SAC. Good times.

~

For the following, why did they use these methods? I figured out the question using the first method specified but I don't understand these. This is q12 of exam 1 2007

[kawfee]

On my old CAS, if this happened I gave it a "hard reboot" by taking the batteries out and putting them back in again - don't think you'll have time for this a rechargeable one, though. Depends on how easy it is to get the batteries out.

If your CAS DOES enter this mode, it's because it's doing a very hard a difficult calculation, and will not respond to anything else. Curiously, what were you trying to do that made it kersplat like that?

Thank you!

Yeah won't have time to take it out, I think I need to unscrew it or something?

I don't remember what I was trying to do :s I think it was a probability question...actually, and ofc I realised I was on the wrong track when that happened.

@ sonasty : thank you!  Hopefully it won't happen in the actual exam... o.O

ALSO, are we allowed to take in x2 CAS calcs??

[Jason12]

when given a matrix and asked to find the image do we multiply the matrix out and make two equations like this. For example,

3x - 1 = y'
2y + 4 = x'

or is it the other way

3x - 1 = x'
2y + 4 = y'

then solve for x and y and substitute into the image of the curve