VCE Methods Question Thread!

[sasa]

cos(x)/(2x+2)
differentiate=>on the top you get -sin(x)*(2x+2)-2cos(x). Plug in pi and you get 2
On the bottom you get (2x+2)^2. Plug in pi and you get 4*(pi+1)^2
Divide=>1/2(pi+1)^2
[/quote]

Nope, sorry. Still don't get it. I did the diff bit correct, but when I plugged in pi I got -2(2pi+2)/(2pi+2)^2, simplifying to -1/pi+1

[achre]

Control 7=> start of line
Control 1=> end of line
at least it works on my TI-nspire

That's awesome man, cheers!

[Henreezy]

vcaa 2011 exam 2

1a)

dv/dt = t^2/5

I know you have to integrate dv/dt to get V, but what about the constant c?

V=( t^3/15) + c

I looked at some solutions and they say V(0) = 0
why?

Integrate so you get
V = t^3/15 + c
given that initially (t=0), V = 6075
V(0) = 0/15 + c
6075 = c
therefore, V = t^3/15 + 6075.
If it says something 'initially' has x amount or whatever, that is an initial condition.
Pay close attention to wording pertaining to initial conditions as they often help you solve diff/integration problems!

Hope this helped. 

[Henreezy]

cos(x)/(2x+2)
differentiate=>on the top you get -sin(x)*(2x+2)-2cos(x). Plug in pi and you get 2
On the bottom you get (2x+2)^2. Plug in pi and you get 4*(pi+1)^2
Divide=>1/2(pi+1)^2


Nope, sorry. Still don't get it. I did the diff bit correct, but when I plugged in pi I got -2(2pi+2)/(2pi+2)^2, simplifying to -1/pi+1

Can you show me all of the steps you did so I can see why (2x+2) is still present?
Since Sin of pi = 0, (2x+2) should have disappeared when you sub in (pi).

[achre]

For the 1b question, is 2/(2pi+2)^2 sufficient for 3 marks? The assessors report just says that some students lost a mark for not 'substituting correctly' but in the bulk of their questions they advise against repeatedly simplifying because you risk errors. 2/(2pi+2)^2 is an equivalent answer

[abcdqdxD]

Hi, can someone explain to me how to work out the value of a combinatoric by hand?

I know the formula is n!/r!(n-r)! but I'm not sure how to work out the values. Thanks!

[sasa]

How is 2/(2pi+1)^2 an equivalent answer? (I did it again and got this answer)

[achre]

How is 2/(2pi+1)^2 an equivalent answer? (I did it again and got this answer)

2/(2pi+2)^2 is, 2/(2pi+1)^2 isn't.
By the quotient rule, your denominator will be (2x+2)^2, I don't know where the +1 is coming from for you

[Daenerys Targaryen]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2011mmcas1-w.pdf

In Qu 10. (c)

Why doesn't it work if you had made the substitution from (b) instead of differentiating first then substituting?

[sasa]

Sorry, I meant +2 

[papertowns]

Part b please! I understand that (x+2)(x+8) has to be greater than or equal to 0 but don't know where to go from there

[ahat]

Random CAS question - is there a way to move back to the start of the line quickly without using the cursor? It feels like about ten seconds of wasted time if you make a typo/error on the start of a long-ish entry.

1. Be at the end of the expression
2. Hit "Caps + Up"
3. Press Left
4. Profit

[Daenerys Targaryen]

Part b please! I understand that (x+2)(x+8) has to be greater than or equal to 0 but don't know where to go from there

draw a lil y=(x+2)(x+8)
and see where it is > 0
in this case x<-8 and x>-2

[Lejn]

if (x+2)(x+8) has to be ≥0, than x≥-2 as that would be both brackets even, or x≤-8, so both brackets are odd, making either 0 or an even number inside the root.

[Henreezy]

Sorry, I meant +2 

2/(2pi+2)^2
expand the denominator
4pi^2 + 8pi + 4
4(pi^2 + 2pi + 1)

This is the same as
4(pi+1)^2 (which becomes 2/4(pi+1)^2 and cancels to give vcaa's answer.)
4(pi^2 + 2pi +1)

v^2 = (2x + 2)^2 = (2(x+1))^2 = 4(x+1)^2
They factorised the denominator and then squared it, whereas I and a few other users on this board approached it by just taking 2x + 2 as it was. Either approach is correct, you just need to realise that these two answers are the same.