VCE Methods Question Thread!

[Zealous]

Is that for nspire cx, I can't find the key

CTRL + del.

[papertowns]

Can someone explain this to me please? looked at assessment reports and itute but still don't get it

[clıppy]

Just a quick question for when we're finding general solutions for trigonometric functions
Lets say we had
If we solve for x we can get
Would we be wrong if I wrote or or do we have to try and write it in the simplest form?

Also, looking at a unit circle while finding general solutions for these. Is there a general rule of thumb when there is one answer with +-,  or when there are two separate answers?

[Lejn]

Can someone explain this to me please? looked at assessment reports and itute but still don't get it

Okay firstly, lets establish what we know:
The graphs intersect at x=m, where m>0, whilst they also intersect at zero.
The shaded area is 64.

We have two variables, we're going to want simultaneous equations.

If we use our first piece of information, the graphs are the same when x=m.

Therefore, f(m)=g(m), and this is our first equations, which is:

m³-am=am
therefore
m³=2am,
a=m²/2 [1], our first equation.

Our second information is the shaded area equalling 64. The area enclosed by g(x) is entirely above the axis, whilst f(x) is some under and some above. The area below the axis enclosed by f(x) must be negative, as it is included in the overall area and is under the axis. The area ABOVE the axis is actually taken away from the shaded area, meaning that we subtract f(x) for this interval too.

Therefore, Area=64=∫g(x)-f(x)dx, where the lower bound is 0 and the upper bound m. Evaluating this allows us to have terms of m and a = 64.

To be precise, a*m²-m⁴/4=64 [2]

As a=m²/2, we can sub a in for this.

giving a²=64, a>0 so a=8

Going back to equation [1] and subbing a=8 in, and m=4.

[Henreezy]

Can someone explain this to me please? looked at assessment reports and itute but still don't get it

Well, I'm pretty bad at explaining, but I'll give it a shot.
You would want to first note that you are given two points, (0,0) which is implied (0 into ax = a(0)) and a (which when substituted gives (m, am).
This gives us our first hint at finding a and m.
Since the point of intersection occurs at the same value of x, we can substitute m into both equations to find m in terms of a.
am = m^3 - am
2am = m^3
2a = m^2
m = sqrt(2a)
Now that we have the 'value' of m, we can use it in our integral that we're going to set up (as we know the /area/ is 64).
We want to do top curve, minus bottom curve, thus we cannot simply do f + g, but we must do an integral of g - f which is 2ax - x^3.
the integral of between 0 and sqrt(2a) 2ax - x^3 = 64
Since we have an integral completely in terms of a, rearrange and solve. Hope this helps friend!

[TrueTears]

Is that for nspire cx, I can't find the key

it's for ti-89

[Henreezy]

Just a quick question for when we're finding general solutions for trigonometric functions
Lets say we had
If we solve for x we can get
Would we be wrong if I wrote or or do we have to try and write it in the simplest form?

Also, looking at a unit circle while finding general solutions for these. Is there a general rule of thumb when there is one answer with +-,  or when there are two separate answers?

My specialist teacher had a talk with us about this, he said that either format is fine, since you are still giving the correct answer, some publishers stick to just +/-, others write two equations. 
General solutions for Sin however, I would not try to find a compact way to write it and just stick to two equations.
e.g. sin (x) = 1/2
pi/6 + n2pi = x
5pi/6 + n2pi = x
Just remember to write that n, k or whatever letter you use to represent the varying number of rotations is an element of Z.

One exception however, would be if they said "express in simplest form", but I doubt they'd ever say that for a 'solve/find the solutions' question.

[Lejn]

Just a quick question for when we're finding general solutions for trigonometric functions
Lets say we had
If we solve for x we can get
Would we be wrong if I wrote or or do we have to try and write it in the simplest form?

Also, looking at a unit circle while finding general solutions for these. Is there a general rule of thumb when there is one answer with +-,  or when there are two separate answers?

They're equal, so they're fine. Just don't forget to write n is an element of Z.

I'm unsure on what you mean about this part, but this is how I interpret gen solutions:

let 'a'=Whatever the special angle in question is,

We know for cos to be positive, it must be first or fourth quad, and so it must be from an even number of pi, or . For Tan, it must be 1st or 3rd, so it's + for any integer of . Sine must be first or second, so it is + from the even and - from the odd. This can remain true for even negative values, as 'a' can be an angle outside of the first quad.

[papertowns]

Thanks guys 

[lzxnl]

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.

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

[Damoz.G]

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

Works for me now, thanks.

How did you figure that out?

[sasa]

Vcaa 2009 ex 1, q 1b. I got -1/(pi+1) but that's incorrect. Help!

[lzxnl]

Works for me now, thanks.

How did you figure that out?

A friend told me that one

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

[Henreezy]

Vcaa 2009 ex 1, q 1b. I got -1/(pi+1) but that's incorrect. Help!

a) y = xlog(x)
u = x
du/dx = 1
v = log(x)
dv(dx) = 1/x
dy/dx = 1 + log(x)


b)
u = cos(x)
du/dx = -sin(x)
v = 2x+2
dv/dx = 2

dy/dx = (-sin(x)(2x+2) - 2cos(x))/(2x+2)^2
f'(pi) = -2(-1)/(2pi+2)^2
= 2/(2pi+2)^2

the answer says 1/2(pi+1)^2

but if you expand both you see that it is actually the same, you'd just need to do this instead
v = 2(x+1)
v^2 = 4(x+1)^2
thus 2/4 = 1/2

[Sanguinne]

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?