My Suggested Solutions to SP2008 Exam 2 [Confirmed]

[eza]

Yeah, sweet thanks that's what I'm hoping for.

[Nelle91]

For the last question you can find the area of the unshaded are of the triangle by taking the area of the triangle minus the intergral of -sqroot(1-xsqrd)+1 between -sqrt(3)/2. This give aprox .5236.
Then take area of circle which is pie - .5236, which gives 2.62

[xers]

Yes i got 2.62 (converted from 5pi/6) too. But majority said 2.53.

[Nelle91]

I dude who got a 49 for methods last year got the same as me too.

[Synesthetic]

Having not attempted that question I can't try and clarify, but compare your working to this:

http://vcenotes.com/forum/index.php/topic,7511.msg93089.html#msg93089

Or wait for itute

[xers]

Also, for the train question, i got 3pi seconds which made the answers that follow wrong too. You think i would get consequential mark if working out and reasoning is correct?

[Synesthetic]

Also, for the train question, i got 3pi seconds which made the answers that follow wrong too. You think i would get consequential mark if working out and reasoning is correct?

In my opinion you will most likely get consequential marks for e)i) and ii) - perhaps 1/2 + 1/2 = 1 of the 2 available.

I'm not too sure about d) though, perhaps 1 / 2 marks again if the examiner is generous.

In short I would expect some marks to be awarded.

[xers]

Well for d, even subbing in 3pi seconds still gave me the answer of root2/3. Think they ll count that as correct.

[orsel]

Also, for the train question, i got 3pi seconds which made the answers that follow wrong too. You think i would get consequential mark if working out and reasoning is correct?

In my opinion you will most likely get consequential marks for e)i) and ii) - perhaps 1/2 + 1/2 = 1 of the 2 available.

I'm not too sure about d) though, perhaps 1 / 2 marks again if the examiner is generous.

In short I would expect some marks to be awarded.

There's no such thing as consequential/half marks in maths. Only method marks.

[riadnicolas]

Yes i got 2.62 (converted from 5pi/6) too. But majority said 2.53.

i think i finished on 2.62 as well

[Mao]

well done guys.

I think I agree with your answers there, never bothered to write mine down.

and a note to VCAA: no benefits from using CAS my ass. I relied on it the whole way.

[eza]

Yeah that licks, wish I did the exam with a CAS instead of lame graphics calc -.-"

[jamestaR`]

INtegrated how? you couldnt use the "top minus bottom" method because the circle went went outside the point of intersection.

And i dunno how to use calc to find 'top minus bottom' areas so..

The circle went outside the point of intersection? The circle _was_ the point of intersection? What you talkin' bout?

You integrate by finding the equation of the circle in terms of x (top half of the circle), which was rt(1-x^2) +1
And you integrated between -rt(3)/2 and 0 with the circle minus the line. The calculator does it for you if you subtract Y2 from Y1...
Then to that you add the pi/2 which was the semi circle...

Yeah i realise the point of intersection is where you integrate from, but part of the circle stuck out to the left of that particular point. Additionally, you are dealing with 2 different equations of the circle, where the top semi circle has a different equation to the bottom. For these 2 reasons you cant simply integrate from -rt(3)/2 to 0 with say, the bottom half, as you are not takin into account that little bit of area to the left of -rt(3)/2, as well as the fact that theres 2 equations of the circle to consider.

So i had no clue.

You only needed the top half of the semi circle as that was the area under it. So you needed the positive sqrt, which was easily found by transposing to find y in terms of x. Then just integrate from -rt(3)/2 to 0. The circle is the circle you used to find the point of intersection of the line, so there can't be a bit "stuck out to the left of that particular point". The circle that was already drawn for you as part of the complex field had nothing to do with it.

No, you need both as the area encompasses both equations of the circle. The area of the whole circle was obviously just pi, but the little bit enclosed by the straight line and the circle which had to be subtracted I could not do by hand. That little bit that is enclosed is not a simple integration thing, it was fucked up coz, as I said, part of the circle sticks out to the left. Do i have to draw you a picture>

[Mao]

I saw a very interesting geometric way of looking at it - the line segment in the circle can form an equilateral triangle circumscribed by the circle. The difference between the area of the circle and the triangle is three of the offending areas we don't want. divide that by 3, then pi minus, gives the correct answer.

[chid]

I got 2.53 as well. I wrote out the intgral and added pi/2 to it. Then I used CAS to evaluate integral coz I couldn't do it by hand.