My Suggested Solutions to SP2008 Exam 2 [Confirmed]

[Blizz]

guessed E haha..mad

[riadnicolas]

hardly any slope fields this year and methods their wasnt one standard normal distribution, funny stuff

[junaman]

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...

[riadnicolas]

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..

i attempted that but lost time and ended up with 2.52 or sumthing funny


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...

[junaman]

2.52 is hardly funny...

[riadnicolas]

2.52 is hardly funny...

nah im just guessing i did some weird things  i remember getting two point something two and integrating about y

[chris]

alrite guys what would g1 c+, g2 b+ and g3 C get me in spec raw, and then standardised,,,,  --- these r following 2007's distribution? 

[Captain]

I did the area of the circle thing this way (picture attached), which didn't require any integration.

Don't know if that's technically correct?  Might have gotten 2.53 as a fluke.

[IMG]http://img91.imageshack.us/img91/5045/1111082203vh5.th.jpg[/img][IMG]http://img91.imageshack.us/images/thpix.gif[/img]

EDIT: Linked to a easier-to-see picture

That looks right.  Good thinking

I've got a friend who by inspecting, decided to try and model a sin wave to fit the area he needed to find, so he could integrate HAHAHA.  I think examiners will see lots of weird methods here.

[junaman]

That looks right.  Good thinking

I've got a friend who by inspecting, decided to try and model a sin wave to fit the area he needed to find, so he could integrate HAHAHA.  I think examiners will see lots of weird methods here.

Fail.

[chris]

****** alrite guys what would g1 c+, g2 b+ and g3 C get me in spec raw, and then standardised,,,,  --- these r following 2007's distribution?   
-- anyoen im desperate to know i cant sleep otherwise,, and i cant figure it out.

[Toothpaste]

****** alrite guys what would g1 c+, g2 b+ and g3 C get me in spec raw, and then standardised,,,,  --- these r following 2007's distribution?   
-- anyoen im desperate to know i cant sleep otherwise,, and i cant figure it out.

http://vcenotes.com/forum/index.php/topic,6930.msg93460/topicseen.html#msg93460

[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.

[junaman]

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.

[eza]

With Q3)b - drawing the path of the train.
Was it required for the max/min points to be labelled? I have a feeling I only chucked in the axis intercept points.

Q1)c - another graph drawing question, specifically asks for these to be labelled. However they both were worth 2 marks lol

[Synesthetic]

With Q3)b - drawing the path of the train.
Was it required for the max/min points to be labelled? I have a feeling I only chucked in the axis intercept points.

Q1)c - another graph drawing question, specifically asks for these to be labelled. However they both were worth 2 marks lol

I'm quite sure 1)c) asked for the two inflection point coordinates, as well as the maximum turning point. You would probably need to label them for both marks.

I think the 'figure eight' sketch [3)b] only required the shape of the curve, though. But because it asked for the 'path' I think arrows on the curve, representing the direction of motion, might be necessary for the second of the two marks. However they may not care...