My Suggested Solutions to SP2008 Exam 2 [Confirmed]

[jamestaR`]

Ye the equilateral traingle is in this thread?

I don't know if you can assume that the line going through the circle is one side of an equilateral triangle in the circle...there's only one length that line can have for it to be true so it seems unlikely...

ASeriously someone show me a proper way to do this question, assuming you have a graphics calc like i did...get fucked vcaa putting questions that advantage ppl with cas.

[polky]

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

You can calculate the side lengths of each side.  The lengths of each side is .  Thus it is a equilateral triangle.

[Mao]

or, you can use the information that the line is , hence the angle between the two lines are 60 degrees, bingo.

[jamestaR`]

Yeah thats a good method, but is there any other method thats actually relevant to thecourse! LIke integration...

As you can see whoever I was talking to before, the intersection is above the halfway point of the circle, so if you were gonna integrate you would need to consider both functions. And, the intersection is to the RIGHT* of some of the enclosed area u have to subtract from pi...so how the fuck do you use a GRAPHIX calc to do that?

The questions should be equally difficult regardless of which calc u have.

[Captain]

Yeah thats a good method, but is there any other method thats actually relevant to thecourse! LIke integration...

Im pretty sure it's relevant to the course.  Area of circles and triangles = pretty basic.

[Mao]

Yeah thats a good method, but is there any other method thats actually relevant to thecourse! LIke integration...

As you can see whoever I was talking to before, the intersection is above the halfway point of the circle, so if you were gonna integrate you would need to consider both functions. And, the intersection is to the RIGHT* of some of the enclosed area u have to subtract from pi...so how the fuck do you use a GRAPHIX calc to do that?

The questions should be equally difficult regardless of which calc u have.

being above the half-way point actually makes it easier, you'll just work out the area on the LHS separately to the RHS



Alternatively, you can take the inverse of everything, so now you have a circle centred at (1,0) with the the intersection below the x axis, in which case

[jamestaR`]

Since the fun is over, I'd like to mention that coblin, Ahmad and I conspired to come up with reasonable-looking yet incorrect answers. Hopefully none of you were bothered too much by it. Kudos to the various people who managed to pinpoint exactly where the errors were

I left the graph, since that's probably the only useful thing from it.
(Image removed from quote.)

Even your graph is wrong. Look at itute its a figure 8 shape, theres no turning points at the origin...unless thats meant to be a figure 8.

And ye I get how to do that cirlce one now..I was just obsessed with finding that enclosed area and minusing from pi

[chid]

After looking at itute solutions ive got a few questions:

For the train question, i wrote the parametric equation as y^2=... instead of y=... (itute did this too so i assume its acceptable)-after exam people were saying that cartesian equation strictly mean y=...though :|

I also got 3pi as the time for one circuit instead of 6pi. I then used this as the value of the upper limit of the integral. Would this generally get a consequential mark (since i've already been penalised for making the mistake). Thing is it's only one mark-so i assume it's an answer mark? Then obviously the next 1 mark question involved evaluating this integral which would again provide the wrong answer!

It seems really harsh to lose 2 or 3 marks for a small oversight, but the marking scheme allows this to hapen.

Does anyone have any thoughts on what might happen/is in a similar situation?

Thanks guys.

[Mao]

for the cartesian equation concern, i did the same as you. though I doubt they will penalise you for that. is a cartesian equation for a circle.... etc