How did we go in Exam 2?

[Special At Specialist]

I didn't do very well unfortunately But oh well, I wouldn't have gotten a 50 anyway because I got 39/40 on exam 1.

I guessed 2 MC, not because I couldn't do them (they were all pretty easy), but because I ran out of time. I also left the last part of questions 4 and 5 blank, as well as about half of question 3...

The space station question used up waaaaay too much time. There were so many big numbers and stuff to type into my calculator. If I'd known this in advance, I would've saved it until last.

The worst one of them all was the last part of the space station question... wtf even was the answer?! I couldn't get the answer to that question. My calculator just said "false". I even got rid of the domain restrictions on t and it still said "false".

OMFG I realised what I did wrong now!!!!! It had x = 6800*cos(pi*(1.3t - 0.1)) and y = 6800*sin(pi*(1.3t - 0.1)) - 6400 and I completely forgot about the -6400 and tried to say that y/x = tan(pi*(1.3t - 0.1)) FFS I spent so long on this question for nothing!!!!!

Another question which I couldn't answer (on question 3) was one where I had v = t^2 / 100 - 9t / 20 + 25 = 0 and we had to find when the particle was at rest and my calculator wouldn't give me a solution

[baddin]

And remember that question which said find values of n for which re(z) = 0, did you guys get n = 6+24k and n = 18 + 24k, where kEZ, k = ....-1,0,1,2....

[Biceps]

And remember that question which said find values of n for which re(z) = 0, did you guys get n = 6+24k and n = 18 + 24k, where kEZ, k = ....-1,0,1,2....

i got that and then used pi and -pi for the question after. is that right?

[Special At Specialist]

MC was easy, questions 1 and 2 of ER were easy, questions 3, 4 and 5 of ER looked easy but were so damn time consuming!

[baddin]

And remember that question which said find values of n for which re(z) = 0, did you guys get n = 6+24k and n = 18 + 24k, where kEZ, k = ....-1,0,1,2....

i got that and then used pi and -pi for the question after. is that right?

I just did 1^(6+24k)cis((pi/12)(6+24k)) and 1^(18+24k)cis((18+24k)(pi/12))

[datfatcat]

I thought the angle have to be (-pi,pi] So n=-6 and 6

[ZanyCat]

I said n = 12k + 6, and then for the next section said (z1)^n = i when k is even and -i when k is odd.
Is this the wrong notation/way to describe it?

[Special At Specialist]

I said n = 12k + 6, and then for the next section said (z1)^n = i when k is even and -i when k is odd.
Is this the wrong notation/way to describe it?

I said k = -3, -1, 1, 3, ... and for the other one k = -4, -2, 0, 2, 4, ...
I think your notation is probably better than mine.

I thought the angle have to be (-pi,pi] So n=-6 and 6

Wow, I didn't even think of this... you could be right.

[baddin]

I thought the angle have to be (-pi,pi] So n=-6 and 6

Yes I think you are correct, damn it LOL I didn't do that and didn't even think of that in the exam FML.

[enwiabe]

Did any of you guys keep a copy to upload?

[scarl54]

It was good. I screwed up a few multi, screwed up the final two pages (made corrections but would have lost 4 or 5 marks) and I wouldn't rule out losing another 10 marks throughout the paper. So I reckon somewhere around the 60 mark out of 80, definitely more than 50, so hopefully I'll get a high good B+

Certainly makes me feel a lot better after my woeful effort on exam one.

[ZanyCat]

But isn't cis(pi/12)^18 = cis(3pi/2) = cis(-pi/2) ?
Yes you might get an angle outside of the Arg restrictions, but by the definition of cis it's still equal to an equivalent term of cis with a correct angle.

[Nick Nack 94]

I thought the angle have to be (-pi,pi] So n=-6 and 6

Wow, I didn't even think of this... you could be right.

The angle was pi/2, 3pi/2 wasn't it?

[InsaneMcFries]

I got n=-6 or n=6 as the angles have to be in the principle range. So that makes pi/2 and -pi/2.

[DJing]

But isn't cis(pi/12)^18 = cis(3pi/2) = cis(-pi/2) ?
Yes you might get an angle outside of the Arg restrictions, but by the definition of cis it's still equal to an equivalent term of cis with a correct angle.

Yep that's right, it wasn't restricted by Arg(z) because it would eventuate into the equivalent angle

so i for k as a positive integer, and -i for k as a negative integer

But I'm pretty sure the general solutions are right