Thank you so much! That was very helpful. I'll definitely keep in mind all that you've said.
A final question: Would you think the best course of action is to overload 6 units of OLE over with in semester 2 and then overload the final 6 in semester 1 of year 2 because it seems hard to overload...
Hello, I'll be studying at the University of Sydney as a first year student this year, and I had a few questions. Firstly, here are my provisional units of study for next year:
CHEM1991
MATH1902
MATH1921
PHYS1903
INFO1110 (?)
CHEM1992B
MATH1905
MATH1923
PHYS1904
INFO1113 (?)
So, I'm doing the...
Really hope I'm right coz this took me forever. Also sorry for bad LaTeX.
I=\int_0^{\pi} \mathrm{ln}(1+2a\mathrm{cos}(x)+a^2)\mathrm{d}x
\frac{\mathrm{d}I}{\mathrm{d}a} =\int_0^{\pi} \frac{2\mathrm{cos}(x)+2a}{1+2a\mathrm{cos}(x)+a^2}\mathrm{d}x = \frac{1}{a}\int_0^{\pi}...
Uhh, my school was weird. I did well in math in year 9 so they were like "yeah screw it you and these other three kids can do accelerated 3U" and then me and my other friend decided to do accelerated 4U.
Correct me if I'm wrong, but here's what I get:
Yeah 8 is B which is just by subbing in (-x) into the equation.
Then 9 is C. So dy/dt = dy/dt * dt/dx = -x/y * 1/y = -x. Then, consider that in the first quadrant, y is positive so dx/dt is positive, and hence x is increasing. So it seems like it...
Yeah, that was my logic as well, hence why I did negative. But, yeah as you said, I don't see any real reason to deduct marks for neglecting the negatives.
I'm guessing, either I'm wrong in doing the negative, or even if you can technically do a negative they won't care about it. Like, I won't pretend to know all the definitions in conics, so you could totally be right. And in that case doing the negative is wrong.
Vertices on the y axis don't exist though. Sub in x = 0 and you get y^2 = -b^2 which unless y = b = 0, is impossible, and b=0 can't be the case either.
I did forget the other case though which I've fixed up now.