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exactly. This inturn makes it a billion times more interesting though

It wasn't too bad lol the good thing about doingg the adv is it can scale very well. I have friends that did it last year getting about 50% in the exam, and that scaling to a final of 90/100. I think i went a bit better than that but it might have been an easier exam, i guess well find out soon

HSC physics is COMPLETELY different than University physics. I cannot emphasis this enough. Alot more mathematical. I had my end of semester adv physics exam yesterday lol

Yeah true... haven't thought auxillary in ages. I do alot of SHM in physics these days... I view it heaps diff than i did at school. You're right though.

x=asin nt + b cos nt dx/dt=nacos(nt)-bnsin(nt) d^2x/dt^2=-n^2asin(nt)-n^2bcos(nt) =-n^2[asin(nt)+bcos(nt)] =-n^2x as requires B) Hint: Velocity = 0 at maximum displacement & T=time for one oscillation ie. Time from a max displacement to another max displacement So that makes...

Yeah i think these would be better at assessing english ability and intellect than the current english course... rote learn and essay

Got the solution now, nah i didn't require calculus (but note that calculus is used VERY often in physics in uni) & thanks man lol

A Marble of mass M and radius r rolls down a track of height h and around a loop-the-loop. Loop- the-loop has radius R. What minimum height h must the track have for the marble to make it around the loop the loop without falling off?

Thanks alot, that really helps. They seem like quite easy questions if you understand the method. Thanks again mate

1) y''+y=cotx 2) y''+y=cosecx Can someone talk me through the solutions, only new to the method. Cheers PS: So it's based on finding: y(x)=yc+yp where yc is the complimentary solution and yp is the particular. This falls under 2 order non-homogenuous differential equations.

3 (Andrew Price, Dana Ma & Aaron Chong) are in my Physics (advanced) class... That's kinda scary

There's probably like 15 girls doing physics this year out of 120 in the Advanced Physics class in Melbourne (First Year).

Wait what? Explain what your saying. lim (x-->0) (sinx)/x = 1. So you try and get it into a fraction of this so that you can use this property. This can be proved by L' Hopitals Rule: lim (x-->0) (sinx)/x Since function is currently indeterminable (0/0) & noting x=/=0 lim(x-->0)...

-512sin(t)e^-sqrt(3)t I believe ( the 'i' wasn't supposed to be in it). It's definitely sin(t) not cos(t) since it's the Imaginary component of the derivative, cos(t) is the real.

Just checking if my answer is correct: Q. Use complex exponential to find the 9th derivative of sint/(e^rt3t). Simpliest form S. Im(e^it).e^-rt3t Im{d^9/dt^9[e^(i-rt3)t]} Im[(i-rt3)^9e^-rt3t cos(t)+i(i-rt3)^9e^-rt3.sint] Im[2^9e^i(45pi/6)e^-rt3cost...

That's how i originally went about solving them. Drongoski, your solution for Q2 is epic! That's very elegant

Thanks mate, makes clear sense now.

Express the following as algebriac functions of x: a) sinh^2(arctanh(x)) b) tanh(arccosh(x)) They might be easy, i might be missing something

http://community.boredofstudies.org/13/mathematics-extension-1/220554/stones-hitting-birds-d.html#post4569969

http://www.youtube.com/watch?v=kqVKvRTmbj0 http://www.youtube.com/watch?v=ZOU8GIRUd_g oMGz IAm SeW alTerNT