HSC 2016 MX1 Marathon (archive) (3 Viewers)

Paradoxica · Apr 12, 2016

Re: HSC 2016 3U Marathon

leehuan said:
There actually is a strong vibe that the other parad0xica is a uni student.

I could easily pass off as a (bad, barely passing) uni student.

porcupinetree · Apr 12, 2016

Re: HSC 2016 3U Marathon

leehuan said:
There actually is a strong vibe that the other parad0xica is a uni student.

Paradoxica said:
I could easily pass off as a (bad, barely passing) uni student.

Especially when you attend usyd physics lectures amirite

Drsoccerball · Apr 12, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
"only"

evidence please.

Well you couldn't do it :L

leehuan · Apr 12, 2016

Re: HSC 2016 3U Marathon

porcupinetree said:
Especially when you attend usyd physics lectures amirite

He attended my math lecture on the day of the ASoc integration bee

Paradoxica · Apr 12, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
Well you couldn't do it :L

Technically I did.

Go back to the thread, I provided the necessary value of delta/epsilon for the limit to exist. The rest was bookwork.

hedgehog_7 · Apr 12, 2016

Re: HSC 2016 3U Marathon

A particle moves in simple harmonic motion with period 8pi. Initially, it is at the point P where x = 4 moving with velocity v =6. Find, correct to three significant figures, how long it takes to return to P:

by expressing the motion in the form x = b sin nt + c cos nt and using the t formulae.

so essentially ive found the displacment formuale to be x = 24 sin (t/4) + 3 cos (t/4) but im not sure what the question is asking me to do nor use the t method. when it asks for how long it returns to P does it want me to let x = 4 in the formulae that i have just found and to solve for t? and if so ,how do i use the t method to do this.

The answer is t = 8 (tan inverse 6) = 11.2 ish

many thanks!!

Nailgun · Apr 12, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
Since when did a person have to be in Uni to know high levels of mathematics?

Plenty of people are self taught.

Not sure about what level of education, but he does have a particularly intense hatred for a certain HS tutoring centre

parad0xica · Apr 12, 2016

Re: HSC 2016 3U Marathon

Nailgun said:
Not sure about what level of education, but he does have a particularly intense hatred for a certain HS tutoring centre

Interesting claim. Could you reveal your reasoning? Or identify the premises.

leehuan · Apr 12, 2016

Re: HSC 2016 3U Marathon

hedgehog_7 said:
A particle moves in simple harmonic motion with period 8pi. Initially, it is at the point P where x = 4 moving with velocity v =6. Find, correct to three significant figures, how long it takes to return to P:

by expressing the motion in the form x = b sin nt + c cos nt and using the t formulae.

so essentially ive found the displacment formuale to be x = 24 sin (t/4) + 3 cos (t/4) but im not sure what the question is asking me to do nor use the t method. when it asks for how long it returns to P does it want me to let x = 4 in the formulae that i have just found and to solve for t? and if so ,how do i use the t method to do this.

The answer is t = 8 (tan inverse 6) = 11.2 ish

many thanks!!

Shouldn't it be x = 24sin(t/4) + 4cos(t/4)?

Nailgun · Apr 12, 2016

Re: HSC 2016 3U Marathon

parad0xica said:
Interesting claim. Could you reveal your reasoning? Or identify the premises.

parad0xica said:
Guys, this is the consequence of being heavily influenced by NGO & SONS.

parad0xica said:
Ayyyy, this guy knows what's up

I had a feeling you go to NGO & SONS based on your past arrogance but it was just intuition...

parad0xica said:
Have you interacted with every single one in your class or enough people to make that claim? Do you think most people at NGO & SONS are easily influenced?

I had Mai and he was a bad person. He knows that and is trying to fix it.

What is the true meaning of life? I would like to know ^_^

Q.E.D

leehuan · Apr 12, 2016

Re: HSC 2016 3U Marathon

Nailgun said:
Q.E.D

This is abuse of Q.E.D.

hedgehog_7 · Apr 12, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Shouldn't it be x = 24sin(t/4) + 4cos(t/4)?

Yes it is sorry my bad... do you know what i have to do afterwards?

parad0xica · Apr 12, 2016

Re: HSC 2016 3U Marathon

Nailgun said:
Q.E.D

lol

leehuan · Apr 12, 2016

Re: HSC 2016 3U Marathon

hedgehog_7 said:
Yes it is sorry my bad... do you know what i have to do afterwards?

Firstly, t = 8 tan^-1(6) is wrong. Try substituting it into the original equation and you get something that's not 0.
I tried this with both 4cos(t/4) AND 3cos(t/4)

So we want when the particle returns to P, which is x=4. Therefore simply let x=4

4 = 24sin(t/4) + 4cos(t/4)
1 = 6sin(t/4) + cos(t/4)

At this point, without the slightest of hesitation bombard with the auxiliary angle method:

1 = sqrt(37)sin(t/4 + tan^-1(1/6))

I'll leave it here, and wait for you to comment first

hedgehog_7 · Apr 13, 2016

Re: HSC 2016 3U Marathon

well the cambridge book stated at the end that the answers was t = 8 tan^-1(6) so became confused as to what i was trying to solve. the question is stating for x=4 right?

leehuan · Apr 13, 2016

Re: HSC 2016 3U Marathon

Should've mentioned earlier that this was a Cambridge question so that I could've had something to refer to.

$Cambridge Yr 12 3U Ex 3D Q21$

$$Given that the first part of the answer is $x=24\sin{\frac{t}{4}}+4\cos{\frac{t}{4}}\\$Since $P$ is the point $x=4$, to find when the particle returns to $P:$
$\\ 4=24\sin{\frac{t}{4}}+4\cos{\frac{t}{4}}\\1=6\sin{\frac{t}{4}}+\cos{\frac{t}{4}}\\$Let $T=\tan{\frac{t}{8}}\\$Then $\sin{\frac{t}{4}}=\frac{2T}{1+{T}^{2}}$ and $\cos{T}=\frac{1-{T}^{2}}{1+{T}^{2}}$

$\begin{align*}1&=\frac{12T}{1+{T}^{2}}+\frac{1-{T}^{2}}{1+{T}^{2}}\\1+{T}^{2}&=-{T}^{2}+12T+1\\2{T}^{2}-12{T}&=0\\2T\left(T-6\right)&=0\text{ note, we reject the case }t=0\\T&=6\\\tan{\frac{t}{8}}&=6\\ t=8\arctan{\frac{t}{6}}\end{align*}$
Note: arctan just means tan^-1

I will leave part b) for you to do yourself first.

kawaiipotato · Apr 16, 2016

Re: HSC 2016 3U Marathon

Prove that a+b< c*sqrt(2) if a,b are two sides of a right angled triangle and c is the hypotenuse.

Paradoxica · Apr 16, 2016

Re: HSC 2016 3U Marathon

kawaiipotato said:
Prove that a+b< c*sqrt(2) if a,b are two sides of a right angled triangle and c is the hypotenuse.

obviously true by inspection

note this is only true if the sides are not the same length.

InteGrand · Apr 16, 2016

Re: HSC 2016 3U Marathon

Yeah, in fact there's equality if and only if the right-triangle is isosceles.

Carrotsticks · Apr 16, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
obviously true by inspection

note this is only true if the sides are not the same length.

For the spirit of this thread, perhaps you'd like to actually answer the questions rather than providing what is more commonly known as 'proof by intimidation'.

Yes, many of these can be done by inspection, good for you. But the reality is that there are several processes that go on in our heads to produce the result. We still follow some train of thought when doing something as menial as factorising quadratics. People want to see and follow this train.

Remember that people read these forums and learn from the solutions. But what can they learn when all they see is you waving around your maths dick all over the Maths forum? You add little to no value to the forum by providing such answers.

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