HSC 2015 MX1 Marathon (archive) (1 Viewer)

Drsoccerball · Jun 8, 2015

Re: HSC 2015 3U Marathon

Not sure what to do to get the marks:

$The rate of change of a quantity Q is given by the equation $ \frac{dQ}{dt} = k(Q-B), $ where t is the time, k and B are constant. Using integration deduce that $ Q=Ae^{kt} +B $ Where A is another constant. 4 Marks$

InteGrand · Jun 8, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Not sure what to do to get the marks:

$The rate of change of a quantity Q is given by the equation $ \frac{dQ}{dt} = k(Q-B), $ where t is the time, k and B are constant. Using integration deduce that $ Q=Ae^{kt} +B $ Where A is another constant. 4 Marks$

$We solve this differential equation using the method of separation of variables. $\frac{\mathrm{d}Q}{\mathrm{d}t}=k(Q-B)\Rightarrow \frac{\mathrm{d}Q}{(Q-B)}= k\text{ d}t$$

$\Rightarrow \int \frac{\mathrm{d}Q}{(Q-B)} =\int k\text{ d}t$

$$\Rightarrow \ln (Q-B) =kt + C $, where $C$ is a constant$

$\Rightarrow Q-B =\mathrm{e}^{kt + C}$

$$\Rightarrow Q =\mathrm{e}^C\mathrm{e}^{kt}+ B=A\mathrm{e}^{kt}+B$, where $A\equiv \mathrm{e}^C$ is a constant.$

(http://en.wikipedia.org/wiki/Separation_of_variables)

Drsoccerball · Jun 8, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$We solve this differential equation using the method of separation of variables. $\frac{\mathrm{d}Q}{\mathrm{d}t}=k(Q-B)\Rightarrow \frac{\mathrm{d}Q}{(Q-B)}= k\text{ d}t$$

$\Rightarrow \int \frac{\mathrm{d}Q}{(Q-B)} =\int k\text{ d}t$

$$\Rightarrow \ln (Q-B) =kt + C $, where $C$ is a constant$

$\Rightarrow Q-B =\mathrm{e}^{kt + C}$

$$\Rightarrow Q =\mathrm{e}^C\mathrm{e}^{kt}+ B=A\mathrm{e}^{kt}+B$, where $A\equiv \mathrm{e}^C$ is a constant.$

(http://en.wikipedia.org/wiki/Separation_of_variables)

Surely my school didnt expect the cohort to know this?...

InteGrand · Jun 8, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Surely my school didnt expect the cohort to know this?...

The HSC is not meant to be allowed to ask you to explicitly solve the differential equation. It's meant to only be allowed to get students to verify that a given equation satisfies the differential equation.

If you have a copy of the 3U Year 12 Cambridge (Pender) maths textbook, check Exercise 7H page 281, the Extension exercise Q.11 there guides you through the method to solve that differential equation.

Drsoccerball · Jun 8, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
The HSC is meant to not be allowed to ask you to explicitly solve the differential equation. It's meant to only be allowed to get students to verify that a given equation satisfies the differential equation.

so this is considered in the sylabus

InteGrand · Jun 9, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
so this is considered in the sylabus

No, as far as I know they're meant to only be able to give you the solution function, and then ask you to show that it satisfies the given differential equation by differentiating it etc.

The HSC isn't meant to be able to just get you to solve that differential equation from scratch, as far as I know.

InteGrand · Jun 9, 2015

Re: HSC 2015 3U Marathon

Check out page 73 of the HSC 2/3U syllabus

http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths23u_syl.pdf

''Derivation of the solution of (3) is not required.''

Drsoccerball · Jun 9, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Check out page 73 of the HSC 2/3U syllabus http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/maths23u_syl.pdf

''Derivation of the solution of (3) is not required.''

But this question was in my schools exam :/?>

InteGrand · Jun 9, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
But this question was in my schools exam :/?>

Yeah, it may be possible that schools can do whatever they like (I'm not sure though)...lol.

Maybe your teachers had set Exercise 7H of the Year 12 3U Pender textbook as homework and told you to look at Extension or something?

Drsoccerball · Jun 9, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Yeah, it may be possible that schools can do whatever they like (I'm not sure though)...lol.

Maybe your teachers had set Exercise 7H of the Year 12 3U Pender textbook as homework and told you to look at Extension or something?

we use fitzpatrick

InteGrand · Jun 9, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
we use fitzpatrick

Well did anyone manage to get out that question? If not, it doesn't make much difference. Or you could show the syllabus to your teacher or whoever wrote the paper.

Drsoccerball · Jun 9, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Well did anyone manage to get out that question? If not, it doesn't make much difference. Or you could show the syllabus to your teacher or whoever wrote the paper.

Wasnt my year anyway

leehuan · Jun 9, 2015

Re: HSC 2015 3U Marathon

Although separation of variable is technically not in the 3U syllabus (a definite integral method is sometimes taught in 4U), you don't need it anyway.

$\frac { dQ }{ dt } =k\left( Q-B \right) \\ \frac { dt }{ dQ } =\frac { 1 }{ k } \frac { 1 }{ Q-B } \\ t=\frac { 1 }{ k } \ln { \left( Q-B \right) } -C\\ t+C=\frac { 1 }{ k } \ln { \left( Q-B \right) } \\ k\left( t+C \right) =\ln { \left( Q-B \right) } \\ Q-B={ e }^{ kt+kC }\\ Q=B+{ e }^{ kC }{ e }^{ kt }\\ Q=B+A{ e }^{ kt }\quad \quad (where\quad { e }^{ kC }=A)$

It just makes the working tidier than what I just did.

And it also helps that this is a first order diff. eqn.

photastic · Jun 9, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Although separation of variable is technically not in the 3U syllabus (a definite integral method is sometimes taught in 4U), you don't need it anyway.

$\frac { dQ }{ dt } =k\left( Q-B \right) \\ \frac { dt }{ dQ } =\frac { 1 }{ k } \frac { 1 }{ Q-B } \\ t=\frac { 1 }{ k } \ln { \left( Q-B \right) } -C\\ t+C=\frac { 1 }{ k } \ln { \left( Q-B \right) } \\ k\left( t+C \right) =\ln { \left( Q-B \right) } \\ Q-B={ e }^{ kt+kC }\\ Q=B+{ e }^{ kC }{ e }^{ kt }\\ Q=B+A{ e }^{ kt }\quad \quad (where\quad { e }^{ kC }=A)$

It just makes the working tidier than what I just did.

And it also helps that this is a first order diff. eqn.

We learnt this method for 3U. It wasn't hard, kinda more interesting but was a waste of time considering the HSC exams only use differentiation and actually provide the equation in a sense

Drsoccerball · Jun 9, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Although separation of variable is technically not in the 3U syllabus (a definite integral method is sometimes taught in 4U), you don't need it anyway.

$\frac { dQ }{ dt } =k\left( Q-B \right) \\ \frac { dt }{ dQ } =\frac { 1 }{ k } \frac { 1 }{ Q-B } \\ t=\frac { 1 }{ k } \ln { \left( Q-B \right) } -C\\ t+C=\frac { 1 }{ k } \ln { \left( Q-B \right) } \\ k\left( t+C \right) =\ln { \left( Q-B \right) } \\ Q-B={ e }^{ kt+kC }\\ Q=B+{ e }^{ kC }{ e }^{ kt }\\ Q=B+A{ e }^{ kt }\quad \quad (where\quad { e }^{ kC }=A)$

It just makes the working tidier than what I just did.

And it also helps that this is a first order diff. eqn.

Didn't see this when i was solving D:

Drsoccerball · Jun 9, 2015

Re: HSC 2015 3U Marathon

$$ From a distance of b above the origin a particle is shot with initial veloicty of V and at an angle of$ $\alpha$

$The particle lands at a horizontal distance of a and the displacement traveled is given by the line which points are (0,b) (a,0)$
$i) $ Show that the displacment traveled is $ =2V \sqrt{\frac{b}{g}}$

leehuan · Jun 9, 2015

Re: HSC 2015 3U Marathon

Please tell me where I went wrong:

$\ddot { x } =0\\ \dot { x } =V\cos { \alpha } \\ x=Vt\cos { \alpha } \\ \ddot { y } =-g\\ \dot { y } =V\sin { \alpha } -gt\\ y=Vt\sin { \alpha } -\frac { g{ t }^{ 2 } }{ 2 } \\ For\quad the\quad time\quad of\quad flight,\quad let\quad y=0\\ \therefore 0=Vt\sin { \alpha } -\frac { g{ t }^{ 2 } }{ 2 } \\ V\sin { \alpha } =\frac { gt }{ 2 } \quad as\quad t=0\quad is\quad the\quad time\quad of\quad projection\\ t=\frac { 2V\sin { \alpha } }{ g } \\ So\quad the\quad range\quad is\\ R=V\cos { \alpha } \left( \frac { 2V\sin { \alpha } }{ g } \right) \\ R=\frac { { V }^{ 2 }\sin { 2\alpha } }{ g }$

$But\quad R=a\quad \therefore a=\frac { { V }^{ 2 }\sin { 2\alpha } }{ g } \\ However\quad \tan { \alpha } =\frac { b }{ a } \\ So\quad \sin { 2\alpha } =\frac { 2\left( \frac { b }{ a } \right) }{ 1+{ \left( \frac { b }{ a } \right) }^{ 2 } } \quad using\quad tangent \quad half-angle \quad substitution \\ =\frac { 2ab }{ { a }^{ 2 }+b^{ 2 } } \\ i.e.\quad a=\frac { { V }^{ 2 } }{ g } \left( \frac { 2ab }{ { a }^{ 2 }+b^{ 2 } } \right) \\ { a }^{ 2 }+{ b }^{ 2 }=\frac { 2{ V }^{ 2 }b }{ g } \\ But\quad \sqrt { { a }^{ 2 }+{ b }^{ 2 } } =d\\ \therefore d=V\sqrt { \frac { 2b }{ g } } \\$

Ekman · Jun 9, 2015

Re: HSC 2015 3U Marathon

I did an alt method:

$x=Vt\cos{\alpha} $ and $ y=\frac{-gt^2}{2} +Vt\sin{\alpha} +b$

$$ The simultaneous equations used: $ a=Vt\cos{\alpha} $ and $ 0=\frac{-gt^2}{2} +Vt\sin{\alpha} +b $ and $ \tan{\alpha}=\frac{b}{a}$

$Vt=a\sec{\alpha}=\sqrt{a^2+b^2} $ $ \therefore $ $ Vt\sin{\alpha}=b$

$\therefore \frac{-gt^2}{2}+2b=0 \rightarrow t=2\sqrt{\frac{b}{g}}$

$\therefore \sqrt{a^2+b^2} = 2V\sqrt{\frac{b}{g}}$

laters · Jun 10, 2015

Re: HSC 2015 3U Marathon

Hey guys, could someone please explain to me the solution to Carrotsticks' 2014 trial question? http://puu.sh/ijqqn/efd1be7108.png

The solution is http://puu.sh/ijqnF/0c9662c363.png but there is no diagram.

I can't get my head around it - I've only used the divider method for doing simple things like assorting rings onto hands/into boxes etc.

InteGrand · Jun 10, 2015

Re: HSC 2015 3U Marathon

laters said:
Hey guys, could someone please explain to me the solution to Carrotsticks' 2014 trial question? http://puu.sh/ijqqn/efd1be7108.png

The solution is http://puu.sh/ijqnF/0c9662c363.png but there is no diagram.

I can't get my head around it - I've only used the divider method for doing simple things like assorting rings onto hands/into boxes etc.

Maybe I've misinterpreted the Q or screwed up some working out, but I'm not sure why the answer is $\binom{n-1}{k-1}\binom{10}{k}$ .

Say we had k = 2 (so two different numbers from 0-9 used), and n = 3 (three-digit password).

Let's just say for now the two distinct digits used were 1 and 2. Since n = 3, these are the possibilities: 112, 121, 211, 221, 212, 122. That's 6 possibilities.

But we could do the same for all possible $\binom{10}{2}$ choices of two distinct digits. So it seems like the total possible ways for n = 3, k = 2 is: $6\binom{10}{2}$ . But this doesn't agree with the answer, since $6\neq \binom{3-1}{2-1}=2$ .

Edit: My bad, Q said combinations, not permutations

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