HSC 2014 MX2 Marathon (archive) (2 Viewers)

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dunjaaa

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Re: HSC 2014 4U Marathon

What I was trying to do was inspect the stationary points, there probably is another way of proving this but it was the best I could do :D
 

Sy123

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Re: HSC 2014 4U Marathon

The way I did the upper bound was:



 

dunjaaa

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Re: HSC 2014 4U Marathon

Can you provide us with a diagram, because I can't translate your instructions into one lol (idk if we have different interpretations of the way a house looks like) :p
 

dunjaaa

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Re: HSC 2014 4U Marathon

Find the volume of the region bounded by the curve y=sin(x) between x=0 and x=pi rotated about the line y=-x.
If anyone else has any good oblique rotation questions, please share! It's in my upcoming assessment task even though its not in the syllabus and there isn't many questions I can source out from past papers or textbooks.
 

Ikki

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Re: HSC 2014 4U Marathon

PHP:
Can you provide us with a diagram, because I can't translate your instructions into one lol (idk if we have different interpretations of the way a house looks like) :p
I know right? Basically inside the house which apparently has nothing in it. At one end of the house it is fired and hits one side of the triangular shaped roof and then reaches its maximum and hits the other side and then falls to finish at the other side of the house. That is why the range is the length of the house.
 

Davo_01

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Re: HSC 2014 4U Marathon

Can you provide us with a diagram, because I can't translate your instructions into one lol (idk if we have different interpretations of the way a house looks like) :p
Here you go, sorry if wording isn't clear enough Untitled.png
 

Davo_01

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Re: HSC 2014 4U Marathon

PHP:
I know right? Basically inside the house which apparently has nothing in it. At one end of the house it is fired and hits one side of the triangular shaped roof and then reaches its maximum and hits the other side and then falls to finish at the other side of the house. That is why the range is the length of the house.
Well maybe its a dog house :p
 

Sy123

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Re: HSC 2014 4U Marathon



Not sure if it belongs in Advanced thread, I thought it was a bit easy to post there
 

Ikki

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Re: HSC 2014 4U Marathon

i) From the condition given, A+B=C.
Multiply everything by e^x. Ae^x+Be^x=Ce^x.
Now the only way to reach the * form is to multiply throughout by the same constant which in this case means that a=b=c must hold true.

ii) As a result of dropping that condition, A, B and C can be any number. Since a,b,c do not have to be the same, each of A, B and C can be manipulated by multiplying by a sufficient constant in the form of e^a/b/c to fullfill the equation. Thus there should be atleast one set of a,b,c so that the identity holds true.
 

Sy123

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Re: HSC 2014 4U Marathon

i) From the condition given, A+B=C.
Multiply everything by e^x. Ae^x+Be^x=Ce^x.
Now the only way to reach the * form is to multiply throughout by the same constant which in this case means that a=b=c must hold true.

ii) As a result of dropping that condition, A, B and C can be any number. Since a,b,c do not have to be the same, each of A, B and C can be manipulated by multiplying by a sufficient constant in the form of e^a/b/c to fullfill the equation. Thus there should be atleast one set of a,b,c so that the identity holds true.
Hmmm yea that looks right I guess, though your wording in the last sentence of your first answer is a little vague.

Rather I would say:





Thus, the only solutions (a,b) such that is when a=c, b=c, thus a=b=c
 

aDimitri

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Re: HSC 2014 4U Marathon

good question! took me a couple of minutes to wrap my head around what the solid looked like but then i realised it said the cross sections were perpendicular not parallel, and i felt like a fool :(
loving the volumes, it's definitely my favourite topic so far :)

EDIT: Started to LaTeX my solution, but cant be bothered, far too long.
 
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mathing

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Re: HSC 2014 4U Marathon

If you want to punish yourself have a go at this inequality:

 

Ikki

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Re: HSC 2014 4U Marathon

If you want to punish yourself have a go at this inequality:

Not too hard:
Begin with expansions of:
(ly-mx)^2>=0
(nx-lz)^2>=0
(ny-mz)^2>=0

Then add them together to obtain:
l^2 * y^2 + l^2 * z^2 + m^2 * x^2 + m^2 * z^2 + n^2 * x^2 + n^2 * y^2 >= 2(lxmy+lxnz+mgnz)

Add l^2 * x^2 + m^2 * y^2 + n^2 * z^2 to both sides and factorise to obtain the required expression.
 

mathing

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Re: HSC 2014 4U Marathon

You got it!
Not too hard:
Begin with expansions of:
(ly-mx)^2>=0
(nx-lz)^2>=0
(ny-mz)^2>=0

Then add them together to obtain:
l^2 * y^2 + l^2 * z^2 + m^2 * x^2 + m^2 * z^2 + n^2 * x^2 + n^2 * y^2 >= 2(lxmy+lxnz+mgnz)

Add l^2 * x^2 + m^2 * y^2 + n^2 * z^2 to both sides and factorise to obtain the required expression.
 
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