Answer to 10b? (1 Viewer)

romanqwerty

New Member
Joined
Mar 26, 2007
Messages
1
Gender
Male
HSC
2007
As the title sugests, does anyone have an answer to 10b that they know is right? I would really like to see if i got it right. Both my friend and i have differing opinions on what the answer is.
 

j3t_pil0t

Getting there....
Joined
Aug 8, 2005
Messages
5
Location
Tuggerah
Gender
Male
HSC
2007
I cant remember much but mine involved...

x = third root of L1/L2 divided by... etc haha thats my two cents.

-Charlie
 

joshuaali

Member
Joined
Oct 17, 2006
Messages
66
Gender
Male
HSC
2007
I couldn't get it, but my friend the equation solver told me this:
 

eppingMCE

Member
Joined
Mar 13, 2007
Messages
49
Gender
Male
HSC
2007
I dont know if mine was correct, but for part (i) i put P= L1/x2 + L2/(m-x)2
 

fishy89sg

Member
Joined
Feb 20, 2006
Messages
674
Gender
Male
HSC
2007
being an extension 2 student who could do Question 8 (the last question) in our exam today, i personally would have got about 6/12 for question 10 in this exam
 

kokodamonkey

Active Member
Joined
Feb 25, 2007
Messages
3,453
Location
Sydney
Gender
Male
HSC
2007
josh what you got was def wrong..
umm i remember mine slightly but not relaly all i remember is that i ended up simplyfying it down to like a cubic equation or something.
 

joshuaali

Member
Joined
Oct 17, 2006
Messages
66
Gender
Male
HSC
2007
Ok.
From part (i), you get N = L1/x2 + L2/(m-x)2, yes?

Once you differentiate with respect to x, you get 2L2/(m-x)3 - 2L1/x3, yes?

Since you are finding a maxima/minima, you equate it to 0. It was at this stage that I could not proceed further in trying to make x the subject (I tried expanding but that led me nowhere).
 

sting24

2 Radical For U
Joined
Oct 20, 2006
Messages
20
Gender
Male
HSC
2007
joshuaali said:
I couldn't get it, but my friend the equation solver told me this:
WTF???? I dont think this would be the answer to a 2 unit Q...
I think that question was too hard, more like a last question for three unit or sumthin
 

mc88

Member
Joined
May 13, 2006
Messages
49
Gender
Male
HSC
2007
Wait, wasn't the loudness supposed to be the subject of the first equation?
 

sting24

2 Radical For U
Joined
Oct 20, 2006
Messages
20
Gender
Male
HSC
2007
justfriedy said:
well i got x=(m*L1)/(L1+L2)
I got the same as u justfriedy, except where u have L1 and L2 i had the cube root of L1, and the cubed root of L2. But i couldnt really prove it was a minimum...
 

joshuaali

Member
Joined
Oct 17, 2006
Messages
66
Gender
Male
HSC
2007
mc88 said:
Wait, wasn't the loudness supposed to be the subject of the first equation?
The subject was the sum of the noise heard, which, I assume, is the sum of the noise from L1 and L2 after you apply the inverse square law (that formula up the top).
 

Steth0scope

Member
Joined
May 23, 2006
Messages
266
Location
Camperdown
Gender
Male
HSC
N/A
nar it was Noise not Loudness.

I go 2L1 / x^3 - 2L2 / (m-x)^3 = 0


2L1 / x^3 = 2L2 / (m-x)^3


(m-x)^3 / x^3 = L2 / L1 (cross multpily sort of thing)

(m-x) / x = cuberoot (L2/L1)

m -x = xcuberoot(L2/L1)

m = x + xcuberoot(L2/L1)

m = x (1 + cuberoot (L2/L1) )\

x = m / (1 + cuberoot(L2/L1))

Then found 2nd deriviate. Sub'd that x into it and formed some fraction with everything to the power of 4 and 4/3 on it which means that the fraction is greater than zero.

Which means theres a min at that x value...hopefully =)

ANDDD i missed 5 marks on question 3 with the AP swimming bullshit coz i didnt even see it .. madddd!

hope that helps
 

cccclaire

Member
Joined
Nov 4, 2006
Messages
660
Gender
Female
HSC
2008
If you put somewhere that to find the minimum value you make y'=0 you should get a mark or two at least.

I got an answer to it, but I can't remember it and I'm not certain its right.
10.a) was easy though, so its ok.
 

mitchbazza

New Member
Joined
Jun 2, 2007
Messages
12
Gender
Male
HSC
2007
markzada said:
nar it was Noise not Loudness.

I go 2L1 / x^3 - 2L2 / (m-x)^3 = 0


2L1 / x^3 = 2L2 / (m-x)^3


(m-x)^3 / x^3 = L2 / L1 (cross multpily sort of thing)

(m-x) / x = cuberoot (L2/L1)

m -x = xcuberoot(L2/L1)

m = x + xcuberoot(L2/L1)

m = x (1 + cuberoot (L2/L1) )\

x = m / (1 + cuberoot(L2/L1))

Then found 2nd deriviate. Sub'd that x into it and formed some fraction with everything to the power of 4 and 4/3 on it which means that the fraction is greater than zero.

Which means theres a min at that x value...hopefully =)

ANDDD i missed 5 marks on question 3 with the AP swimming bullshit coz i didnt even see it .. madddd!

hope that helps
Your the man!

I got the same as my answer for x lad, pretty sure i'm right cause i just did it again.
 

hodarrenkm

New Member
Joined
Aug 20, 2006
Messages
25
Gender
Male
HSC
2008
i got x = m*cube root L1 / (cube root L1 + cube root L2), which is basically the same thing as x = m / (1 + cuberoot(L2/L1))

pretty sure it was right cos i actually proved that its noise level would be at minimum

what i did was i sub L1 = 8, L2 = 27, and m = 10...since there are all constant, which wont affect the result

substituting those values i got x = 4....then calculate N
try x = 3.9 and 4.1, the N value was greater than the one using x = 4
hence its a minimum
 

Fish Sauce

Active Member
Joined
Feb 11, 2007
Messages
1,051
Gender
Undisclosed
HSC
N/A
hahahaha shit

I did this absolutely MASSIVE expansion when I was differentiating and realised I made a tiny mistake earlier so I just gave up.

I should not have made a common denominator in part i, haha crap it would've been so much easier. I have been a fool.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top