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Answer to 10b? (1 Viewer)

romanqwerty

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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

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I cant remember much but mine involved...

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

-Charlie
 

joshuaali

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I couldn't get it, but my friend the equation solver told me this:
 

eppingMCE

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I dont know if mine was correct, but for part (i) i put P= L1/x2 + L2/(m-x)2
 

fishy89sg

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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

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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

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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

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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

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Wait, wasn't the loudness supposed to be the subject of the first equation?
 

sting24

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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

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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

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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

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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

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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

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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

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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.
 

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