WTF???? I dont think this would be the answer to a 2 unit Q...joshuaali said:I couldn't get it, but my friend the equation solver told me this:
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...justfriedy said:well i got x=(m*L1)/(L1+L2)
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).mc88 said:Wait, wasn't the loudness supposed to be the subject of the first equation?
Your the man!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