Answer to 10b? (1 Viewer)

[romanqwerty]

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]

I cant remember much but mine involved...

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

-Charlie

 

[joshuaali]

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

 

[eppingMCE]

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

 

[Originality]

Should dat even be 2u question -_- 4 marks Sheesh.

 

[fishy89sg]

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]

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]

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).

 

[justfriedy]

well i got x=(m*L1)/(L1+L2)

 

[sting24]

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]

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

 

[sting24]

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...

 

[mc88]

Stupid me and my stupid rushing.

 

[joshuaali]

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]

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]

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]

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.

 

[joshuaali]

 

[hodarrenkm]

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]

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.