Answers (1 Viewer)

[redom]

3 a) yeh
c) 2 pi
4c yeh, but i had 1/2 + ...
6 b) lol i wrote something like "cause its proportional to the velocity" LOL

ohhh i had these answers too....but i had for 6b) that the rain drop is falling from the sky? thus the negative sign....and then what u had!

yay! i hope your right!....

 

[gaoOO]

3 a) yeh
c) 2 pi
4c yeh, but i had 1/2 + ...
6 b) lol i wrote something like "cause its proportional to the velocity" LOL

crap, i can't believe i got the last question right only to make a numerical error on 4c)

btw, i'm pretty sure the answer they were looking for on 6 b) was air resistance. it's like, g is to gravity as 0.2v^2 is to air resistance.



if anyone is willing to take the time to post up answers to the numerical questions, that would be greatly appreciated.

It seems i got the proof questions but i made some incredibly stupid numerical errors.

 

[roadrage75]

well i didnt get that for 4b), i got something like 4e - 1 - e^2/2....

 

[roadrage75]

well i didnt get that for 4c), i got something like 4e - 1 - e^2/2....

 

[roadrage75]

well i didnt get that for 4c), i got something like 4e - 1 - e^2/2....

 

[roadrage75]

well i didnt get that for 4c), i got something like 4e - 1 - e^2/2....
lol, and i hated how they did the y axis horizontally and teh x axis vertically!

 

[roadrage75]

well i didnt get that for 4c), i got something like 4e - 1 - e^2/2....
lol, and i hated how they did the y axis horizontally and teh x axis vertically!

 

[gaoOO]

well i didnt get that for 4b), i got something like 4e - 1 - e^2/2....

AHAH, that's not what i got either. I got 2e -1/2-e^2

 

[Golbez]

if anyone is willing to take the time to post up answers to the numerical questions, that would be greatly appreciated.

Which ones?

 

[gaoOO]

Which ones?

the definite integrals and volume questions thanks

 

[Golbez]

Ehh... I cbf doing the integrals right now...

but for the volumes

3c

each cylinder has height y, radius x

NOTE: A cylinder is 2pi(r) [The circumference of the circle] * h [The height of the cylinder]

therefore each is 2pi(x)(y)

2pi e-INT-1 (x)(y) dx

2pi e-INT-1 (x)(lnx/x) dx

2 pi e-INT-1 ln(x) * 1 dx

IBP-

u= loge (x)
u'= 1/x

v'=1
v=x

2pi[xlnx - INT x/x dx ] e->1

2pi[xlnx - x] e->1

2pi[e(1)-e - ((1)ln(1) - 1)]

2pi[ 0 - (0 - 1)]

=2pi

 

[roadrage75]

actually, gao, i didnt type it right, i got 2e - e^2/2 - 1/2, which is almost what you got

 

[JasonNg1025]

omg how can i forget to put a number like n=1000 into the equation

NOOOOOOOOOOOOOOO

I FORGOT TO SQUARE =/

 

[sphelein]

For the probability question part (i), I think I get about 0.02. But I'm not really sure of it.

 

[gaoOO]

actually, gao, i didnt type it right, i got 2e - e^2/2 - 1/2, which is almost what you got

oops, i typed it wrong too. I think i got what you got *looks hopeful*

anyway, thanks golbez, i think i got 2pi. I guess i'm just waiting on the integrals and the probability questions. The rest were from memory, just proofs?

 

[roadrage75]

well, for the probability question, for the first part, i got something like .36 ish...

 

[pires_w]

5di was kind of annoying...

Basically, BCP and BAP are similar coz they have equal sides/angles whatever.

And for a 4 sided polygon each interior angle is (n-2)180/5 degrees = 3pi/5

Therefore Angle BAP = 3pi/5 but bap and ead are pi/5 therefore angle cad = pi/5.

Also, Side AC = AP+PC

And cos (pi/5) = AP/1

So AP = u

Similarly, PC = u

And so AC = 2u

Similarly, AD = 2u


Using the cos rule,

CD^2 = AC^2 + AD^2 - 2(AC)(AD)Cos (CAD)

CD=1

1 = (2u)^2 + (2u)^2 - 2(2u)(2u)u
1= 8u^2 - 8u^3

Rearraging

8u^3-8u^2+1 = 0

As required.

thz... what a stupid mistake... when i tried to fig out what < CAD is... i found it was pi/10=.=

 

[Dumsum]

Can someone scan and upload the exam plz. Ta

 

[pires_w]

well, for the probability question, for the first part, i got something like .36 ish...

part i: 0.02 (0.018xxx)
partii: 0.03 (0.03452)

 

[Hikari Clover]

回复: Re: Answers

for the probability one , i got
0.3x for part i
0.6x for part 2