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Who did BOX + 2 Triangles + p + q question? (1 Viewer)

Who did the QUESTION with the BOX and p, g, triangles

  • YES

    Votes: 9 27.3%
  • NO

    Votes: 24 72.7%

  • Total voters
    33

Toodulu

werd!
Joined
Apr 15, 2003
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1,335
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Sydney
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HSC
2003
ohhhhhhhhhhhhhhhhhhhhhhhh

i took out the 1/2 ahah, it's not integration silly :/
 

Ats

New Member
Joined
Oct 20, 2003
Messages
3
Location
NE NSW
I got the 1st bit, and I just realised how to do the second when the time ran out... another 20 seconds and 2 more marks, bah...
 

walla

Satisfied Customer
Joined
Nov 9, 2002
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285
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2003
Originally posted by Constip8edSkunk
tired of physics so:

A=1-p/2+(p-1)/2(1+p)
dA/dp = -1/2 + [2(1+p)-2(p-1)]/4(1+p)^2
=1/2[ (1+p-p+1)/(1+p)^2 - (1+p)^2/(1+p)^2]
=1/2[(2-1+2p-p^2)/(1+p)^2]
=0 when
-1+2p-p^2=0
by using quadrtic formula
p={-2+/-sqrt[4+4]}/2 = -1+/-sqrt2
but p>0
so p=sqrt2-1

(show this is a maximum)

sub into original eqn

A=1-[sqrt2-1]/2+[sqrt2-2]/[2sqrt2]
=1-1/sqrt2 +1/2+1/2-1/sqrt2
=2-2sqrt2
typo on the last line?
1/sqrt2 + 1/sqrt2 = 2/sqrt2
=sqrt2 not 2sqrt2
 
Last edited:
N

ND

Guest
Originally posted by Constip8edSkunk

(show this is a maximum)
I just wrote: 'Since this is the only stationary pt, and the question is asking for a max, sqrt2-1 is a max pt.'
They better not take marks off...

Btw, you didn't need to exapnd it all out:

-1/2 + [2(1+p)-2(p-1)]/4(1+p)^2 = 0
(1+p)^2 = 2
1+p = sqrt2
p = sqrt2 - 1.
 

Constip8edSkunk

Joga Bonito
Joined
Apr 15, 2003
Messages
2,398
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HSC
2003
Originally posted by walla
typo on the last line?
1/sqrt2 + 1/sqrt2 = 2/sqrt2
=sqrt2 not 2sqrt2
yep... i was typing 2/sqrt2 then realised it was sqrt2 but didnt cancel the 2 off lol
Originally posted by ND
I just wrote: 'Since this is the only stationary pt, and the question is asking for a max, sqrt2-1 is a max pt.'
They better not take marks off...

Btw, you didn't need to exapnd it all out:

-1/2 + [2(1+p)-2(p-1)]/4(1+p)^2 = 0
(1+p)^2 = 2
1+p = sqrt2
p = sqrt2 - 1.
nice,,, wish i saw that in the exam... haha
 

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