typo on the last line?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
I just wrote: 'Since this is the only stationary pt, and the question is asking for a max, sqrt2-1 is a max pt.'Originally posted by Constip8edSkunk
(show this is a maximum)
Thats what i got as well.... As i was doing that in the exam, i was praying that i would not make a silly mistake.Originally posted by walla
area is 2 - sqrt(2)
which when you think about it is close to 0.5
you have to sub in p
yep... i was typing 2/sqrt2 then realised it was sqrt2 but didnt cancel the 2 off lolOriginally posted by walla
typo on the last line?
1/sqrt2 + 1/sqrt2 = 2/sqrt2
=sqrt2 not 2sqrt2
nice,,, wish i saw that in the exam... hahaOriginally 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.