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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Please could anyone explain how to do question 2aii? Thanks :smile:image.jpg
Reply 1021
Avatar for kennz
kennz
8
Original post by economicss
Please could anyone explain how to do question 2aii? Thanks :smile:image.jpg


arguments work like logs
rearrange to make z=(w/2)^(1/3)
therefore arg((w/2)^(1/3))= pi/6
so arg((w/2))/3 = pi/6
so arg(o.5w) = pi/2
so half line, angle pi/2
Original post by Seytonic
I hope this is clear enough.

IMG_20160606_121704.jpg


why do you multiply -e^-t by dt/dx when you have written that e^-t = dt/tx? bit confused
Original post by Patrick2810
why do you multiply -e^-t by dt/dx when you have written that e^-t = dt/tx? bit confused


You have to use the chain rule. dy/dx of something in this case is = dy/dt * dt/dx So dy/dx of e^-t is -e^-t * dt/dx.
Original post by kennz
arguments work like logs
rearrange to make z=(w/2)^(1/3)
therefore arg((w/2)^(1/3))= pi/6
so arg((w/2))/3 = pi/6
so arg(o.5w) = pi/2
so half line, angle pi/2


Couldn't you find y in terms of x (y/x = tan pi/6) and then write z = x+ root3/3 xi and then sub that into w = ...
Use complex conjugates and then isolate real and imaginary and bit of algebraic manipulation to get something in terms of u and v?

Also in your working, how did you go from the third to the 4th line?
(edited 7 years ago)
Original post by kennz
arguments work like logs
rearrange to make z=(w/2)^(1/3)
therefore arg((w/2)^(1/3))= pi/6
so arg((w/2))/3 = pi/6
so arg(o.5w) = pi/2
so half line, angle pi/2

Thank you :smile:
Original post by EricPiphany
Sorry if this is a little off topic, someone sent me this differential equation I couldn't make sense of:IMG-20160606-WA0003.jpg


No takers? Is it corrupt?
Original post by EricPiphany
No takers? Is it corrupt?


Is that even internally consistent? Surely you'd want the sum of the RHS's to be zero?

I suspect the last equation may have an erroneous minus, since if the water is draining into C you'd expect the volume to be rising, surely.


Posted from TSR Mobile
Reply 1028
Avatar for kennz
kennz
8
Original post by coolguy123456
Couldn't you find y in terms of x (y/x = tan pi/6) and then write z = x+ root3/3 xi and then sub that into w = ...
Use complex conjugates and then isolate real and imaginary and bit of algebraic manipulation to get something in terms of u and v?

Also in your working, how did you go from the third to the 4th line?


I wouldnt use your method personally, as its a lot longer and obscure.
I multiplied by 3 in the 3rd line
Original post by kennz
I wouldnt use your method personally, as its a lot longer and obscure.
I multiplied by 3 in the 3rd line


I meant the line where you have it to ^1/3 and then the next line you divide by 3?
Original post by Krollo
Is that even internally consistent? Surely you'd want the sum of the RHS's to be zero?

I suspect the last equation may have an erroneous minus, since if the water is draining into C you'd expect the volume to be rising, surely.


Posted from TSR Mobile


Ye, thought so myself. Possibly Vc=Va+Vb. first equation is easily solvable, (2)-(3) can then be solved. then you might say Va+Vb+Vc=100, to get them all, but I'm not sure if the last statement hold the whole time. Basically was messing up.
Reply 1031
Avatar for kennz
kennz
8
Original post by coolguy123456
I meant the line where you have it to ^1/3 and then the next line you divide by 3?


arguments work like logs
arg(w/z)=arg(w)-arg(z)
Original post by kennz
arguments work like logs
arg(w/z)=arg(w)-arg(z)


yeah i get that but you went from arg((0.5W)^1/3) to arg (0.5W/3) ??
Original post by coolguy123456
yeah i get that but you went from arg((0.5W)^1/3) to arg (0.5W/3) ??


What does DeMoivre tell you about the argument of a complex number raised to a power? Or, using the exponential definition of a complex number and raising it to a power and using basic index laws, what does that tell you? You should really be able to do this by now.
Original post by coolguy123456
yeah i get that but you went from arg((0.5W)^1/3) to arg (0.5W/3) ??


arg( (0.5W)^1/3 ) = arg( 0.5W ) / 3

by De Moivre's theorem, W = cos(theta) + isin(theta)
W^1/3 = ( cos(theta) + isin(theta) )^1/3 = cos(theta/3) + isin(theta/3)

from this, its obvious that if
arg(W) = theta
then
arg(W^1/3) = theta/3 = arg(W) / 3

works like logarithms (e.g. log(a^b) = blog(a); similarly arg(a^b) = b * arg(a) )
(edited 7 years ago)
Original post by Zacken
What does DeMoivre tell you about the argument of a complex number raised to a power? Or, using the exponential definition of a complex number and raising it to a power and using basic index laws, what does that tell you? You should really be able to do this by now.


I think I'm just missing something obvious, the FP2 book doesn't apply DeMoivre to complex number arguments in the form arg(..) etc (unless it does and i've missed it?)
Original post by coolguy123456
I think I'm just missing something obvious, the FP2 book doesn't apply DeMoivre to complex number arguments in the form arg(..) etc (unless it does and i've missed it?)


Surely you know that if

z = r(cos x + i sin x)

then

z^n = r^n cos (nx) + i sin (nx)

i.e: where the argument has become the power multiplied by the old argument.
Original post by kennz
arguments work like logs
rearrange to make z=(w/2)^(1/3)
therefore arg((w/2)^(1/3))= pi/6
so arg((w/2))/3 = pi/6
so arg(o.5w) = pi/2
so half line, angle pi/2

How did u go from second line to third line
Original post by Zacken
Surely you know that if

z = r(cos x + i sin x)

then

z^n = r^n cos (nx) + i sin (nx)

i.e: where the argument has become the power multiplied by the old argument.


Yeah that's all fine
Original post by coolguy123456
Yeah that's all fine


So what's the problem? The argument of a complex number raised to a power is the power multiplied by the argument.

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