Turn on thread page Beta
    Offline

    3
    ReputationRep:
    Please could anyone explain how to do question 2aii? Thanks Name:  image.jpg
Views: 199
Size:  500.9 KB
    Offline

    3
    ReputationRep:
    (Original post by economicss)
    Please could anyone explain how to do question 2aii? Thanks Name:  image.jpg
Views: 199
Size:  500.9 KB
    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
    Offline

    11
    ReputationRep:
    (Original post by Seytonic)
    I hope this is clear enough.

    Attachment 544391
    why do you multiply -e^-t by dt/dx when you have written that e^-t = dt/tx? bit confused
    Offline

    2
    ReputationRep:
    (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.
    Offline

    2
    ReputationRep:
    (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?
    Offline

    3
    ReputationRep:
    (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
    Offline

    15
    (Original post by EricPiphany)
    Sorry if this is a little off topic, someone sent me this differential equation I couldn't make sense of:Attachment 544761
    No takers? Is it corrupt?
    Online

    17
    ReputationRep:
    (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
    Offline

    3
    ReputationRep:
    (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
    Offline

    2
    ReputationRep:
    (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?
    Offline

    15
    (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.
    Offline

    3
    ReputationRep:
    (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)
    Offline

    2
    ReputationRep:
    (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) ??
    Offline

    22
    ReputationRep:
    (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.
    Offline

    2
    ReputationRep:
    (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) )
    Offline

    2
    ReputationRep:
    (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?)
    Offline

    22
    ReputationRep:
    (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.
    Offline

    2
    ReputationRep:
    (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
    Offline

    2
    ReputationRep:
    (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
    Offline

    22
    ReputationRep:
    (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.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 24, 2017

5,265

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.