• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

knowitall10

knowitall10

Messages
2,865
Reaction score
13,723
Points
523
Alright.. i don't know if it's right though.. like i said, i'm taking Edexcel, not Cambridge:)
I'm sorry i'm late, the internet stopped working :/
 

Attachments

  • DSC05191.JPG
    DSC05191.JPG
    2 MB · Views: 7
knowitall10

knowitall10

Messages
2,865
Reaction score
13,723
Points
523
oops.. wait, i just checked the marking scheme.. i think it's wrong. then i don't know how to do it, i'm sorry:(
 
iKhaled

iKhaled

Messages
870
Reaction score
374
Points
73
knowitall10 said:
btw, when ur done, please explain it to me too.. its bothering me now, i can't study anything else
Click to expand...
Farru here u go!

dy/dx = (y^2 +4)/xy
dx/dy = xy/(y^2 +4)
1/x dx/dy = y/(y^2+4)
1/x dx = y/(y^2+4) dy
∫1/x dx = ∫y/(y^2+4) dy

i think u know how to integeate..if not tell me and i will teach u too but here is the rest
ln x = 1/2ln(y^2+4) + c
sub the values of x and y and u will get c = -1/2ln4

ln x = 1/2ln(y^2+4) -1/2ln4
ln x + 1/2ln 4 = 1/2 ln(y^2+4)
2ln x + ln 4 = ln (y^2+4)
ln (4x^2) = ln (y^2+4)
4x^2 = y^2 + 4
y^2 = 4x^2-4

thats it!!
 
knowitall10

knowitall10

Messages
2,865
Reaction score
13,723
Points
523
iKhaled said:
Farru here u go!

dy/dx = (y^2 +4)/xy
dx/dy = xy/(y^2 +4)
1/x dx/dy = y/(y^2+4)
1/x dx = y/(y^2+4) dy
∫1/x dx = ∫y/(y^2+4) dy

i think u know how to integeate..if not tell me and i will teach u too but here is the rest
ln x = 1/2ln(y^2+4) + c
sub the values of x and y and u will get c = -1/2ln4

ln x = 1/2ln(y^2+4) -1/2ln4
ln x + 1/2ln 4 = 1/2 ln(y^2+4)
2ln x + ln 4 = ln (y^2+4)
ln (4x^2) = ln (y^2+4)
4x^2 = y^2 + 4
y^2 = 4x^2-4

thats it!!
Click to expand...
but iKhaled what i learnt was to bring all the x values on one side all the y values on one side...u guys learnt a different method or is it a totally different topic?
 
knowitall10

knowitall10

Messages
2,865
Reaction score
13,723
Points
523
Oh wait, i take that reply back.. iget what you're doing:)
Thanks a lot!! Jazak Allahu Khair brother!!!:)
and Farru i'm really sorry for being useless...:notworthy:
 
iKhaled

iKhaled

Messages
870
Reaction score
374
Points
73
knowitall10 said:
but iKhaled what i learnt was to bring all the x values on one side all the y values on one side...u guys learnt a different method or is it a totally different topic?
Click to expand...
nope thats the method..u put x on one side and y on the other side thats the separate integrals and thats what i did see..

1/x dx = y/(y^2+4) dy

u still dont get it ?
 
A

A star

Messages
2,703
Reaction score
3,939
Points
273
iKhaled said:
Farru here u go!

dy/dx = (y^2 +4)/xy
dx/dy = xy/(y^2 +4)
1/x dx/dy = y/(y^2+4)
1/x dx = y/(y^2+4) dy
∫1/x dx = ∫y/(y^2+4) dy

i think u know how to integeate..if not tell me and i will teach u too but here is the rest
ln x = 1/2ln(y^2+4) + c
sub the values of x and y and u will get c = -1/2ln4

ln x = 1/2ln(y^2+4) -1/2ln4
ln x + 1/2ln 4 = 1/2 ln(y^2+4)
2ln x + ln 4 = ln (y^2+4)
ln (4x^2) = ln (y^2+4)
4x^2 = y^2 + 4
y^2 = 4x^2-4

thats it!!
Click to expand...
looked like french for a non p3 student :p
 
Farru

Farru

Messages
37
Reaction score
9
Points
18
iKhaled said:
Farru here u go!

dy/dx = (y^2 +4)/xy
dx/dy = xy/(y^2 +4)
1/x dx/dy = y/(y^2+4)
1/x dx = y/(y^2+4) dy
∫1/x dx = ∫y/(y^2+4) dy

i think u know how to integeate..if not tell me and i will teach u too but here is the rest
ln x = 1/2ln(y^2+4) + c
sub the values of x and y and u will get c = -1/2ln4

ln x = 1/2ln(y^2+4) -1/2ln4
ln x + 1/2ln 4 = 1/2 ln(y^2+4)
2ln x + ln 4 = ln (y^2+4)
ln (4x^2) = ln (y^2+4)
4x^2 = y^2 + 4
y^2 = 4x^2-4

thats it!!
Click to expand...

Thank you so much..
I know the integration part, I was stuck in making ysq the subject.. :/
and Jazak Allah.. :) :) :)
 
Esme

Esme

Messages
681
Reaction score
438
Points
73
Pie-man said:
bother? you aren't bothering me :)

so first thing is writing the forces so that you wouldn't get confused by them, as the 5.5 N and the W N are pulling the particle upwards.
then draw a line in the middle of the right angle to make 2 letter Z (one from AP1X to the line and the other from CP2X to the line) and label the angles AP1X with a symbol and CP2X with another symbol, so now you have the angle AP1X equal to the left side of the line you drew and CP2X is the same as the right side of the line.

so lami says that 7.3/sin(90) = 5.5/sin(AP1X) = W/sin(CP2X)

so using cross-multiplication (5.5 x sin(90))/7.3 = 48.9 (angle CP2X)

angle AP1X = 90 - 48.9 = 41.1

so angle AP1X= 41.1


then to get W you do the same but you use the angle 41.1 so
(sin(41.1) x 7.3)/sin(90) = 4.8

so W= 4.8

did you get it now? if you want me to scan the paper I solved it on i don't mind ;)
Click to expand...

But I thought we use Lami's rule when we have coplanar forces.
And how can you find angle CP2X first since we have 7.3/sin90 = 5.5/sinAP1X = W/sinCP2X
you did sinCP2X= (5.5xsin90)/7.3 right ? But this should be for angle AP1X. because 5.5 is its opposite force.

I'd be grateful if you showed me your workings possibly with a diagram. Thanks a lot :)
 
You must log in or register to reply here.
Top