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Mathematics: Post your doubts here!

iKhaled

iKhaled

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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

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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

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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

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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 :)
 
5

50sum

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Pie-man said:
the link from the rule "Final Energy = Initial Energy + Driving force - Resistance force" ? or the one "Driving force - Resisting force = mass x acceleration"?

so you get the final and initial energy using "K.E. [1/2 mass x velocity(squared) ] + P.E. (mass x gravity [10] x height)" and substituting the mass, height and velocity as given at each point.

if given a driving force or resistant force you put them in the equation "F.E. = I.E. + DF - RF"
Click to expand...



What if it's just a particle(with no driving force) going upwards or downwards a rough slope? Can you explain the conditions for both motions. For example sometimes while going down a rough slope the speed increases, but in some cases it also decrease and comes to rest, like in w_12 qp 41, question 6
 
Pie-man

Pie-man

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Esme said:
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 :)
Click to expand...
all i know is if you have 3 different forces on an object you can use lami
i'm not exactly sure why i got CP2X it just made more sense to me and after getting all the other angles i was sure it was right but it will make more sense if you turn it into a triangle (well for most people i know)
 

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Pie-man

Pie-man

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50sum said:
What if it's just a particle(with no driving force) going upwards or downwards a rough slope? Can you explain the conditions for both motions. For example sometimes while going down a rough slope the speed increases, but in some cases it also decrease and comes to rest, like in w_12 qp 41, question 6
Click to expand...
well in 6 i) you use the F.E. = I.E. as there is no driving force (he didn't mention one) nor resisting force (as it is a smooth plane)

in ii) i'm thinking of getting the I.E. of C (which get you the work done against friction) and F.E. at the mid point where the V is will be the only unknown

if like you say the speed increases while going down a rough slope then there is a driving force that is greater than the resisting force unlike the question above if there is a driving force then he shall give you its value, or the value of the resistance and ask you to obtain the driving force.
 
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