Mathematics: Post your doubts here!

[Iffat]

Since u got k = 2 , the eq:
y = 2x^2 -3x +5
y = 5x-3

By equating (simultaneous eq):
2x^2 -3x +5 = 5x-3
2x^2 -3x -5x +5 +3 =0
2x^2 - 8x + 8 =0
2 (x^2 - 4x + 4) =0
2(x-2)(x-2) = 0
Hence, x= 2

When x=2 , y = 5(2) - 3 = 7
(2 , 7)

thanx

 

[georges]

can anyone help with the friction in mechanics m1?
it has been a pain in the ass
so plz if there are any revision sheets provide me with them

 

[whitecorp]

can anyone help with the friction in mechanics m1?
it has been a pain in the ass
so plz if there are any revision sheets provide me with them

Just remember the basic idea of frictional force as one opposing the motion of an object.

However, if you are looking at a stationary object, make sure you consider the static coefficient of friction. Do also note
the direction in which the object has a tendency to move so as to determine how the frictional force would act.

I have solved some problems on my site related to mechanics, you may wish to take a look:

http://www.whitegroupmaths.com/2011/10/problem-2-mechanics.html

http://www.whitegroupmaths.com/2011/10/problem-6-double-incline-aka-wedge.html

Hope these help. Peace.

 

[rafayet]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_12.pdf


I need help with Q.9(i). i got coor of M, but how can i get coor of D? can you please help me?

 

[anonymous123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_12.pdf


I need help with Q.9(i). i got coor of M, but how can i get coor of D? can you please help me?

(Coordinates of B + Coordinates of D)/2 = Coordinates of M

 

[rafayet]

(Coordinates of B + Coordinates of D)/2 = Coordinates of M

thanks

 

[Ammirul Shafiq]

The complex number 2/-1+i is denoted by u.

1) Find the modulus and argument of u and u^2.

Can someone help me to solve this question?

 

[iKhaled]

The complex number 2/-1+i is denoted by u.

1) Find the modulus and argument of u and u^2.

Can someone help me to solve this question?

ok, when you have a question in that form u need to change it to the form x+ yi and to do this u multiply the denominator by the conjugate (1-i) and u multiply the nominator by the conjugate so..

2/1+i = 2(1-i)/(1+i)(1-i) = 2-2i/(1^2 + 1^2) = 1-i

the mod of |1-i| = √(1^2 + 1^2) = √2
arg (z) = tan^-1 (-1/1)
u^2 = (1-i)(1-i) = -2i

that's all, hope u got it bro!

 

[IbtiCool]

Hey. I need some help in Finding the area bounded by a curve and the y-axis and Volume of revolution about the y-axis. I study using examsolutions.net and this topic isn't there yet. I have the CIE A-Level Mathematics Pure 1 book with a black cover and some sea shell thingy, and I cant find this topic there either. Any help would be appreciated.

 

[anonymous123]

Hey. I need some help in Finding the area bounded by a curve and the y-axis and Volume of revolution about the y-axis. I study using examsolutions.net and this topic isn't there yet. I have the CIE A-Level Mathematics Pure 1 book with a black cover and some sea shell thingy, and I cant find this topic there either. Any help would be appreciated.

i assume u have no problem with rev about x-axis..also, I can only give a brief explanation here but i hope u get it:

volume around x-axis = pi int(y^2) dx, and then put the limits
if the question says find volume about y-axis, simply make 'x' the subject and then V= pi int(x^2) dx
Now, if you are given the limits for y, simply put the values and get the result..If you are only given the values of x, then put them in the equation and get the limits for y.

e.g
y=x^2 ; Find the volume of rev when the region between x=1 and x=5 and the curve 'y' is rotated about y-axis.

> make x subject: x= √y
> we are only given x-coordinates but to revolve around y-axis we nede y-coordinates: when x=1, y=1..when x= 5, y=25.....so our limits are 1 and 25
> square and integrate: V= pi int(√y)^2 dy =>V=pi int dy ==> V=pi (y^2 /2) with limits 1 , 25.
the rest is simple calculation...hope i helped

 

[whitecorp]

i assume u have no problem with rev about x-axis..also, I can only give a brief explanation here but i hope u get it:

volume around x-axis = pi int(y^2) dx, and then put the limits
if the question says find volume about y-axis, simply make 'x' the subject and then V= pi int(x^2) dx
Now, if you are given the limits for y, simply put the values and get the result..If you are only given the values of x, then put them in the equation and get the limits for y.

e.g
y=x^2 ; Find the volume of rev when the region between x=1 and x=5 and the curve 'y' is rotated about y-axis.

> make x subject: x= √y
> we are only given x-coordinates but to revolve around y-axis we nede y-coordinates: when x=1, y=1..when x= 5, y=25.....so our limits are 1 and 25
> square and integrate: V= pi int(√y)^2 dy =>V=pi int dy ==> V=pi (y^2 /2) with limits 1 , 25.
the rest is simple calculation...hope i helped

I have written extensive material regarding volume of revolution on my website here:

http://www.whitegroupmaths.com/2010/02/understanding-matters-5.html

Hope it helps. Peace.

 

[anonymous123]

I have written extensive material regarding volume of revolution on my website here:

http://www.whitegroupmaths.com/2010/02/understanding-matters-5.html

Hope it helps. Peace.

you quoted the wrong person but i hope he sees it

 

[whitecorp]

you quoted the wrong person but i hope he sees it

Thanks for pointing it out. Peace.

 

[7devilsallaround]

Can someone help me out with this integration sum?

Use a substitution of the form ax+b=u to find the integral for x/(2x+3)

The answer's x/2 - 3 ln |2x+3| +k
But I got x/2 +3/4 - 3 ln |2x+3| +k

I'd very much appreciate if you showed all the steps please.

 

[anonymous123]

Can someone help me out with this integration sum?

Use a substitution of the form ax+b=u to find the integral for x/(2x+3)

The answer's x/2 - 3 ln |2x+3| +k
But I got x/2 +3/4 - 3 ln |2x+3| +k

I'd very much appreciate if you showed all the steps please.

taking sub u=2x+3:
du/dx = 2
dx=du/2

x=(u-3)/2

insert values in original eq:
=> ⌡{u-3)/2}/u du/2
take 1/4 outside: 1/4 ⌡(u-3)/u du
simplify: 1/4 ⌡1 - (3/u) du
integrate: 1/4 [u - 3lnu] + k
replace u by x: 1/4[(2x+3) - 3ln|2x+3|] +k
simplify: x/2 + 3/4 -3/4 ln|2x+3| + k

mine includes one more term too..are u sure that answer is correct??

 

[IbtiCool]

i assume u have no problem with rev about x-axis..also, I can only give a brief explanation here but i hope u get it:

volume around x-axis = pi int(y^2) dx, and then put the limits
if the question says find volume about y-axis, simply make 'x' the subject and then V= pi int(x^2) dx
Now, if you are given the limits for y, simply put the values and get the result..If you are only given the values of x, then put them in the equation and get the limits for y.

e.g
y=x^2 ; Find the volume of rev when the region between x=1 and x=5 and the curve 'y' is rotated about y-axis.

> make x subject: x= √y
> we are only given x-coordinates but to revolve around y-axis we nede y-coordinates: when x=1, y=1..when x= 5, y=25.....so our limits are 1 and 25
> square and integrate: V= pi int(√y)^2 dy =>V=pi int dy ==> V=pi (y^2 /2) with limits 1 , 25.
the rest is simple calculation...hope i helped

Thank you very much. Can you please tell me about the AREA bounded by a curve and the y-axis too?

 

[anonymous123]

Thank you very much. Can you please tell me about the AREA bounded by a curve and the y-axis too?

Its the same thing except that you don't have to square and don't need to put pi....but there are certain cases where u have to be careful e.g the area bounded by the sine curve and x-axis..u have to calculate the negative and positive part separately..u will have to do the same for y-axis

 

[IbtiCool]

i assume u have no problem with rev about x-axis..also, I can only give a brief explanation here but i hope u get it:

volume around x-axis = pi int(y^2) dx, and then put the limits
if the question says find volume about y-axis, simply make 'x' the subject and then V= pi int(x^2) dx
Now, if you are given the limits for y, simply put the values and get the result..If you are only given the values of x, then put them in the equation and get the limits for y.

e.g
y=x^2 ; Find the volume of rev when the region between x=1 and x=5 and the curve 'y' is rotated about y-axis.

> make x subject: x= √y
> we are only given x-coordinates but to revolve around y-axis we nede y-coordinates: when x=1, y=1..when x= 5, y=25.....so our limits are 1 and 25
> square and integrate: V= pi int(√y)^2 dy =>V=pi int dy ==> V=pi (y^2 /2) with limits 1 , 25.
the rest is simple calculation...hope i helped

What's funny is that ( y ) turned into

 

[IbtiCool]

Its the same thing except that you don't have to square and don't need to put pi....but there are certain cases where u have to be careful e.g the area bounded by the sine curve and x-axis..u have to calculate the negative and positive part separately..u will have to do the same for y-axis

Can you do an example too?

 

[anonymous123]

What's funny is that ( y ) turned into

maths is fun, no?

Can you do an example too?

you want a simple one or that special case?? coz i think u know how to work out for curves like y=x^2