Mathematics: Post your doubts here!

[iKhaled]

got it

oops sorry i guess i was too late to answer ur question..u already got it -_-

 

[Alice123]

oops sorry i guess i was too late to answer ur question..u already got it -_-

da/dt = KV
da/dt = k(4/3πr^3)
da/dt = 4/3πkr^3

A=4πr^2 V = 4/3πr^3

da/dr = 8πr dv/dr = 4πr^3

dr/dt = dr/da x da/dt
dr/dt = (1/8πr) x 4/3πkr^3 substitute r and dr/dt to find K then u will end up with the given equation

did u get it ?

u givin p3 this may?

 

[iKhaled]

u givin p3 this may?

inshAllah yes i will.

 

[Alice123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
N0 10i Dug PhyZac or any1 having p3 please help

cant do 10iii iKhaled

 

[ahmed abdulla]

find f inverse(x) and put x^2-2 in place of x

well .. i did that ... and mark scheme shows a different answerr!

 

[iKhaled]

cant do 10iii iKhaled

we have r = 5/ 1-0.4t

find t when the denomerator is equal to 0 so 1-0.4t = 0 t = 2.5 now any value higher than 2.5 will give u a negative value and we cant have radius in negative so the maximum value of t is just below 2.5 and the lowest one is t = 0 so 0 < t < 2.5​

 

[Shreeram]

hi can anyone give me a detailed explanation for q3 may june 2012 p13 thanks
Alice123 or abruzzi

 

[Alice123]

well .. i did that ... and mark scheme shows a different answerr!

is it (x^2-3)/2????

 

[abruzzi]

sin 3x + 2cos 3x = 0

divid the equation by cos x and u will get tan 3x +2 = 0
tan 3x = -2
3x = -64.3 and 116.5
x= 158.9 and 38.9 but i have no idea how the mark scheme got 98.9 as one of x values sorry :/

Thanks I tried using the same method in a different way. This is what I did.. hope it'll help you:

tan 3x = -2
Let 3x = x,
tan x = -2
(Ignoring negative); x = tan-1 (2)
x = 63.435

2nd quadrant:
180 - 63.435 = 116.565
3x = 116.565
x = 38.8

4th quadrant:
360 - 63.435 = 296.565
3x = 296.565
x = 98.8

2nd revolution:
360 + 116.565 = 476.565
3x = 476.565
x = 158.8
Angles => 38.8, 98.8, 158.8

 

[salvatore]

Then we add 3, since -2tan(1/2x) + 3 , so when y is 0 becomes y 3
View attachment 24379


And that is the final shape!

I was going through this again and failed to understand something.. does it matter where the curve cuts the x-axis in the final shape? Like in your case, it cuts slightly after π/2..

 

[PANDA-]

I was going through this again and failed to understand something.. does it matter where the curve cuts the x-axis in the final shape? Like in your case, it cuts slightly after π/2..

According to the ms, no. But it would make your graph more accurate I guess.... and it's very simple, just replace y=0 and solve the eqn.

 

[Rutzaba]

Aslamoalikum
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_12.pdf
Q9 (ii) adn Q 11 (i) and (iii)

this is so very easy.. al you have to kp in mind that the gradient of tangent is also gradient og curve at aprticular point they meet. curve =9/2x+3
differentiate you will get -9/ (2x+3)^2
since its tangent at A x=3
-9 /( 81)
grad = -1/9
now u make an equation with coords 3,1 and grad = -1/9
then u get eq of line 9y= -x+12
notice that C lies on this line... hence to take out coords of C put x=0 as it lies on the y axis
u will get points (0, 4/3)
now take distance of C (0, 4/3) and O ( 0, 0)
and distance C(0, 4/3) and B (0, 3)
the lesser the distance the closer the points are...
the formula of distance is root of (((x1-x2)^2 +(y1- y2)^2))

 

[Rutzaba]

hey guys check 11 part i
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_12.pdf

 

[PhyZac]

I was going through this again and failed to understand something.. does it matter where the curve cuts the x-axis in the final shape? Like in your case, it cuts slightly after π/2..

Well, I usually pay attention to x-axis, and as PANDA- said, easily by replacing y=0, the thing is, i forgot to do that when I replied to you.
To be safe side, do consider the x-axis.

 

[A star]

hey guys check 11 part i
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_12.pdf

easy
we can see that OQ= 20 hence Oc= 20-x
sin(0.6)=x/OC
equate both and get the answer

 

[PhyZac]

hey guys check 11 part i
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_12.pdf

You made tensed seeing you post.

 

[PANDA-]

hey guys check 11 part i
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_12.pdf

OR is a tangent to the circle inside at point R... So angle ORC = 90. That's something, lol still thinking

 

[Rutzaba]

ahahahah i was tensed k i cant solve such an easy questions... actually i have phy practicl on my mind... not my best subject :'(

 

[Rutzaba]

OR is a tangent to the circle inside at point R... So angle ORC = 90. That's something, lol still thinking

i tried that frst xD

 

[A star]

ahahahah i was tensed k i cant solve such an easy questions... actually i have phy practicl on my mind... not my best subject :'(

well i have both practicals and computing on my mind