Mathematics: Post your doubts here!

[sma786]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 7 part i !

 

[falcon678]

You didn't tell me if you understood my explanation for your Mechanics question

dude i liked tht post? meaning i got it?

 

[falcon678]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 7 part i !

it simple....using the equation...find the sum for two diff terms.....use tht term n its sum in the original sum of terms eqation...u will get two equations...solve them simultaneously!
hope u got it?

 

[daredevil]

when scaler product less then 1 we sayy acute if greater then 1 then obtuse
other is simple integration

ok thanks.. igot the integration one by myself already but thanks all the samee
and yeah our teacher never told us this ..... thanks

 

[syed1995]

help please! I am always confused with rates questions ! if you can please tell me the method of doing these questions

dA/dt = DA/dr * dr/dt

where t is time
r is radius and
A is area ..
dA/dt is change in Area and
dr/dt is change in radius .. and
DA/dr is the diffrentiation of Area with respect to radius..

A= πr^2
DA/dr = 2πr
DA/dr at start = 2π*50 = 100π


DA/dt = DA/dr * dr/dt
DA/dt = 100π*3
DA/dt = 300π

The Area is increasing at a rate of 300π per second

 

[sma786]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 8, ii and iii , lol wtf is this?

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_12.pdf
Question 8, ii and iii , lol wtf is this?

This is CIE making life difficult for us...

To get the vectors we will find the unit vector and multiply by the magnitude..

OA = Unit Vector .. 3i − 4k/√25

15/5(3i-4k)
=3(3i-4k)
=9i-12k

Unit Vector of OB = 2i + 3j − 6k / √(2^2)+3^2+(-6^2)
Unit Vector of OB = 2i+3j-6k/7

OB = 14/7(2i+3j-6k)
OB = 2(2i+3j-6k)
OB = 4i+6j-12k

 

[shafayat]

shafayat , iKhaled i found a much easier way, thanks

:O what much easier ? tell me also ..

 

[shafayat]

dA/dt = DA/dr * dr/dt

where t is time
r is radius and
A is area ..
dA/dt is change in Area and
dr/dt is change in radius .. and
DA/dr is the diffrentiation of Area with respect to radius..

A= πr^2
DA/dr = 2πr
DA/dr at start = 2π*50 = 100π


DA/dt = DA/dr * dr/dt
DA/dt = 100π*3
DA/dt = 300π

The Area is increasing at a rate of 300π per second

thanku , this helped me a lot !...

 

[MustafaMotani]

syed1995 long time no see..

 

[falcon678]

guys question 8 part (i) n (ii) please!

 

[MustafaMotani]

falcon678 u just have to double derivate

d2y/dx2 = 2 x 3/2 (3x+4)^(3/2-1) x 3 - 6

just insert x = -1 in that dy/dx and it should be zero if stationary point lies at x = -1
2(3x-1 + )^1.5 - 6x-1 +8 = 0 (shown)

for nature just insert x = -1 in d2y/dx2 and if its less than zero that point is max and if greater than zero than min

if yu facing prob in derivating 2(3x + 4)^3/2
then remember power gets multiplied to the coefficient, then 1 is subtracted from the power and the function is multiplied by the derivative of whatver is inside the bracket..

i know its difficult .. reread it again and agin and see what i have done....

Hope it helps

 

[falcon678]

falcon678 u just have to double derivate

d2y/dx2 = 2 x 3/2 (3x+4)^(3/2-1) x 3 - 6

just insert x = -1 in that dy/dx and it should be zero if stationary point lies at x = -1
2(3x-1 + )^1.5 - 6x-1 +8 = 0 (shown)

for nature just insert x = -1 in d2y/dx2 and if its less than zero that point is max and if greater than zero than min

if yu facing prob in derivating 2(3x + 4)^3/2
then remember power gets multiplied to the coefficient, then 1 is subtracted from the power and the function is multiplied by the derivative of whatver is inside the bracket..

i know its difficult .. reread it again and agin and see what i have done....

Hope it helps

it waz tht easy? wat is wrong with me??

thanks alot

 

[falcon678]

ummm guys..... question 9 (i) n (ii)

 

[Sarah22]

it was a tough question.c by putting the value of k=10 u will get anohter point q amd frm that point find the gradient
and u need to find the gradient of the curve also which is dy/dx
dy/dx=12/x
dy/dx=-12/x2
by putting x =2 u will get 3
then to find the angle difference tan theta2m-tan(theta1m)
so for the 1st by calculating u will get m=-2
so tan(3)-tan(2) and thats the answer

Thank you.. I understood the gradient part, but I dont quite get the angle difference part.. Only and only if u dont mind, can u explain that part

 

[MustafaMotani]

falcon678 WelcOme..

if OAB is a straight line then the angle between them has to be 180.. just dot the both vectors and equate it the product of their magnitudes and cos180

ii) is easier if they are perpendicular then angle between them is 90... just dot them again and equate it to zero..

If yu didnt get it just tell me.... i like explained it quite quickly..

 

[MustafaMotani]

Sarah22 whats the question..?

 

[falcon678]

falcon678 WelcOme..

if OAB is a straight line then the angle between them has to be 180.. just dot the both vectors and equate it the product of their magnitudes and cos180

ii) is easier if they are perpendicular then angle between them is 90... just dot them again and equate it to zero..

If yu didnt get it just tell me.... i like explained it quite quickly..

dude i tried tht...it gets realy tricky n more then the allotted worth of 3 marks....questions regarding scalar product are around 5-6 marks.....n if thtz the only way....can u please solve? thanks!

 

[shafayat]

Thank you.. I understood the gradient part, but I dont quite get the angle difference part.. Only and only if u dont mind, can u explain that part

tan inverse of gradient .. is the angle the line makes with x-axis , so when ever u need to find the angle of two lines intersecting ! just find their gradients and do tan^-1(gradient)
the one with larger angle minus the smaller angle so u get the angle ... this angle can be acute or obtuse ..
some time cie makes it thougher for us .. wen we get acute angle ! they might ask for an obtuse angle, so at that time you need to 180-(what ever angle u got) to get the obtuse angle

 

[Rutzaba]

see..

dy/dx = tangent
tangent = y/x which is same as tan theta = y/x

so tan theta = dy/dx....find the gradient at point p and q by substituting the x values of p and q in the dy/dx u got (if u don't know how to differentiate tell me)

tan theta = 1.5 at point p and tan theta = 0.75 at point q
theta = 56.3 and 36.87
find the difference between the 2 56.3 - 36.87 = 19.4 and thats out answer!

i hope you got it

this^


or this

well we did it in one go if u see the previous post but since u dint get it ... ah well
we have two lines thus two gradients ryt?
so put in formula tan x = gradieant of line 1
tan y= gradient of line two
y-x will give you the answer kapish?

SalmanPakRocks