Mathematics: Post your doubts here!

[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

 

[syed1995]

syed1995 long time no see..

And the Maths legend is here How are ya man?? What subjects you got this year? You planning on being active again on XPF ?

If you've got time I have some maths questions for ya =)

 

[MustafaMotani]

shafayat pardon me bro..but i think yu are saying little wrong... tan of angle gives yu gradient... therefre angle = tan inverse of gradient

 

[shafayat]

shafayat pardon me bro..but i think yu are saying little wrong... tan of angle gives yu gradient... therefre angle = tan inverse of gradient

yeah, That's right ! I am wrong ! tan inverse .. thanks for rectifying my mistake bro

 

[MustafaMotani]

syed1995 yeah i think so... i just remeberd lst year when we were on xpf for more than 12 hrs and so came again..
my subs are accel maths, physix, chem and eco

 

[gary221]

guys... help pls!!!
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_3.pdf
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_ms_3.pdf

ques 2
thnx!!

 

[MustafaMotani]

falcon678 i) can be little hectic wait i might try.. ii) will be very easy because cos 90=0 and therefore whole product will be zero

 

[daredevil]

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

this is just how i might work through the question so just look at it and then its ur call totally.,..
well here goes:

i) if OAB is a straight line take them lines to be parellel

[ if vector a is parellel to vector b then
a=kb where k is a constant]

find out that constant here
as the j values are both given we use them and from there we can determine that OB = 2(OA)
so OA = 2i + j + k
so p=2
and unit vector will be OA/|OA|
|OA| = (4)^1/2
= 2

so unit vector in the direction of OA = i + (1/2)j + (1/2)k

i hope u got it....
will post the next part in the nxt reply because this is already looking at me like me with big round glasses a.k.a. NERD!! B)

 

[syed1995]

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!

MustafaMotani

AH YEAH.. I remember that question! It won't be solved with the dot product method.. try it .. it will blow your brains out and you won't be able to solve it.

One of our Add maths concepts is what is going to be used here

If Two Vector A and B are collinear (on the same line).. then Vector B = k(Vector A)

OA = p,1,1 OB = 4,2,p

Comparing the j coefficents ..

OB = k(OA)
2 = k(1)
k=2

now compare either of the first or last to get p... Like comparing the first one we get

OB = k(OA)
OB = k(OA)
4= 2(p)
p=2 Answer

 

[MustafaMotani]

syed1995 daredevil