• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Pure Mathematics 1.... AS students only

Messages
1,189
Reaction score
863
Points
123
Some confusions in Integration and vextors.. when you intergrate it, you increase the power and divide it by the increased power, sometimes we have to multiply the first number of the Nominator also,, aswell as in integration. Help?
 
Messages
946
Reaction score
1,144
Points
153
Both the questions are related to differentiation.In the 1st one we have been given dr/dt and he radius 50m. It is a circular patch and we have to find the rate at which the area of oil is increasing.Area of a circle is 3.142r^2.Differentiate it you will get 2pi*r.Substitute the value of r in 2pir. Than multiply it by 3 .
dA/dt=dA/dr *dr/dt
 
Messages
91
Reaction score
177
Points
28
Question7 O/N12 paper 11 ,part ii
Easy hy
For finding n first
Remember cos sq theota =1-sinsq theota
Apply it and make the equation
2sinsq ntheota + 3sin ntheota - 2 =0
Solve it quadratic
U get sinn theota =0.5. And =-2
Ignore -2 as it will give math error
Ok now when u solve for sin theotA=0.5 u get ntheota= (1st and 2n quad) 30,150
It says for the minimum POSITIVE value theota =10
Ntheota=30 , 10n=30 , n=3
This n will remain constant

Now the next part it says the largest value in 0-360
For sin 3theota find the values they will be
30,150,390,510,750,870 ( add 360 2 times as there will be three rotations as its 3 theota)
Now divide by 3 largest number in 360 range is 870/3=290
:)
No likes needed just pray we all get straight a*
 
Messages
91
Reaction score
177
Points
28
Some confusions in Integration and vextors.. when you intergrate it, you increase the power and divide it by the increased power, sometimes we have to multiply the first number of the Nominator also,, aswell as in integration. Help?
When u differetiate for example 2(3x+3)^3
U multiply the power and reduce one and multiply the derivative in the brackets thats whys confusing u
In this it will be 6(3x+3)^2 multiply by 3

For integration of same
2(3x+3)^3
Increase power by one , divide it and multiply the derivate by denomatior!!
It will go like this
2(3x+3)^4 /4 multiply 3

I dont need likes pray we all get straight a*
 
Messages
91
Reaction score
177
Points
28

First of all remember if A=bk ( k constant )
Then rate formula is
DA/DT = DB/DT multiply DB/DA
IN FIRST QUESTION FORMULA WILL BE A=pie r^2
Find DR/DA
2pie r =Da/dr
Now the formula for this is
Da/dt = dr/dt multiply dr/da
Radius increasing at 3 when initial is 50
Da/dt=3 x 2 pie (50)
=300 pie is answer


For second question apply the same with that formula find the value of k by putting values
M=kr^3
3.2=10^3k
K=o.oo32
If u dont get any part do let me know before the paper :p
 
Messages
20
Reaction score
0
Points
11
Can u help me with this ? 9709_w12_qp_12

9) iii) Why cant you expand (2x+3)^2 first and then integrate the expanded terms. ALso on mark scheme says lower limit of zero can't be igonored, but why ?

11) i) Why Angle CRO is a right angle triangle?

Thx!
 
Messages
946
Reaction score
1,144
Points
153
J
Easy hy
For finding n first
Remember cos sq theota =1-sinsq theota
Apply it and make the equation
2sinsq ntheota + 3sin ntheota - 2 =0
Solve it quadratic
U get sinn theota =0.5. And =-2
Ignore -2 as it will give math error
Ok now when u solve for sin theotA=0.5 u get ntheota= (1st and 2n quad) 30,150
It says for the minimum POSITIVE value theota =10
Ntheota=30 , 10n=30 , n=3
This n will remain constant

Now the next part it says the largest value in 0-360
For sin 3theota find the values they will be
30,150,390,510,750,870 ( add 360 2 times as there will be three rotations as its 3 theota)
Now divide by 3 largest number in 360 range is 870/3=290
:)
No likes needed just pray we all get straight a*
JazakAllah :)
 
Messages
23
Reaction score
9
Points
13
Hey its easy bro just remember the concept of sin graph
the equation y=asin(bx)+c in every sin graph will mean
a=amplitude , in this case 9-3 =6
b=no of complete wave in 2pie (360 degrees) , in this case 2
c=the mean point where wave is starting , in this case 3
so it becomes
6sin2x+3=y


for the second part put y=0
u will get sin2 inverse = -0.5
ignore - and take sin in the opposite quadrants (2,3 )
now for sin2 inverse 0.5 u get 30
2nd quad
180-30=150

3rd quad
180+30=210

so you have two values 150,210 WAIT , remember it was sin2theota ? , divide it by 2
u get 75 and 105 least is 75 , thats ur answer if u dont understand ask me ,


bro isnt the 4 quadrant thing like this S | A
T | C

With S being taking Sin as positive, A as all being positive, C being cos postive and T being tan positive. SO when you worked out the quadrant, why did you use S and T - 180 + x (s) and 180 - X (t) . Shouldnt it be 180 + x (T) and 360 - x (C) since you get a negative value??
 
Messages
21
Reaction score
1
Points
13
Both the questions are related to differentiation.In the 1st one we have been given dr/dt and he radius 50m. It is a circular patch and we have to find the rate at which the area of oil is increasing.Area of a circle is 3.142r^2.Differentiate it you will get 2pi*r.Substitute the value of r in 2pir. Than multiply it by 3 .
dA/dt=dA/dr *dr/dt
First of all remember if A=bk ( k constant )
Then rate formula is
DA/DT = DB/DT multiply DB/DA
IN FIRST QUESTION FORMULA WILL BE A=pie r^2
Find DR/DA
2pie r =Da/dr
Now the formula for this is
Da/dt = dr/dt multiply dr/da
Radius increasing at 3 when initial is 50
Da/dt=3 x 2 pie (50)
=300 pie is answer


For second question apply the same with that formula find the value of k by putting values
M=kr^3
3.2=10^3k
K=o.oo32
If u dont get any part do let me know before the paper :p

Thanks :)
 
Messages
91
Reaction score
177
Points
28
bro isnt the 4 quadrant thing like this S | A
T | C

With S being taking Sin as positive, A as all being positive, C being cos postive and T being tan positive. SO when you worked out the quadrant, why did you use S and T - 180 + x (s) and 180 - X (t) . Shouldnt it be 180 + x (T) and 360 - x (C) since you get a negative value??
100% right , i mixed it with cosine , my bad
Answer same hy ittefak se :D
 
Messages
1,189
Reaction score
863
Points
123
I am doing variant two paper, so just solving variant 12 papers is enough?
 
Top