Mathematics: Post your doubts here!

[ashcull14]

Thank you for your explantions. May I know why WD BY DRIVING FORCE= WD by Kinetic energy + WD BY FRICTION + Gain in gpe? I can't imagine it.

according to law of conservation of energy is tht energy is converted from one form to another WD is energy
so when an object travels up an inclined plane the total WD by driving force is utilised in various forms
while travelling up the plane some WD will be done in gaining height u see an inclined plane always has some height,...so it will be equal to WD by GPE
some energy will be utilized for motion so WD is done by Kinetic energy.....
if there are resistive forces present like friction ...WD will be done against Friction (WD = force*distance) nd friction is a force
This formula is the base of energy conversions for inclined plane ques in our syllabus
it is altered according to the situation given in the question...
there can b conditions where either one of the factors will be eliminated for eg
on smooth surfaces no WD against Fr will be done
or if speed is constant no change in speed will occur thus Ke will be 0J etc
Hope u got it now

 

[sara kamal]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s12_qp_41.pdf
please explain q4 part 2,i dont get why limit is taken as 0 to 12 why not 0 to 8 as found in first part..???

 

[qwertypoiu]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s12_qp_41.pdf
please explain q4 part 2,i dont get why limit is taken as 0 to 12 why not 0 to 8 as found in first part..???

Note that reversing direction means the VELOCITY changes sign from positive to negative, NOT acceleration! I've plotted a graph, where you can see the velocity becomes negative when t=12, not t=8 which is simply a local maxima.

 

[TheJDOG]

http://onlineexamhelp.com/wp-content/uploads/2014/02/9709_w13_qp_31.pdf
pls help Q2, Q4.. i cant get it

Hi, for Q2, let 3^x= u
You now have 2|u-1|=u
Now |2u-2|=u
Square both sides, (2u-2)^2=u^2
4u^2 -2(2u)(2)+ 2^2= u^2
3u^2 -8u +4=0
Solve on calculator
u= 2 u= 2/3
But, 3^x = u
So 3^x=2 and 3^x=2/3
So xln3=ln2 and xln3= ln(2/3)
x= 0.631 and x= -0.369

Q4
x=e^-t.cost and y= e^-t.sint
Differentiate each one with respect to t, and then divide dy/dt by dx/dt to get dy/dx
Use product rule,
Dx/dt= (e^-t)(sint) - (cost)(e^-t)
Dy/dt= (e^-t)(cost) - (sint)(e^-t)
Now dy/dx = (sint - cost)/(sint + cost) cancel all (e^-t)
(tant - 1)/(tant + 1) divide by cost
But expand tan(t - 1/4pie)= (tant- tan1/4pie)/(1 + tant.tan1/4pie)
= (tant - 1)/(1 + tant)
And so they're both equal
If I have any mistake tell me

 

[TheJDOG]

Can someone please help me with this question?

Here you go if there's any more doubt tell me

Justification : Drop down a perpendicular to get AB or AC in terms of r and use simple trigonometry , then cross out r^2 and simplify , change (Costheta)^2 and 2sintheta.costheta into their identities . Then make cos2theta the subject

 

[Jewel504]

Hey guys can someone pls tell me the formula for the perimeter of a segment of a circle

 

[DeadlYxDemon]

Variant 42 - O/N - 2014
Q1)

I understood that when the Particle P reaches it's highest point it meets with partical Q and then they both come down together and hit at the saame time..

 

[DeadlYxDemon]

Variant 42 - O/N - 2014
Q4 - ii) Need explanation!

 

[papajohn]

http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_33.pdf
Q9 part 1 please anyone. I can`t getting the ans..
MS ---> http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_ms_33.pdf

 

[Ram97]

Hey guys can someone pls tell me the formula for the perimeter of a segment of a circle

Length of chord + Arc length

 

[Bhaijan]

9709_w07_qp4 (M1)
what would be the solution to part(ii) of this question?
pls help

 

[Dr Death]

How to do this?
We get x and y on two sides.

 

[sara kamal]

Note that reversing direction means the VELOCITY changes sign from positive to negative, NOT acceleration! I've plotted a graph, where you can see the velocity becomes negative when t=12, not t=8 which is simply a local maxima. View attachment 52494

thanks alot!!!

please can u tell me that in the question 6 part 2 of following paper why the direction of line is taken as 'i' even tough its parallel to x axis,why not 7j??
http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_31.pdf

thanks in advance.

 

[qwertypoiu]

thanks alot!!!

please can u tell me that in the question 6 part 2 of following paper why the direction of line is taken as 'i' even tough its parallel to x axis,why not 7j??
http://maxpapers.com/wp-content/uploads/2012/11/9709_s13_qp_31.pdf

thanks in advance.

I haven't taken A2, but I can tell you what i know:
Unit vector to the right side, parallel to x-axis is i.
And unit vector going UP, parallel to y-axis is j.
Hope that makes sense.

 

[mak kings]

http://maxpapers.com/wp-content/uploads/2012/11/9709_w13_qp_3.pdf pls sm1 telll me q6 /... plsss!!

 

[qwertypoiu]

Seems like A2 math is a big jump from AS

 

[TheJDOG]

Seems like A2 math is a big jump from AS

I can tell you no, not much. Tbh, I actually see A2 math easier than AS math I don't know why xD it's ironic

 

[TheJDOG]

http://maxpapers.com/wp-content/uploads/2012/11/9709_w13_qp_3.pdf pls sm1 telll me q6 /... plsss!!

Here you go, if you have any more doubts tell me

Justification : Drop down a perpendicular to get AB or AC in terms of r and use simple trigonometry , then cross out r^2 and simplify , change (Costheta)^2 and 2sintheta.costheta into their identities . Then make cos2theta the subject

https://www.xtremepapers.com/community/attachments/image-jpg.52500/
For part ii) just iterate it

 

[qwertypoiu]

I can tell you no, not much. Tbh, I actually see A2 math easier than AS math I don't know why xD it's ironic

haha lol thanks for the encouragement

 

[Mayisha M]

How to do this?
We get x and y on two sides.View attachment 52533

Well you differentiate it like you normally would with an implicit function, until you get an expression which is probably x^2+y^2 = 1. I'm not sure, did this a few days ago haha. Then rearrange to get x^2 = y^2 - 1, and substitute this equation in the initial equation that you'd differentiated.
Hope that helps!