Mathematics: Post your doubts here!

[Copy Cat]

dx/dt = 1/x - x/4
Take L.C.M
dx/dt = (4 - x^2)/4x
Cross multiply
dx*4x = 4dt - x^2dt
dx*4x = dt(4-x^2)
(dx*4x)/(4-x^2) = dt
Switch sides and integrate.
t = -2ln(4-x^2) + c

t = o
x = 1

>> 0=-2ln(4-1) + c
0 = -2ln3
c = 2ln3
Substitute.
t = -2ln(4-x^2) + 2ln3
t - 2ln3 = -2ln(4-x^2)
-t + 2ln3 = 2ln(4-x^2)
-t = 2ln(4-x^2) - 2ln3
-t = 2ln[(4-x^2)/3]
-t/2 = ln[(4-x^2)/3]
exp(-t/2) = (4 - x^2)/3
3exp(-t/2) = 4 - x^2
3exp(-t/2) - 4 = -x^2
switch sides and change sign
x^2 = 4 - 3exp(-t/2)

I hope I'm right tho :3

Thanks a bunch....

 

[Copy Cat]

last part

 

[DaniyalK]

View attachment 58034
last part

take lcm
dh/dt = (100-h^2)/20h^2
(20h^2/(100-h^2)) dh = dt
Use result from part 2.
-20 + 2000/(10-h)(10+h) dh = dt
use partial fractions
-20 + 100/(10-h) + 100/(10+h) dh = dt
then integrate

 

[Anum96]

View attachment 58034
last part

Did you try partial fractions?

 

[happyoner_5]

Mass = 1100
Fd = 1800
Fr = 700N
Distance = x
Speed = v
i) K.E = 1/2(1100)v^2 = 550v^2
P.E = (1100)(10)(160) = 1760000 Now we have to find it in terms of x which is distance. So it will become 1760000/1760 = 1000x

kv^2 = x
1800x = 700x+ 550v^2+ 1000x
^This is the WD equation. I just substituted everything in terms of x and v
Solve it further you will get
100x = 550v^2
x = 550/100v^2
x = 5.5v^2
Hence k = 5.5 Ans.

ii) 5.5v^2 = 1760
v^2 = 320

1800(x-1760) = 700(x-1760)+ 550(v^2-320) [we subtracted 320 because it was our initial velocity and so to find gain in K.E we already have v^2(initial velocity)
1800x-3168000 = 700x - 1232000+ 550v^2 - 176000
1800x-700-3168000 = -1408000+ 550v^2
1100x-3168000+ 1408000 = 550v^2
1100x+ 1760000 = 550v^2
(1100x+ 176000)/550 = v^2
v^2 = 2x - 3200 (shown)

thnx a lot : )

 

[Copy Cat]

take lcm
dh/dt = (100-h^2)/20h^2
(20h^2/(100-h^2)) dh = dt
Use result from part 2.
-20 + 2000/(10-h)(10+h) dh = dt
use partial fractions
-20 + 100/(10-h) + 100/(10+h) dh = dt
then integrate

Did you try partial fractions?

Thaks a lot
Yes i did but my C is zero just wanted to confirm because in ms there's no correct answer given.

Btw in 2 nd part can we just say assume h=1 anwers are 2/100 and 2/99 so since the answers are close enough hence we can say that they are eqivalent

 

[DaniyalK]

Thaks a lot
Yes i did but my C is zero just wanted to confirm because in ms there's no correct answer given.

Btw in 2 nd part can we just say assume h=1 anwers are 2/100 and 2/99 so since the answers are close enough hence we can say that they are eqivalent

Both of those things are in the examiner report.

 

[happyoner_5]

can u plz suggest a good book for further math

 

[DaniyalK]

can u plz suggest a good book for further math

Further Pure Mathematics by Brian and Mark Gaulter
A Concise Course in Advanced Level Statistics by Crawshaw and Chambers
Mechanics for A Level by Bostock and Chandler

They don't cover all the topics, but you can find the majority of them in these books.

 

[happyoner_5]

Further Pure Mathematics by Brian and Mark Gaulter
A Concise Course in Advanced Level Statistics by Crawshaw and Chambers
Mechanics for A Level by Bostock and Chandler

They don't cover all the topics, but you can find the majority of them in these books.

thnx

 

[Wâlèé Atèéq]

https://drive.google.com/file/d/0B6rlbmTCHUFZRm0tN2RkZEJVNW8/view?usp=sharing

Q8..! Mechanics..? need help..!

 

[***amd***]

https://drive.google.com/file/d/0B6rlbmTCHUFZRm0tN2RkZEJVNW8/view?usp=sharing

Q8..! Mechanics..? need help..!

consider points A and C, u have...
time = 6s
displacement = 60m
initial velocity = 4 m/s

use s = ut + 1/2 at^2, and u get a = 2 m/s^2

now that u have acceleration,
consider points A and B,
time = 3s
initial velocity = 4 m/s
acc = 2

use again s = ut + 1/2 at^2
s = 21 m

 

[Wâlèé Atèéq]

consider points A and C, u have...
time = 6s
displacement = 60m
initial velocity = 4 m/s

use s = ut + 1/2 at^2, and u get a = 2 m/s^2

now that u have acceleration,
consider points A and B,
time = 3s
initial velocity = 4 m/s
acc = 2

use again s = ut + 1/2 at^2
s = 21 m

Thanx bro..!

 

[extremesuser]

https://mathway.com/

 

[The Sarcastic Retard]

(iii)

 

[Anum96]

View attachment 58090
(iii)

http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-rcostheta-alpha-2009-1.pdf
Use these notes for these type of questions. You will find what u need in the last heading that's all about maximum and minimum values

 

[The Sarcastic Retard]

http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-rcostheta-alpha-2009-1.pdf
Use these notes for these type of questions. You will find what u need in the last heading that's all about maximum and minimum values

Thanks

 

[The Sarcastic Retard]

http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-rcostheta-alpha-2009-1.pdf
Use these notes for these type of questions. You will find what u need in the last heading that's all about maximum and minimum values

Can u please post the solution. I already refered to this site.. nahi aa raha meko

 

[Anum96]

Can u please post the solution. I already refered to this site.. nahi aa raha meko

Sure. Wait up
Whats the answer ? I'll post if I get it correct

 

[The Sarcastic Retard]

Sure. Wait up
Whats the answer ? I'll post if I get it correct

IDK no solution with me thats why m confused.
its a homework :3