Mathematics: Post your doubts here!

[Wolfgangs]

Can someone please give me the final answer and working for this question.

 

[Anum96]

View attachment 57664

View attachment 57664

y = e^(-x) sinx

For stationary point of any curve put dy/dx equal to 0

For dy/dx

Let u = e^(-x)

v = sinx

u’ = -e^(-x)

v’ = cosx

dy/dx = vu’ + uv’

You have all the values now. Just plug in.

0 = (sinx)(-e^(-x)) + (e^(-x))(cosx)

0 = e^(-x) {-sinx + cosx}

e^(-x) {-sinx + cosx} = 0

-sinx +cosx = 0

-sinx=-cosx

sinx=cosx

sinx/cosx = 0

tanx = 0

x = 45® or π/4 Ans.



b) For nature determination, Differentiate again. That will give you d2y/dx2

e^(-x) {-sinx + cosx}

d2y/dx2 = vu’ + uv’

u = e^(-x)

v = {-sinx + cosx}

u’ = -e^(-x)

v’ = -cosx –sinx

d2y/dx2 = vu’ + uv’

= [{-sinx + cosx}*-e^(-x) ]+[( e^(-x))*( -cosx –sinx)]

e^(-x)[sinx -cosx – cosx –sinx] = 0

[sinx -cosx – cosx –sinx] = 0

-2cosx = 0 Put x = 45 or π/4

-2(π/4) = 0

-π/2 < 0 Therefore this point is maximum

 

[Anum96]

Can someone please give me the final answer and working for this question.

First find the acceleration.
Acceleration = (forces aiding acceleration - forces opposing acceleration)/ sum of masses
Here, Block A has higher mass so acceleration will be in the direction of A. In other words A will move down.
Therefore.
After resolving forces you will get acceleration = (40sin53 - 20sin37)/(4 + 2) = 10/3
Now, For tension. Remember when acceleration is downwards the equation to find tension is F - T = ma
So, just plug in the values. You can take any one block. A or B. Ill take A
40sin53 - T = (4)(10/3)
- T = 13.33 - 40sin53
- T = - 18.61
T = 18.61
Rounding off will give you 18.7 That is your option B

 

[Wolfgangs]

How to find the average velocity for this question?

 

[Anum96]

How to find the average velocity for this question?

If you sketch a velocity time graph. you will note that the slope of the curve will be constant since acceleration is constant. You will start from 0 (rest) and till 30m/s(final velocity)
Now, Average speed = total distance/total time.
Let time = t
Total distance will be the area under the graph
=[1/2*(t)*(30)]
Therefore,
Average speed = [1/2*(t)*(30)]/t
t and t cancels out and you will get 15 as your answer. Choice C

 

[Wolfgangs]

If you sketch a velocity time graph. you will note that the slope of the curve will be constant since acceleration is constant. You will start from 0 (rest) and till 30m/s(final velocity)
Now, Average speed = total distance/total time.
Let time = t
Total distance will be the area under the graph
=[1/2*(t)*(30)]
Therefore,
Average speed = [1/2*(t)*(30)]/t
t and t cancels out and you will get 15 as your answer. Choice C

Thanks alot. I have a doubt in one more question, if you are free.

 

[Anum96]

Thanks alot. I have a doubt in one more question, if you are free.

First step. Acceleration. As I stated in the previous question.
Acceleration = (forces aiding acceleration - forces opposing acceleration)/ sum of masses
Now mass of m1 is greater than the mass of m2 so we'll just assume(for now) that m1 will move down since its heavier. (If true the resule should be positive)
Lets see.
Acceleration = (100sin37 - 80)/(10 + 8) = (60.2 - 80)/18 = -19.8/18 = -1.1
Now, The result is negative, It shouldve been positive therefore m1 will NOT move down even though its heavier. Acceleration came out negative because of the force, suggesting that m1 will move up the incline Choice A

 

[Sarosh Jameel]

PLZ help with this !

 

[Anum96]

PLZ help with this !

U = sin4x

Find the new limits.

1) X = π/24

U = sin4(π/24) = sin(π/6) = ½

2) X =0

U = sin4(0) = sin(0) = 0

Differentiate the substitution.

U = sin4x

Du/dx = 4cos4x

Find dx

Du = 4cos4x (dx)

Dx = du/4cos4x

Now re-write the expression

= ʃcos^3(4x) dx

You have value of dx

= ʃcos^3(4x) * (du/4cos4x)

Cancelling out will give you:

= ʃcos^2(4x)/4 du

= ¼ ʃ(1 – sin^2(4x) ) du

= ¼ ʃ(1 – u^2)

= ¼ [ u – (u^(3)/3) ]

= ¼ [(3u – u^3)/3)]

Put the limits

= ¼ {(3(0.5) – (0.5^(3))/3) – (3(0) – (0^3)/3)}

= ¼ {[(1.5 – 0.125)/3] – (0)}

= ¼ (1.375/3)

= ¼ (11/24)

= 11/96 Ans

 

[Sarosh Jameel]

U = sin4x

Find the new limits.

1) X = π/24

U = sin4(π/24) = sin(π/6) = ½

2) X =0

U = sin4(0) = sin(0) = 0

Differentiate the substitution.

U = sin4x

Du/dx = 4cos4x

Find dx

Du = 4cos4x (dx)

Dx = du/4cos4x

Now re-write the expression

= ʃcos^3(4x) dx

You have value of dx

= ʃcos^3(4x) * (du/4cos4x)

Cancelling out will give you:

= ʃcos^2(4x)/4 du

= ¼ ʃ(1 – sin^2(4x) ) du

= ¼ ʃ(1 – u^2)

= ¼ [ u – (u^(3)/3) ]

= ¼ [(3u – u^3)/3)]

Put the limits

= ¼ {(3(0.5) – (0.5^(3))/3) – (3(0) – (0^3)/3)}

= ¼ {[(1.5 – 0.125)/3] – (0)}

= ¼ (1.375/3)

= ¼ (11/24)

= 11/96 Ans

Thanks a lot !!

 

[Sarosh Jameel]

in part ii i am stuck at this step .. ʃ -5sin^2(x) cos^2(x) du

 

[Anum96]

in part ii i am stuck at this step ..
Use the substitution now

ʃ5sin^3x*cos^2x dx

You have dx and u substitution

= ʃ 5sin^3x(u^2) (du/-sinx)

Cancelling out will give you,

=ʃ-5sin^2x(u^2)

Use trig identity : sin^2x = 1 – cos^2x

=-ʃ5(1 – cos^2x)(u^2)du

=-5ʃ(1 – u^2)(u^2)du

= -5ʃu^2 – u^4 du

=- 5 [(u^3/3) – (u^5/5)]

Put limits in.

-5 {[(1/3 – 1/5) – (0/3 – 0/5)]}

= -5 (2/15)

= -10/15

= -2/3

Area = 2/3 Ans.
*I think I messed up a sign somewhere or its the way it is

 

[Sarosh Jameel]

thanks !

 

[Hassan Ali Abid]

I'm using this book :
http://www.amazon.com/Concise-Course-Advanced-Level-Statistics/dp/074875475X/ref=oosr
covers both S1 and S2
As for pastpapers, when I bought solved past papers, I had chosen S1 and M1 but later I changed my mind and found Statistics much more bearable than Mechanics, so I've solved papers for S1 but not for S2..But it doesn't matter, I'll print out unsolved ones and I can check them through mark schemes! What's your plan?

So aren't you having this book in pdf form or is it available for free online ?
Are there any solved papers for S2 ??? ...if you have them then i need them urgently.
I have tried to use to use ms for S2 but it didnt work out for me bcz the method is not given in the ms :/

 

[Hassan Ali Abid]

I'm using this book :
http://www.amazon.com/Concise-Course-Advanced-Level-Statistics/dp/074875475X/ref=oosr
covers both S1 and S2
As for pastpapers, when I bought solved past papers, I had chosen S1 and M1 but later I changed my mind and found Statistics much more bearable than Mechanics, so I've solved papers for S1 but not for S2..But it doesn't matter, I'll print out unsolved ones and I can check them through mark schemes! What's your plan?

and one thing more ...does that book contains the topics such as cdf ,pdf , chi square etc ???

 

[Wolfgangs]

Can someone please help me with this question (projectile motion)?

 

[Anum96]

Can someone please help me with this question (projectile motion)?

Use this. Pretty similar example

 

[bakhita]

So aren't you having this book in pdf form or is it available for free online ?
Are there any solved papers for S2 ??? ...if you have them then i need them urgently.
I have tried to use to use ms for S2 but it didnt work out for me bcz the method is not given in the ms :/

Yes, the book is available online, check this website:
http://www.aliensservices.com/E-books.html

As for the solved papers, it's difficult, a mere chance, that they'll be available for S2, very few students choose S2, but I'll search and let you know if I find!

 

[bakhita]

and one this more ...does that book contains the topics such as cdf ,pdf , chi square etc ???

Yes, it does cover these, the probability density function and cumulative distribution function are covered in 6th unit, chi square must be there too and I can assure you, it covers ALL of the syllabus, and even more than that, soyou must keep checking the syllabus, like there is the one "Regression and Corrlation", that I haven't really found in syllabus so I skipped that.

 

[Wolfgangs]

Can someone tell me the final answer and working for this question please?