Mathematics: Post your doubts here!

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Give me five minutes. Ill help you out :D
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How to solve this question please??? I'm completely lost :cry:
Dm /dt = K(M) cos (0.02t)
dM / m = kcos0.02t dt
ʃ dM / m = ʃ kcos0.02t dt
(M ^(1/2)) / (1/2) = ksin0.02t/0.02 + c
Rearrange:
2M =(ksin0.02t)/0.02 + C
Now we have t = 0 and M = 100 so just substitute inorder to find the constant C
Therefore,
2100 = (ksin0.02(0))/0.02 + C
2(10) = 0 + C
C = 20
Now : substitute the value of C to find the relationship.
2M = (ksin0.02t)/0.02 +20
**1/0.02 = 50
So, it will become: 2M = 50ksin0.02t + 20
b) M = 196 and t = 50
Just plug in the values.
2196 = 50ksin0.02(50) +20
14*2 = 50ksin0.02(50) + 20
28 = 50ksin0.02(50) + 20
28 – 20 = 50ksin0.02(50)
8 = 50ksin(1)
8/50 = ksin(1)
K = 0.19 Ans.
c) 2M = 50ksin0.02t + 20
Make m the subject
You will get:
M = ((50ksin0.02t + 20)/2)^2
Plus in the values.
M =((50(0.19)sin0.02t + 20)/2)^2
Solve the square. And you will get around 27 or 28 as your answer
If you still have problem, Inbox me :)
I dont own a cellphone otherwise a picture wouldve cleared it out well. Sorry :p
 
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Dm /dt = K(M) cos (0.02t)
dM / m = kcos0.02t dt
ʃ dM / m = ʃ kcos0.02t dt
(M ^(1/2)) / (1/2) = ksin0.02t/0.02 + c
Rearrange:
2M =(ksin0.02t)/0.02 + C
Now we have t = 0 and M = 100 so just substitute inorder to find the constant C
Therefore,
2100 = (ksin0.02(0))/0.02 + C
2(10) = 0 + C
C = 20
Now : substitute the value of C to find the relationship.
2M = (ksin0.02t)/0.02 +20
**1/0.02 = 50
So, it will become: 2M = 50ksin0.02t + 20
b) M = 196 and t = 50
Just plug in the values.
2196 = 50ksin0.02(50) +20
14*2 = 50ksin0.02(50) + 20
28 = 50ksin0.02(50) + 20
28 – 20 = 50ksin0.02(50)
8 = 50ksin(1)
8/50 = ksin(1)
K = 0.19 Ans.
c) 2M = 50ksin0.02t + 20
Make m the subject
You will get:
M = ((50ksin0.02t + 20)/2)^2
Plus in the values.
M =((50(0.19)sin0.02t + 20)/2)^2
Solve the square. And you will get around 27 or 28 as your answer
If you still have problem, Inbox me :)
I dont own a cellphone otherwise a picture wouldve cleared it out well. Sorry :p
Thankyou soooooooo much!!
This is great! Your workings and steps are clear!!
Thanks a ton for your effort and time!!! (y):D
 
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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 (y)
 
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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 :)
 
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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 :)

 
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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.
 

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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 :)
 
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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
 
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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 !! :)
 
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