Mathematics: Post your doubts here!

[mohmed ahmed soliman]

help in II

 

[mohmed ahmed soliman]

HELP IN
III
IV
V

 

[Copy Cat]

 

[The Sarcastic Retard]

View attachment 59834

Knowing that u have idea about rules of differentiation. You might have reached to this step :
(i)
dy/dx = -1/[(1 + x)^(3/2).(1 - x)^(1/2)]
Then split dy/dx = (1 + x)^(3/2).(1 - x)^(1/2)
you will obtain result similar to this : (1+x)(1+x)^(1/2).(1-x)^(1/2)
We know that,
(1+x)^(1/2).(1-x)^(1/2) = (1 - x^2)^(1/2)
So now dy/dx = -1/[(1 + x)^(3/2).(1 - x)^(1/2)] = -1 / (1+x).(1 - x^2)^(1/2)
Normal to it is -ve reciprocal of it = (1+x).(1 - x^2)^(1/2)

(ii)
Take y = (1+x).(1 - x^2)^(1/2)
differentiate it.
equate it to zero
obtain x = 1/2

 

[The Sarcastic Retard]

View attachment 59796

Here :

 

[Copy Cat]

Knowing that u have idea about rules of differentiation. You might have reached to this step :
(i)
dy/dx = -1/[(1 + x)^(3/2).(1 - x)^(1/2)]
Then split dy/dx = (1 + x)^(3/2).(1 - x)^(1/2)
you will obtain result similar to this : (1+x)(1+x)^(1/2).(1-x)^(1/2)
We know that,
(1+x)^(1/2).(1-x)^(1/2) = (1 - x^2)^(1/2)
So now dy/dx = -1/[(1 + x)^(3/2).(1 - x)^(1/2)] = -1 / (1+x).(1 - x^2)^(1/2)
Normal to it is -ve reciprocal of it = (1+x).(1 - x^2)^(1/2)

(ii)
Take y = (1+x).(1 - x^2)^(1/2)
differentiate it.
equate it to zero
obtain x = 1/2

Thanks.

 

[Copy Cat]

 

[Eugene99]

2Ф (b/0.15) = 1.80
Ф (b/0.15) = 0.90

Using Normal Distribution tables,

b/0.15 = 1.281
b = 0.192 (3sf)

Could you please explain this part a little, how you got 1.281 from the table?

 

[mistique_bee]

View attachment 59840

Just use quadratic equation:

a = 1
b = 2√3
c = 4
-->

b² = 4*3 = 12
and
4ac = 4*1*4 = 16
and
2a = 2*1 = 2
-->

(-2√3 ± √(12 - 16)) / 2
-->

√(12 - 16) = √(-4) = i√4 = 2i
-->

(-2√3 ± 2i) / 2
--> divide out the 2

√3 ± i
-->

the two roots are:

z = √3 + i
z = √3 - i

I'm not going to multiply that out to the 6th degree, it's easier if you convert to polar form, which I don't know if you've seen or not:

z = r * exp(iθ) = r * (cos(θ) + i * sin(θ))

So in this case the first root is in the first quadrant and we have:

tan(θ) = 1/√3
--> that's a 30-60-90 triangle and it's the smaller angle (i.e. 30° or π/6 radians)

the r is just the magnitude which is the square of the real and imaginary parts added up (then square rooted):

(√3)² + 1² = 3 + 1 = 4 --> √4 = 2
-->

r = 2

So we have:

z = √3 + i = 2 * exp(πi/6)

For the other one, it's the same angle, just in the 4th quadrant, so we get either:

-π/6 or 11π/6

Now raising to the 6th power is easy:

(2 * exp(±πi/6))^6 = 2^6 * exp(±πi/6 * 6) = 64 * exp(±πi)

But exp(±πi) = -1 (because we have cos(±π) + i * sin(±π) = -1 + 0 = -1
-->

So those roots do indeed satisfy the other equation (since you get 64 * -1 = -64)

 

[Rizwan Javed]

Could you please explain this part a little, how you got 1.281 from the table?

I looked up this value from the Normal Distribution Table which is provided with the paper.

 

[Eugene99]

I looked up this value from the Normal Distribution Table which is provided with the paper.

You used value for 0.90 or what?

 

[Eugene99]

I looked up this value from the Normal Distribution Table which is provided with the paper.

it's 0.8159 at 0.90

 

[Rizwan Javed]

You used value for 0.90 or what?

Yep! I looked up for the value of z which gives 0.90.

 

[Eugene99]

Yep! I looked up for the value of z which gives 0.90.

OH! I got that!!
thanks

 

[farhan141]

Here : View attachment 59837

Thanks. Can u explain me how is cos x equal to (1/2 AB)/R

 

[Copy Cat]

Just use quadratic equation:

a = 1
b = 2√3
c = 4
-->

b² = 4*3 = 12
and
4ac = 4*1*4 = 16
and
2a = 2*1 = 2
-->

(-2√3 ± √(12 - 16)) / 2
-->

√(12 - 16) = √(-4) = i√4 = 2i
-->

(-2√3 ± 2i) / 2
--> divide out the 2

√3 ± i
-->

the two roots are:

z = √3 + i
z = √3 - i

I'm not going to multiply that out to the 6th degree, it's easier if you convert to polar form, which I don't know if you've seen or not:

z = r * exp(iθ) = r * (cos(θ) + i * sin(θ))

So in this case the first root is in the first quadrant and we have:

tan(θ) = 1/√3
--> that's a 30-60-90 triangle and it's the smaller angle (i.e. 30° or π/6 radians)

the r is just the magnitude which is the square of the real and imaginary parts added up (then square rooted):

(√3)² + 1² = 3 + 1 = 4 --> √4 = 2
-->

r = 2

So we have:

z = √3 + i = 2 * exp(πi/6)

For the other one, it's the same angle, just in the 4th quadrant, so we get either:

-π/6 or 11π/6

Now raising to the 6th power is easy:

(2 * exp(±πi/6))^6 = 2^6 * exp(±πi/6 * 6) = 64 * exp(±πi)

But exp(±πi) = -1 (because we have cos(±π) + i * sin(±π) = -1 + 0 = -1
-->

So those roots do indeed satisfy the other equation (since you get 64 * -1 = -64)

Is there any other method?

 

[mistique_bee]

Is there any other method?

Well i have no idea bt tht.. This is hw i do it.. So i did...
Ask may be Rizwan Javed Anum96

 

[Rizwan Javed]

Well i have no idea bt tht.. This is hw i do it.. So i did...
Ask may be Rizwan Javed Anum96

z^6 = (-√3 +i)^6
= (2(cos(5pi/6) + i sin(5pi/6)))^6
= 2^6 (cos 5pi + i sin5pi)
= -64

z^6 = (-√3 - i)^6
= (2(cos(-5pi/6) + i sin(-5pi/6)))^6
= 2^6 (cos(-5pi) + i sin (-5pi))
= -64

 

[mistique_bee]

z^6 = (-√3 +i)^6
= (2(cos(5pi/6) + i sin(5pi/6)))^6
= 2^6 (cos 5pi + i sin5pi)
= -64

z^6 = (-√3 - i)^6
= (2(cos(-5pi/6) + i sin(-5pi/6)))^6
= 2^6 (cos(-5pi) + i sin (-5pi))
= -64

Wah xD thn wth im up to :/

 

[Copy Cat]

2^6 (cos 5pi + i sin5pi)

From this = (2(cos(5pi/6) + i sin(5pi/6)))^6
to this = 2^6 (cos 5pi + i sin5pi)
How?
An explanation would help.....Thanks