₦airaland Forum

Laplacian:
w= (z+2j)/(z+j), w(z+j)=z+2j or

z=(2j+wj)/(w-1)=[(u+2)j-v]/[(u-1)+vj] or
taking conjugates;
z=[(u+2)j-v]*[(u-1)-vj]/[(u-1)^2+v^2]

={[(u-1)(u+2)+v^2]j+[-uv+v(u+2)]}/[(u-1)^2+v^2]
so
x=[-uv+v(u+2)]/[(u-1)^2+v^2]
and
y=[(u-1)(u+2)+v^2]/[(u-1)^2+v^2]

if u make this substitution u 'll get the centre

tartar9:
the one we use for mechanics of materials is Strength of Material by R.K RAJPUT you can get the PDF online don't know bout the fluid mechanics.

Drniyi4u:
yoų çāŋ ţŗy Engineering Mechanics - Statics, by Meriam, Kraige
OR
Beer & Johnston: Statics and Dynamics

ţђęy'ŗę woŋdęŗƒųl įŋ męçђāŋįçş!

akpos4uall:
I'm not 100% sure but I think there is a Schaum's Outline Series textbook on Mechanics of Materials.

efficiencie:
Given that:
arcsina+arcsinb+arcsinc=π

arcsina=π-arcsinb-arcsinc

Take the cosine of both sides:

cos(arcsina)=cos(π-arcsinb-arcsinc)

cos(arcsina)=cos(π-(arcsinb+arcsinc))

cos(arcsina)=
-cos(arcsinb+arcsinc)

cos(arcsina)=
-cos(arcsinb)cos(arcsinc)+ sin(arcsinb)sin(arcsinc)

√(1-sin^2(arcsina))=
-(√(1-sin^2(arcsinb))
√(1-sin^2(arcsinc)))+
sin(arcsinb).sin(arcsinc)

a√(1-a^2)=-a.√(1-b^2).√(1-c^2)+abc...1

and similarly the following is derived:

b√(1-b^2)=-b√(1-a^2).√(1-c^2)+abc...2

c√(1-c^2)=-c√(1-b^2).√(1-a^2)+abc...3

On adding up:

a√(1-a^2)+b√(1-b^2)+c√(1-c^2) =-a.√(1-b^2).√(1-c^2)
-b.√(1-a^2).√(1-c^2)
-c.√(1-b^2).√(1-a^2)+3abc ...4

Going back to the original equatn:
arcsina+arcsinb+arcsinc=π

arcsina+arcsinb=π-arcsinc
On taking the sine of both sides

sin(arcsina+arcsinb)=
sin(π-arcsinc)

sin(arcsina)cos(arcsinb)+ sin(arcsinb)cos(arcsina) =c

a.√(1-b^2)+b.√(1-a^2) =c ...5

and similarly the followin is derivd
a.√(1-c^2)+c.√(1-a^2) =b ...6

c.√(1-b^2)+b.√(1-c^2) =a ...7

And hence multiply 5 by
√(1-c^2), 6 by √(1-b^2) and 7 by √(1-a^2) the followin results:

a.√(1-b^2).√(1-c^2)+b.√(1-a^2).√(1-c^2)=c.√(1-c^2) ...8

a.√(1-c^2).√(1-b^2) +c.√(1-a^2).√(1-b^2) =b√(1-b^2) ...9

c.√(1-b^2).√(1-a^2)+b.√(1-c^2). √(1-a^2)=a.√(1-a^2) ...10

On adding up

2a.√(1-b^2).√(1-c^2)+
2b.√(1-a^2).√(1-c^2)+
2c.√(1-b^2).√(1-a^2) =
a.√(1-a^2)+b.√(1-b^2)+
c.√(1-c^2) ...11

On multiplyn equatn 4 by 2 we av:

2a√(1-a^2)+2b√(1-b^2)+2c√(1-c^2) =-2a.√(1-b^2).√(1-c^2)
-2b.√(1-a^2).√(1-c^2)
-2c.√(1-b^2).√(1-a^2)+6abc

on substituting the value of:
2a.√(1-b^2).√(1-c^2)+
2b.√(1-a^2).√(1-c^2)+
2c.√(1-b^2).√(1-a^2) from equatn 11

The following results:
2a√(1-a^2)+2b√(1-b^2)+2c√(1-c^2) =-a.√(1-a^2)-b.√(1-b^2)-
c.√(1-c^2) +6abc

3a√(1-a^2)+3b√(1-b^2)+3c√(1-c^2) =6abc

a√(1-a^2)+b√(1-b^2)+c√(1-c^2) =2abc

QED!

Too long! Any shortcut solutions!?

bolkay47:
thanks so much sir...but why did you assume "a" to be 20?

bolkay47:
this is just excellent.... Although I don't know the answer yet cos its an assignment but WAO.#respect..

bolkay47:
the question is on GP....it should somehow form a sequence. That's what I don't understand.

jackpot:
hmmm. I can see the mathematics. But. . .

Let's take another look.

Given y'+y=0, differentiating both sides w.r.t. x gives a new differential equation
y''+y'=0.

It is really a heresy to say that both differential equations are equivalent since y'+y=0 has the solution y=C e^-x, whereas y''+y'=0 has the solution y=C₁+C₂e^-x.


Because y'+y=0 is easier to handle and it's solution satisfies y''+y'=0, I can not conclude that both differential equations has the same (general) solution.

So, I will say that the two differential equations you compared are not equivalent.

Or, what do you think?

agentofchange1:
@ PatEinstEin ( m ), AlphaMaximus

this's. benbuks. speaking ...

greet thee. men & brethren..

happy solving...

PatEinstEin:
I thought as much

V/T₁ + V/T₂ = V/T₃ .............(i)

Where V/T is the rate of flow

T₂ = T₁ - 5 = T₃ + 18
T₂ = T₁ - 5
T₃ + 18 = T₁ - 5
T₃ = T₁ - 23

Substituting into (i)

V/T₁ + V/(T₁ - 5) = V/(T₁ - 23)
1/T₁ + 1/(T₁ - 5) = 1/(T₁ - 23)
(2T₁ - 5)/(T₁² - 5T₁) = 1/(T₁ - 23)

T₁² - 46T₁ + 115 = 0

T₁ = 43.347 hrs or 2.653 hrs

PatEinstEin:
Hailings Sir @AlphaMaximus
I got the latest Oxford Advanced Learners Dictionary for your comments

Drniyi4u:
r u sure equation (i) is correct??

tohero:
Your solution might be right. It is not compulsory to have a whole no as an answer. Just convert your decimal part to terms of Minutes and Seconds

sorbentcrooner:
Hello house , kindly help out with ds. Thanks in anticipation:
A tank is filled by three pipes with uniform flow. The
first two pipes operating simultaneously fill the tank
in the same time during which the tank is filled by
the third pipe alone. The second pipe fills the tank 5
hours faster than the first pipe and 18 hours slower
than the third pipe. The time required by the first is?

sorbentcrooner:
Hello house , kindly help out with ds. Thanks in anticipation:
A tank is filled by three pipes with uniform flow. The
first two pipes operating simultaneously fill the tank
in the same time during which the tank is filled by
the third pipe alone. The second pipe fills the tank 5
hours faster than the first pipe and 18 hours slower
than the third pipe. The time required by the first is?