+0  
 
0
892
6
avatar

 Dec 31, 2014

Best Answer 

 #3
avatar+128083 
+10

Here's (c)

u(x,t) = cos(x + ct )

ut (x, t)  = - sin(x  ct) (c) = - (c)sin (x + ct)

So

utt = -(c)cos(x + ct)(c) = -(c2) cos (x + ct)    

And

ux (x, t) = -sin(x + ct)(1)= -sin(x + ct)

So

uxx = -cos( x + t)(1) = -cos( x + ct)

So

(c2) uxx = (c2)[-cos( x + ct) ] = -(c2) cos (x + ct)

And ......we have shown that.....

utt = (c2) uxx

 

 Jan 1, 2015
 #1
avatar+128083 
+10

∫√(4 - x^2) dx       

Let x = 2sinΘ       dx = 2cosΘ dΘ  .....    when x =0, Θ = 0  and when x = 2, Θ = pi/2 ... so we have

∫√(4 - 4sin^2 Θ) 2cosΘ dΘ =

2∫√[(4(1 - sin^2  Θ) ] cos Θ  dΘ =

2 ∫2 √[ 1 - sin^2  Θ) ] cos Θ  dΘ =

4 ∫ cos Θ * cos Θ  dΘ =

4 ∫ cos ^2 Θ  dΘ          and using  cos^2 Θ = (1/2)(1 + cos2 Θ)  , we have

2 ∫ [1 + cos2 Θ]  dΘ  =

2Θ   + sin 2Θ  ....and substituting in the limits of integration, we have

2 [ pi/2 - 0 ] + [sin 2 (pi/2) - sin 2(0)] =

pi + [ sin (pi) - sin(0)] =

[pi ] +  [0]  =

pi =  about 3.1416

 

 Dec 31, 2014
 #2
avatar+128083 
+10

Here's (b)

 q =  [P1P2 + 2P1 ] / [P1P2 - 2P2 ].......we can use the quotient rule here.....

∂q/∂P1 =  [ [ (P2 + 2)(P1P2 - 2P2) ] - [ (P1P2 + 2P1) (P2) ] ] /  [P1P2 - 2P2 ]2   =

[ P1P22 + 2P1P2 - 2P22 - 4P2  - P1P22 - 2P1P2 ] / [P1P2 - 2P2 ]2 =

- [ 2P22 + 4P2 ] /  [P1P2 - 2P2 ]2 =

-2P2 [ P2 + 2 ] / [P1P2 - 2P2 ]2

-2P2 [ P2 + 2 ] / [P2 (P1 - 2)]2 =

-2 [ P2 + 2 ] / [P2 (P1 - 2)2 ]

 

 

 Jan 1, 2015
 #3
avatar+128083 
+10
Best Answer

Here's (c)

u(x,t) = cos(x + ct )

ut (x, t)  = - sin(x  ct) (c) = - (c)sin (x + ct)

So

utt = -(c)cos(x + ct)(c) = -(c2) cos (x + ct)    

And

ux (x, t) = -sin(x + ct)(1)= -sin(x + ct)

So

uxx = -cos( x + t)(1) = -cos( x + ct)

So

(c2) uxx = (c2)[-cos( x + ct) ] = -(c2) cos (x + ct)

And ......we have shown that.....

utt = (c2) uxx

 

CPhill Jan 1, 2015
 #4
avatar+808 
+5

I know why your profile picture reads "666", because you're a devil with numbers ;)

 Jan 1, 2015
 #5
avatar+128083 
0

LOL!!!! (I'm really not that good....more lucky....!!! )

 

 Jan 1, 2015
 #6
avatar+118587 
+5

Thank you Chris  

Please anon - put one question per post and turn your picture around the right way before you upload it.

If you are capable of calculus then you are able to learn how to rotate a picture!

 Jan 2, 2015

3 Online Users

avatar
avatar