Author Topic: Need help!!! how to solve tis!!!  (Read 1645 times)

Offline guowie88

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Need help!!! how to solve tis!!!
« on: October 20, 2010, 12:25:54 pm »
Q9 onli pls help file at below  ^^
thk 4 any help
« Last Edit: October 20, 2010, 12:31:11 pm by guowie88 »

Offline guowie88

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Re: Need help!!! how to solve tis!!!
« Reply #1 on: October 20, 2010, 12:32:30 pm »
juz edited ^^ help me solve pls... think for 1 hour alrdy..

Offline ashish

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Re: Need help!!! how to solve tis!!!
« Reply #2 on: October 21, 2010, 06:06:15 am »
y=(1-x)(1+x)-1

using product rule
let U= 1-x
du/dx = -1

V=(1+x)-1

dv/dx = -1(1+x)-2 *1

dy/dx = U dv/dx + V du/dx
     =(1-x)(-1/(1+x)2 +(-1/(1+x))
 
= (-1+x)/(1+x)2  - 1/(1+x)

after combining the 2 fraction we get
dy/dx = -2/(1+x)2


from the question y = ((1-x)/(1+x))1/2
square on both side

y2 = (1-x)(1+x)-1

diff wrt x

2y dy/dx = -2/(1+x)2

dy/dx =  -1/y(1+x)2

replace y by ((1-x)/(1+x))1/2

dy/dx =  -1/((1-x)/(1+x))1/2(1+x)2
         
= -(1+x)1/2/(1-x)1/2(1+x)2

rationalize now

dy/dx = -1/ (1+x2)1/2 (1+x)

 normal gradient = negative reciprocal of of dy/dx]

dy/dx= (1-x2)1/2 (1+x)

Offline Deadly_king

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Re: Need help!!! how to solve tis!!!
« Reply #3 on: October 21, 2010, 06:22:39 am »
Good job dude :D

I would have +rep you but I need to spread the love ;)

Offline ashish

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Re: Need help!!! how to solve tis!!!
« Reply #4 on: October 21, 2010, 06:36:45 am »
Good job dude :D

I would have +rep you but I need to spread the love ;)

thanks again but i am stuck on the second part , i think i didn't quite understand the question ! can you help how can normal have maximum value?

Offline guowie88

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Re: Need help!!! how to solve tis!!!
« Reply #5 on: October 21, 2010, 06:44:02 am »
thk u very much !!! ^^ appreciate it *thumbs up* for u  XD

Offline Deadly_king

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Re: Need help!!! how to solve tis!!!
« Reply #6 on: October 21, 2010, 06:53:29 am »
thanks again but i am stuck on the second part , i think i didn't quite understand the question ! can you help how can normal have maximum value?

The graph represents a curve and as you move along the curve gradient of the tangent changes. This will cause gradient of normals to change as well. The question stated that the gradient of the normal has its maximum value at P which implies the d/dx(gradient of normal) = 0.

The result you obtained in the first part is actually the gradient of the normal in terms of x.

Maximum value ---> stationary value ----> d/dx = 0

Hence you need to differentiate your answer to part 1 and equate it to zero to obtain the required solution ;)

Hope it helps :)

Offline ashish

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Re: Need help!!! how to solve tis!!!
« Reply #7 on: October 21, 2010, 07:09:23 am »
The graph represents a curve and as you move along the curve gradient of the tangent changes. This will cause gradient of normals to change as well. The question stated that the gradient of the normal has its maximum value at P which implies the d/dx(gradient of normal) = 0.

The result you obtained in the first part is actually the gradient of the normal in terms of x.

Maximum value ---> stationary value ----> d/dx = 0

Hence you need to differentiate your answer to part 1 and equate it to zero to obtain the required solution ;)

Hope it helps :)

ok i got it :D thanks ++rep 4 u

here is the answer

d((1-x2)1/2(1+x))/dx  = using quotient rule

U= (1-x2)1/2

du/dx = 0.5* -2x *(1-x2)-1/2


V= 1+x

dv/dx= 1

using the formula

Udv/dx + Vdu/dx =   (1-x2) + (1+x) ((-x)/(1-x2)1/2
                       = (1 - x2-x- x2)/ (1-x2)1/2

since dy/dx = 0

0= 1-2x2-x
2x2+ x - 1=0

(2x-1)(x+1) = 0

x = 1/2 since it is in the right (postive x - axis)


thk u very much !!! ^^ appreciate it *thumbs up* for u  XD
my pleasure  ;D
« Last Edit: October 21, 2010, 07:16:26 am by ashish »

Offline guowie88

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Re: Need help!!! how to solve tis!!!
« Reply #8 on: October 21, 2010, 08:03:26 am »
part 1: the rationalize part i dont understand
can show step by step??
^^ thk

Offline ashish

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Re: Need help!!! how to solve tis!!!
« Reply #9 on: October 21, 2010, 08:23:38 am »
part 1: the rationalize part i dont understand
can show step by step??
^^ thk

dy/dx = -(1+x)1/2 / (1-x)1/2 (1+x)2

rationalize

-(1+x)1/2 / (1-x)1/2 (1+x)2  * ((1+x)1/2/(1+x)1/2)

= -(1+x)/ (1-x2)(1+x)2

(1+x) is common in numerator and denominator
 
= -1/ (1-x2)(1+x)

Offline guowie88

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Re: Need help!!! how to solve tis!!!
« Reply #10 on: October 21, 2010, 08:39:59 am »
thank again ^^