Need help!!! how to solve tis!!!

[guowie88]

Q9 onli pls help file at below  ^^
thk 4 any help

[guowie88]

juz edited ^^ help me solve pls... think for 1 hour alrdy..

[ashish]

y=(1-x)(1+x)^-1

using product rule
let U= 1-x
^du/_dx = -1

V=(1+x)<sup>-1

dv</sup>/_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)

[Deadly_king]

Good job dude

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

[ashish]

Good job dude

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?

[guowie88]

thk u very much !!! ^^ appreciate it *thumbs up* for u  XD

[Deadly_king]

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

[ashish]

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 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 

[guowie88]

part 1: the rationalize part i dont understand
can show step by step??
^^ thk

[ashish]

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)

[guowie88]

thank again ^^