change of parameter prob. .pls help!

[Lost souls]

hey guys. .can someone please tell me what would the next steps b?

x + p / (1+p²)^1/2 = a ...............................................where a =constant, p = dy/dx. .

=> x= a - p / (1+p²)<sup>1/2

=></sup> dx/dy = 0 - d/dy(p / (1+p²)^1/2)

then?

i'm pretty new to this stuff so excuse me if this seems quite easy. .

 

[mmm4444bot]

In general, I do not bother looking at text speak. Others may.

(These boards do not comprise a chat room.)

Cheers :cool:

 

[Lost souls]

In general, I do not bother looking at text speak. Others may.

(These boards do not comprise a chat room.)

Cheers :cool:

how do you want it buddy?

 

[Lost souls]

English is nice

sorry people. .i didn't realise the mistake. .now it's fixed. can someone please help now??

 

[Deleted member 4993]

hey guys. .can someone please tell me what would the next steps b?

x + p / (1+p²)^-2 = a ...............................................where a =constant, p = dy/dx. .

=> x= a - p / (1+p²)<sup>-2

=></sup> dx/dy = 0 - d/dy(p / (1+p²)^-2)

then?

i'm pretty new to this stuff so excuse me if this seems quite easy. .

What do you need to find?

I f you substitute \(\displaystyle p \ = \ \tan(\theta)\) .................in.................. x + p / (1+p²)^-2 = a

What do you get?

 

[Lost souls]

What do you need to find?

I f you substitute \(\displaystyle p \ = \ \tan(\theta)\) .................in.................. x + p / (1+p²)^-2 = a

What do you get?

i need to solve the differential equation stated above in terms of 'x' and 'y' where p=(dy/dx). .cant put \(\displaystyle p \ = \ \tan(\theta)\), see?

 

[Lost souls]

I cannot. This is beyond me; I mostly help out kids with their algebra.

Here is what I suggest. Wait a bit for someone more competent than me to show up. Everyone here is a volunteer, and some have day jobs so you cannot expect immediate answers. There is a problem though. If people see a lot of responses to a post, they may assume that the post has been answered. So if no one answers this thread by this evening, repost as a new question and apologize for reposting and blame me for the need to repost.

i've read the forum rules n i know what u r talking about. .i'm not rushing and pushing for anything. .i have all the time in the world. .

 

[Lost souls]

basically if someone can tell me how to differentiate p/(1+p²)^{-2. .}

 

[srmichael]

basically if someone can tell me how to differentiate p/(1+p²)^{-2. .}

Do you know the Quotient Rule? Or, even better, the Product Rule as p/(1+p²)^(-2) = p(1+p²)²

 

[Lost souls]

Do you know the Quotient Rule? Or, even better, the Product Rule as p/(1+p²)^(-2) = p(1+p²)²

sorry ,my mistake.. .its p/(1+p²)^1/2

 

[Deleted member 4993]

i need to solve the differential equation stated above in terms of 'x' and 'y' where p=(dy/dx). .cant put \(\displaystyle p \ = \ \tan(\theta)\), see?

You can!!

put \(\displaystyle p = tan(\theta)\) → solve for tan(Θ) = f(x) → p = f(x) → dy/dx = f(x) → simple integration and solve for 'y'

Nothing can be simpler than that !!!

 

[Deleted member 4993]

Oooops .. did not see that the problem changed.

But the strategy remains basically same - please show effort...

 

[Deleted member 4993]

Few steps...

I'll start-off

a = x + p/(1+p²)^1/2

Replace p = tan(Θ)

a - x = sin(Θ)

\(\displaystyle tan(\theta) \ = \ \pm\dfrac{a-x}{\sqrt{1 - (a-x)^2}}\)

\(\displaystyle \dfrac{dy}{dx} \ = \ \pm\dfrac{a-x}{\sqrt{1 - (a-x)^2}}\)

Now can you finish it.....

 

[Deleted member 4993]

\(\displaystyle \dfrac{dy}{dx} \ = \ \pm\dfrac{a-x}{\sqrt{1 - (a-x)^2}}\)

Now can you finish it.....

You can get here by using algebra - but trigonometric substitution is easier.

 

[Lost souls]

You can get here by using algebra - but trigonometric substitution is easier.

taking (a-x)² = z

int(dy/dz)= int[ (-dz/2)/(1-z)^1/2]

y = ((-1/2)((1-z)^(-1/2)+1)/((-1/2)+1) + c

y = -(1-z)^1/2 + c

y = -[1-(a-x)²]^1/2 + c


is this correct?

 

[Deleted member 4993]

taking (a-x)² = z

int(dy/dz)= int[ (-dz/2)/(1-z)^1/2]

y = ((-1/2)((1-z)^(-1/2)+1)/((-1/2)+1) + c

y = -(1-z)^1/2 + c

y = -[1-(a-x)²]^1/2 + c


is this correct?

What happened to \(\displaystyle \pm\) in front of the integral?

Otherwise correct....

 

[Lost souls]

What happened to \(\displaystyle \pm\) in front of the integral?

Otherwise correct....

my bad! i always mess up the signs n end up with wrong answers. .btw, thanx for the help buddy. .much appreciated. .