1. ## Algebraic Simplification

Show that $\displaystyle \frac{(-1+(a)^2)}{(2-y)}=\frac{17}{(2+y)^3}$ where

$\displaystyle a=\frac{(1-x)}{(2-y)}$ and also $\displaystyle x^2 - y^2 - 2x +4y -20 =0$

Can anyone do? lol
p.s. This is the end simplification of another problem which I just don't know how to do.

Looking in particular to learn useful methods to do this efficiently for future reference.

Edit: Still need help...

3. ## Re: Algebraic Simplification

Sorry, it's

Show that (-1+(a)^2)/(2-y)=17/(x+3)^3 where a=(1-x)/(2-y)

4. ## Re: Algebraic Simplification

Originally Posted by Umair
Sorry, it's

Show that (-1+(a)^2)/(2-y)=17/(x+3)^3 where a=(1-x)/(2-y)

This is manifestly untrue. Simply pick x = 1, y = 1. Then a = 0. The LHS does not equal the RHS.

5. ## Re: Algebraic Simplification

Simplify $\displaystyle \displaystyle -1+\left(\frac{1-x}{2-y}\right)^2$

Do you get the RHS?

6. ## Re: Algebraic Simplification

Oh man, I'm so sorry, let me re-type it LOL. I got the RHS denominator bit wrong...

It's supposed to be:

Show that (-1+(a)^2)/(2-y)=17/(2+y)^3 where a=(1-x)/(2-y)

7. ## Re: Algebraic Simplification

Originally Posted by Umair
Oh man, I'm so sorry, let me re-type it LOL. I got the RHS denominator bit wrong...

It's supposed to be:

Show that (-1+(a)^2)/(2-y)=17/(2+y)^3 where a=(1-x)/(2-y)
Still not true. I'm always suspicious when I see x's on one side of an equation that don't seem to have any way of canceling, and then I don't see them on the RHS of the equation. I do get the following:

$\displaystyle \frac{a^{2}-1}{2-y}=-\frac{(1+x-y)(x+y-3)}{(2-y)^{3}}.$

8. ## Re: Algebraic Simplification

Nup, this should definitely be true... unless my University maths professor made the mistake lol

Urm, its actually the end bit of another problem. Maybe you could do the full problem?:

Equation of a circle centre (1, -2) and radius 5 is given by
x^2-y^2-2x+4y-20=0

Show that (d^2.y)/(d.x^2 )=17/(2+y)^3

At the original post, the LHS is d^2y/dx^2, and a = dy/dx.

9. ## Re: Algebraic Simplification

Originally Posted by Ackbeet
Still not true. I'm always suspicious when I see x's on one side of an equation that don't seem to have any way of canceling, and then I don't see them on the RHS of the equation. I do get the following:

$\displaystyle \frac{a^{2}-1}{2-y}=-\frac{(1+x-y)(x+y-3)}{(2-y)^{3}}.$
The Bottom part of the RHS is $\displaystyle (2+y)^3$, not $\displaystyle (2-y)^3$ :S

10. ## Re: Algebraic Simplification

Originally Posted by Umair
Nup, this should definitely be true... unless my University maths professor made the mistake lol

Urm, its actually the end bit of another problem. Maybe you could do the full problem?:

Equation of a circle centre (1, -2) and radius 5 is given by
x^2-y^2-2x+4y-20=0

Show that (d^2.y)/(d.x^2 )=17/(2+y)^3

At the original post, the LHS is d^2y/dx^2, and a = dy/dx.
not a circle ...

Equation of a circle centre (1, -2) and radius 5 is given by
$\displaystyle (x-1)^2 + (y+2)^2 = 25$

start from here ...

11. ## Re: Algebraic Simplification

Both of those equations are the same circle, lol

12. ## Re: Algebraic Simplification

Hmm. I've moved this thread to Calculus, now that we're talking about second derivatives obtained implicitly. I get the following:

$\displaystyle x^{2}-y^{2}-2x+4y-20=0$

$\displaystyle y'=\frac{x-1}{y-2},$

which appears to be your 'a'.

Using the quotient rule now, we get

$\displaystyle y''=\frac{1-(y')^{2}}{y-2}=\frac{(1+x-y)(x+y-3)}{(2-y)^{3}}.$

As skeeter pointed out, the equation

$\displaystyle x^{2}-y^{2}-2x+4y-20=0$

is not a circle. It is an hyperbola, since the signs of the $\displaystyle x^{2}$ and $\displaystyle y^{2}$ terms are opposite.

13. ## Re: Algebraic Simplification

Ohh quite a few things make sense now! Thank you very much! One question though, is your value of y' supposed to be that? I've re-done my calculation for dy/dx many times and checked that you didn't get my answer for a?

14. ## Re: Algebraic Simplification

I did my calculations in Post # 12 based on the (incorrect) hyperbola equation, not the (correct) circle equation exhibited by skeeter in post # 10. For the correct circle equation, I get the following:

$\displaystyle y'=\frac{1-x}{2+y},$

and

$\displaystyle y''=\frac{(2+y)(-1)-(1-x)(y')}{(2+y)^{2}}=-\frac{y+2-(x-1)y'}{(2+y)^{2}}=-\frac{5-2x+x^{2}+4y+y^{2}}{(y+2)^{3}}$

$\displaystyle =-\frac{x^{2}-2x+1+y^{2}+4y+4}{(y+2)^{3}}=-\frac{(x-1)^{2}+(y+2)^{2}}{(y+2)^{3}}.$

The x's don't cancel, and I didn't think they would.

15. ## Re: Algebraic Simplification

Originally Posted by Ackbeet
I did my calculations in Post # 12 based on the (incorrect) hyperbola equation, not the (correct) circle equation exhibited by skeeter in post # 10. For the correct circle equation, I get the following:

$\displaystyle y'=\frac{1-x}{2+y},$

and

$\displaystyle y''=\frac{(2+y)(-1)-(1-x)(y')}{(2+y)^{2}}=-\frac{y+2-(x-1)y'}{(2+y)^{2}}=-\frac{5-2x+x^{2}+4y+y^{2}}{(y+2)^{3}}$

$\displaystyle =-\frac{x^{2}-2x+1+y^{2}+4y+4}{(y+2)^{3}}=-\frac{(x-1)^{2}+(y+2)^{2}}{(y+2)^{3}}.$

The x's don't cancel, and I didn't think they would.

The fact it may be a circle or not is trivial in the question I think, but it's the "incorrect" equation we need to use.

For the equation of $\displaystyle x^2-y^2-2x+4y-20=0$,

I got:

$\displaystyle 2x -2y\frac{dy}{dx} - 2 +4\frac{dy}{dx} = 0$

which simplifies to: $\displaystyle (2-y)\frac{dy}{dx}=1-x$

and finally $\displaystyle y'=\frac{dy}{dx}=\frac{1-x}{2-y}$

I'm not sure if this is correct or not ,but it's how I did it... please elaborate

Edit:

For $\displaystyle d^2y/dx^2$, I got the following:

$\displaystyle (1) \frac{d}{dx}((2-y)\frac{dy}{dx}+x-1)=0$

$\displaystyle (2) -(\frac{dy}{dx})^2+(2-y)\frac{d^2y}{dx^2}+1=0$

$\displaystyle (3) \frac{d^2y}{dx^2}=\frac{-1+\(dy/dx)^2}{2-y}$

SOMEHOW simplifies to:

$\displaystyle (4) \frac{d^2y}{dx^2}=\frac{x^2-y^2-2x+4y-3}{(2+y)^3} = \frac{17}{(2+y)^3}$

Any idea how (3) simplifies to Middle section of (4)?

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