# Algebraic Simplification

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Aug 4th 2011, 05:49 PM
Umair
Algebraic Simplification
Show that $\frac{(-1+(a)^2)}{(2-y)}=\frac{17}{(2+y)^3}$ where

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

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...
• Aug 4th 2011, 05:56 PM
pickslides
re: Algebraic Simplification
• Aug 4th 2011, 06:01 PM
Umair
Re: Algebraic Simplification
Sorry, it's

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

• Aug 4th 2011, 06:04 PM
Ackbeet
Re: Algebraic Simplification
Quote:

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.
• Aug 4th 2011, 06:04 PM
pickslides
Re: Algebraic Simplification
Simplify $\displaystyle -1+\left(\frac{1-x}{2-y}\right)^2$

Do you get the RHS?
• Aug 4th 2011, 06:08 PM
Umair
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)
• Aug 4th 2011, 06:11 PM
Ackbeet
Re: Algebraic Simplification
Quote:

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:

$\frac{a^{2}-1}{2-y}=-\frac{(1+x-y)(x+y-3)}{(2-y)^{3}}.$
• Aug 4th 2011, 06:20 PM
Umair
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.
• Aug 4th 2011, 06:33 PM
Umair
Re: Algebraic Simplification
Quote:

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:

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

The Bottom part of the RHS is $(2+y)^3$, not $(2-y)^3$ :S
• Aug 4th 2011, 06:34 PM
skeeter
Re: Algebraic Simplification
Quote:

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 ...

Quote:

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

start from here ...
• Aug 4th 2011, 06:37 PM
Umair
Re: Algebraic Simplification
Both of those equations are the same circle, lol
• Aug 4th 2011, 06:42 PM
Ackbeet
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:

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

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

which appears to be your 'a'.

Using the quotient rule now, we get

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

As skeeter pointed out, the equation

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

is not a circle. It is an hyperbola, since the signs of the $x^{2}$ and $y^{2}$ terms are opposite.
• Aug 4th 2011, 06:51 PM
Umair
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?
• Aug 4th 2011, 07:14 PM
Ackbeet
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:

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

and

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

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

The x's don't cancel, and I didn't think they would.
• Aug 4th 2011, 07:22 PM
Umair
Re: Algebraic Simplification
Quote:

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:

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

and

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

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

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 $x^2-y^2-2x+4y-20=0$,

I got:

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

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

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

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

Edit:

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

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

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

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

SOMEHOW simplifies to:

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

Any idea how (3) simplifies to Middle section of (4)?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last