1. ## Re: Algebraic Simplification!

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

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

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

$\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)?
I get $\displaystyle d^2y/dx^2=-\frac{x^2-2 x+y^2-4 y+5}{(y-2)^3}\,.$

Which gives: $\displaystyle d^2y/dx^2=\frac{2y^2-8 y+25}{(2-y)^3}\,.$

2. ## Re: Algebraic Simplification!

Originally Posted by SammyS
I get $\displaystyle d^2y/dx^2=-\frac{x^2-2 x+y^2-4 y+5}{(y-2)^3}\,.$

Which gives: $\displaystyle d^2y/dx^2=\frac{2y^2-8 y+25}{(2-y)^3}\,.$
How did you manage to derive $\displaystyle d^2y/dx^2=-\frac{x^2-2 x+y^2-4 y+5}{(y-2)^3}\,$?

3. ## Re: Algebraic Simplification!

You know: $\displaystyle \frac{dy}{dx}=\frac{1-x}{2-y}$ (1)
and you got:
$\displaystyle \frac{d^2y}{dx^2}=\frac{-1+\left(\frac{dy}{dx}\right)^2}{2-y}$ (2)

Substituting (1) in (2) gives:
$\displaystyle \frac{d^2y}{dx^2}=\frac{-1+\left(\frac{1-x}{2-y}\right)^2}{2-y}$
$\displaystyle =\frac{-1+\frac{(1-x)^2}{(2-y)^2}}{2-y}$
$\displaystyle =\frac{\frac{-1\cdot(2-y)^2+(1-x)^2}{(2-y)^2}}{2-y}$
$\displaystyle =\frac{-1(4-4y+y^2)+1-2x+x^2}{(2-y)^3}$
$\displaystyle =\frac{-4+4y-y^2+1-2x+x^2}{(2-y)^3}$
$\displaystyle =\frac{-y^2+4y+x^2-2x-3}{(2-y)^3}$

4. ## Re: Algebraic Simplification

Originally Posted by Umair
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.
That statement makes no sense. If the original problem statement is talking about the equation of a circle centered at (1, -2) and of radius 5, then you are NOT talking about the equation x^2-y^2-2x+4y-20=0, because that simply isn't a circle at all, much less the circle specified.

However, if you work from the circle equation given by skeeter in post # 10, you don't get to the result indicated in the OP. If you work from the incorrect hyperbola equation, you still don't get to the result indicated in the OP. In summary: you have been asked to do the impossible. There is a mistake in the problem.

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
That's correct for the hyperbola equation.

Edit:

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

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

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

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

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)?
Here you plug in the value for the derivative that you got before:

$\displaystyle \frac{d^2y}{dx^2}=\frac{-1+\left(dy/dx\right)^2}{2-y}=\frac{-1+\left(\frac{1-x}{2-y}\right)^2}{2-y},$

and then you get the common denominator, add the fractions, etc., etc., etc.

5. ## Re: Algebraic Simplification

Originally Posted by Siron
You know: $\displaystyle \frac{dy}{dx}=\frac{1-x}{2-y}$ (1)
and you got:
$\displaystyle \frac{d^2y}{dx^2}=\frac{-1+\left(\frac{dy}{dx}\right)^2}{2-y}$ (2)

Substituting (1) in (2) gives:
$\displaystyle \frac{d^2y}{dx^2}=\frac{-1+\left(\frac{1-x}{2-y}\right)^2}{2-y}$
$\displaystyle =\frac{-1+\frac{(1-x)^2}{(2-y)^2}}{2-y}$
$\displaystyle =\frac{\frac{-1\cdot(2-y)^2+(1-x)^2}{(2-y)^2}}{2-y}$
$\displaystyle =\frac{-1(4-4y+y^2)+1-2x+x^2}{(2-y)^3}$
$\displaystyle =\frac{-4+4y-y^2+1-2x+x^2}{(2-y)^3}$
$\displaystyle =\frac{-y^2+4y+x^2-2x-3}{(2-y)^3}$
Thank you!! This was exactly what I was looking for! However it seems your denominator is $\displaystyle (2-y)^3 instead of (2+y)^3$

Although your numerator is exactly what was needed, I think the solutions I got from my professor were wrong lol. No wonder I killed off a few brain cells trying to work this one out, lol. Everyone was getting something different...

Originally Posted by Ackbeet
That statement makes no sense. If the original problem statement is talking about the equation of a circle centered at (1, -2) and of radius 5, then you are NOT talking about the equation x^2-y^2-2x+4y-20=0, because that simply isn't a circle at all, much less the circle specified.

However, if you work from the circle equation given by skeeter in post # 10, you don't get to the result indicated in the OP. If you work from the incorrect hyperbola equation, you still don't get to the result indicated in the OP. In summary: you have been asked to do the impossible. There is a mistake in the problem.

That's correct for the hyperbola equation.

Here you plug in the value for the derivative that you got before:

$\displaystyle \frac{d^2y}{dx^2}=\frac{-1+\left(dy/dx\right)^2}{2-y}=\frac{-1+\left(\frac{1-x}{2-y}\right)^2}{2-y},$

and then you get the common denominator, add the fractions, etc., etc., etc.
I think I discovered quite a few errors in the sheet, lol. Either way what I meant to say was the equation for the hyperbola had to be used.

More importantly, your expression for dy/dx for the hyperbola equation was different to mine, although you still said I did it correctly?

6. ## Re: Algebraic Simplification

Originally Posted by Umair
I think I discovered quite a few errors in the sheet, lol. Either way what I meant to say was the equation for the hyperbola had to be used.

More importantly, your expression for dy/dx for the hyperbola equation was different to mine, although you still said I did it correctly?
In my post # 12, multiply top and bottom by -1, thus not changing the fraction, and you get your expression for the derivative.

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