- #1

Tabiri

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- 0

## Homework Statement

If ##xs^2 + yt^2 = 1## (1) and ##x^2s + y^2t = xy - 4,## (2) find ##\frac{\partial x}{\partial s}, \frac{\partial x}{\partial t}, \frac{\partial y}{\partial s}, \frac{\partial y}{\partial t}## at ##(x,y,s,t) = (1,-3,2,-1)##.

## Homework Equations

Pretty much those just listed above.

## The Attempt at a Solution

Alright so I spent quite a while on this one. First of all, I took the differentials of both the equations, (1) and (2), and came up with $$s^2\,dx + 2sx\,ds + t^2\,dy + 2ty\,dt = 0 $$ for (1) and $$ 2xs\,dx + x^2\,ds + 2yt\,dy + y^2\,dt = y\,dx + x\,dy$$ for (2).

Then, substituting in ##(x,y,s,t) = (1,-3,2,-1)## for the respective variables I came up with $$4\,dx + 4\,ds + \,dy + 6\,dt = 0 $$ for (1) and $$4\,dx + \,ds + 6\,dy + 9\,dt = -3\,dx + \,dy$$ for (2).

Simplifying and moving things around a bit, I got $$ 4\,dx + 4\,ds = -\,dy - 6\,dt $$ for (1) and $$ 7\,dx + \,ds = -5\,dy - 9\,dt $$ for (2).

Then I used Cramer's Rule to solve these two equations for ##\,dx## and ##\,ds.##. For ##\,dx## I got $$ \,dx = \frac{

\begin{vmatrix}

-\,dy - 6\,dt & 4 \\

-5\,dy - 9\,dt & 1

\end{vmatrix}

}{

\begin{vmatrix}

4 & 4 \\

7 & 1

\end{vmatrix}

}

= \frac{-\,dy - 6\,dt + 4(5\,dy + 9\,dt)}{-24} = \frac{-\,dy - 6\,dt + 20\,dy + 36\,dt}{-24} = \frac{19\,dy + 30\,dt}{-24} = \frac{-19\,dy - 30\,dt}{24}$$

$$ \,ds = \frac{

\begin{vmatrix}

4 & -\,dy - 6\,dt \\

7 & -5\,dy - 9\,dt

\end{vmatrix}

}{

\begin{vmatrix}

4 & 4 \\

7 & 1

\end{vmatrix}

} = \frac{4(-5\,dy - 9\,dt) + 7(\,dy + 6\,dt)}{-24} = \frac{-20\,dy - 36\,dt + 7\,dy + 42\,dt}{-24} = \frac{-13\,dy + 6\,dt}{-24} = \frac{13\,dy - 6\,dt}{24}.

$$

Then I changed the equations around to $$\,dy + 6\,dt = -4\,dx - 4\,ds$$ for (1) and $$5\,dy + 9\,dt = -7\,dx - \,ds$$ for (2). Using Cramer's Rule again to solve for ##\,dy## and ##\,dt## I got

$$\,dy = \frac{-6\,dx + 30\,ds}{21}$$

$$\,dt = \frac{-13\,dx - 19\,ds}{21}.$$

Then I wanted to find ##\frac{\partial x}{\partial s}##, so I took what I got for ##\,dy## and ##\,dt## plugged it into (1), ##4\,dx + 4\,ds + \,dy + 6\,dt = 0,## and got $$4\,dx + 4\,ds + \frac{-6\,dx + 30\,ds}{21} + 6(\frac{-13\,dx - 19\,ds}{21}) = 0.$$ Multiplying this all out, everything just cancels to zero and I can't find ##\frac{\partial x}{\partial s}##. I've checked the math for the applications of Cramer's Rule and can't find anything wrong there, so... what am I doing wrong?

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