Partial differentiation and partial derivatives

[Tabiri]

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?

 

[Ray Vickson]

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?

I cannot understand what you are trying to do, or why you want to do it. The easiest way is just to differentiate directly. Letting $x_s = \partial x/\partial s$, etc, we have:
(1) \Longrightarrow s^2 x_s + 2 s x + t^2 y_s = 0 \\<br /> (2) \Longrightarrow x^2 + 2 s x x_s + 2 t y y_s = x y_s + y x_s<br />
which you can solve for $x_s,y_s$. It is much easier if you first substitute in the numerical values of $x,y,s,t$.

 

[Tabiri]

As in solve for $y_s$ in (1) and then plug that value for $y_s$ into (2)?

Edit: Also, I know that the answer for $\frac{\partial x}{\partial s}$ is $\frac{-19}{13}$

 

[Ray Vickson]

As in solve for $y_s$ in (1) and then plug that value for $y_s$ into (2)?

Edit: Also, I know that the answer for $\frac{\partial x}{\partial s}$ is $\frac{-19}{13}$

You have two equations in the two unknowns $x_s, y_s$, and these equations are linear in these unknowns. They happen to be nonlinear in $s,t,x,y$, but that does not matter, since we are already given numerical values for them. The things we don't know are $x_s, y_s$ and $x_t,y_t$.

 

[Tabiri]

Yeah, I just did it that way and got the right answer. Thanks!