Finding partial derivatives using differentials in a system of equations

Question

This homework exercise comes from M. Boas, Mathematical Methods in the Physical Sciences:

Typed out: If $xs^2+yt^2=1$ and $x^s+y^2t=xy-4$, find $\partial x/\partial s$, $\partial x/\partial t$, $\partial y/\partial s$, $\partial y/\partial t$, at $(x,y,s,t)=(1,-3,2,-1)$. Hint: To simplify the work, substitute the numerical values just after you have taken differentials.

So, if I understand the question right there are two shapes, given by the two equations, and they intersect somewhere in 4-D space. We want to find just how much e.g. d$x$ changes when we change d$s$ while keeping $y$ and $t$ fixed.

The differentials I find to be: \begin{align} s^2 (dx)+2xs(ds) + 2yt(dt) + t^2(dy) & = 0,\\ 2sx(dx) + x^2 (ds) + y^2(dt)+2ty(dy) & = x(dy) + y(dx). \end{align}

I substitute the given values and simplify, \begin{align} 4(dx) + 4(ds) + 6(dt) + (dy) & = 0,\\ 4(dx) + (ds) + 9(dt) + 5(dy) & = 0. \end{align}

Initially I assumed that, to find $\partial x/\partial s$, I could just set $(dt)=(dy)=0$, but that system doesn't have a solution. So it appears as if I need to solve the system fully... Using Cramer's rule I obtain \begin{align} (dx) = \frac{\left\lvert\begin{array}{cc} -6(dt)-(dy) & 4 \\ -9(dt)-5(dy)& 1 \end{array} \right\rvert}{ \left\lvert \begin{array}{cc} 4 & 4 \\ 7 & 1 \end{array} \right\rvert } = \frac{30(dt)+19(dy)}{-24}. \end{align} Similarly for $(ds)$ I obtain \begin{align} (ds) = \frac{\left\lvert\begin{array}{cc} 4 & -6(dt)-(dy) \\ 7 & -9(dt)-5(dy) \end{array} \right\rvert}{ \left\lvert \begin{array}{cc} 4 & 4 \\ 7 & 1 \end{array} \right\rvert } = \frac{6(dt)-13(dy)}{-24}. \end{align} Taking the ratio of the two gives $$ \frac{dx}{ds}=\frac{30(dt)+19(dy)}{6(dt)-13(dy)} $$ The correct answer, according to the answer sheet, is $\partial x/\partial s=-19/13$.

I repeated the procedure for all the four desired partial derivatives, to find: \begin{align} \frac{dx}{ds}&=\frac{30(dt)+19(dy)}{6(dt)-13(dy)},\\ \frac{dx}{dt}&=\frac{-30(ds)+21(dy)}{24(ds)-13(dy)},\\ \frac{dy}{ds}&=\frac{-24(dx)+2(dt)}{-13(dx)-21(dt)},\\ \frac{dy}{dt}&=\frac{-6(dx)+30(ds)}{-13(dx)-19(ds)}. \end{align} Comparing to the answers, I would obtain the right solution if I would set $(ds)$ or $(dt)$ to zero on the right-hand side each time. But I don't see anything in the question that motivatives why such an assumption is appropriate. Was there anything in the question/equations that suggests that I could set $(dt)$ or $(ds)$ to zero? Did I take the wrong route to obtain my answers?

Answer 1

Either bad notations, or the books explained the notations elsewhere. This is why partial derivative is confusing, especially when used in physics without clear explanation. The simple reason is that $x,y$ are being treated as a functions in term of $s,t$ only. So the partial derivative? They are keeping only one other variable constant. For example $\frac{\partial x}{\partial s}$ keep $t$ constant, so you can only set $dt=0$, $dy$ is not affected.

Which makes sense. $2$ equations in $4$-dimensional space should cut out a surface, which is $2$-dimensional, so you can only use $2$ variables as coordinate.