I did an exercise in the book Vector Calculus[Marsden & Tromba] and I check my answer in the book Vector Calculus Study Guide and Solutions Manual[Karen Pao, Frederick Soon] but my answer is not the same and I don't understand why.
- Is it possible to solve the system of equations \begin{align} xy^2+xzu+yv^2 & = 3 \\ u^3yz+2xv-u^2v^2 &= 2 \end{align} for $u(x,y,z)$, $v(x,y,z)$ near $(x,y,z)=(1,1,1)$, $(u,v)=(1,1)$? Compute $\partial u/\partial y$ at $(x,y,z)=(1,1,1)$.
This is my answer: \begin{align} f_1(x,y,z,u,v) &= xy^2+xzu+yv^2-3 \\ f_2(x,y,z,u,v) &= u^3yz+2xv-u^2v^2-2 \end{align} $$\Delta = \frac{\partial(f_1,f_2)}{\partial (u,v)} = \begin{vmatrix} xz & 2vy \\ -2uv^2+3u^2yz & -2u^2v+2x\end{vmatrix}$$ $$\left.\Delta\right|_{(x,y,z,u,v)=(1,1,1,1,1)} = \begin{vmatrix} 1 & 2 \\ 1 & 0\end{vmatrix} = -2 \neq 0$$ Thus, it is possible to solve in terms of $x$, $y$, $z$ at the given point.
To compute $\partial v/\partial y$ I derive implicitly: $$ \begin{align} \frac{\partial f_1}{\partial y} = 2xy+xz\frac{\partial u}{\partial y}+v^2+2yv\frac{\partial v}{\partial y} &= 0 \\ \frac{\partial f_2}{\partial y} = 3u^2yz\frac{\partial u}{\partial y}+u^3z+2x\frac{\partial v}{\partial y}-2uv^2\frac{\partial u}{\partial y}-2u^2v\frac{\partial v}{\partial y} &= 0 \end{align} $$ For $(x,y,z)=(1,1,1)$ we have: $$ \begin{align} 2+\frac{\partial u}{\partial y} + v^2+2v\frac{\partial v}{\partial y} &= 0 \\ 3u^2\frac{\partial u}{\partial y}+u^3+2\frac{\partial v}{\partial y}-2uv^2\frac{\partial u}{\partial y}-2u^2v\frac{\partial v}{\partial y} &= 0 \end{align} \tag{1}\label{1} $$
And, after a boring time, I get: $$\frac{\partial v}{\partial y} = \frac{6u^2+3u^2v^2-u^3-4uv^2-2uv^4}{2+4uv^3-2u^2v-6u^2v}$$
But in the Solutions Manual, they substitute also $(u,v)=(1,1)$ in $\eqref{1}$(which is much faster) and their answer is $$ \frac{\partial v}{\partial y} = -1 $$
If that's correct, I don't understand it because in the exercise they ask us to compute $\partial v/\partial y$ at $(x,y,z)=(1,1,1)$