Computing partial derivatives using three implicitly defined equations

Question

The three equations

$x^2-y\operatorname{cos}(uv)+z^2=0$

$x^2+y^2-\operatorname{sin}(uv)+2z^2=0$

$xy-\operatorname{sin}u\operatorname{cosv}+z=0$

define $x,y,z$ as functions of $u,v$. Compute the partial derivatives $\partial x/\partial u$ and $\partial x/\partial v$ at the point $x=y=1, u=\pi/2, v=0, z=0$.

I think I need to set three functions $F[X(u,v),Y(u,v),Z(u,v)],G[X(u,v),Y(u,v),Z(u,v)],H[X(u,v),Y(u,v),Z(u,v)]$, and use chain rule and solve the system of three linear equations. However, the mixture of $u$ and $v$ in the equations make it complicated, and I don't know how to progress with the partial differentiations.

I would greatly appreciate it if anyone could help me solve this problem.

Answer 1

Since $x=x(u,v)$ and $y=y(u,v)$, you can differentiate your equations by $u$ and $v$. $$2x\frac{\partial x}{\partial u}-\frac{\partial y}{\partial u}\cos(uv) + yv\sin(uv)+2z\frac{\partial z}{\partial u}=0$$ $$2x\frac{\partial x}{\partial v}-\frac{\partial y}{\partial v}\cos(uv) + yu\sin(uv)+2z\frac{\partial z}{\partial v}=0$$ $$2x\frac{\partial x}{\partial u}+2y\frac{\partial y}{\partial u}-v\cos(uv)+4z\frac{\partial z}{\partial u}=0$$ $$2x\frac{\partial x}{\partial v}+2y\frac{\partial y}{\partial v}-u\cos(uv)+4z\frac{\partial z}{\partial v}=0$$ $$y\frac{\partial x}{\partial u} + x\frac{\partial y}{\partial u}-\cos u\cos v+\frac{\partial z}{\partial u}=0$$ $$y\frac{\partial x}{\partial v} + x\frac{\partial y}{\partial v}+\sin u\sin v+\frac{\partial z}{\partial v}=0$$

Let's substitute $x=y=1$, $z=0$, $u=\pi/2$, $v=0$; denote $\frac{\partial x}{\partial u}=a$, $\frac{\partial x}{\partial v}=b$, $\frac{\partial y}{\partial u}=c$, $\frac{\partial y}{\partial v}=d$, $\frac{\partial z}{\partial u}=e$, $\frac{\partial z}{\partial v}=f$ (at this points). Then $$2a-c=0$$ $$2b-d=0$$ $$2a+2c=0$$ $$2b+2d-\frac\pi2=0$$ $$a+c+e=0$$ $$b+d+f=0$$ Solution is $a=c=0$, $b=\pi/12$, $d=\pi/6$, $e=0$, $f=-\pi/4$.

Answer 2

Denote the equations by $(1), (2)$ and $(3)$ respectively. Plugging $(3)$ into $(1)$ and $(2)$:

$$ x^2 - y\cos (uv) + (\sin u \cos v - xy)^2 =0 \ (4)$$ $$ x^2+y^2 - \sin (uv) +2(\sin u \cos v - xy)^2=0 \ (5)$$

Let $a$ be the point in $(x,y,z,u,v)$ space with $x=y=1, \ u=\pi /2, \ v=z=0 $.

$$\dfrac{\partial}{\partial u}\bigg|_{a}(4): \ 2\dfrac{\partial x}{\partial u}\bigg|_a -\dfrac{\partial y}{\partial u}\bigg|_a =0$$

$$\dfrac{\partial }{\partial u}\bigg|_a (5): \ 2\dfrac{\partial x}{\partial u}\bigg|_a+2\dfrac{\partial y}{\partial u}\bigg|_a =0$$

$$\implies \boxed{\dfrac{\partial x}{\partial u}\bigg|_a =0}$$

Similarly

$$\dfrac{\partial }{\partial v}\bigg|_a (4): \ 2\dfrac{\partial x}{\partial v}\bigg|_a -\dfrac{\partial y}{\partial v}\bigg|_a=0$$

$$ \dfrac{\partial }{\partial v}\bigg|_a(5): \ 2\dfrac{\partial x}{\partial v}\bigg|_a + 2\dfrac{\partial y}{\partial v}\bigg|_a-\dfrac{\pi}{2} = 0$$

$$\implies \boxed{\dfrac{\partial x}{\partial v}\bigg|_a = \dfrac{\pi}{12}}$$