1.14 Implicit Function Theorem
For \(i = 1,2,\cdots \cdots , m\) , let the functions \(F_i\big (x_1,\cdots \cdots ,x_n\,,\ y_1,\cdots \cdots ,y_m\big )\) be defined in a neighbourhood of a point \(P_0\big (x_1^0,\cdots \cdots ,x_n^0\,,\ y_1^0,\cdots \cdots , y_m^0\big )\) and have continuous first partial
derivatives in this neighbourhood.
Let the equations \(F_i\big (x_1,\cdots \cdots ,x_n\,,\ y_1,\cdots \cdots ,y_m\big ) = 0\), \(i= 1,\cdots \cdots ,m\), be satisfied at \(P_0\), and let \[\frac {\partial \big (F_1,\cdots \cdots ,F_m\big )} {\partial \big (y_1,\cdots \cdots ,y_m\big )} \neq 0 \hspace {0.4cm}\text {at}\hspace {0.4cm} P_0 .\]
Then in an appropriate neighbourhood of \((x^0_1,\cdots \cdots , x^0_n)\) there is a unique set of continuous functions
\[y_i = f_i (x_1, \cdots \cdots ,x_n) \hspace {0.5cm} i = 1,\cdots \cdots \cdots , m\]
such that
\[y_i^0 = f_i (x^0_1,\cdots \cdots ,x^0_n)\hspace {0.4cm} \text {for}\hspace {0.4cm} i = 1,2,\cdots \cdots , m,\]
and such that, for every \(i\),
\[F_i\big (f_1(x_1,\cdots \cdots ,x_n),\cdots \cdots , f_m(x_1,\cdots \cdots ,x_n)\,,\
x_1,\cdots \cdots , x_n\big ) = 0\]
throughout that neighbourhood.
Let \(F(x,y,z) = x^2y^3 + 4xyz^2 - 5xyz^4 + z^2 + 1 = 0\) Show that \(z\) can be written as a function of \(x\) and \(y\) in a neighbourhood of the point \((2,-1,1)\) and compute \(\dfrac {\partial z}{\partial x}\) and \(\dfrac {\partial z}{\partial y}\) at the point.
Solution.
\(F(2,-1,1) = -4-8+10+2 =0\)
So that the hypothesis of the IFT are verified.
\[\frac {\partial F}{\partial z} = 32xyz^7 - 20xyz^3 + 2z\]
\[F_z(2,-1,1) = -22 \neq 0\]
So that for \((x,y)\) in a neighbourhood \((2,-1)\) there is a unique function \(f\) such that \(z=f(x,y)\) \[F(x,y,f(x,y)) =0\]
N.B \(f(2,-1) = 1\)
\begin {align*} \text {Then}\hspace {0.5cm} \frac {\partial z}{\partial x} & = \frac {-\partial F}{\partial x}\Bigg / \frac {\partial F}{\partial z}\\ & =\frac {-\big (2xy^3 + 4yz^8 -5yz^4\big )}{-22}\\ & = \frac {-3}{22}\\ \end {align*}
\begin {align*} \frac {\partial z}{\partial y} & = \frac {-F_y}{F_z}\\ & = \frac {-\big (3x^2y^2 + 4xz^8 - 5xz^4\big )}{-22}\\ & = \frac {5}{11}\\ \end {align*}
\(2x^2 + y^2 + z^2 -zw = 0 \hspace {1.7cm} F\)
\(x^2 + y^2 +2z^2 + zw- 8 = 0 \hspace {1cm} G\)
Find \(\dfrac {\partial z}{\partial x}, \dfrac {\partial w}{\partial x}\)
- 1.
- Show that \(w\) and \(z\) can be written as functions of \(x\) and \(y\) in a neighbourhood of \((1,1,1,4)\).
\(F(1,1,1,4) = 0\)
\(G(1,1,1,4) = 0\) \begin {align*} \frac {\partial \big (F,G\big )}{\partial (z,w)} & = \begin {vmatrix} F_z & F_w\\ G_z & G_w\\ \end {vmatrix}=\begin {vmatrix} 2z - w & -z\\ 4z + w & z\\ \end {vmatrix}\\\\ & = 6z^2\\ \end {align*}
at \(\dfrac {\partial \big (F,G\big )}{\partial (z,w)} = 6 \neq 0\)
- 2.
- \begin {align*} \frac {\partial z}{\partial x} & = \frac {- \partial \big (F,G\big )}{\partial (x,y)}\Bigg / \frac {\partial \big (F,G\big )}{\partial (z,w)}\\\\ & = \frac {-\begin {vmatrix} F_x & F_w\\ G_x & G_w\\ \end {vmatrix} }{6} = \frac {- \begin {vmatrix} 4x & -z\\ 2x & z\\ \end {vmatrix} }{6}\\ & = \frac {-6xz}{6}\\\\ & = -xz \end {align*}
\[\implies \hspace {1cm} \frac {\partial z}{\partial x}(1,1,1,4) = -1\]
\begin {align*} \frac {\partial w}{\partial x} & = \frac {- \partial \big (F,G\big )}{\partial (z,x)}\Bigg / \frac {\partial \big (F,G\big )}{\partial (z,w)}\\\\ & = \frac {-\begin {vmatrix} F_z & F_x\\ G_z & G_x\\ \end {vmatrix} }{6} = \frac {- \begin {vmatrix} 2z-w & 4x\\ 4z + w & 2x\\ \end {vmatrix} }{6}\\ & = \frac {-(-12xz -6xw)}{6}\\ & = 2xz + xw \end {align*}
\[\frac {\partial w}{\partial x}(1,1,1,4) =6\]
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.