Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2/Implicit Method

From ProofWiki
Jump to navigation Jump to search

Example of Partial Derivative

Consider the simultaneous equations:

\(\ds u^2 + v^2\) \(=\) \(\ds x^2\)
\(\ds 2 u v\) \(=\) \(\ds 2 x y + y^2\)

Then:

$\map {u_1} {1, -2} = 1$

at $u = 1$, $v = 0$.


Proof

By definition of partial derivative:

$\map {u_1} {1, -2} = \valueat {\dfrac {\partial u} {\partial x} } {x \mathop = 1, y \mathop = -2}$

hence the motivation for the abbreviated notation on the left hand side.


Lemma

Consider the simultaneous equations:

\(\ds u^2 + v^2\) \(=\) \(\ds x^2\)
\(\ds 2 u v\) \(=\) \(\ds 2 x y + y^2\)

Then:

$x = 1$, $y = -2$ is a solution at $u = 1$, $v = 0$.

$\Box$


We have:

\(\ds u^2 + v^2\) \(=\) \(\ds x^2\)
\(\ds \leadsto \ \ \) \(\ds 2 u \dfrac {\partial u} {\partial x} + 2 v \dfrac {\partial v} {\partial x}\) \(=\) \(\ds 2 x\) Power Rule for Derivatives, Chain Rule for Derivatives
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds u \dfrac {\partial u} {\partial x} + v \dfrac {\partial v} {\partial x}\) \(=\) \(\ds x\) simplifying


Then we have:

\(\ds 2 u v\) \(=\) \(\ds 2 x y + y^2\)
\(\ds \leadsto \ \ \) \(\ds 2 v \dfrac {\partial u} {\partial x} + 2 u \dfrac {\partial v} {\partial x}\) \(=\) \(\ds 2 y\) Power Rule for Derivatives, Chain Rule for Derivatives, keeping $y$ constant
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds v \dfrac {\partial u} {\partial x} + u \dfrac {\partial v} {\partial x}\) \(=\) \(\ds y\)


Combining $(1)$ and $(2)$ into matrix form:

$\begin {pmatrix} u & v \\ v & u \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial x} \\ \dfrac {\partial v} {\partial x} \end {pmatrix} = \begin {pmatrix} x \\ y \end {pmatrix}$


Hence by Cramer's Rule:

\(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds \dfrac {\begin {vmatrix} x & v \\ y & u \end {vmatrix} } {\begin {vmatrix} u & v \\ v & u \end {vmatrix} }\)
\(\ds \) \(=\) \(\ds \dfrac {x u - v y} {u^2 - v^2}\) Definition of Determinant

Hence:

\(\ds \map {u_1} {1, -2}\) \(=\) \(\ds \dfrac {1 \times 1 - \paren {-2} \times 0} {1^2 - 0^2}\) substituting $\tuple {1, -2, 1, 0}$ for $\tuple {x, y, u, v}$
\(\ds \) \(=\) \(\ds 1\) simplifying

$\blacksquare$


Sources