Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2/Implicit Method
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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
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $15$