Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2
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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 $u = 1$, $v = 0$.
$\Box$
Explicit Method
We have:
\(\ds u^2 + v^2\) | \(=\) | \(\ds x^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds u\) | \(=\) | \(\ds \sqrt {x^2 - v^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds \dfrac 1 {2 \paren {\sqrt {x^2 - v^2} } } \paren {2 x - 2 v \dfrac {\partial v} {\partial x} }\) | Power Rule for Derivatives, Chain Rule for Derivatives | ||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds \dfrac {x - v \dfrac {\partial v} {\partial x} } {\sqrt {x^2 - v^2} }\) | simplifying |
At this point we investigate the sign of $\sqrt {x^2 - v^2}$ which is needed.
We see that:
\(\ds u\) | \(=\) | \(\ds \sqrt {x^2 - v^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(=\) | \(\ds +\sqrt {x^2 - v^2}\) |
Hence when we are at the point of plugging in numbers we will need to take the positive square root.
Then we have:
\(\ds 2 u v\) | \(=\) | \(\ds 2 x y + y^2\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \dfrac {2 x y + y^2} {2 u}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial v} {\partial x}\) | \(=\) | \(\ds \dfrac {2 u \map {\dfrac \partial {\partial x} } {2 x y + y^2} - \paren {2 x y + y^2} \map {\dfrac \partial {\partial x} } {2 u} } {\paren {2 u}^2}\) | Quotient Rule for Derivatives | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 u \times 2 x - \paren {2 x y + y^2} 2 \dfrac {\partial u} {\partial x} } {4 u^2}\) | Power Rule for Derivatives, Chain Rule for Derivatives | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds \dfrac {2 u x - \paren {2 x y + y^2} \dfrac {\partial u} {\partial x} } {2 u^2}\) | Power Rule for Derivatives, Chain Rule for Derivatives |
Substituting $\dfrac {\partial v} {\partial x}$ from $(2)$ into $(1)$ gives:
\(\ds \dfrac {\partial u} {\partial x}\) | \(=\) | \(\ds \dfrac {x - v \paren {\dfrac {2 u x - \paren {2 x y + y^2} \dfrac {\partial u} {\partial x} } {2 u^2} } } {\sqrt {x^2 - v^2} }\) | |||||||||||||
Could rearrange and simplify, but easier to substitute the values of $x$ and $y$ at this point: | |||||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1 - 0 \times \paren {\cdots} } {+\sqrt {1^2 - 0^2} }\) | $v$ vanishes, hence the big parenthesis is irrelevant | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | simplifying |
$\blacksquare$
Implicit Method
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 $15$