Partial Derivative/Examples/2 u + 3 v = sin x, u + 2 v = x cos y/Explicit Method

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Example of Partial Derivative

Consider the simultaneous equations:

$\begin {cases} 2 u + 3 v & = \sin x \\ u + 2 v & = x \cos y \end {cases}$


Then:

$\map {u_1} {\dfrac \pi 2, \pi} = 3$


Proof

By definition of partial derivative:

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

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


We have:

$2 u + 3 v = \sin x$

Thus:

\(\ds u\) \(=\) \(\ds \dfrac {\sin x - 3 v} 2\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds \dfrac {\cos x} 2 - \dfrac 3 2 \dfrac {\partial v} {\partial x}\)


Then we have:

$u + 2 v = x \cos y$

Thus:

\(\ds v\) \(=\) \(\ds \dfrac {x \cos y - u} 2\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {\partial v} {\partial x}\) \(=\) \(\ds \dfrac {\cos y} 2 - \dfrac 1 2 \dfrac {\partial u} {\partial x}\)


Substituting $\dfrac {\partial v} {\partial x}$ from $(2)$ into $(1)$ gives:

\(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds \dfrac {\cos x} 2 - \dfrac 3 2 \paren {\dfrac {\cos y} 2 - \dfrac 1 2 \dfrac {\partial u} {\partial x} }\)
\(\ds \) \(=\) \(\ds \dfrac {\cos x} 2 - \dfrac {3 \cos y} 4 + \dfrac 3 4 \dfrac {\partial u} {\partial x}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial x} \paren {1 - \dfrac 3 4}\) \(=\) \(\ds \dfrac {\cos x} 2 - \dfrac {3 \cos y} 4\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\partial u} {\partial x}\) \(=\) \(\ds 2 \cos x - 3 \cos y\)
\(\ds \leadsto \ \ \) \(\ds \map {u_1} {\dfrac \pi 2, \pi}\) \(=\) \(\ds 2 \map \cos {\dfrac \pi 2} - 3 \cos \pi\)
\(\ds \) \(=\) \(\ds 2 \times 0 - 3 \times \paren {-1}\) Cosine of Right Angle, Cosine of Straight Angle
\(\ds \) \(=\) \(\ds 3\) simplifying

$\blacksquare$


Sources

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