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

Lemma for Example: $u^2 + v^2 = x^2$, $2 u v = 2 x y + y^2$

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

Proof

We make sure that $\tuple {1, -2}$ is actually a solution at $u = 1$, $v = 0$:

\(\ds u^2 + v^2\)

\(=\)

\(\ds \paren {-1}^2 + 0^2\)

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

\(\ds \)

\(=\)

\(\ds x^2\)

at $\tuple {1, -2}$

\(\ds 2 u v\)

\(=\)

\(\ds 2 \times \paren {-1} \times 0\)

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

\(\ds \)

\(=\)

\(\ds 2 \times 1 \times \paren {-2} + \paren {-2}^2\)

at $\tuple {1, -2}$

\(\ds \)

\(=\)

\(\ds 2 x y + y^2\)

at $\tuple {1, -2}$

and it is seen that this is the case.

$\blacksquare$

Sources

• 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $15$