Ejercicios

5.1 Sea

f (x, y) = \frac{xy}{x^{2} - y^{2}} .

Calcule

\frac{∂f}{∂y},

\frac{∂f}{∂x}

f_{y} (2, 1) .

Usando la regla para la derivada del cociente,

\begin{matrix} \frac{∂f}{\partial y} & = & \frac{\frac{\partial}{∂y} [xy] \cdot (x^{2} - y^{2}) - \frac{\partial}{∂y} [x^{2} - y^{2}] \cdot xy}{{(x^{2} - y^{2})}^{2}} \\ = & \frac{x \cdot (x^{2} - y^{2}) + 2 y \cdot xy}{{(x^{2} - y^{2})}^{2}} \end{matrix}

\begin{matrix} \frac{∂f}{\partial x} & = & \frac{\frac{\partial}{∂x} [xy] \cdot (x^{2} - y^{2}) - \frac{\partial}{∂x} [x^{2} - y^{2}] \cdot xy}{{(x^{2} - y^{2})}^{2}} \\ = & \frac{y \cdot (x^{2} - y^{2}) - 2 x \cdot xy}{{(x^{2} - y^{2})}^{2}} \end{matrix}

f_{y} (2, 1) = \frac{10}{9}

5.2 Sea

f (x, y) = {ln}^{5} (x^{y} + x^{2} + 2^{y})

Calcule

\frac{∂f}{∂y},

\frac{∂f}{∂x} .

Se debe usar la regla de la cadena para funciones de una variable,

\begin{matrix} \frac{∂f}{\partial y} & = & 5 {ln}^{4} (x^{y} + x^{2} + 2^{y}) \cdot \frac{\partial}{∂y} [ln (x^{y} + x^{2} + 2^{y})] \\ = & 5 {ln}^{4} (x^{y} + x^{2} + 2^{y}) \cdot \frac{1}{x^{y} + x^{2} + 2^{y}} \cdot (x^{y} ln x + 2^{y} ln 2) \\ = \\ \frac{∂f}{\partial x} & = & 5 {ln}^{4} (x^{y} + x^{2} + 2^{y}) \cdot \frac{\partial}{∂x} [ln (x^{y} + x^{2} + 2^{y})] \\ = & 5 {ln}^{4} (x^{y} + x^{2} + 2^{y}) \cdot \frac{1}{x^{y} + x^{2} + 2^{y}} \cdot (y \cdot x^{y - 1} + 2 x) \end{matrix}

5.3 Sea

z (x, y) = 2 {(ax + by)}^{2} - (x^{2} + y^{2})

con

a^{2} + b^{2} = 1 .

Verifique que

\frac{\partial 2 z}{∂x 2} + \frac{\partial 2 z}{∂y 2} = 0 .

•: $\frac{\partial 2 z}{∂x 2} = 4 a^{2} - 2$
•: $\frac{\partial 2 z}{∂y 2} = 4 b^{2} - 2$
•: $\frac{\partial 2 z}{∂x 2} + \frac{\partial 2 z}{∂y 2} = 4 (a^{2} + b^{2}) - 4 = 0 . \sqrt$

5.4 Sea

z = f (\frac{x^{2}}{y})

con

f

derivable. Verifique que

x \frac{∂z}{∂x} + 2 y \frac{∂z}{∂y} = 0 .

Sea

u = \frac{x^{2}}{y},

entonces

z = f (u)

•: $\frac{∂z}{∂x} = f^{'} (u) \cdot \frac{2 x}{y}$
•: $\frac{∂z}{∂y} = f^{'} (u) \cdot \frac{- x^{2}}{y^{2}}$
•: $x \frac{∂z}{∂x} + 2 y \frac{∂z}{∂y} = f^{'} (u) [\frac{2 x^{2}}{y} - \frac{2 x^{2}}{y}] = 0 \sqrt$

5.5 Sea

z = \sqrt{xy + arctan (\frac{y}{x})}

. Demuestre que

zx \frac{∂z}{∂x} + zy \frac{∂z}{∂y} = xy

•: $\frac{∂z}{∂x} = \frac{y - \frac{y}{x^{2} + y^{2}}}{2 z}$
•: $\frac{∂z}{∂x} = \frac{x + \frac{x}{x^{2} + y^{2}}}{2 z}$
•: Ahora sustituimos, $\begin{array}{rcl} zx \frac{∂z}{∂x} + zy \frac{∂z}{∂y} & = & zx \frac{y - \frac{y}{x^{2} + y^{2}}}{2 z} + zy \frac{x + \frac{x}{x^{2} + y^{2}}}{2 z} \\ = & \frac{2 xy - \frac{xy}{x^{2} + y^{2}} + \frac{xy}{x^{2} + y^{2}}}{2} = xy \end{array}$

5.6 Sea

k

una constante y

C (x, t) = t^{- 1 ∕ 2} e^{^{- x^{2} ∕ kt}}

. Verifique que esta función satisface la ecuación (de difusión)

\frac{k}{4} \cdot \frac{\partial^{2} C}{\partial x^{2}} = \frac{∂C}{∂t}

Pongamos

C (x, t) = \frac{e^{^{- x^{2} ∕ kt}}}{\sqrt{t}} .

•: $\frac{∂C}{∂t} = \frac{(\sqrt{t} \frac{- 2 x}{kt} - \frac{1}{\sqrt{t}}) e^{^{- x^{2} ∕ kt}}}{t} = e^{^{- x^{2} ∕ kt}} (\frac{x^{2}}{k t^{5 ∕ 2}} - \frac{1}{2 t^{3 ∕ 2}})$
•: $\frac{∂C}{∂x} = \frac{1}{\sqrt{t}} \frac{- 2 x}{kt} e^{^{- x^{2} ∕ kt}}$
•: $\frac{\partial^{2} C}{\partial x^{2}} = e^{^{- x^{2} ∕ kt}} \frac{1}{\sqrt{t}} (\frac{4 x^{2}}{k^{2} t^{2}} - \frac{2}{kt}) = e^{^{- x^{2} ∕ kt}} (\frac{4 x^{2}}{k^{2} t^{5 ∕ 2}} - \frac{2}{k t^{3 ∕ 2}})$
•: Luego, multiplicando $\frac{\partial^{2} C}{\partial x^{2}}$ por $\frac{k}{4}$ se obtiene la identidad.

5.7 Sea

z = f (x^{2} y + y) \cdot \sqrt{x + y^{2}} .

Calcule

\frac{∂f}{∂y},

\frac{∂f}{∂x} .

$z$ es una función de dos variables pero $f$ es una función de un solo argumento y como tal, se deriva de la manera ordinaria. Aquí es conveniente hacer el cambio de variable $u = x^{2} y + y$ de tal manera que $z = f (u) \cdot \sqrt{x + y^{2}} .$

\begin{matrix} \frac{∂z}{\partial y} & = & f^{'} (u) \cdot \frac{\partial}{∂y} [u] \cdot \sqrt{x + y^{2}} + f (u) \cdot \frac{\partial}{∂y} [\sqrt{x + y^{2}}] \\ = & f^{'} (u) \cdot (x^{2} + 1) \cdot \sqrt{x + y^{2}} + f (u) \cdot \frac{y}{\sqrt{x + y^{2}}} \\ = \\ \frac{∂z}{\partial x} & = & f^{'} (u) \cdot \frac{\partial}{∂x} [u] \cdot \sqrt{x + y^{2}} + f (u) \cdot \frac{\partial}{∂x} [\sqrt{x + y^{2}}] \\ = & f^{'} (u) \cdot (2 xy) \cdot \sqrt{x + y^{2}} + f (u) \cdot \frac{1}{2 \sqrt{x + y^{2}}} \end{matrix}

5.8 Verifique que

u (x, y) = e^{y} sen x

satisface la ecuación de Laplace

\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0

•: $u_{x} = e^{y} cos x$
•: $u_{y} = e^{y} sen x$
•: $u_{xx} = - e^{y} sen x$
•: $u_{xy} = e^{y} sen x$
•: $\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = - e^{y} sen x + e^{y} sen x = 0$ $\sqrt$

5.9 Sea

a \in ℝ

una constante. Verifique que

u (x, t) = sen (x - at) + ln (x + at)

es solución de la ecuación de onda

u_{tt} = a^{2} u_{xx} .

•: $u_{t} = - a cos (x - at) + \frac{a}{x + at}$
•: $u_{tt} = - a^{2} sen (x - at) - \frac{a^{2}}{{(x + at)}^{2}}$
•: $u_{x} = cos (x - at) + \frac{1}{x + at}$
•: $u_{xx} = - sen (x - at) - \frac{1}{{(x + at)}^{2}}$
•: $u_{tt} = - a^{2} sen (x - at) - \frac{a^{2}}{{(x + at)}^{2}} = a^{2} \cdot (- sen (x - at) - \frac{1}{{(x + at)}^{2}}) = a^{2} \cdot u_{xx} .$ $\sqrt$

5.10 Sea

a \in ℝ

una constante y

f

g

funciones dos veces derivables. Verifique que

u (x, t) = f (x - at) + g (x + at)

es solución de la ecuación de onda

u_{tt} = a^{2} u_{xx} .

Sea

A = x - at

B = x + at,

entonces

u (x, t) = f (A) + f (B) .

•: $u_{t} = - a f^{'} (A) + a g^{'} (B)$
•: $u_{tt} = a^{2} f^{″} (A) + a^{2} g^{″} (B)$
•: $u_{x} = f^{'} (A) + g^{'} (B)$
•: $u_{xx} = f^{″} (A) + g^{″} (B)$
•: $u_{tt} = a^{2} f^{″} (A) + a^{2} g^{″} (B) = a^{2} \cdot u_{xx} .$ $\sqrt$

5.11 Verifique que

z = ln (e^{x} + e^{y})

es solución de las ecuación diferencial

\frac{∂z}{∂x} + \frac{∂z}{∂y} = 1

y de la ecuación diferencial

\frac{\partial 2 z}{∂x 2} \cdot \frac{\partial 2 z}{∂y 2} - {(\frac{\partial 2 z}{∂x∂y})}^{2} = 0 .

Satisface

\frac{∂z}{∂x} + \frac{∂z}{∂y} = 1 .

•: $z_{x} = \frac{e^{x}}{e^{x} + e^{y}}$
•: $z_{y} = \frac{e^{y}}{e^{x} + e^{y}}$
•: $\frac{∂z}{∂x} + \frac{∂z}{∂y} = \frac{e^{x}}{e^{x} + e^{y}} + \frac{e^{y}}{e^{x} + e^{y}} = 1$ $\sqrt$

Satisface $\frac{\partial 2 z}{∂x 2} \cdot \frac{\partial 2 z}{∂y 2} - {(\frac{\partial 2 z}{∂x∂y})}^{2} = 0 .$

•: $z_{xx} = \frac{e^{x} \cdot (e^{x} + e^{y}) - e^{x} \cdot e^{x}}{{(e^{x} + e^{y})}^{2}}$
•: $z_{xy} = \frac{e^{y} \cdot (e^{x} + e^{y}) - e^{y} \cdot e^{y}}{{(e^{x} + e^{y})}^{2}}$
•: $\frac{\partial 2 z}{∂x∂y} = \frac{\partial}{∂x} [\frac{e^{y}}{e^{x} + e^{y}}] = \frac{- e^{y} \cdot e^{x}}{{(e^{x} + e^{y})}^{2}} .$
•: $\begin{matrix} \frac{\partial 2 z}{∂x 2} \cdot \frac{\partial 2 z}{∂y 2} - {(\frac{\partial 2 z}{∂x∂y})}^{2} & = & \frac{e^{x} \cdot (e^{x} + e^{y}) - e^{x} \cdot e^{x}}{{(e^{x} + e^{y})}^{2}} \cdot \frac{e^{y} \cdot (e^{x} + e^{y}) - e^{y} \cdot e^{y}}{{(e^{x} + e^{y})}^{2}} - {(\frac{- e^{y} \cdot e^{x}}{{(e^{x} + e^{y})}^{2}})}^{2} \\ = & \frac{e^{x} \cdot e^{y}}{{(e^{x} + e^{y})}^{2}} \cdot \frac{e^{y} \cdot e^{x}}{{(e^{x} + e^{y})}^{2}} - {(\frac{- e^{y} \cdot e^{x}}{{(e^{x} + e^{y})}^{2}})}^{2} = 0 \sqrt \end{matrix}$

5.12 Sea

f

una función derivable en todo

ℝ

y sea

w (x, y) = f (y sen x) .

Verifique que

cos (x) \frac{∂w}{∂x} + y sen (x) \frac{∂w}{∂y} = y f^{'} (y sen x)

Sea $u = y sen (x),$ entonces $w = f (u) .$

•: $w_{x} = f^{'} (u) \cdot y cos (x)$
•: $w_{y} = f^{'} (u) \cdot sen (x)$
•: $cos (x) w_{x} + y sen (x) w_{y} = {cos}^{2} (x) \cdot y \cdot f^{'} (u) + {sen}^{2} (x) \cdot y \cdot f^{'} (u) = ({cos}^{2} x + {sen}^{2} x) y f^{'} (u) = y f^{'} (u)$

5.13 Sea

g (x, y) = x^{2} sen (3 x - 2 y) .

Verifique la identidad

x \cdot \frac{\partial 2 g}{∂y∂x} = 2 \frac{∂g}{∂y} + 6 x \cdot g (x, y) .

$\frac{∂g}{∂x} = 2 x sen (3 x - 2 y) + 3 x^{2} cos (3 x - 2 y),$ $\frac{∂g}{∂y} = - 2 x^{2} cos (3 x - 2 y)$ y $\frac{\partial 2 g}{∂y∂x} = - 4 x cos (3 x - 2 y) - 6 x^{2} sen (3 x - 2 y) .$ La identidad se verifica de manera directa.

5.14 La resistencia total

R

producida por tres conductores con resistencias

R_{1}, R_{2}

R_{3}

conectadas en paralelo en un circuito eléctrico está dado por la fórmula

\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} .

Calcule

\frac{∂R}{\partial R_{1}} .

Sugerencia: derive a ambos lados respecto a

R_{1}

Derivamos a ambos lados respecto a

R_{1},

$\begin{matrix} \frac{\partial}{\partial R_{1}} [\frac{1}{R}] & = & \frac{\partial}{\partial R_{1}} [\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}] \\ \frac{- 1 \cdot \frac{∂R}{\partial R_{1}}}{R^{2}} & = & \frac{- 1}{R_{1}^{2}} ⟹ \frac{∂R}{\partial R_{1}} = \frac{R^{2}}{R_{1}^{2} .} \end{matrix}$

5.15 La ley de gases para un gas ideal de masa fija

m,

temperatura absoluta

T,

presión

P

y volumen

V

PV = mRT

donde

R

es la constante universal de los gases ideales. Verifique que

\frac{∂P}{∂V} \frac{∂V}{∂T} \frac{∂T}{∂P} = - 1 .

•: $\frac{∂P}{∂V} = - \frac{P}{V}$
•: $\frac{∂V}{∂T} = \frac{mR}{P}$
•: $\frac{∂T}{∂P} = \frac{V}{mR}$
•: $\frac{∂P}{∂V} \frac{∂V}{∂T} \frac{∂T}{∂P} = - \frac{P}{V} \frac{mR}{P} \frac{V}{mR} = - 1 . \sqrt$

5.16 La energía cinética de un cuerpo de masa

m

y velocidad

v

K = \frac{1}{2} m v^{2} .

Verifique que

\frac{∂K}{∂m} \frac{\partial 2 K}{∂v 2} = K .

\frac{∂K}{∂m} \frac{\partial 2 K}{∂v 2} = \frac{1}{2} v^{2} \cdot m = K . \sqrt

5.17 Sea

f

g

funciones dos veces derivables. Sea

u = x^{2} + y^{2}

w (x, y) = f (u) \cdot g (y) .

Calcule

\frac{∂w}{∂x}, \frac{\partial 2 w}{∂x 2}, \frac{\partial 2 w}{∂y∂x}

\frac{∂w}{∂y} .

•: $\frac{∂w}{∂x} = f^{'} (u) \cdot 2 x \cdot g (y)$
•: $\frac{\partial 2 w}{∂x 2} = 2 g (y) \cdot [f^{″} (u) \cdot 2 x^{2} + f^{'} (u)]$
•: $\frac{\partial 2 w}{∂y∂x} = 2 x [f^{″} (u) \cdot 2 y \cdot g (y) + g^{'} (y) \cdot f^{'} (u)]$
•: $\frac{∂w}{∂y} = f^{'} (u) \cdot 2 y \cdot g (y) + g^{'} (y) \cdot f (u)$

5.18 Sea

f

g

funciones dos veces derivables. Sea

w (x, y) = f (u) + g (v)

donde

u = \frac{x}{y}

v = \frac{y}{x} .

Calcule

\frac{∂w}{∂x}, \frac{\partial 2 w}{∂y∂x} .

•: $\frac{∂w}{∂x} = f^{'} (u) \cdot \frac{1}{y} + g^{'} (v) \cdot \frac{- y}{x^{2}}$
•: $\frac{\partial 2 w}{∂y∂x} = f^{″} (u) \cdot \frac{- x}{y^{2}} \cdot \frac{1}{y} - \frac{1}{y^{2}} \cdot f^{'} (u) + g^{″} (v) \cdot \frac{1}{x} \cdot \frac{- y}{x^{2}} - \frac{1}{x^{2}} \cdot g^{'} (v) .$

5.19 Sea

w = e^{3 x} \cdot f (x^{2} - 4 y^{2}),

donde

f

es una función dos veces diferenciable. Calcule

\frac{\partial 2 w}{∂x∂y} .

Sea

u = x^{2} - 4 y^{2},

•: $\frac{∂w}{∂y} = - e^{3 x} f^{'} (u) \cdot 8 y$
•: $\frac{\partial 2 w}{∂x∂y} = 8 y [- 3 e^{3 x} f^{'} (u) - e^{3 x} f^{″} (u) \cdot 2 x]$

5.20Sea

u (r, 𝜃) = r^{n} cos (n 𝜃)

con

n \in ℕ

una constante. Verfique que

u

satisface la ecuacón

u_{rr} + \frac{u_{r}}{r} + \frac{u_{𝜃𝜃}}{r^{2}} = 0

\frac{∂u}{\partial rr} = n (n - 1) r^{n - 2} cos (n𝜃) y \frac{∂u}{\partial 𝜃𝜃} = - n^{2} r^{n} cos (n 𝜃) .

Sustituyendo y simplificando se verifica la ecuación.