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UNIVERSIDAD PEDAGÓGICA NACIONAL
FACULTAD DE CIENCIA Y TECNOLOGÍA
LICENCIATURA EN ELECTRÓNICA
MATEMÁTICAS IV
DIEGO ARMANDO BOLAÑOS MORENO
1. Cambio de dimensiones de una caja. Las longitudes 𝒂, 𝒃 y 𝒄 de las aristas de una caja rectangular cambian
con el tiempo. En un instante en cuestión
𝑎 = 1 𝑚, 𝑏 = 2 𝑚, 𝑐 = 3 𝑚
𝑑𝑎
𝑑𝑡
=
𝑑𝑏
𝑑𝑡
= 1 𝑚/𝑠
𝑑𝑐
𝑑𝑡
= −3 𝑚/𝑠
a. ¿Qué valores tienen las tasas de cambio instantáneas en ese instante?
i. Volumen 𝑉
ii. Área 𝑆
b. ¿La longitud de las diagonales interiores de la caja crece o decrece?
PARA EL VOLUMEN 𝑉
Tenemos que el volumen se define por la siguiente ecuación
𝑉 = 𝑎 ∙ 𝑏 ∙ 𝑐
Ahora, expresamos la ecuación en término de las derivadas parciales
𝑑𝑉
𝑑𝑡
=
𝜕𝑉
𝜕𝑎
∙
𝑑𝑎
𝑑𝑡
+
𝜕𝑉
𝜕𝑏
∙
𝑑𝑏
𝑑𝑡
+
𝜕𝑉
𝜕𝑐
∙
𝑑𝑐
𝑑𝑡
resolvemos las derivas parciales
𝑑𝑉
𝑑𝑡
= (𝑏𝑐) ∙
𝑑𝑎
𝑑𝑡
+ (𝑎𝑐) ∙
𝑑𝑏
𝑑𝑡
+ (𝑎𝑏) ∙
𝑑𝑐
𝑑𝑡
Evaluamos con las siguientes condiciones planteadas en el ejercicio
𝑎 = 1 𝑚, 𝑏 = 2 𝑚, 𝑐 = 3 𝑚
Remplazamos
𝑑𝑉
𝑑𝑡
= (2 𝑚 ∙ 3 𝑚) ∙
𝑑𝑎
𝑑𝑡
+ (1 𝑚 ∙ 3 𝑚) ∙
𝑑𝑏
𝑑𝑡
+ (1 𝑚 ∙ 2 𝑚) ∙
𝑑𝑐
𝑑𝑡
Remplazamos las siguientes condiciones que nos fueron dadas en el problema
𝑑𝑎
𝑑𝑡
=
𝑑𝑏
𝑑𝑡
= 1 𝑚/𝑠
𝑑𝑐
𝑑𝑡
= −3 𝑚/𝑠
Resolvemos
𝑑𝑉
𝑑𝑡
= (6 𝑚) ∙ (1 𝑚/𝑠) + (3 𝑚) ∙ (1 𝑚/𝑠) + (2 𝑚) ∙ (−3 𝑚/𝑠)
Resolvemos y obtenemos la respuesta
𝒅𝑽
𝒅𝒕
= 𝟑 𝒎𝟑
/𝒔
𝑬𝒍 𝒗𝒐𝒍𝒖𝒎𝒆𝒏 𝒂𝒖𝒎𝒆𝒏𝒕𝒂 𝒂: 𝟑 𝒎𝟑
/𝒔
PARA EL ÁREA 𝑺
Tenemos que del área se define por la siguiente ecuación
𝑆 = 2𝑎𝑏 + 2𝑎𝑐 + 2𝑏𝑐
Ahora, expresamos la ecuación en término de las derivadas parciales
𝑑𝑆
𝑑𝑡
=
𝜕𝑆
𝜕𝑎
∙
𝑑𝑎
𝑑𝑡
+
𝜕𝑆
𝜕𝑏
∙
𝑑𝑏
𝑑𝑡
+
𝜕𝑆
𝜕𝑐
∙
𝑑𝑐
𝑑𝑡
resolvemos las derivas parciales
𝑑𝑆
𝑑𝑡
= (2(𝑏 + 𝑐)) ∙
𝑑𝑎
𝑑𝑡
+ (2(𝑎 + 𝑐)) ∙
𝑑𝑏
𝑑𝑡
+ (2(𝑎 + 𝑏)) ∙
𝑑𝑐
𝑑𝑡
Evaluamos con las siguientes condiciones planteadas en el ejercicio
𝑎 = 1 𝑚, 𝑏 = 2 𝑚, 𝑐 = 3 𝑚
Remplazamos
𝑑𝑆
𝑑𝑡
= (2(2 𝑚 + 3 𝑚)) ∙
𝑑𝑎
𝑑𝑡
+ (2(1 𝑚 + 3 𝑚)) ∙
𝑑𝑏
𝑑𝑡
+ (2(1 𝑚 + 2 𝑚)) ∙
𝑑𝑐
𝑑𝑡
Remplazamos las siguientes condiciones que nos fueron dadas en el problema
𝑑𝑎
𝑑𝑡
=
𝑑𝑏
𝑑𝑡
= 1 𝑚/𝑠
𝑑𝑐
𝑑𝑡
= −3 𝑚/𝑠
Resolvemos
𝑑𝑆
𝑑𝑡
= (10 𝑚) ∙ (1 𝑚/𝑠) + (8 𝑚) ∙ (1 𝑚/𝑠) + (6 𝑚) ∙ (−3 𝑚/𝑠)
Resolvemos y obtenemos la respuesta
𝒅𝑺
𝒅𝒕
= 𝟎 𝒎𝟑
/𝒔
𝑬𝒍 á𝒓𝒆𝒂 𝒏𝒐 𝒄𝒂𝒎𝒃𝒊𝒂 𝒅𝒂𝒅𝒂 𝒍𝒂 𝒊𝒏𝒇𝒐𝒓𝒎𝒂𝒄𝒊ó𝒏 𝒅𝒆: 𝟎 𝒎𝟑
/𝒔
PARA LAS DIAGONALES 𝑫
Tenemos que del área se define por la siguiente ecuación
𝐷 = √𝑎2 + 𝑏2 + 𝑐2
Ahora, expresamos la ecuación en término de las derivadas parciales
𝑑𝐷
𝑑𝑡
=
𝜕𝐷
𝜕𝑎
∙
𝑑𝑎
𝑑𝑡
+
𝜕𝐷
𝜕𝑏
∙
𝑑𝑏
𝑑𝑡
+
𝜕𝐷
𝜕𝑐
∙
𝑑𝑐
𝑑𝑡
resolvemos las derivas parciales
𝑑𝐷
𝑑𝑡
= (
𝑎
√𝑎2 + 𝑏2 + 𝑐2
) ∙
𝑑𝑎
𝑑𝑡
+ (
𝑏
√𝑎2 + 𝑏2 + 𝑐2
) ∙
𝑑𝑏
𝑑𝑡
+ (
𝑐
√𝑎2 + 𝑏2 + 𝑐2
) ∙
𝑑𝑐
𝑑𝑡
Al simplificar la expresión nos queda
𝑑𝐷
𝑑𝑡
=
1
√𝑎2 + 𝑏2 + 𝑐2
∙ (𝑎 ∙
𝑑𝑎
𝑑𝑡
+ 𝑏 ∙
𝑑𝑏
𝑑𝑡
+ 𝑐 ∙
𝑑𝑐
𝑑𝑡
)
Evaluamos con las siguientes condiciones planteadas en el ejercicio
𝑎 = 1 𝑚, 𝑏 = 2 𝑚, 𝑐 = 3 𝑚
Remplazamos
𝑑𝐷
𝑑𝑡
=
1
√12 + 22 + 32
∙ (1 ∙
𝑑𝑎
𝑑𝑡
+ 2 ∙
𝑑𝑏
𝑑𝑡
+ 3 ∙
𝑑𝑐
𝑑𝑡
)
Remplazamos las siguientes condiciones que nos fueron dadas en el problema
𝑑𝑎
𝑑𝑡
=
𝑑𝑏
𝑑𝑡
= 1 𝑚/𝑠
𝑑𝑐
𝑑𝑡
= −3 𝑚/𝑠
Resolvemos
𝑑𝐷
𝑑𝑡
=
1
√14 𝑚
∙ (1 𝑚 ∙ (1 𝑚/𝑠) + 2 ∙ (1 𝑚/𝑠) + 3 ∙ (−3 𝑚/𝑠))
Resolvemos y obtenemos la respuesta
𝒅𝑫
𝒅𝒕
= −
𝟑√𝟏𝟒
𝟕
𝒎𝟑
/𝒔
𝒅𝑫
𝒅𝒕
≈ −𝟏. 𝟔𝟎𝟒 𝒎𝟑
/𝒔
𝑳𝒂 𝒅𝒊𝒂𝒈𝒐𝒏𝒂𝒍 𝒆𝒔𝒕á 𝒅𝒆𝒄𝒓𝒆𝒄𝒊𝒆𝒏𝒅𝒐 𝒆𝒏 𝒔𝒖 𝒍𝒐𝒏𝒈𝒖𝒊𝒕𝒖𝒅 𝒂: −
𝟑√𝟏𝟒
𝟕
𝒎𝟑
/𝒔

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Cambio de dimensiones. Matemática.

  • 1. UNIVERSIDAD PEDAGÓGICA NACIONAL FACULTAD DE CIENCIA Y TECNOLOGÍA LICENCIATURA EN ELECTRÓNICA MATEMÁTICAS IV DIEGO ARMANDO BOLAÑOS MORENO 1. Cambio de dimensiones de una caja. Las longitudes 𝒂, 𝒃 y 𝒄 de las aristas de una caja rectangular cambian con el tiempo. En un instante en cuestión 𝑎 = 1 𝑚, 𝑏 = 2 𝑚, 𝑐 = 3 𝑚 𝑑𝑎 𝑑𝑡 = 𝑑𝑏 𝑑𝑡 = 1 𝑚/𝑠 𝑑𝑐 𝑑𝑡 = −3 𝑚/𝑠 a. ¿Qué valores tienen las tasas de cambio instantáneas en ese instante? i. Volumen 𝑉 ii. Área 𝑆 b. ¿La longitud de las diagonales interiores de la caja crece o decrece?
  • 2. PARA EL VOLUMEN 𝑉 Tenemos que el volumen se define por la siguiente ecuación 𝑉 = 𝑎 ∙ 𝑏 ∙ 𝑐 Ahora, expresamos la ecuación en término de las derivadas parciales 𝑑𝑉 𝑑𝑡 = 𝜕𝑉 𝜕𝑎 ∙ 𝑑𝑎 𝑑𝑡 + 𝜕𝑉 𝜕𝑏 ∙ 𝑑𝑏 𝑑𝑡 + 𝜕𝑉 𝜕𝑐 ∙ 𝑑𝑐 𝑑𝑡 resolvemos las derivas parciales 𝑑𝑉 𝑑𝑡 = (𝑏𝑐) ∙ 𝑑𝑎 𝑑𝑡 + (𝑎𝑐) ∙ 𝑑𝑏 𝑑𝑡 + (𝑎𝑏) ∙ 𝑑𝑐 𝑑𝑡 Evaluamos con las siguientes condiciones planteadas en el ejercicio 𝑎 = 1 𝑚, 𝑏 = 2 𝑚, 𝑐 = 3 𝑚 Remplazamos 𝑑𝑉 𝑑𝑡 = (2 𝑚 ∙ 3 𝑚) ∙ 𝑑𝑎 𝑑𝑡 + (1 𝑚 ∙ 3 𝑚) ∙ 𝑑𝑏 𝑑𝑡 + (1 𝑚 ∙ 2 𝑚) ∙ 𝑑𝑐 𝑑𝑡 Remplazamos las siguientes condiciones que nos fueron dadas en el problema 𝑑𝑎 𝑑𝑡 = 𝑑𝑏 𝑑𝑡 = 1 𝑚/𝑠 𝑑𝑐 𝑑𝑡 = −3 𝑚/𝑠
  • 3. Resolvemos 𝑑𝑉 𝑑𝑡 = (6 𝑚) ∙ (1 𝑚/𝑠) + (3 𝑚) ∙ (1 𝑚/𝑠) + (2 𝑚) ∙ (−3 𝑚/𝑠) Resolvemos y obtenemos la respuesta 𝒅𝑽 𝒅𝒕 = 𝟑 𝒎𝟑 /𝒔 𝑬𝒍 𝒗𝒐𝒍𝒖𝒎𝒆𝒏 𝒂𝒖𝒎𝒆𝒏𝒕𝒂 𝒂: 𝟑 𝒎𝟑 /𝒔 PARA EL ÁREA 𝑺 Tenemos que del área se define por la siguiente ecuación 𝑆 = 2𝑎𝑏 + 2𝑎𝑐 + 2𝑏𝑐 Ahora, expresamos la ecuación en término de las derivadas parciales 𝑑𝑆 𝑑𝑡 = 𝜕𝑆 𝜕𝑎 ∙ 𝑑𝑎 𝑑𝑡 + 𝜕𝑆 𝜕𝑏 ∙ 𝑑𝑏 𝑑𝑡 + 𝜕𝑆 𝜕𝑐 ∙ 𝑑𝑐 𝑑𝑡 resolvemos las derivas parciales 𝑑𝑆 𝑑𝑡 = (2(𝑏 + 𝑐)) ∙ 𝑑𝑎 𝑑𝑡 + (2(𝑎 + 𝑐)) ∙ 𝑑𝑏 𝑑𝑡 + (2(𝑎 + 𝑏)) ∙ 𝑑𝑐 𝑑𝑡 Evaluamos con las siguientes condiciones planteadas en el ejercicio 𝑎 = 1 𝑚, 𝑏 = 2 𝑚, 𝑐 = 3 𝑚 Remplazamos 𝑑𝑆 𝑑𝑡 = (2(2 𝑚 + 3 𝑚)) ∙ 𝑑𝑎 𝑑𝑡 + (2(1 𝑚 + 3 𝑚)) ∙ 𝑑𝑏 𝑑𝑡 + (2(1 𝑚 + 2 𝑚)) ∙ 𝑑𝑐 𝑑𝑡 Remplazamos las siguientes condiciones que nos fueron dadas en el problema 𝑑𝑎 𝑑𝑡 = 𝑑𝑏 𝑑𝑡 = 1 𝑚/𝑠 𝑑𝑐 𝑑𝑡 = −3 𝑚/𝑠
  • 4. Resolvemos 𝑑𝑆 𝑑𝑡 = (10 𝑚) ∙ (1 𝑚/𝑠) + (8 𝑚) ∙ (1 𝑚/𝑠) + (6 𝑚) ∙ (−3 𝑚/𝑠) Resolvemos y obtenemos la respuesta 𝒅𝑺 𝒅𝒕 = 𝟎 𝒎𝟑 /𝒔 𝑬𝒍 á𝒓𝒆𝒂 𝒏𝒐 𝒄𝒂𝒎𝒃𝒊𝒂 𝒅𝒂𝒅𝒂 𝒍𝒂 𝒊𝒏𝒇𝒐𝒓𝒎𝒂𝒄𝒊ó𝒏 𝒅𝒆: 𝟎 𝒎𝟑 /𝒔 PARA LAS DIAGONALES 𝑫 Tenemos que del área se define por la siguiente ecuación 𝐷 = √𝑎2 + 𝑏2 + 𝑐2 Ahora, expresamos la ecuación en término de las derivadas parciales 𝑑𝐷 𝑑𝑡 = 𝜕𝐷 𝜕𝑎 ∙ 𝑑𝑎 𝑑𝑡 + 𝜕𝐷 𝜕𝑏 ∙ 𝑑𝑏 𝑑𝑡 + 𝜕𝐷 𝜕𝑐 ∙ 𝑑𝑐 𝑑𝑡 resolvemos las derivas parciales 𝑑𝐷 𝑑𝑡 = ( 𝑎 √𝑎2 + 𝑏2 + 𝑐2 ) ∙ 𝑑𝑎 𝑑𝑡 + ( 𝑏 √𝑎2 + 𝑏2 + 𝑐2 ) ∙ 𝑑𝑏 𝑑𝑡 + ( 𝑐 √𝑎2 + 𝑏2 + 𝑐2 ) ∙ 𝑑𝑐 𝑑𝑡 Al simplificar la expresión nos queda 𝑑𝐷 𝑑𝑡 = 1 √𝑎2 + 𝑏2 + 𝑐2 ∙ (𝑎 ∙ 𝑑𝑎 𝑑𝑡 + 𝑏 ∙ 𝑑𝑏 𝑑𝑡 + 𝑐 ∙ 𝑑𝑐 𝑑𝑡 ) Evaluamos con las siguientes condiciones planteadas en el ejercicio 𝑎 = 1 𝑚, 𝑏 = 2 𝑚, 𝑐 = 3 𝑚 Remplazamos 𝑑𝐷 𝑑𝑡 = 1 √12 + 22 + 32 ∙ (1 ∙ 𝑑𝑎 𝑑𝑡 + 2 ∙ 𝑑𝑏 𝑑𝑡 + 3 ∙ 𝑑𝑐 𝑑𝑡 )
  • 5. Remplazamos las siguientes condiciones que nos fueron dadas en el problema 𝑑𝑎 𝑑𝑡 = 𝑑𝑏 𝑑𝑡 = 1 𝑚/𝑠 𝑑𝑐 𝑑𝑡 = −3 𝑚/𝑠 Resolvemos 𝑑𝐷 𝑑𝑡 = 1 √14 𝑚 ∙ (1 𝑚 ∙ (1 𝑚/𝑠) + 2 ∙ (1 𝑚/𝑠) + 3 ∙ (−3 𝑚/𝑠)) Resolvemos y obtenemos la respuesta 𝒅𝑫 𝒅𝒕 = − 𝟑√𝟏𝟒 𝟕 𝒎𝟑 /𝒔 𝒅𝑫 𝒅𝒕 ≈ −𝟏. 𝟔𝟎𝟒 𝒎𝟑 /𝒔 𝑳𝒂 𝒅𝒊𝒂𝒈𝒐𝒏𝒂𝒍 𝒆𝒔𝒕á 𝒅𝒆𝒄𝒓𝒆𝒄𝒊𝒆𝒏𝒅𝒐 𝒆𝒏 𝒔𝒖 𝒍𝒐𝒏𝒈𝒖𝒊𝒕𝒖𝒅 𝒂: − 𝟑√𝟏𝟒 𝟕 𝒎𝟑 /𝒔