Ejercicios de demostraciones

Ejercicio 3:

a) Es negativo, debido a que la dirección de la curva está en contra del campo vectorial

b) Es positiva, debido a que la magnitud de los vectores que salen del punto es mayor a la magnitud de los vectores entran


Ejercicio 4:

5 lines Line 1: d i v F equals d F over d x i plus d F over d y j plus d F over d z k Line 2: d i v F equals negative e to the xth power s e n of y i plus e raised to the negative y power s e n of z j plus e raised to the negative z power s e n of x k Line 3: r o t F equals x F Line 4: r o t F equals open paren 0 minus e raised to the negative y power cosine z close paren times i minus open paren e raised to the negative z power cosine x minus 0 close paren times j plus open paren 0 minus e raised to the negative x power cosine y close paren times k Line 5: r o t F equals negative e raised to the negative y power cosine z i minus e raised to the negative z power cosine x j minus e raised to the negative x power cosine y k

Ejercicio 5:

2 lines Line 1: d i v of open paren r o t G close paren equals is less than the fraction with numerator and denominator x comma the fraction with numerator and denominator y comma the fraction with numerator and denominator z is greater than period is less than 2 x comma 3 y z comma negative x z squared is greater than equals is less than 2 comma 3 z comma negative 2 x z squared is greater than Line 2: the of F sub open paren x comma y comma z close paren slash d i v of open paren r o t G close paren equals 0 period Por lo tanto no existe campo vectorial cap G

Ejercicio 6:

Ejercicio 7:





Ejercicio 8:






Ejercicio 9:





Ejercicio 10:




18) Si C es una curva plana, cerrada, simple y suave por segmentos y f y g son funciones derivables, desmuestre que

6 lines Line 1: the integral from c to of f of x plus g of y d y equals 0 Line 2: Por las caracter modifying above i with acute sticas de la curva comma podemos aplicar el teorema de Green colon Line 3: the integral from c to of f of x plus g of y d y equals the integral from to of the integral from D to of open bracket g of y over x minus f of x over x close bracket Line 4: the integral from c to of f of x plus g of y d y equals the integral from to of the integral from D to of open bracket 0 minus 0 close bracket Line 5: the integral from c to of f of x plus g of y d y equals 0


19) Si f  y g son funciones dos veces derivables, desmuestre que:

8 lines Line 1: squared f of g equals f of squared g plus g of squared f plus 2 f period g Line 2: squared f of g equals f of g Line 3: P o r l a. p r o p i e d a. d f of g equals f of g plus g of f Line 4: equals the of open paren f of g plus g of f close paren Line 5: equals f of g plus g of f Line 6: equals f of squared g plus g of f plus g of squared f plus g of f Line 7: equals f of squared g plus g of squared f plus 2 times g of f

20) Si f es una función armónica, es decir, ⊽² f = 0 demuestre que la integral de línea ∫fydx – fxdy es independiente de la trayectoria.

Si la integral es independiente de la trayectoria, entonces: 7 lines Line 1: the integral from to of f sub y of d x minus f sub x of d y equals 0 Line 2: squared f equals f minus is greater than the of open paren f over x plus f over y close paren equals 0 minus is greater than the fraction with numerator squared f and denominator the of x squared plus the fraction with numerator squared f and denominator the of y squared equals 0 Line 3: P o d e m o s v e r i f of i c a. r c o n e l t e o r e m a. d e G of r e e n colon Line 4: the integral from c to of f sub y of d x minus f sub x of d y equals the integral from to of the integral from D to of open paren the fraction with numerator the of open paren negative f sub x close paren and denominator x minus the fraction with numerator the of open paren f sub y close paren and denominator y close paren Line 5: the integral from c to of f sub y of d x minus f sub x of d y equals negative the integral from to of the integral from D to of open paren the fraction with numerator squared f and denominator the of x squared minus the fraction with numerator squared f and denominator the of y squared close paren Line 6: the integral from c to of f sub y of d x minus f sub x of d y equals the integral from to of the integral from D to of 0 Line 7: the integral from c to of f sub y of d x minus f sub x of d y equals 0

Por lo tanto, la integral de línea es efectivamente independiente de la trayectoria.



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