Demonstre que

a ) grad r^n = n r^(n - 2) r

b ) grad f (r) = (r)/(r) (d f)/(d r)

c ) grad^2 (ln r) = (1)/(r^2)

Passo 1

E aí pessoal, belezinha? Vamos trabalhar agora em algumas questões de matrizes, vetores e cálculo vetorial.

Item a

grad r^n = sum_(i = 1)^3 e_i (del r^n)/(del x_i) = sum e_i (del)/(del x_i) [(underset(j)(sum) x_j^2)^((1)/(2))]^n = underset(i)(sum) e_i 2 x_i (n)/(2) (underset(j)(sum) x_j^2)^((n)/(2) - 1) =

= underset(i)(sum) e_i x_i n (underset(j)(sum) x_j^2)^((n)/(2) - 1) = underset(i)(sum) e_i x_i n r^(n - 2)

Portanto:

grad r^n = n r^(n - 2) r

Item b

grad f (r) = sum_(i = 1)^3 e_i (del f (r))/(del x_i) = sum_(i = 1)^3 e_i (del f (r))/(del r) (del r)/(del x_i) = underset(i)(sum) e_i (del)/(del x_i) (underset(j)(sum) x_j^2)^(- (1)/(2)) (del f (r))/(del r) =

= underset(i)(sum) e_i x_i (underset(j)(sum) x_j^2)^(- (1)/(2)) (del f (r))/(del r) = underset(i)(sum) e_i (x_i)/(r) (del f (r))/(del r)

Portanto:

grad f (r) = (r)/(r) (d f)/(d r)

Item c

grad^2 (ln r) = underset(i)(sum) (del^2 ln r)/(del x_i^2) = sum (del^2)/(del x_i^2) [ln (underset(j)(sum) x_j^2)^((1)/(2))] = underset(i)(sum) (del)/(del x_i) [((1)/(2) * 2 x_i (underset(j)(sum) x_j^2)^(- (1)/(2)))/((underset(j)(sum) x_j^2)^((1)/(2)))] =

= underset(i)(sum) (del)/(del x_i) [x_i (underset(j)(sum) x_j^2)^(- 1)] = underset(i)(sum) (- x_i) (2 x_i) (underset(j)(sum) x_j^2)^(- 2) + underset(i)(sum) (del x_i)/(del x_i) (underset(j)(sum) x_j^2)^(- 1) =

= underset(i)(sum) (- 2 x_j^2) (r^2)^(- 2) + 3 ((1)/(r^2)) = - (2 r^2)/(r^4) + (3)/(r^2) = (1)/(r^2)

Portanto:

grad^2 (ln r) = (1)/(r^2)

Resposta

Ver passo a passo.

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