Respuestas
Hola.
Solution.
r''(t) = 12t i - 3t^(-1/2)j + 2k (Integrando con respecto a t)
r' (t) = (12t²/2 +c1)i + ([- 3t^(1/2)]/(1/2) +c2)j + (2t + c3)k (simplificando)
r' (t) = (6t² +c1)i + (- 6t^(1/2) +c2)j + (2t + c3)k Ec.1
Con t=1
r' (1)= (6(1)² +c1)i + (-6(1)^(1/2) +c2)j + (2(1) + c3)k
j = (6 +c1)i + (-6 +c2)j + (2 + c3)k
con j = 0i + j +0k
Comparando cada componente.
(6 +c1) = 0 (-6 +c2) = 1 (2 + c3) = 0
c1= - 6 c2= 7 c3 = -2
Reemplazando valores en Ec.1
r' (t) = (6t² +c1)i + (- 6t^(1/2) +c2)j + (2t + c3)k
r' (t) = (6t² - 6)i + (- 6t^(1/2) + 7)j + (2t - 2)k Ec. 2 (integrand. y simplif.)
r(t) = (2t³ - 6t + k1)i + (- 4t^(3/2) + 7t+k2)j + (t² - 2t + k3)k Ec. 2
k1, k2, k3 constantes de integración.
Cuando t= 1 en Ec. 2
r(t) = (2t³ - 6t + k1)i + (- 4t^(3/2) + 7t+k2)j + (t² - 2t + k3)k
r(1) = (2(1)³ - 6(1) + k1)i + (- 4(1)^(3/2) + 7(1)+k2)j + ((1)² - 2(1) + k3)k
r(1) = (- 4 + k1)i + (3+k2)j + (- 1+ k3)k
Comparando componente a componente con r(1) = 2i – k = 2i + 0j – k
r(1) = (- 4 + k1)i + (3+k2)j + (- 1+ k3)k = 2i + 0j – k
(- 4 + k1) = 2 (3+k2) = 0 (- 1+ k3) = -1
k1 = 6 k2 = -3 k3 = 0
Reemplazando los valores en Ec. 2
r(t) = (2t³ - 6t + k1)i + (- 4t^(3/2) + 7t+k2)j + (t² - 2t + k3)k
r(t) = (2t³ - 6t + 6)i + (- 4t^(3/2) + 7t- 3)j + (t² - 2t )k (respuesta)
Gracias por tu pregunta.