Respuesta :
Answer:
[tex]r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k[/tex]
Explanation:
a(t)=20t i+sin(t) j +cos(2t) k
v(t)=[tex]\int\limits^a_b {a(t)} \, dt[/tex]
=[tex](10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k[/tex]---------------eqn 1
given v(0)=i
i=[tex]c_1i+(-1+c_2)j+(0+c_3)k[/tex]
[tex]c_1=1[/tex] [tex]c_2=1[/tex] [tex]c_3=0[/tex]
from equation 1
V(t)=[tex](10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k[/tex]----------eqn 2
now r(t)=[tex]\int\limits^a_b {v (t)} \, dt[/tex]
[tex](\frac{10t^3}{3}+t+c_1)i+(-sint+t+c_2)j+(\frac{-cos2t}{4}+c_3)k[/tex]
given r(0)=j
0i+1j+ok = [tex]c_1i+c_2j+(\frac{-1}{4}+c_3)k[/tex]
[tex]c_1=0[/tex] [tex]c_2=0[/tex] [tex]c_3 = \frac{1}{4}[/tex]
[tex]r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k[/tex]
