## Calculus 8th Edition

$r(t)=(t-\sin t) i+(3-2 \cos t)j+(t^3-t-4)k$
As we know $v(t)=\int a(t)$ and $r(t)=\int v(t)$ $v(t)=\int (\sin t i+2 \cos tj+6tk) dt$ and $v(t)=(1-\cos t) i+2 \sin tj+(3t^2-1)k$ Solve $r(t)=\int [(1-\cos t) i+2 \sin tj+(3t^2-1)k] dt=(t-\sin t) i-2 \cos tj+(t^3-t)k+c$ Also, $r(0)=j-4k$ and $c= j-4k$ Therefore, $r(t)=(t-\sin t) i-2 \cos tj+(t^3-t)k+j-4k$ Hence, the result. $r(t)=(t-\sin t) i+(3-2 \cos t)j+(t^3-t-4)k$