## Calculus: Early Transcendentals 8th Edition

$r(t)=(t-\sin t) i+(3-2 \cos t)j+(t^3-t-4)k$
As we are given that $a(t)=\sin t i+2 \cos tj+6tk$ and $v(0)=-k$ Since, $v(t)=\int a(t)$ and $r(t)=\int v(t)$ Thus, $v(t)=\int (\sin t i+2 \cos tj+6tk) dt$ $\implies v(t)=(1-\cos t) i+2 \sin tj+(3t^2-1)k$ Now, $r(t)=\int [(1-\cos t) i+2 \sin tj+(3t^2-1)k] dt=(t-\sin t) i-2 \cos tj+(t^3-t)k+c$ Here, $c$ represents a constant of integration. That is, $r(0)=j-4k \implies c= j-4k$ $r(t)=(t-\sin t) i-2 \cos tj+(t^3-t)k+j-4k$ Hence, $r(t)=(t-\sin t) i+(3-2 \cos t)j+(t^3-t-4)k$