Find \(\vec {v}(t)\) so that \(\vec {v}(\pi ) = \vector {1,2,3}\) and \(\vec {v}'(t) = \vec {f}(t)\).
\[ \vec {v}(t) = \vector {\answer {1 + \sin t},\answer {1 - \cos t},\answer {3 + \sin t}}. \]
By antidifferentiating, \(\vec {v}(t) = \vector {\sin t,-\cos t,\sin t} + \vec {C}\) for some constant \(\vec {C}\).
When \(t = \pi \), we find \(\vec {v}(\pi ) = \vector {\sin \pi ,-\cos \pi ,\sin \pi } + \vec {C}\) which is \(\vector {0,1,0} + \vec {C}\).
Since we are told that \(\vec {v}(\pi ) = \vector {1,2,3}\), it must be that \(\vec {C} = \vector {1,1,3}\).
Therefore \(\vec {v}(t) = \vector {\sin t,-\cos t,\sin t} + \vector {1,1,3}\) is the desired function.
In this case, the \(x\)-coordinate is \(1 + \sin t\).
The \(y\)-coordinate is \(1 - \cos t\).
The \(z\)-coordinate is \(3 + \sin t\).
We can check our work by differentiating: note that the derivative of \(\vector {{1 + \sin t},{1 - \cos t},{3 + \sin t}}\) with respect
to \(t\) is \(\vector {\cos t,\sin t,\cos t}\) as desired. Moreover, \(\vector {{1 + \sin \pi },{1 - \cos \pi },{3 + \sin \pi }}\) equals \(\vector {1,2,3}\).