5f7641…: “Out of the amazing village”

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Let $\vec {f}(t) = \vector {\cos t,\sin t,\cos t}$ . 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 $\vector {1,2,3}$ .