Use the given velocity vectors $v\left(t\right)=r\text{'}\left(t\right)$ and initial positions in exercise to find the position function $r\left(t\right)$.

Consider

The objective is to find the position function $r\left(t\right)$.

Since $v\left(t\right)=r\text{'}\left(t\right)$,

we have

where c is a vector constant

$=c_{1}i+(sect+c_2)j+(\tan t+c_3)k$where $c_1,c_2$and $c_3$are scalars.

So

$$\begin{gathered} \mathrm{r}\left(\mathrm{t}\right)={\mathrm{c}}_1\mathrm{i}+(\sec \mathrm{t}+{\mathrm{c}}_2)\mathrm{j}+(\tan \mathrm{t}+{\mathrm{c}}_3)\mathrm{k} \\ r\left(\mathrm{\pi }\right)={\mathrm{c}}_1\mathrm{i}+(\sec \mathrm{\pi }+{\mathrm{c}}_2)\mathrm{j}+(\tan \mathrm{\pi }+{\mathrm{c}}_3)\mathrm{k} \\ \mathrm{r}\left(\mathrm{\pi }\right)={\mathrm{c}}_1\mathrm{i}+(-1+{\mathrm{c}}_2)\mathrm{j}+{\mathrm{c}}_3\mathrm{k}.......\left(1\right) \end{gathered}$$

But the given initial position

Comparing (1) and (2) equations,

$$\begin{gathered} c_1=-2,-1+c_2=4,c_3=3 \\ {c}_2=5 \end{gathered}$$

Thus$r(t0=-2i+(sect+5)j+(\tan t+3)k$

Check: taking the derivative of$r\left(t\right)=-2i+(sect+5)j+(\tan t+3)k$

We obtain $r\text{'}\left(t\right)=sect\tan tj+se{c}^{2}tk$which is the correct tangent vector function.

Next, we evaluate

Which is true according to the problem.

Thus$r\left(t\right)=-2i+(sect+5)j+(\tan t+3)k$