Question

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

localid="1650740363372" $$\begin{gathered} v\left(t\right)=\cos ti+\sin tj+tk, \\ r\left(0\right)=-i+5j \end{gathered}$$

Short answer

$\mathrm{r}\left(\mathrm{t}\right)=\left(\mathrm{sint}-1\right)\mathrm{i}+\left(-\mathrm{cost}+6\right)\mathrm{j}+\frac{{\mathrm{t}}^2}{2}\mathrm{k}$

Step-by-step solution

Given Information

Consider $v\left(t\right)=\cos ti+\sin tj+tk,r\left(0\right)=-i+5j$

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

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

We have,

$$\begin{gathered} r\left(t\right)=\int v\left(t\right)dt \\ =\int \left(\mathrm{costi}+\mathrm{sintj}+\mathrm{tk}\right)\mathrm{dt} \\ =i\int \cos tdt+j\int \sin tdt+k\int tdt \\ =\sin ti-\cos tj+\frac{t^2}{2}k+c \\ =\left(\sin t+c_1\right)i+\left(-\cos t+c_2\right)j+\left(\frac{t^2}{2}+c_3\right)k \end{gathered}$$

Where $c_1,c_2$and $c_3$are scalar constants.

Calculation

$$\begin{gathered} r\left(t\right)=\left(\sin t+c_1\right)i+\left(-\cos t+c_2\right)j+\left(\frac{t^2}{2}+c_3\right)k \\ r\left(0\right)=\left(\sin 0+c_1\right)i+\left(-\cos 0+c_2\right)j+\left(\frac{0^2}{2}+c_3\right)k \\ r\left(0\right)=c_{1}i+\left(-1+c_2\right)j+c_{3}k..............\left(1\right) \\ r\left(0\right)=-i+5j................\left(2\right) \end{gathered}$$

comparing (1) and (2) equations,

$c1=-1.-1+c2=5,c3=0c2=6$

$\mathrm{r}\left(\mathrm{t}\right)=\left(\mathrm{sint}-1\right)+\left(-\mathrm{cost}+6\right)\mathrm{j}+\left(\frac{{\mathrm{t}}^2}{2}\right)\mathrm{k}$

Calculation

Check: Taking the derivatives of $r\left(t\right)$, we obtain

$r^2\left(t\right)=\cos ti+\sin tj+tk,$

Which is the correct tangent vector function

Next we evaluate

$$\begin{gathered} r\left(0\right)=\left(\sin 0-1\right)i+\left(-\cos 0+6\right)j+\left(\frac{0^2}{2}\right)k \\ =-i+5j \end{gathered}$$

Which is true according to the problem

Thus $r\left(t\right)=\left(\sin t-1\right)i+\left(-\cos t+6\right)j+\frac{t^2}{2}k$