Use the given velocity vectors v(t)=r'(t) and initial positions din Exercises 45-48 to find the position function r(t).

consider

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

Since localid="1651746230157" $v\left(t\right)=r^2\left(t\right)$we have localid="1651746236416" $r\left(t\right)=\int v\left(t\right)dt$

localid="1651746076251" $=\frac{t^2}{2}i+\frac{t^3}{3}j+c$, where c is a vector constant.

localid="1651746082066" $=\left(\frac{t^2}{2}+c_1\right)i+\left(\frac{t^3}{3}+c_2\right)j$where$c_1$and$c_2$are scalars.

So

But the given initial position

Equating $\left(1\right)$and $\left(2\right)$equations,

$c_{1}i+c_{2}j=3i-4j$

we obtain $c_1=3,c_2=-4$thus

$r\left(t\right)=\left(\frac{t^2}{2}+3\right)i+\left(\frac{t^3}{3}-4\right)j$

Check: taking the derivative of $r\left(t\right),\left(\frac{t^2}{2}+3\right)i+\left(\frac{t^3}{3}-4\right)j$we obtain

$r^1\left(t\right)=ti+t^{2}j,$which is the correct tangent vector function.

Next, we evaluate

localid="1651743714640" $$\begin{gathered} r\left(0\right)=\left(\frac{0^2}{2}+3\right)i+\left(\frac{0^3}{3}-4\right)j \\ =3i-4j \end{gathered}$$

which is true according to the problem.

Thus localid="1651746030104" $r\left(t\right)=\left(\frac{t^2}{2}+3\right)i+\left(\frac{t^3}{3}-4\right)j$is the required position function.