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 $r\left(t\right)$.

Short answer

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

Step-by-step solution

Given Information

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 $r\left(t\right)=\int v\left(t\right)dt$

where $c$is a vector constant.

$=\left(\frac{t^2}{2}+c_1\right)i+\left(\frac{t^3}{3}+c_2\right)j$

Where $c_1$and $c_2$are scalars.

Calculation

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 $r\text{'}\left(t\right)=ti+t^{2}j$,

which is the correct tangent vector function.

Next, we evaluate

,

which is true according to the problem.

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