Answer to Question #209257 in Differential Geometry | Topology for eyes

Question

if v=2ti+t^2j+tk,find the position vector r of the particle at t=1 given that the particle initially was at i+2j+2k

Answer 1

$$\vec v=\dfrac{d\vec r}{dt}=2t\vec i+t^2\vec j+t\vec k$$

$$\vec r(t)=t^2\vec i+\dfrac{t^3}{3}\vec j+\dfrac{t^2}{2}\vec k+\vec r_0$$

$$\vec r(0)=\vec r_0=\vec i+2\vec j+2\vec k$$

$$\vec r(t)=(t^2+1)\vec i+(\dfrac{t^3}{3}+2)\vec j+(\dfrac{t^2}{2}+2)\vec k$$

$$\vec r(1)=(1+1)\vec i+(\dfrac{1}{3}+2)\vec j+(\dfrac{1}{2}+2)\vec k$$

$$\vec r(1)=2\vec i+\dfrac{7}{3}\vec j+\dfrac{5}{2}\vec k$$