A particle moves in the xy-plane so that its velocity vector at time t is $v(t)=\langle t^2, \sin(\pi t)\rangle$ and the particle's position vector at time $t = 0$ is $\langle 1,0 \rangle$. What is the position vector of the particle when t = 3?

Question

anonymous · Answer

Integrate $v(t)$ to get our position vector $s(t)$.$$s(t)=\int(t^2,\sin\pi t)\,dt=\left(\frac13t^3+C_x,-\frac1\pi\cos\pi t+C_y\right)$$Now we merely verify solve for $C_x,C_y$ so that our initial condition holds:$$s(0)=\left(0+C_x,-\frac1\pi+C_y\right)=\left(0,1\right)$$so $C_x=0,C_y=1+\frac1{\pi}\pi$

The question is then $s(3)=\left(\frac133^3,-\frac1\pi\cos3\pi+1+\frac1\pi\right)=\left(9,\frac2\pi+1\right)$