Question

The position vector of a particle is r= (t^2 - 1) i + 2t j. How can I find out the trajectory of the particle in the XY-plane ?

Answer 1

Task. The position vector of a particle is $r(t)=(t^{2}-1)i+2tj$. How can I find out the trajectory of the particle in the XY-plane ?

Solution. Let $x(t)=t^{2}-1$ and $y(t)=2t$ be the coordinate functions of the vector, so

$r(t)=(t^{2}-1)i+2tj=x(y)i+y(t)j.$

Notice that we can express $t$ via $y(t)$:

$y(t)=2t,$

whence

$t=y/2.$

Substituting this formula into the equation of $x(t)$ we obtain

$x=t^{2}-1$

$x=(y/2)^{2}-1$

$x=\frac{y^{2}}{4}-1$

$4x=y^{2}-4$

$y^{2}-4x=4.$

The trajectory of the particle is given by the following equation

$y^{2}-4x=4.$