I have to find the parameter $t$ of a vector position $r(t)$, but don't get how since $r(t)$ return a vector.
my vector position is $$r(t) = \frac{1}{2}t - \frac{1}{4} \sin(2t)i - \frac{1}{2} \sin^{2}(t)k + C$$ with the points of the extremities $(0,1,0)$ and $(\pi/2, 1, 0)$.
So for the first point $(0,1,0)$, I just plugged in the function and I find out that the $t$ in that point is $t=0$ because $j = C = 1$ in this case. So $t$ needs to be equal to $0$ to return $1$.
But when I try the same thing for the second point, it just doesn't make sense because of $C = j = 1.$ I would've taken $t = \pi/2$ but the $j$ composant break everything so I don't understand.
Can someone Can enlighten me? Am I doing the wrong thing? Thanks in advance.
As Andrew says, it looks like your expression for $r(t)$ involves a sum of scalars with vectors. This is mathematically undefined, which makes me think you may have transcribed the equation wrong. Thus I can't provide a specific answer unless you rewrite it, but I can tell you the general technique for solving an equation of the form $$ r(t) := \Big(x(t),\ y(t),\ z(t) \Big) = (a, b, c) $$ The first step is to determine the components of $r(t)$ for each $t$ (that is, determine $x(t)$, $y(t)$, and $z(t)$) and set each one separately equal to the corresponding component of the right-hand side. That is, set \begin{align} x(t) &= a \\ y(t) &= b \\ z(t) &= c \end{align} and then solve each equation for $t$. Remember that these three equations are supposed to be simultaneously satisfied for a single $t$. If you solve each of the three equations and get multiple solutions for each, then the solutions to the entire vector equation are whatever $t$ solutions are shared among the three equations. If the three equations have no solutions in common, then it means your original vector equation has no solution.