I have the following problem.
A particle moves along an ellipse $$3x^2 + y^2 = 1$$ with a position vector $$\vec{r} (t)= (f(t), g(t))$$ the displacement is such that the horizontal component of the velocity vector in each instant $ t $ is $ -g(t)$.
Is the particle having a negative or positive displacement?
Show that the vertical component of the velocity vector in instant $ t $ is proportional to $ f(t)$. Find the constant of proportionality.
This is what I have:
$$\vec{r} (t)= ( \cos(t)/\sqrt{3} , \sin(t))$$
$$\vec{v} (t) = ( -\sin(t)/\sqrt{3} , \cos(t))$$
Velocity vector with horizontal component $ -g(t)$:
$$\vec{v} (t) = ( -\sin(-sin(t))/\sqrt{3} , \cos(t))$$
This is equal to:
$$\vec{v} (t) = ( \sin(sin(t))/\sqrt{3} , \cos(t))$$
My conclusion, if I am not wrong, is that since both components of the velocity vector are positive then the particle displacement is positive.
I am still not sure about what to do regarding if the vertical component of the velocity vector in instant $ t $ is proportional to $ f(t)$, and about finding the constant of proportionality.
Any tips on how to solve that problem?
Edit:
Constant of proportionality
Please excuse my english. Thanks.
Best Answer
Particle moves along $3x^2+y^2=1$
So we define $\vec r(t) = (\frac{1}{\sqrt3} \cos (at), \sin (at))$ where $a$ is a constant. For $a = 1$, it will take $2\pi$ unit of time for the particle to go along the ellipse and return to its starting position.
$\vec v(t) = (- \frac{a}{\sqrt3} \sin (at), a \cos (at))$
We also know that $\vec r(t) = (f(t), g(t))$ such that $x$ component of velocity vector at any given $t$ is $-g(t)$.
So what is the value of $a$? Can you take it from here?