How to calculate the time to travel around the ellipse?

A particle moves along the ellipse $$3x^2 + y^2 = 1$$ with positions vector $$\vec{r(t)} = f(t)\vec i + g(t) \vec j$$. The motion is such that the horizontal component of the velocity vector at time $$t$$ is $$-g(t)$$. How much time is required for the particle to go once around the ellipse?

Now I found that the particle travels counterclockwise, though I suspect it doesn't matter. I also found that the position vector is $$(x, ±\sqrt{1-3x^2} )$$ and hence the velocity vector is $$(\mp \sqrt{1-3x^2} , 3x)$$.

I am not supposed to use arc length, so I am quite confused as to how to solve this problem. I am guessing that I am supposed to integrate something, but what, how, and why....

• Integrate velocity from $0$ to $t$ to get displacement. You require zero displacement for the particle to go around the ellipse hence set this equal to zero. This should give $f(t)=0$ which can be solved to give $t$ - the time required. – Peter Foreman Aug 14 at 8:15
• Why does integrating velocity not give me the $(x, ±\sqrt{1-3x^2} )$? Is position different than displacement? In that case, why can we write $f'(t)$ and $g'(t)$ as the components of velocity? – John Arg Aug 14 at 8:40
• You need only consider one component of the motion to determine the time for one orbit. – amd Aug 14 at 9:03

Lets paremertize the elipse

$$f(t) = \frac {1}{\sqrt 3}\cos (\omega(t))\\ g(t) = \sin (\omega (t))$$

and it is given that: $$f'(t) = -g(t)$$

$$f'(t) = -(\frac {1}{\sqrt 3}\cos \omega)\omega' = \sin (\omega (t))\\ \omega' = \sqrt 3\\ \omega(t) = \sqrt 3 \ t$$

It will take $$\frac {2\pi}{\sqrt 3}$$ to complete one orbit.

Or you could say

$$(x,y) = (\omega(t), \sqrt {1-3\omega^2})$$

But this only gets you half way around.... then you need to get back along the path

$$(x,y) = (\omega(t), -\sqrt {1-3\omega^2})$$

$$\omega' = -\sqrt {1-3\omega^2}$$

and we solve the differential equation and get the same trig equations we have above.

Hint: The examiner wants you to use a parametrisation of the ellipse such that the derivative has a horizontal component which satisfies the given condition. That is, such that $$f'(t)=-g(t).$$

Well, one such parametrisation is got by setting $$x =f(t)=\frac{1}{\sqrt 3}\cos(\sqrt 3 t)$$ and $$y=g(t)=\sin( \sqrt 3 t).$$

• I was able to parametrize the equation, and the parametrization matches what is given in the solution booklet. However, I do not understand how to find the time taken. – John Arg Aug 14 at 8:43
• @PeterForeman This is essentially what the accepted answer amounts to. OP was expected by examiner to know that the time is a multiple of $2π.$ They are only required to find a parametrisation satisfying the given conditions and adjusting the standard time based on the period, etc. – Allawonder Aug 14 at 18:02
• @JohnArg The time is a multiple of $2π$ since it's a simple closed curve. – Allawonder Aug 14 at 18:03