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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....

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  • $\begingroup$ 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. $\endgroup$ – Peter Foreman Aug 14 at 8:15
  • $\begingroup$ 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? $\endgroup$ – John Arg Aug 14 at 8:40
  • $\begingroup$ You need only consider one component of the motion to determine the time for one orbit. $\endgroup$ – amd Aug 14 at 9:03
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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.

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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 $y=g(t)=\sin( \sqrt 3 t).$

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  • $\begingroup$ 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. $\endgroup$ – John Arg Aug 14 at 8:43
  • $\begingroup$ @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. $\endgroup$ – Allawonder Aug 14 at 18:02
  • $\begingroup$ @JohnArg The time is a multiple of $2π$ since it's a simple closed curve. $\endgroup$ – Allawonder Aug 14 at 18:03

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