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....

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

August 14, 2019 08:23 AM

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.

August 14, 2019 09:09 AM

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