# How to calculate the time to travel around the ellipse?

by John Arg   Last Updated August 14, 2019 09:20 AM - source

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

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.