I Show that a particle moves over a circumference

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1. Mar 7, 2016

Thales Costa

The problem asks me to show that a particle moves over a circumference with its center at the origin.

The position vector of a moving particle is:

I've tried using the x2+y2=r2 formula of the circumference, squaring both components of the vector function but I couldn't figure out what to do with the result.

How should I go in order to show this?

2. Mar 8, 2016

Svein

Simple algebra: $x^{2}+y^{2}=(\frac{1-t^{2}}{1+t^{2}})^{2}+(\frac{2t}{1+t^{2}})^{2}=\frac{1-2t^{2}+t^{4}+4t^{4}}{(1+t^{2})^{2}}=\frac{1+2t^{2}+t^{4}}{1+2t^{2}+t^{4}}=1$.

3. Mar 8, 2016

andrewkirk

Those formulas are from the 't formula' that is sometimes useful in trigonometry. If you read the first part of the link, you'll see why the particle travels around a circle.

4. Mar 8, 2016

Thales Costa

Oh my god, that's it. I did the math but for some reason I had 4t2 instead of 2t2. I should double check my math more often. Thanks for the reply.

The problem also asks about the direction the particle moves as t increases, if it's counter or clockwise, and if there are points in the circumference that are not occupied by the particle when t goes from negative infinity to infinity.

Do I just plug in t values and see how the function behaves to find out the direction?

5. Mar 8, 2016

Staff: Mentor

There's a typo in one of the expressions - $\frac{1-2t^{2}+t^{4}+4t^{4}}{(1+t^{2})^{2}}$ should be $\frac{1-2t^{2}+t^{4}+4t^{2}}{(1+t^{2})^{2}}$

6. Mar 8, 2016

zinq

Looks to me that the original formula has a typo: The numerator of the y-coordinate should be 2t2, not 2t.

Then you actually get

x(t)2 + y(t)2 = 1​

for all t.

Hint: You might notice that

y(t) / x(t) = tan(θ),​

so

θ(t) = arctan(y(t) / x(t))​

will tell you the angle θ(t) (or θ(t) ± π) around the circumference. In either case dθ/dt is the rate the angle is increasing.

7. Mar 8, 2016

Samy_A

How so?
Looking only at the numerators:
$(1-t²)²+(2t)²=1-2t²+t^4+4t²=1+2t²+t^4=(1+t²)²$, and that is equal to the denominator squared.

Last edited: Mar 8, 2016
8. Mar 8, 2016

andrewkirk

If you use the 't formula' link from post 3, you'll see that the formulas for x and y can be expressed as sin and cos of an angle (which in the link they call $x$ but for this purpose would be better thought of as $\theta$), which is the angle travelled around the circle in the conventional direction for polar coordinates.

9. Mar 8, 2016

zinq

Samy, thanks again for the correction. It seems I ought to join up so I have more time to delete my stupid posts.

10. Mar 8, 2016

Math_QED

These are indeed the t-formulas: Let tan(x/2) = t
=>
(1-t^2)/(1+t^2) = cos x
2t/(1+t^2) = sin x

11. Mar 8, 2016

Thales Costa

I see that now. Thank you for your help!!!

The problem also asks if there are points that are not occupied by the particle as t goes from negative infinity to positive infinity.
Since the vector function (cosθ, sinθ) is defined at every t from -infinity to +infinity, is it safe to say that all the points are occupied by the particle as t changes?

The answer says that the point (-1,0) is not occupied, by I couldn't figure out why. I think it has something to do with the r2 x2+y2=r2 being ±1

12. Mar 8, 2016

andrewkirk

The function is defined for every $t\in(-\infty,\infty)$. But $t$ must be real, so cannot take on a value of $\pm\infty$ as that is not a real number. Hence points on the circle that would require an infinite $t$ are never attained - only asymptotically approached.

13. Mar 8, 2016

Thales Costa

And why is the point (-1,0) not occupied by the particle if the circle is always defined for all real numbers?

Is it because making x = (1-t^2)/(1+t^2) = -1 would give me 2 = 0, therefore not possible.
And plugging x= -1 in x2+y2 = -1 would give me y = 0?

Sorry if it's confusing, my english is a bit bad

14. Mar 8, 2016

andrewkirk

Yes

15. Mar 8, 2016

Thales Costa

Got it. Thank you so much for your help!