# 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!