Show that a particle moves over a circumference

[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?

 

[Svein]

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

 

[andrewkirk]

Those formulas are from the http://www.qc.edu.hk/math/Certificate%20Level/t%20method.htm 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.

 

[Thales Costa]

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

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?

 

[Mark44]

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

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

 

[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.

 

[Samy_A]

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.

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.

 

[andrewkirk]

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

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 traveled around the circle in the conventional direction for polar coordinates.

 

[zinq]

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

 

[member 587159]

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

 

[Thales Costa]

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 traveled around the circle in the conventional direction for polar coordinates.

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

 

[andrewkirk]

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 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.

 

[Thales Costa]

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.

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

 

[andrewkirk]

Is it because making x = (1-t^2)/(1+t^2) = -1 would give me 2 = 0, therefore not possible.

Yes

 

[Thales Costa]

Yes

Got it. Thank you so much for your help!