Can somebody help me with the following problems? For each of the following parametric curves, describe the motion of the particle. That is, tell the type of curve, the starting point of the particle, the direction of motion, and the behavior of the particle.

1. x=sect
y=tant
t= [0, pi/2)U(pi/2, 3*pi/2)

2. x^2=(sint)^2
y^2=(cost)^2
0 less than or equal to t less than or equal to 2pi

3. x=sint
y=(sint)^2
0 less than or equal to t less than or equal to 2pi

4. x=(1-t^2)/(1+t^2)
y=(2t)/(1+t^2)
t is greater than or equal to zero

2. Originally Posted by riverjib
Can somebody help me with the following problems? For each of the following parametric curves, describe the motion of the particle. That is, tell the type of curve, the starting point of the particle, the direction of motion, and the behavior of the particle.

1. x=sect
y=tant
t= [0, pi/2)U(pi/2, 3*pi/2)
This is a partial hyperbola. You can see that all point on this curve satisfy the hyperbola equation if you elimate the variable 't'.
2. x^2=(sint)^2
y^2=(cost)^2
There are 4 possibilities here:

x=sin t
y=cos t

x=-sin t
y=-cos t

x=-sin t
y=cos t

x=sin t
y=-cos t

All are the same curve (except some can the orientation of the closed curve).
Which is a circle since,
x^2+y^2=(sin t)^2+(cos t)^2=1 [Note this is not a correct proof, because I only shown that the point lie on a circle rather than all points are on a circle].

3. x=sint
y=(sint)^2
Substitute,
y=x*x=x^2
A parabola

4. x=(1-t^2)/(1+t^2)
y=(2t)/(1+t^2)
See that x^2+y^2=1
So all points lie on a circle, again this does not show that all the points are the circle (which in fact is untrue). But since t>0 you see that 'y' only have positive terms so it seems to be a positive circle, that is, a semicircle.

3. Hello, riverjib!

Here's #4 . . . I happen to be familiar with the problem.

Describe the motion of the particle.
That is, tell the type of curve, the starting point of the particle,
the direction of motion, and the behavior of the particle.

$4)\; x\:= \: \frac{1 - t^2}{1 + t^2}$
. . $y \:= \:\frac{2t}{1 + t^2}$
. . $t \geq 0$

The parametric equations describe a circle.

Square the two equations: . $\begin{cc}x^2\:=\:\frac{(1-t^2)^2}{(1+t^2)^2} \\ y^2 \:=\:\frac{(2t)^2}{(1+t^2)^2}$

The first equation gives us: . . .x(1 + t²) .= .1 - t²
The second equation gives us: .y(1 + t²) .= .2t

Square the equations: .x²(1 + t²)² .= .(1 - t²)²
. . . . . . . . . . . . . . . . y²(1 + t²)² .= . .(2t)²

Add: . x²(1 + t²)² + y²(1 + t²)² . = . (1 - t²)² + (2t)²

The left side factors: .(1 + t²)²(x² + y²)

The right side is: .1 - 2t² + t^4 + 4t² .= .1 + 2t² + t^4 .= .(1 + t²)²

So we have: .(1 + t²)²(x² + y²) . = . (1 + t²)²

Divide by (1 + t²)²: . x² + y² .= .1 . . . a unit circle!

When t = 0: x = 1, y = 0 . . . We have the point (1,0).
. . The particle starts at the "3 o'clock position".

When t = 1: x = 0, y = 1 . . . We have the point (0,1).
. . The particle moves counterclockwise.

As t →
: x → -1, y → 0 . . . We approach the point (-1,0)
. . The particle heads towards (-1,0), the "9 o'clock position".