243 parametric equations and motion direction

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karush
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11.1 Parametric equations and a parameter interval for the motion of a particle in the xy-plane given. Identify the paritcals path by finding a Cartestian equation for it $x=2\cos t, \quad 2 \sin t, \quad \pi\le t \le 2\pi$
(a) Identify the particles path by finding a Cartesian Equation the Cartesian equation is
$$x^2+y^2=4$$
(b) Indicate the portion of the graph traced by the particle and the direction of motion
so if $x=2cos{(\pi)}=-2$ and $x=2cos{(2\pi)}=2$ then
$$-2\le t \le 2$$
and the particle moves in a clockwise direction


View attachment 9217

ok, I think this is correct but I got the carresian equation just by ploting the parametric into desmos and saw that it was a circle with a radius of 2. the examples didn't the normal steps

also obviously I just pluged into see the direction of motion so...

I was going to try to use tikx on this but didn't how to use an interval be cute to put an arrow also
 

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  • #2
HallsofIvy
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Did you really need to use "desmos"? If [tex]x= 2 cos(t)[/tex] and [tex]y= 2 sin(t)[/tex] then [tex]x^2= 4 cos^2(t)[/tex] and [tex]y^2= 4 sin^2(t)[/tex] so [tex]x^2+ y^2= 4cos^2(t)+ 4sin^2(t)= 4(cos^2(t)+ sin^2(t))= 4[/tex], the equation of a circle of radius 2. When [tex]t= \pi[/tex], [tex]x= 2 cos(\pi)= -2[/tex], [tex]y= 2 sin(\pi)= 0[/tex] and when [tex]t= 2\pi[/tex], [tex]x= 2 cos(2\pi)= 2[/tex], [tex]y= 2 sin(2\pi)= 0[/tex] so the particle moves counter-clock wise from (-2, 0) to (2, 0). You say "[tex]-2\le t\le 2[/tex]". I am sure you mean "[tex]-2\le x\le 2[/tex]".
 
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karush
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Ok I didn't understand how the square got there

View attachment 9218

Why is the text encroaching
 

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  • #4
HallsofIvy
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Do you mean [tex]x^2[/tex] and [tex]y^2[/tex]? They "got there" because I put them there!

And I put them there because I wanted an equation in x and y only. I wanted to eliminate "t" and I knew that, for any [tex]t[/tex], [tex]sin^2(t)+ cos^2(t)= 1[/tex].
 

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