curvesparametric
Find parametric equations for a particle moving two full revolutions clockwise around a circle of radius $2$ centered at $(3,-1)$. In other words give equations for $x(t)$ and $y(t)$, and specify the time interval.
Is $x=3+2cost$
and $\ \ $$y=-1+2sint$
correct?
I only have several examples from class from which I can study .
it would be great if someone could provide detailed basic steps to solving this kind of problems
Thanks in advance .
The equation $x^2+y^2=1$ works. How can we tell? Experience: if you have a sine and a cosine of the same quantity (in this case, of $\frac{1}{2}\theta$), then you can always "get rid" of the dependence on the quantity by squaring and adding; no matter what the quantity is, you always get $\sin^2(\text{quantity}) + \cos^2(\text{quantity}) = 1$.
But let's say you don't want to try things to see how the parameter can be eliminated. In this case, you can simply solve for the parameter in each equation: $$\begin{align*} x&=\sin\left(\frac{1}{2}\theta\right)\\ \arcsin(x) &= \frac{1}{2}\theta\\ 2\arcsin(x) &=\theta;\\ y &= \cos\left(\frac{1}{2}\theta\right)\\ \arccos(y) &= \frac{1}{2}\theta\\ 2\arccos(y) &= \theta. \end{align*}$$ Therefore, $x$ and $y$ will satisfy $$2\arcsin(x) = 2\arccos(y)$$ or equivalently, $$\arcsin(x) = \arccos(y).$$
The problem is that this equation is ugly; arcsine and arccosine are annoying functions. Better to try to get rid of them. One way to do that is to first take sines on both sides, to get $$x = \sin(\arccos(y)).$$ Now, what is $\sin(\arccos(y))$? Well, since $\sin^2z + \cos^2z = 1$ (there is again!), then $\sin^2 z = 1-\cos^2 z$, so $|\sin z| = \sqrt{1-\cos^2(z)}$. Letting $z=\arccos(y)$, we get $$\sin(\arccos(y)) = \pm\sqrt{1 - \cos^2(\arccos(y))} = \pm\sqrt{1-y^2}.$$ So our equation becomes $$x = \pm\sqrt{1-y^2}.$$ Again, the $\pm$ is annoying. We can get rid of it by squaring both sides, so we get $$x^2 = 1-y^2$$ which can then be rewritten as $$x^2 + y^2 = 1.$$ Now, this is a nice equation; we know exactly what it is (circle of radius $1$ centered at the origin), and it does not involve any annoying functions like inverse trigonometric functions. So this is a good stopping point.
The parametric equation of a cycloid generated by rolling a wheel of radius $a$ at a constant rate by an angle $\theta$ is
$$x = a (\theta-\sin{\theta}) \quad y=a (1-\cos{\theta})$$
Rotating the $(x,y)$ coordinate system to a new coordinate system $(x',y')$ by an angle $\phi$ is accomplished by the tranformation
$$x'=x \cos{\phi}+y \sin{\phi} \quad y'=-x \sin{\phi}+y \cos{\phi}$$
so that the equation of the cycloid here is, after some simplification:
$$x'=a (\sin{\phi}+\theta \cos{\phi}) - a \sin{(\theta+\phi)} \quad y' = a(\cos{\phi}-\theta \sin{\phi}) - a \cos{(\theta+\phi)}$$
Note this assume that the point $P$ begins on the ramp. Note also that $\phi = \pi/2-\alpha$, the angle of incline.
Here is a plot for a constant rate at the specified angle of incline, $\alpha=\pi/3$.
Best Answer
Facts:
Q: What is the equation of that circle using Cartesian coordinates?
A: As it centered in $(+3,-1)$ with radii $2$ so it is $$(x-3)^2+(y+1)^2=4$$
Q: Do the new relations I've got, satisfy the equation above? Will we have a right statement the?
A: As we have $x=3+2\cos(t),~~y=-1+2\sin(t)$ so $$(x-3)^2+(y+1)^2=4\cos^2(t)+4\sin^2(t)=4$$
Q: So, I have done it right?
A: Yes. you did it right. :-)