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Question

particle starts at $(0,-3)$ and moves clockwise around origin on graph $x^2+y^2=9$, revolve in $9$ seconds find parametric equation in term of $t$.

What I've done so far:

I first thought that the graph ought to be $x^2+y^2=9$

so then I say that $x=\frac{9}{2}\cos{t}=x$ and $y=\frac{9}{2}\sin{t}$

but then this particle in this graph travels CCW so then I change it to :

$x=\frac{9}{2}\cos{t}=x$ and $y=\frac{9}{2}\sin{-t}$

but then I found out that when I plug in $t=0$, I do not get -3

How do I phase shift this parametric equation so that it satisfies the fact that the particle starts at $(0.-3)$?

  • 1
    The factor $\frac92$ shouldn't be here ... A parametric representation of this circle is given by $x=3\cos(t)$ and $y=3\sin(t)$. This because the rhs in the cartesian equation is the *square* of the radius (and not twice the radius !). – 2017-02-13

4 Answers 4

2

Let's start with the standard clockwise parametrization,

$$x=3\cos (t)$$

$$y=-3\sin (t)$$

With $t \in [0,2\pi]$. This starts at $(3,0)$ and then moves clockwise.

We want it to start at $(0,-3)$. If we rotate $(3,0)$ clockwise $90$ degrees we get $(0,-3)$. So shift $\frac{\pi}{2}$ radians clockwise. Let,

$$x=3\cos (t+\frac{\pi}{2})$$

$$y=-3\sin (t+\frac{\pi}{2})$$

Now we need to deal with the fact that it should take $9$ seconds to do a full revolution. Right now it takes $2\pi$ seconds because the period of both is $2\pi$. If the new period is $9$ then $2\pi$ over the horizontal shift $a$ should be $9$.

$$\frac{2\pi}{a}=9$$

$$a=\frac{2\pi}{9}$$

So,

$$x=3\cos (\frac{2\pi}{9}t+\frac{\pi}{2})$$

$$y=-3\sin (\frac{2\pi}{9}t+\frac{\pi}{2})$$

2

You have two problems. As you say, the graph is $x^2+y^2=9.$ That means $x,y$ range from $-3$ to $+3$. As $\sin$ and $\cos$ range from $-1$ to $1$ you need to scale by a factor $3$, not $\frac 92$. Your idea of using $-t$ in the $y$ equation to reverse the direction of rotation is spot on. Now you just need to get the phase right. You want $x=3\cos (t + \phi), y=3\sin(-t+\phi)$ You just need to find $\phi$ and you are home. Good work.

0

Let's start with the value of $\theta$, in radians, as a function of time. Because the particle is travelling at a constant velocity, revolving in $9$ seconds, $\dfrac{d\theta}{dt} = -\dfrac{2\pi}{9}$. The minus sign is because the particle is travelling clockwise. Then $\theta(t) = -\dfrac{2\pi}{9}t + \theta_0$. Since the angle of the point $(-3,0)$ is $-\pi$, we must have

$-\pi = \theta(0) = \theta_0$

So

$\theta(t) = -\dfrac{2\pi}{9}t - \pi$.

So $(x,y) = \left(3 \cos\left( -\dfrac{2\pi}{9}t - \pi\right), 3 \sin\left( -\dfrac{2\pi}{9}t - \pi\right) \right) = \left( -3\cos\left( \dfrac{2\pi}{9}t\right), 3\sin\left( \dfrac{2\pi}{9}t\right) \right)$

-1

$$ x=-3\sin\left(\dfrac{2\pi}{9}t\right), \quad y=-3\cos\left(\dfrac{2\pi}{9}t\right) $$

  • 2
    Can you explain how you did this? I cannot understand how you got this answer. – 2017-02-13