Describing the motion of a particle (sphere)

Question

If I have the following position at time t : $\hat{r}(t) = 3\cos(t)\hat{i} + 4\cos(t)\hat{j} + 5\sin(t)\hat{k}$ , then how can I tell if the particle's path lies on a sphere or not? If e.g. the second term was simply $t$ and not a trigonometric function, I know that it would be a spiral, since $t$ will change along (in this case) y-axis on this curve, but if it was a constant then we would simply have an ellipse; but I don't know how to explain or derive a sphere from the given position vector, I tried to use spherical coordinates but to no avail. Could someone explain how one can see if the path of the particle lies on the sphere? If it indeed is a sphere, then from the norm of the position we know that the radius will be 5 units.

Answer 1

In your case spherical coordinates should work I think? with $\sin (a) = \frac{3}{5}$ ,$$r=5(\sin (a)\cos (t), \cos (a)\cos (t), \sin (t))$$ which is indeed a spherical coordinate with $a$ is the angle around $z$ axis and $t$ is the angle away from $z$ axis. Hope this could help. I think one can try to show $$\frac{d}{dt}\left( (r_x-a)^2+(r_y-b)^2+(r_z-c)^2\right)=0$$ with (a,b,c) can be derived by set $t=t_0,t_1,t_2,t_4$ and use this 4 points to find the sphere across them and (a,b,c) is the center of the sphere. Hope this can help.

Answer 2

Let $$r=x\hat i+y\hat j+z\hat k$$ comparing, we get $$\frac{x}{3}=\cos t,\frac{y}{4}=\cos t, \frac{z}{5}=\sin t $$ $$\implies \cos t= \frac{1}{2}(\frac{x}{3}+\frac{y}{4})$$and $$\sin t=\frac{z}{5}$$ $$\implies \frac{1}{4}(\frac{x}{3}+\frac{y}{4})^2+\frac{z^2}{25}=1$$ Can you proceed?

Answer 3

Let $t = \theta + \frac{\pi}{2}$, so that $\hat{r}(t)$ gives $x = 3\cos(\theta + \frac{\pi}{2}) = -3\sin\theta, y= -4\sin \theta, z = 5\cos\theta$.

Comparing to the spherical coordinates: $x = r\sin\theta \cos\phi, y = r\sin\theta \sin\phi, z = r\cos\theta$.

Then radius of sphere $r = 5$, but we need to find the fixed azimuth $\phi$ such that $5\cos \phi = -3$, and $5 \sin\phi = -4$.

Answer 4

You can easily check that every point has distance $5$ from $(0,0,0)$. Moreover, all the points lie on the plane $4x=3y$, so $\hat r$ is a circle.

Answer 5

The locus is a sphere of radius 5 centered at origin. Since y/x is constant, there is a cutting plane though the poles. You are on meridians of longitude 180*arctan (4/3) and another on the other side of sphere.