Elliptical Motion Under Uniform Angular Speed

Introduction

A point can move around an ellipse while its parameter increases at a constant rate. That regular change in parameter may suggest that the point travels at a steady pace. On a circle, it does; on a noncircular ellipse, it does not.

The reason is geometric: equal changes in the parameter correspond to different distances along different parts of the ellipse. Differentiating the position makes this difference precise. It also gives the velocity and acceleration, so we can identify where the motion is fastest and slowest.

Position on an Ellipse

Let \(a>0\) and \(b>0\). The curve \(x=a\cos\theta,\ y=b\sin\theta\) is an ellipse centered at the origin, with semiaxis lengths \(a\) along the horizontal direction and \(b\) along the vertical direction. A point moving on it can be described by the position vector \(\mathbf{r}(t)=(a\cos\theta(t),b\sin\theta(t))\).

Here, “angular rate” refers to the rate of this parameter. It does not generally mean that the direction of the position vector from the center turns at a constant rate. For the calculations below, take \(\omega>0\); if the motion runs in the reverse direction, its speed uses \(|\omega|\).

Velocity, Speed, and Acceleration

Velocity is the derivative of position with respect to time. Applying the chain rule to each coordinate gives

Speed is the magnitude of velocity, so it is nonnegative:

Differentiate the velocity once more to obtain acceleration:

This last identity holds because the parameter rate is constant. It says that acceleration points toward the center and is proportional to position. However, on a noncircular ellipse, that does not mean acceleration is perpendicular to velocity or that speed stays constant. Indeed, the velocity and acceleration need not be perpendicular.

Why the Speed Changes

The squared speed can be rewritten using \(\sin^2\theta+\cos^2\theta=1\):

If \(a\ne b\), the coefficient \(a^2-b^2\) is nonzero and \(\sin^2\theta\) changes as the point moves. Therefore the speed changes. If \(a=b\), the curve is a circle and the expression reduces to \(v^2=\omega^2a^2\), a constant.

The speed’s rate of change follows by differentiating \(v^2\). Since \(v>0\) for \(a,b,\omega>0\), we can divide by \(2v\):

The denominator is positive. Thus, except at the axis endpoints or when \(a=b\), the sign of \(dv/dt\) is determined by the numerator. The point speeds up on some portions of the path and slows down on others, even though \(\theta'(t)\) remains constant.

To verify the result, suppose first that \(a\ge b\). Since \(0\le\sin^2\theta\le1\), the squared speed lies between \(\omega^2b^2\) and \(\omega^2a^2\). It equals its minimum when \(\sin^2\theta=0\), at \(\theta=0,\pi\) modulo \(2\pi\); those positions are \((a,0)\) and \((-a,0)\), the ends of the longer horizontal semiaxis. It equals its maximum when \(\sin^2\theta=1\), at \(\theta=\pi/2,3\pi/2\); those positions are \((0,b)\) and \((0,-b)\), the ends of the shorter vertical semiaxis. If \(b>a\), interchange the roles of sine and cosine: the maximum occurs at the ends of the shorter horizontal semiaxis, and the minimum at the ends of the longer vertical semiaxis. This proves the claim.

Worked Example: Speed at Points on an Ellipse

Worked Example: Including a Different Uniform Rate

Acceleration and the Changing Pace

For the motion in the previous example, the velocity and acceleration have dot product \[ (-5\sqrt{2})(-10\sqrt{2})+(3\sqrt{2})(-6\sqrt{2})=100-36=64. \] For a nonzero velocity, the rate of change of speed equals \((\mathbf{v}\cdot\mathbf{a})/v\). Here it is \(64/(2\sqrt{17})=32/\sqrt{17}>0\), so the point is speeding up at that instant.

This agrees with the general formula: with \(a=5\), \(b=3\), \(\omega=2\), and \(\theta=\pi/4\), it gives \[ \frac{4(25-9)(1/2)}{\sqrt{25(1/2)+9(1/2)}}=\frac{32}{\sqrt{17}}. \] The two calculations match. The acceleration identity \(\mathbf{a}=-\omega^2\mathbf{r}\) also checks directly: multiplying the position in this example by \(-4\) gives \((-10\sqrt{2},-6\sqrt{2})\).

Common Pitfall: Using Position Components for Speed

Speed is the magnitude of velocity, not the magnitude of position. For \((x,y)=(a\cos t,b\sin t)\), the velocity components are \((-a\sin t,b\cos t)\). Using the position components in their place swaps where the maximum and minimum appear.

For example, with \(a=4\) and \(b=2\), the incorrect expression \(\sqrt{a^2\cos^2t+b^2\sin^2t}\) gives \(4\) at \(t=0\). That would incorrectly assign the maximum speed to \((4,0)\). The correction is to differentiate first: \[ \mathbf{v}(t)=(-4\sin t,2\cos t), \qquad v(t)=\sqrt{16\sin^2t+4\cos^2t}. \] At \(t=0\), this gives speed \(2\); at \(t=\pi/2\), it gives speed \(4\). The point moves fastest at the ends of the shorter semiaxis and slowest at the ends of the longer one.

Worked Example: Identifying the Speed Range

Why This Motion Is Useful to Understand

This ellipse is a clear example of the difference between a parameter’s rate of change and distance traveled per unit time. The parameter advances uniformly, but the coordinate derivatives have different sizes because they are scaled by \(a\) and \(b\). Consequently, the linear speed varies unless the semiaxes are equal.

The speed formula also provides a direct check on calculations: differentiate the position to obtain velocity, take its magnitude, and evaluate at the parameter values of interest. Keep the semiaxis associated with each coordinate in place throughout; swapping \(a\) and \(b\), or using position where velocity is required, can reverse the conclusions about where the motion is fastest.