Speed From Parametric Derivatives

From Coordinate Derivatives to Speed

A parametrized path describes position by giving coordinate functions in terms of a parameter, often time. If the position is \((x(t),y(t))\), then \(x'(t)\) and \(y'(t)\) describe the instantaneous changes in the two coordinates. Neither coordinate derivative alone measures how quickly the point is traveling along the path: the point can change both coordinates at once.

The two coordinate changes combine through the ordinary distance formula. The resulting quantity is the magnitude of the derivative vector. To interpret it as the instantaneous rate of change of distance traveled, however, we need a regularity condition: continuous coordinate derivatives near the instant in question are sufficient.

Here, distance traveled over a parameter interval means the length of the path traced over that interval, not merely the distance between its endpoints. Those two distances need not agree when the path bends. The speed formula will follow by comparing path length with endpoint displacement over a short interval.

The Speed Formula

Write the coordinate-derivative vector at \(t\) as \((x'(t),y'(t))\). Its magnitude is the ordinary Euclidean length of that vector. Therefore the formula for speed is

Speed is nonnegative. A negative value of \(x'(t)\) or \(y'(t)\) means that the corresponding coordinate is decreasing; it does not mean that speed is negative. Squaring the coordinate derivatives before adding accounts for changes in either direction.

In physical applications, if \(x\) and \(y\) are measured in metres and \(t\) in seconds, then both coordinate derivatives are measured in metres per second, and the speed is also measured in metres per second. More generally, speed has units of distance divided by units of the parameter.

Why the Formula Gives a Distance Rate

We justify the formula using the definition of curve length as the supremum of the sums of lengths of successive chords. The proof needs continuous derivatives near \(t_0\), not just derivatives at that single instant. Let \(L(a,b)\) denote the length of the path traced as the parameter runs from \(a\) to \(b\), with \(a\le b\).

More explicitly, let \(v(u)=\sqrt{(x'(u))^2+(y'(u))^2}\), which is continuous near \(t_0\). For \(h\ne 0\), the endpoint chord and the integral bound give

For the upper bound, take any partition of the interval. On each subinterval \([a,b]\), the displacement vector is \(\int_a^b (x'(u),y'(u))\,du\). Its magnitude is at most \(\int_a^b v(u)\,du\), by the triangle inequality for integrals. Summing these inequalities bounds every partition sum by the displayed integral; taking the supremum preserves that bound.

Divide the inequalities by \(|h|\). By differentiability of both coordinates, the leftmost quotient tends to \(\sqrt{(x'(t_0))^2+(y'(t_0))^2}\). By continuity of \(v\), its average over the shrinking interval tends to \(v(t_0)\). The two bounds therefore have the same limit, so the distance-traveled quotient has that limit as well. This proves that the speed is the magnitude of the coordinate-derivative vector.

Worked Examples

A Common Mistake: Using a Derivative at One Instant Alone

A tempting shortcut is to assume that differentiability at \(t_0\) by itself guarantees that the formula gives the rate of distance traveled there. It does not. Differentiability controls the change in position between two nearby parameter values, but it does not rule out a path with infinite length accumulating near \(t_0\).

For example, set \(y(t)=0\), \(x(0)=0\), and \(x(t)=t^2\sin(1/t^2)\) for \(t>0\). The derivative \(x'(0)\) exists and equals \(0\), since

Yet the path has infinite variation on every interval \([0,h]\) with \(h>0\). At successive extrema approaching zero, the values of \(x\) alternate in sign and have magnitudes comparable to \(1/n\); the sum of the lengths between these successive values therefore diverges like the harmonic series. Thus its distance traveled from \(0\) is infinite, not a quantity with rate \(0\). This example is not continuously differentiable near \(0\). The continuity hypothesis in the proof rules out this behavior.

Another frequent error is to add coordinate derivatives without taking their magnitude. In the third worked example, that incorrect calculation would give \(5+(-12)=-7\) metres per second. Speed cannot be negative, and the correct calculation is \(\sqrt{5^2+(-12)^2}=13\) metres per second.

Why the Distinction Matters

Coordinate derivatives describe changes in separate directions, while speed describes the rate at which distance along the path accumulates. Combining the derivatives by the Euclidean magnitude accounts for both coordinate changes without allowing opposite signs to cancel. The regularity condition matters because the speed formula is a statement about distance traveled, not merely a calculation involving two derivatives.

For practical calculations, the method is consistent: differentiate each coordinate with respect to the same parameter, evaluate both derivatives at the instant, square them, add, and take the nonnegative square root. Check that the coordinate functions have continuous derivatives nearby when interpreting the result as an instantaneous distance rate, and preserve units throughout.