Accumulate speed along a path

Try

A point moves according to x=3t,y=4t from t=0 to t=2. Each second its displacement vector is (3,4), whose magnitude is 5. Predict the traveled length and compare it with the displacement magnitude. They agree here because the path is a straight segment traveled without reversal.

Predict

Now consider x=t²,y=0 from t=−1 to t=1. Both endpoint positions are (1,0), but the point travels to the origin and back. Predict the displacement and distance separately. A curve’s image, its endpoint displacement and the distance counted by its parameterization are different objects.

Experiment

For the returning path, x′=2t and y′=0, so speed is |2t|. Build a directed-motion ledger.

Parameter interval	Position change	Distance
[−1,0]	from (1,0) to (0,0)	1
[0,1]	from (0,0) to (1,0)	1
Whole interval	zero displacement	2

Sketch the same line segment with two opposite arrows. Integrating 2t would give signed horizontal change zero, while integrating its magnitude gives traveled length 2. The signed motion lab supplies a one-dimensional counterpart to this vector distinction.

Understand

For a continuously differentiable path (x(t),y(t)) on [a,b], traveled length is ∫ₐᵇ√[(x′)²+(y′)²]dt. Piecewise smooth paths can be split into suitable intervals and their lengths added. The integrand is speed, the nonnegative magnitude of velocity. A derivation approximates the path by short chords and uses the Pythagorean length of each displacement; refinement produces the integral under the regularity assumptions.

If t represents time, velocity components have length-per-time units and speed times dt has length units. If t is merely a parameter, the same formula measures length without a physical clock. A change of parameter with appropriate regularity and one-to-one traversal preserves the length, while retracing portions counts them again.

For a circle x=Rcos t,y=Rsin t, speed is √(R²sin²t+R²cos²t)=R for R>0. On [0,2π] the length is 2πR. On [0,4π] it is 4πR because the circle is traveled twice. The geometric image has only one circumference; the parameterized integral records traversal multiplicity. State which quantity the problem requests.

For x=t,y=(2/3)t^(3/2) on [0,3], derivatives are 1 and √t. Length is ∫₀³√(1+t)dt=(2/3)[(1+t)^(3/2)]₀³=14/3. The speed is continuous including t=0, so the ordinary length formula applies. A zero component at an endpoint does not automatically invalidate the parameterization.

Speed differs from the slope y′/x′. Slope can be negative, infinite or undefined while speed remains finite and nonnegative. A vertical straight motion x=0,y=t has speed 1 even though no finite dy/dx exists. Conversely, a stationary point may have speed zero without ending the path or making its total length undefined.

Apply

For each practice task, state the parameter interval and calculate the velocity magnitude before integrating. Check for retracing and separate displacement from distance. In the circle problem, explain the factor of two geometrically rather than treating the larger integral as an algebraic error.

Continue

Next use polar coordinates to describe regions by angle and radius. The tracing interval becomes especially important when radii change sign or a closed curve is represented more than once.