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Speed along a path

A parametric representation gives both xx and yy as functions of the time tt. The velocity vector is the derivative, coordinate by coordinate, and it points along the tangent. The speed is the length of the velocity vector, not one of the coordinates alone. You differentiate first, insert the time, and use Pythagoras. The acceleration is the derivative of the velocity vector.

v⃗(t)=[x′(t),y′(t)]\vec v(t) = [x'(t), y'(t)]velocity vector
∣v⃗(t)∣=x′(t)2+y′(t)2|\vec v(t)| = \sqrt{x'(t)^2 + y'(t)^2}speed

Symbols

ttthe parameter, often time
v⃗\vec vthe velocity vector

Example

r⃗(t)=[2t,t2]\vec r(t) = [2t, t^2] gives v⃗(t)=[2,2t]\vec v(t) = [2, 2t]. At t=2t = 2 the velocity vector is [2,4][2, 4], and the speed is 20≈4.47\sqrt{20} \approx 4.47.

Speed is a positive number. The velocity vector has a direction and can have negative coordinates.
Practise parametric curves for free →

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Part of Mathematics R1: Parametric curves.