Vector-Valued Functions and Motion Along a Curve
Describe a path in space by a vector that depends on one parameter. Differentiating gives velocity and acceleration, and the arc length is an integral of speed.
What a learner can do afterwards
- Differentiate a vector-valued function componentwise to get velocity, speed, and acceleration
- Find the unit tangent at a point on a space curve
- Set up the arc length of a curve as an integral of speed
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A vector-valued function takes one parameter t and returns a position vector r(t) = (x(t), y(t), z(t)). When t is time, r(t) is the position of a moving point. Differentiation gives the instantaneous rate of change. The derivative of a constant is 0, of ct is c, of ct² is 2ct, and of ct³ is 3ct². Sums differentiate term by term. Apply these rules to each component separately.
For r(t) = (t, t², t³), differentiate each component. The derivative of t is 1, of t² is 2t, and of t³ is 3t². So the velocity is v(t) = (1, 2t, 3t²). When t is time, v(t) is the velocity vector. For r(t) = (3t, 4t, 0), the velocity is (3, 4, 0), constant.
Speed is the magnitude of velocity. Square each component, add the squares, and take the square root. This extends Pythagoras to three dimensions. Speed is a scalar, always nonnegative. It tells how fast the point moves, not which way.
The unit tangent vector is T(t) = v(t) / |v(t)|, defined when the speed |v(t)| is nonzero. For v = (3, 4, 0), the speed is √(9 + 16) = 5, so T = (3/5, 4/5, 0). This keeps the direction of motion and drops the size. At a stationary instant where v = 0, this formula is undefined. That does not mean no geometric tangent exists; it means this particular formula cannot be used there.
Acceleration is the derivative of velocity, or the second derivative of position. For r(t) = (t, t², t³), velocity is (1, 2t, 3t²). Differentiate again: the derivative of 1 is 0, of 2t is 2, and of 3t² is 6t. So a(t) = (0, 2, 6t). Acceleration is a vector. It differs from the scalar rate of change of speed.
Arc length is the integral of speed over the interval. For a <= b and a path that is continuously or piecewise continuously differentiable, the arc length is the integral from a to b of |v(t)| dt. The integral accumulates small distances along the path. It gives distance traveled, counting repeated traversal, not merely endpoint displacement. Constant speed gives speed times (b - a).
A particle moves with speed |v(t)| = 3t², which is nonnegative on [0, 2]. An antiderivative is F(t) = t³, since F'(t) = 3t². The distance traveled is F(2) - F(0) = 8 - 0 = 8. The integral from 0 to 2 of 3t² dt equals the same 8.
Differentiate component by component for velocity, divide by nonzero speed for the unit tangent, differentiate again for acceleration, and integrate speed for arc length.
Eight extra tasks were freshly authored under original:mt_Xcjt6exMy3:x1 through x8 with null license. Four readings are fresh original prose under original:mt_Xcjt6exMy3:reading1 through reading4 with null license. Two Khan Academy video objects are unchanged with null licenses and metadata-only verification. The OpenStax publisher pages state NC-SA, not CC0, and the PG corpus CC0 labels are unsupported, so no source text was carried over. The curvature video is optional extension and not a prerequisite for core tasks.
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