Space Curves, Velocity and Acceleration

Differentiate each component

A space curve $r(t)=(x(t),y(t),z(t))$ records position. Velocity is $v=r'$ , speed is $\|v\|$ , and acceleration is $a=r''$ . Speed is nonnegative, but each velocity component can have either sign. A constant speed does not imply zero acceleration because direction can change.

On a smooth interval, distance traveled is $\int_a^b\|r'(t)\|dt$ . Displacement is $r(b)-r(a)$ . The magnitude of displacement is at most the traveled distance. Repeated traversal increases distance even when it returns to the same point.

Visual guide

VISUAL GUIDEA helix rises while circling

Helix

The projected curve r(t) = (2 cos t, 2 sin t, t) completes one turn as t runs from 0 to 2π. Equal parameter increments have equal 3D arc length because speed is √5. Their projected lengths need not be equal.

Worked example: a helix

For $r(t)=(2\cos t,2\sin t,t)$ ,

v=(-2\sin t,2\cos t,1),

a=(-2\cos t,-2\sin t,0).

Speed is $\sqrt5$ at every time. From $0$ to $2\pi$ , distance is $2\pi\sqrt5$ , while displacement is $(0,0,2\pi)$ : one complete turn gains height but ends directly above the starting point.

At $t=0$ , the tangent line is $(2,0,0)+s(0,2,1)$ . Its parameter $s$ describes points on the tangent, not generally the original motion at time $t=s$ .

Worked example: reconstruct motion

Suppose acceleration is constant $a=(0,0,-g)$ , initial position is zero, and initial velocity is $(u,0,w)$ . Integrating componentwise gives $v=(u,0,w-gt)$ and $r=(ut,0,wt-gt^2/2)$ .

For $u,w>0$ , return to ground occurs at $t=2w/g$ , giving horizontal range $2uw/g$ . This ideal model ignores air resistance, terrain and Earth's curvature. These assumptions define where its exact mathematical solution is physically useful.

For an arbitrary constant-speed motion, differentiating $v\cdot v$ yields $2v\cdot a=0$ . Thus acceleration is perpendicular to velocity whenever speed is constant and the derivatives exist.

Practice

Find velocity and speed for $r(t)=(3t,4t,0)$ .
Find displacement and distance from $t=0$ to $2$ for that curve.
For the unit circle $(\cos t,\sin t,0)$ , verify velocity is perpendicular to acceleration.

Show worked solutions

$v=(3,4,0)$ and speed is $5$ .
Displacement is $(6,8,0)$ and distance is $10$ .
The dot product $(-\sin t,\cos t,0)\cdot(-\cos t,-\sin t,0)$ is zero.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.