Difficulty: Intermediate to Advanced | Prerequisites: parametric curves, derivatives of vector-valued functions, dot product, cross product, curvature
Once you can describe a curve parametrically, the natural next questions are: how fast is a particle moving along it, which way is it heading, and how is its motion changing? This topic pulls together velocity, speed, the unit tangent and unit normal vectors, and the decomposition of acceleration into tangential and normal components. It is one of the most formula-heavy sections of the course, but every formula has a clean geometric meaning. You should be comfortable with curvature and the cross product before diving in.
The velocity vector r'(t) tells you the direction and speed of motion. The unit tangent T points along the curve; the unit normal N points toward the centre of the curve's turning. Acceleration decomposes into two pieces: a tangential component aT (how fast the speed is changing) and a normal component aN (how fast the direction is changing).
Velocity vector v(t)
v(t) = r'(t). The derivative of the position vector. Its direction is the direction of motion; its magnitude is the speed.
Think of it as the arrow showing which way and how fast the particle is going at time t.
Speed
The scalar v = |r'(t)| = |v(t)|. Always non-negative.
In simple terms, how fast, with no regard to direction.
Unit tangent vector T(t)
T(t) = r'(t) / |r'(t)| = v(t) / |v(t)|. A unit vector pointing in the direction of motion.
In simple terms, the direction of travel, normalised to length 1.
Unit normal vector N(t)
N(t) = T'(t) / |T'(t)|. Points in the direction the curve is turning (toward the centre of curvature).
In simple terms, the direction in which the curve is bending.
Osculating plane
The plane spanned by T and N at a given point. The acceleration vector always lies in this plane.
Tangential component of acceleration (aT)
aT = comp_T(a) = (a . v) / |v|, or equivalently d/dt(|r'(t)|). Measures the rate of change of speed.
Think of it as the part of acceleration that speeds you up or slows you down.
Normal component of acceleration (aN)
aN = comp_N(a). Measures the rate of change of the direction of the velocity vector. Can be computed as sqrt(|a|^2 - aT^2), or as |r' x r''| / |r'|, or as kappa * v^2.
Think of it as the part of acceleration that turns you.
The velocity vector is v(t) = r'(t)
Its direction is given by the unit tangent vector T
Its magnitude is the scalar speed v = |r'(t)|
Unit tangent T(t) = r'(t) / |r'(t)|: shows the direction the curve is heading
Unit normal N(t) = T'(t) / |T'(t)|: shows the direction the curve is turning (perpendicular to T, pointing inward)
Together, T and N span the osculating plane
The acceleration vector a(t) = r''(t) lies in the osculating plane. It decomposes into:
a(t) = aT * T + aN * N
Tangential component aT:
Measures how the speed (magnitude of velocity) is changing
Computed as: aT = (r'' . r') / |r'| = (a . v) / |v|
Alternatively: aT = d/dt(|r'(t)|), the derivative of the speed function
Normal component aN:
Measures how the direction of velocity is changing
Computed as: aN = sqrt(|a|^2 - aT^2)
Or: aN = |r' x r''| / |r'|
Or: aN = kappa(t) * v(t)^2
The easiest method depends on what you already have computed. The sqrt formula is often the quickest if you already know aT and |a|.
\vec{v}(t) = \vec{r}\,'(t), \quad v = |\vec{r}\,'| = |\vec{v}|\vec{T}(t) = \frac{\vec{r}\,'(t)}{|\vec{r}\,'(t)|}, \quad \vec{N}(t) = \frac{\vec{T}\,'(t)}{|\vec{T}\,'(t)|}\vec{a}(t) = \vec{r}\,''(t) = a_T\,\vec{T} + a_N\,\vec{N}a_T = \frac{\vec{r}\,'' \cdot \vec{r}\,'}{|\vec{r}\,'|} = \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}a_N = \sqrt{|\vec{a}|^2 - a_T^2} = \frac{|\vec{r}\,' \times \vec{r}\,''|}{|\vec{r}\,'|} = \kappa\, v^2This decomposition is exactly how physicists and engineers analyse forces on a particle moving along a curved path. The tangential component tells you about the engine or brakes (speeding up or slowing down). The normal component tells you about the centripetal force needed to maintain the curve, which is why cars skid on icy bends: the road cannot supply enough normal force.
Students often confuse velocity (a vector) with speed (a scalar). Speed is the magnitude of velocity and is always non-negative.
The unit normal N does not point "up" or in any fixed direction. It points toward the centre of curvature, which changes along the curve.
Students sometimes think aT and aN are always positive. aT can be negative (the particle is slowing down). aN is always non-negative.
Forgetting that a(t) = aT * T + aN * N, not aT + aN. The components are multiplied by their respective unit vectors before adding.
⚠️ Be able to compute T(t) and N(t) from a given r(t). This is a standard multi-step problem.
⚠️ Know at least two methods for computing aT and aN. The dot product formula for aT and the sqrt formula for aN are usually the fastest pair.
⚠️ Understand the physical interpretation: aT = rate of change of speed, aN = rate of change of direction.
⚠️ The relationship aN = kappa * v^2 ties this topic back to curvature. Expect exam questions that link both concepts.
True or false: Speed can be negative.
False. Speed = |v(t)| >= 0 always.
Fill in the blank: The unit tangent vector T(t) is obtained by dividing r'(t) by ___.
|r'(t)| (its magnitude).
True or false: The normal component of acceleration aN can be negative.
False. aN >= 0.
Fill in the blank: The acceleration vector lies in the ___ plane.
Osculating.
True or false: If a particle moves at constant speed along a curve, the tangential component of acceleration is zero.
True. Constant speed means d/dt(|r'|) = 0, so aT = 0.
Q: Given r(t) =
A: v(t) = r'(t) = <-sin(t), cos(t), 1>. Speed = |v(t)| = sqrt(sin^2(t) + cos^2(t) + 1) = sqrt(2).
Q: For the same curve r(t) =
A: T(t) = v(t)/|v(t)| = <-sin(t)/sqrt(2), cos(t)/sqrt(2), 1/sqrt(2)>.
Q: Find aT and aN for r(t) =
A: r'(t) = <1, 2t, 0>, r''(t) = <0, 2, 0>. |r'| = sqrt(1 + 4t^2). aT = (r'' . r')/|r'| = 4t / sqrt(1 + 4t^2). |r''| = 2, so |a|^2 = 4. aN = sqrt(4 - 16t^2/(1 + 4t^2)) = sqrt(4/(1 + 4t^2)) = 2/sqrt(1 + 4t^2).
Q: If a particle moves at constant speed along a circle, what are aT and aN?
A: aT = 0 (constant speed means no tangential acceleration). aN = kappa * v^2 = v^2 / R, where R is the radius. All the acceleration is centripetal.
This topic directly uses curvature (kappa appears in the aN formula) and the cross product (needed for the |r' x r''| / |r'| formula for aN). The tangent and normal vectors reappear in the Frenet-Serret frame, which adds the binormal vector B = T x N. In physics, this decomposition of acceleration is fundamental to Newton's second law applied to curvilinear motion.
velocity vector, speed, unit tangent vector, unit normal vector, acceleration, tangential acceleration, normal acceleration, centripetal acceleration, osculating plane, Frenet-Serret frame, binormal vector, curvature, parametric curve, MATH 126, Calculus III, vector-valued functions, kinematics