Tangential and Normal Components of Acceleration
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 3.4: “Motion in Space” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/3-4-motion-in-space |
| Textbook used in class | Stewart, Calculus, Section 13.4: “Motion in Space: Velocity and Acceleration” (Examples 4, 5) |
Quick Reference
Decomposition: $$\mathbf{a} = a_T \mathbf{T} + a_N \mathbf{N}$$
| Component | Meaning | Formula |
|---|---|---|
| $a_T$ (tangential) | rate of change of speed | $\dfrac{d|\mathbf{v}|}{dt} = \dfrac{\mathbf{r}' \cdot \mathbf{r}''}{|\mathbf{r}'|}$ |
| $a_N$ (normal) | centripetal (direction change) | $\kappa |\mathbf{v}|^2 = \dfrac{|\mathbf{r}' \times \mathbf{r}''|}{|\mathbf{r}'|}$ |
Magnitude check: $|\mathbf{a}|^2 = a_T^2 + a_N^2$ (since $\mathbf{T} \perp \mathbf{N}$).
Motivation
When a car rounds a curve, the driver feels two distinct forces: one pushing into the seat (acceleration in the direction of travel -- speeding up or braking) and one pushing sideways (centripetal -- the direction change). These come from the same acceleration vector $\mathbf{a}(t)$, split into two perpendicular components along $\mathbf{T}$ and $\mathbf{N}$.
The decomposition $\mathbf{a} = a_T\mathbf{T} + a_N\mathbf{N}$ makes this separation precise. It also shows that curvature and speed together determine how much the curve bends a particle’s path: $a_N = \kappa |\mathbf{v}|^2$ grows both with sharper turns (large $\kappa$) and with greater speed.
Key Concepts
1. The Decomposition
Because $\mathbf{v} = |\mathbf{v}|\mathbf{T}$, differentiating gives: $$\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d|\mathbf{v}|}{dt}\mathbf{T} + |\mathbf{v}|\mathbf{T}'.$$
Now $\mathbf{T}' = |\mathbf{T}'|\mathbf{N} = \kappa|\mathbf{r}'|\mathbf{N}$ (from the definition of curvature), so: $$\mathbf{a} = \underbrace{\frac{d|\mathbf{v}|}{dt}}_{a_T}\mathbf{T} + \underbrace{\kappa|\mathbf{v}|^2}_{a_N}\mathbf{N}.$$
The result: acceleration has NO component in the binormal direction $\mathbf{B}$. The acceleration vector always lies in the osculating plane.
2. Computing $a_T$ and $a_N$ in Practice
Direct differentiation of speed can be cumbersome. The dot and cross product formulas are cleaner: $$a_T = \frac{\mathbf{r}'(t) \cdot \mathbf{r}''(t)}{|\mathbf{r}'(t)|}, \qquad a_N = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|}.$$
If $|\mathbf{a}|^2$ is easy to compute, you can also find $a_N = \sqrt{|\mathbf{a}|^2 - a_T^2}$ without a cross product.
3. Physical Meaning
$a_T > 0$ means the particle is speeding up; $a_T < 0$ means it is slowing down; $a_T = 0$ means the speed is momentarily constant. $a_N \geq 0$ always: it is zero only when the curve is straight (no direction change).
Worked Example
Find $a_T$ and $a_N$ for $\mathbf{r}(t) = \langle t^2, t, t^3\rangle$ at $t = 1$. (Adapted from Stewart 13.4, Example 4.)
Step 1. Compute derivatives at $t = 1$. $$\mathbf{r}'(t) = \langle 2t, 1, 3t^2\rangle \implies \mathbf{r}'(1) = \langle 2, 1, 3\rangle.$$ $$\mathbf{r}''(t) = \langle 2, 0, 6t\rangle \implies \mathbf{r}''(1) = \langle 2, 0, 6\rangle.$$
Step 2. Compute $|\mathbf{r}'(1)|$. $$|\mathbf{r}'(1)| = \sqrt{4+1+9} = \sqrt{14}.$$
Step 3. Compute $a_T$. $$a_T = \frac{\mathbf{r}' \cdot \mathbf{r}''}{|\mathbf{r}'|} = \frac{(2)(2) + (1)(0) + (3)(6)}{\sqrt{14}} = \frac{4 + 0 + 18}{\sqrt{14}} = \frac{22}{\sqrt{14}}.$$
Step 4. Compute $a_N$ via the cross product. $$\mathbf{r}' \times \mathbf{r}'' = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 3 \\ 2 & 0 & 6\end{vmatrix} = \mathbf{i}(6-0) - \mathbf{j}(12-6) + \mathbf{k}(0-2) = \langle 6, -6, -2\rangle.$$ $$|\mathbf{r}' \times \mathbf{r}''| = \sqrt{36+36+4} = \sqrt{76} = 2\sqrt{19}.$$ $$a_N = \frac{2\sqrt{19}}{\sqrt{14}}.$$
Check: $|\mathbf{a}(1)| = |\langle 2,0,6\rangle| = \sqrt{40} = 2\sqrt{10}$.
$a_T^2 + a_N^2 = \dfrac{484}{14} + \dfrac{76}{14} = \dfrac{560}{14} = 40 = |\mathbf{a}|^2$. Correct.
$a_T$ is the component of $\mathbf{a}$ in the direction of motion, computed by projecting onto $\mathbf{r}'$ (not $\mathbf{T}$). The formula $a_T = (\mathbf{r}' \cdot \mathbf{r}'')/|\mathbf{r}'|$ IS the projection of $\mathbf{r}''$ onto the unit tangent $\mathbf{T} = \mathbf{r}'/|\mathbf{r}'|$. If you mistakenly project onto the un-normalized $\mathbf{r}'$, you get $(\mathbf{r}'\cdot\mathbf{r}'')/|\mathbf{r}'|^2$, which is wrong. Always divide by $|\mathbf{r}'|$, not $|\mathbf{r}'|^2$.
Common Misconceptions
the tangential component of acceleration is the dot product of acceleration with the un-normalized velocity vector $\mathbf{r}'$.
This is the multiplicative-not-additive error in projections. The formula $a_T = (\mathbf{r}' \cdot \mathbf{r}'')/|\mathbf{r}'|$ is the projection of $\mathbf{r}''$ onto the unit tangent $\mathbf{T} = \mathbf{r}'/|\mathbf{r}'|$. Projecting onto the un-normalized $\mathbf{r}'$ gives $(\mathbf{r}' \cdot \mathbf{r}'')/|\mathbf{r}'|^2$, which is the scalar projection divided by $|\mathbf{r}'|$ and has different units and magnitude. Always divide by $|\mathbf{r}'|$ once, not twice.
Leveled Practice
Problem 1. A particle moves along a circle of radius 3 at constant speed 6 m/s. Find $a_T$ and $a_N$.
Show answer
Constant speed means $d|\mathbf{v}|/dt = 0$, so $a_T = 0$.
Curvature of a circle of radius 3 is $\kappa = 1/3$.
$a_N = \kappa |\mathbf{v}|^2 = (1/3)(36) = 12$ m/s$^2$.
All acceleration is centripetal.