MATH-283 (Calculus III) Curriculum

Session 06 - Arc Length and Curvature

Handout Worksheet

Fall Semester 2026

Session Objectives & Overview

By the end of this session, you will be able to:

Compute the total arc length $s = \int_a^b \|\vec{r}'(t)\|\,dt$ of smooth 2D and 3D space curves, derive the accumulated arc length function $s(t)$ , and reparametrize space curves in terms of the intrinsic arc length parameter $s$ to guarantee unit speed traversal ( $\|\vec{r}'(s)\| = 1$ ).
Calculate the curvature $\kappa$ and radius of curvature $\rho = \frac{1}{\kappa}$ of space curves and planar curves using the unit tangent derivative formula $\kappa = \frac{\|\hat{T}'(t)\|}{\|\vec{r}'(t)\|}$ , the cross-product formula $\kappa = \frac{\|\vec{r}'(t) \times \vec{r}''(t)\|}{\|\vec{r}'(t)\|^3}$ , and the intrinsic formula $\kappa(s) = \|\vec{r}''(s)\|$ , and geometrically interpret the osculating circle.
Decompose the kinematic acceleration vector into orthogonal tangential and normal components ( $\vec{a}(t) = a_T \hat{T}(t) + a_N \hat{N}(t)$ ) using $a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}$ and $a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|} = \sqrt{\|\vec{a}\|^2 - a_T^2}$ , connecting scalar throttle/braking dynamics and centripetal turning forces to Newton's Second Law and structural engineering safety limits.

Micro-Lecture

1. Arc Length & Intrinsic Spatial Parametrization

Mathematical Foundation: When a particle moves along a 3D trajectory $\vec{r}(t) = \langle x(t), y(t), z(t) \rangle$ , its scalar speed is the magnitude of its velocity vector: $\|\vec{r}'(t)\| = \sqrt{(x')^2 + (y')^2 + (z')^2}$ . Integrating speed over time yields the Arc Length of a Space Curve: $s = \int_a^b \|\vec{r}'(t)\|\,dt$ . By defining the accumulated distance function $s(t) = \int_a^t \|\vec{r}'(u)\|\,du$ and inverting it to solve for $t(s)$ , we obtain the Arc Length Parametrization $\vec{r}(s)$ . Under this intrinsic parametrization, the traversal speed is identically unity ( $\|\vec{r}'(s)\| = 1$ ), isolating pure spatial geometry from arbitrary time parameterizations.
Engineering Context: In automated manufacturing, CNC milling spindles and laser cutting nozzles must maintain a strictly constant physical feed rate across complex curves to prevent localized melting, thermal stress, or tool chatter.

2. Curvature ( $\kappa$ ), Turning Sharpness, and the Osculating Circle

Mathematical Foundation: Curvature of a Space Curve measures the instantaneous rate at which a curve changes its directional heading per unit distance traveled: $\kappa = \|d\hat{T}/ds\|$ . For generic time-parameterized space curves, curvature is evaluated via the cross-product formula $\kappa(t) = \frac{\|\vec{r}'(t) \times \vec{r}''(t)\|}{\|\vec{r}'(t)\|^3}$ . The reciprocal of curvature defines the Radius of Curvature and Osculating Circles $\rho = \frac{1}{\kappa}$ , representing the radius of the tangent osculating circle nestled against the curve along the Principal Unit Normal Vector $\hat{N}(t) = \frac{\hat{T}'(t)}{\|\hat{T}'(t)\|}$ .
Engineering Context: Civil and transportation engineers design highway interchange ramps and railway transitions with bounded curvature ( $\rho \ge \rho_{\min}$ ) to prevent vehicle rollovers, lateral tire slippage, and passenger discomfort at design speeds.

3. Orthogonal Decomposition of Acceleration ( $\vec{a} = a_T \hat{T} + a_N \hat{N}$ ) & Dynamic Forces

Mathematical Foundation: Kinematic acceleration $\vec{a}(t) = \vec{r}''(t)$ lies in the osculating plane and decomposes into two mutually perpendicular components: $\vec{a}(t) = a_T \hat{T}(t) + a_N \hat{N}(t)$ (detailed in Tangential and Normal Components of Acceleration). The tangential component $a_T = \frac{d}{dt}\|\vec{v}\| = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}$ measures the rate of change of scalar speed (throttle or braking), while the normal component $a_N = \kappa \|\vec{v}\|^2 = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}$ measures the centripetal turning acceleration directed toward the center of curvature.
Engineering Context: Applying Newton's Second Law ( $\vec{F}_{\text{net}} = m a_T \hat{T} + m a_N \hat{N}$ ), automotive and roller coaster engineers isolate the longitudinal traction forces ( $m a_T$ ) from the lateral cornering forces ( $m a_N$ ) to verify tire grip limits and ensure passenger G-force biomechanical safety.

Skill Block 1

Worked Example A: Computing Arc Length and Intrinsic Parametrization

Given the 2D path described by the position vector:

\vec{r}(t) = (3t^2 - 7)\hat{i} + (-4t^2 - 10)\hat{j} \quad \text{for } 0 \le t \le 4

Find the total arc length $s$ traversed from $t = 0$ to $t = 4$ .
Solve for the arc length function $s(t)$ starting at $t = 0$ , and evaluate $s(4)$ .
Determine the intrinsic arc length parametrization $\vec{r}(s)$ and specify its valid parameter domain.

Worked Solution:

Differentiate Position Vector and Compute Scalar Speed: State the Vector Derivative Formula:
$\vec{r}'(t) = \frac{d}{dt}[3t^2 - 7]\hat{i} + \frac{d}{dt}[-4t^2 - 10]\hat{j} = 6t\hat{i} - 8t\hat{j} = \langle 6t, -8t \rangle$

State the Vector Magnitude Formula:
$\|\vec{r}'(t)\| = \sqrt{(6t)^2 + (-8t)^2} = \sqrt{36t^2 + 64t^2} = \sqrt{100t^2} = 10|t|$ Since $t \ge 0$ , $\|\vec{r}'(t)\| = 10t$ . (This speed function is defined in Arc Length of a Space Curve.)
Compute Total Arc Length ( $0 \le t \le 4$ ): State the Definite Arc Length Formula:
$s = \int_{a}^{b} \|\vec{r}'(t)\|\,dt$ Substitute limits $a = 0$ and $b = 4$ : $s = \int_{0}^{4} 10t\,dt = \left[ 5t^2 \right]_{0}^{4} = 5(4)^2 - 5(0)^2 = 5(16) = 80\text{ units}$
Derive the Arc Length Function $s(t)$ : State the Accumulated Arc Length Function Formula:
$s(t) = \int_{0}^{t} \|\vec{r}'(u)\|\,du = \int_{0}^{t} 10u\,du = \left[ 5u^2 \right]_{0}^{t} = 5t^2$ Evaluating at $t = 4$ : $s(4) = 5(4)^2 = 80\text{ units}$
Invert $s(t)$ and Formulate $\vec{r}(s)$ : State the Arc Length Parametrization Inversion Step:
$s = 5t^2 \implies t^2 = \frac{s}{5} \implies t(s) = \sqrt{\frac{s}{5}} \quad (\text{since } t \ge 0)$ Substitute $t(s)$ into $\vec{r}(t)$ : $\vec{r}(s) = \left(3\left(\sqrt{\frac{s}{5}}\right)^2 - 7\right)\hat{i} + \left(-4\left(\sqrt{\frac{s}{5}}\right)^2 - 10\right)\hat{j}$ $\vec{r}(s) = \left(\frac{3}{5}s - 7\right)\hat{i} + \left(-\frac{4}{5}s - 10\right)\hat{j} = \left\langle \frac{3}{5}s - 7, -\frac{4}{5}s - 10 \right\rangle$ Since $t \in [0, 4]$ , the valid domain for $s$ is $[0, 5(4)^2] = [0, 80]$ . (Notice that $\|\vec{r}'(s)\| = \sqrt{(3/5)^2 + (-4/5)^2} = \sqrt{25/25} = 1$ , confirming unit speed traversal as established in Arc Length Parametrization.)

Practice Problem 1: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain why the scalar differential of arc length is written as $ds = \|\vec{r}'(t)\|\,dt$ and how this connects instantaneous speed to accumulated distance.
(b) In Worked Example A, explain why the relation $s(t) = 5t^2$ is invertible on the domain $t \ge 0$ and how this leads to the unit speed property $\|\vec{r}'(s)\| = 1$ .
(c) In the linked engineering application Robotic Manufacturing & CNC Toolpaths (Mechanical & Manufacturing Engineering), why must the CNC controller program the toolhead using the arc length parametrization $\vec{r}(s)$ instead of raw time parameter $t$ ?

Solution Key:

(a) In 3D space, an infinitesimal position change is $d\vec{r} = \langle x'(t), y'(t), z'(t) \rangle dt$ . Its scalar length is $ds = \|d\vec{r}\| = \|\vec{r}'(t)\|dt = v(t)dt$ . The arc length integral $s = \int_a^b v(t)dt$ represents the accumulation of scalar speed over time, exactly analogous to distance in single-variable kinematics.
(b) Because the speed $\|\vec{r}'(t)\| = 10t > 0$ for all $t > 0$ , the arc length function $s(t)$ has a strictly positive derivative $\frac{ds}{dt} > 0$ . By the Inverse Function Theorem, $s(t)$ is strictly monotonic and bijective, admitting a single-valued inverse $t(s) = \sqrt{s/5}$ . Applying the chain rule to $\vec{r}(s) = \vec{r}(t(s))$ yields $\frac{d\vec{r}}{ds} = \vec{r}'(t)\frac{dt}{ds} = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$ , which is the unit tangent vector $\hat{T}(t)$ , possessing constant magnitude $\|\vec{r}'(s)\| = 1$ .
(c) In CNC laser cutting and milling, toolpaths defined with non-uniform speed profiles cause the cutter to decelerate around sharp contours, concentrating laser energy and overheating the metal. Using the intrinsic parametrization $\vec{r}(s)$ , the machine firmware commands servo velocities directly per unit length of cut, ensuring a uniform energy flux and dimensional tolerance.

Show your steps and calculations here.

Practice Problem 2

Given the position vector for a circular path:

\vec{r}(t) = 3\sin(t)\hat{i} + 3\cos(t)\hat{j} \quad \text{for } 0 \le t \le 4\pi

Find the total arc length $s$ for $0 \le t \le 4\pi$ .
Solve for the arc length function $s(t)$ starting from $t = 0$ , and evaluate $s(4\pi)$ .
Determine the intrinsic arc length parametrization $\vec{r}(s)$ .

Solution Key:

Differentiate Position Vector and Compute Speed: State the Vector Derivative Formula:
$\vec{r}'(t) = \frac{d}{dt}[3\sin(t)]\hat{i} + \frac{d}{dt}[3\cos(t)]\hat{j} = 3\cos(t)\hat{i} - 3\sin(t)\hat{j} = \langle 3\cos(t), -3\sin(t) \rangle$

State the Scalar Speed Formula:
$\|\vec{r}'(t)\| = \sqrt{(3\cos(t))^2 + (-3\sin(t))^2} = \sqrt{9\cos^2(t) + 9\sin^2(t)} = \sqrt{9(\cos^2(t) + \sin^2(t))} = \sqrt{9(1)} = 3$ (Defined in Arc Length of a Space Curve.)
Compute Total Arc Length ( $0 \le t \le 4\pi$ ): State the Definite Arc Length Formula:
$s = \int_{0}^{4\pi} \|\vec{r}'(t)\|\,dt = \int_{0}^{4\pi} 3\,dt = [3t]_{0}^{4\pi} = 3(4\pi) - 3(0) = 12\pi\text{ units} \approx 37.70$
Derive Arc Length Function $s(t)$ : State the Accumulated Arc Length Formula:
$s(t) = \int_{0}^{t} \|\vec{r}'(u)\|\,du = \int_{0}^{t} 3\,du = [3u]_{0}^{t} = 3t$ Evaluating at $t = 4\pi$ : $s(4\pi) = 3(4\pi) = 12\pi$
Invert $s(t)$ and Substitute to Find $\vec{r}(s)$ : Invert $s(t) = 3t \implies t(s) = \frac{s}{3}$ . Substitute into $\vec{r}(t)$ :
$\vec{r}(s) = 3\sin\left(\frac{s}{3}\right)\hat{i} + 3\cos\left(\frac{s}{3}\right)\hat{j} \quad \text{for } 0 \le s \le 12\pi$

Show your steps and calculations here.

Practice Problem 3

Given the 3D position vector for a circular helix:

\vec{r}(t) = 3\cos(t)\hat{i} + 3\sin(t)\hat{j} + 4t\hat{k} \quad \text{for } 0 \le t \le \pi

Solve for the total arc length $s$ over the interval $0 \le t \le \pi$ .
Explain whether or not a closed domain such as $0 \le t \le \pi$ permits an arc length parametrization $\vec{r}(s)$ , and formulate $\vec{r}(s)$ with its explicit parameter domain.

Solution Key:

Differentiate Position Vector and Compute Scalar Speed: State the 3D Space Curve Derivative Formula:
$\vec{r}'(t) = \frac{d}{dt}[3\cos(t)]\hat{i} + \frac{d}{dt}[3\sin(t)]\hat{j} + \frac{d}{dt}[4t]\hat{k} = -3\sin(t)\hat{i} + 3\cos(t)\hat{j} + 4\hat{k}$

State the 3D Speed Formula:
$\|\vec{r}'(t)\| = \sqrt{(-3\sin(t))^2 + (3\cos(t))^2 + 4^2} = \sqrt{9\sin^2(t) + 9\cos^2(t) + 16} = \sqrt{9(1) + 16} = \sqrt{25} = 5$
Compute Total Arc Length ( $0 \le t \le \pi$ ): State the Definite Arc Length Formula:
$s = \int_{0}^{\pi} \|\vec{r}'(t)\|\,dt = \int_{0}^{\pi} 5\,dt = [5t]_{0}^{\pi} = 5\pi\text{ units} \approx 15.71$
Domain Validity and Intrinsic Parametrization:
- Monotonicity & Bijective Mapping: The arc length function $s(t) = \int_0^t 5\,du = 5t$ is smooth, continuous, and strictly increasing on the closed interval $[0, \pi]$ . Because its derivative $\frac{ds}{dt} = 5 > 0$ is strictly positive, $s(t)$ is one-to-one, guaranteeing a unique continuous inverse $t(s) = \frac{s}{5}$ .
- Bounded Parameter Domain: The closed time domain $t \in [0, \pi]$ maps directly to the bounded distance domain $s \in [0, 5\pi]$ .
- Parametrization Formulation: $\vec{r}(s) = 3\cos\left(\frac{s}{5}\right)\hat{i} + 3\sin\left(\frac{s}{5}\right)\hat{j} + 4\left(\frac{s}{5}\right)\hat{k} = 3\cos\left(\frac{s}{5}\right)\hat{i} + 3\sin\left(\frac{s}{5}\right)\hat{j} + \frac{4}{5}s\hat{k} \quad \text{for } 0 \le s \le 5\pi$ (Defined in Arc Length Parametrization.)

Show your steps and calculations here.

Regroup 1

Review the algebraic simplification of perfect square trinomials under square roots ( $u^4 + 2ku^2 + k^2 = (u^2 + k)^2$ ).
Clarify why parametrizing by arc length forces $\|\vec{r}'(s)\| = 1$ and how domain boundaries transform from $t \in [a, b]$ to $s \in [0, s(b)]$ .

Skill Block 2

Worked Example A: Finding Curvature with the Cross-Product Formula

Given the 3D position vector:

\vec{r}(t) = \langle 4t^2, -4t - 3, -2t \rangle

Solve for the principal unit tangent vector $\hat{T}(t)$ .
Calculate the curvature $\kappa(1)$ at parameter value $t = 1$ .

Worked Solution:

Compute Velocity Vector and Scalar Speed: State the Vector Derivative Formula:
$\vec{r}'(t) = \frac{d}{dt}\langle 4t^2, -4t - 3, -2t \rangle = \langle 8t, -4, -2 \rangle$

State the Vector Magnitude Formula:
$\|\vec{r}'(t)\| = \sqrt{(8t)^2 + (-4)^2 + (-2)^2} = \sqrt{64t^2 + 16 + 4} = \sqrt{64t^2 + 20} = \sqrt{4(16t^2 + 5)} = 2\sqrt{16t^2 + 5}$
Formulate the Unit Tangent Vector $\hat{T}(t)$ : State the Unit Tangent Vector Formula:
$\hat{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|}$ $\hat{T}(t) = \frac{\langle 8t, -4, -2 \rangle}{2\sqrt{16t^2 + 5}} = \left\langle \frac{4t}{\sqrt{16t^2 + 5}}, \frac{-2}{\sqrt{16t^2 + 5}}, \frac{-1}{\sqrt{16t^2 + 5}} \right\rangle$ (Defined in Unit Tangent Vector.)
Compute Curvature $\kappa(1)$ via Cross-Product Formula: State the Cross-Product Curvature Formula:
$\kappa(t) = \frac{\|\vec{r}'(t) \times \vec{r}''(t)\|}{\|\vec{r}'(t)\|^3}$
(Defined in Curvature of a Space Curve.)
- Compute second derivative (acceleration): $\vec{r}''(t) = \frac{d}{dt}\langle 8t, -4, -2 \rangle = \langle 8, 0, 0 \rangle$
- Evaluate $\vec{r}'(1)$ and $\vec{r}''(1)$ : $\vec{r}'(1) = \langle 8(1), -4, -2 \rangle = \langle 8, -4, -2 \rangle$ $\vec{r}''(1) = \langle 8, 0, 0 \rangle$
- Compute speed at $t = 1$ : $\|\vec{r}'(1)\| = 2\sqrt{16(1)^2 + 5} = 2\sqrt{21}$
- Compute cross product $\vec{r}'(1) \times \vec{r}''(1)$ : $\vec{r}'(1) \times \vec{r}''(1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8 & -4 & -2 \\ 8 & 0 & 0 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(0 - (-16)) + \hat{k}(0 - (-32)) = \langle 0, -16, 32 \rangle$
- Compute magnitude of cross product: $\|\vec{r}'(1) \times \vec{r}''(1)\| = \sqrt{0^2 + (-16)^2 + 32^2} = \sqrt{256 + 1024} = \sqrt{1280} = \sqrt{256 \times 5} = 16\sqrt{5}$
- Substitute into curvature equation: $\kappa(1) = \frac{16\sqrt{5}}{(2\sqrt{21})^3} = \frac{16\sqrt{5}}{8 \times 21\sqrt{21}} = \frac{2\sqrt{5}}{21\sqrt{21}} = \frac{2\sqrt{105}}{441} \approx 0.0464$

Practice Problem 4: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain why a straight line has zero curvature ( $\kappa = 0$ ), whereas a circle of radius $R$ has constant curvature $\kappa = 1/R$ .
(b) In Worked Example A, explain why the cross-product formula $\kappa = \frac{\|\vec{r}' \times \vec{r}''\|}{\|\vec{r}'\|^3}$ was chosen over the unit tangent derivative formula $\kappa = \frac{\|\hat{T}'\|}{\|\vec{r}'\|}$ .
(c) In the linked engineering application Civil Engineering & Highway Exit Design (Civil & Transportation Engineering), why is the radius of curvature $\rho = 1/\kappa$ used to verify vehicle safety compliance rather than curvature $\kappa$ directly?

Solution Key:

(a) Curvature is defined as $\kappa = \|d\hat{T}/ds\|$ . For any straight line path $\vec{r}(t) = \vec{r}_0 + t\vec{v}$ , the unit tangent vector $\hat{T} = \vec{v}/\|\vec{v}\|$ is constant for all $t$ . Thus $d\hat{T}/ds = \vec{0}$ , yielding $\kappa = 0$ . For a circular path of radius $R$ , $\vec{r}(t) = \langle R\cos(t), R\sin(t) \rangle$ , the unit tangent is $\hat{T}(t) = \langle -\sin(t), \cos(t) \rangle$ and speed is $ds/dt = R$ . Differentiating gives $\|d\hat{T}/ds\| = \|\hat{T}'(t)\| / (ds/dt) = 1/R$ , showing that smaller circles bend more sharply (higher curvature) and larger circles bend more gently (lower curvature).
(b) The unit tangent vector $\hat{T}(t)$ contains a radical denominator $\sqrt{16t^2 + 5}$ . Differentiating $\hat{T}(t)$ requires three separate quotient rule expansions with chain rule derivatives of the radical, increasing the risk of algebraic errors. The cross-product formula $\kappa = \frac{\|\vec{r}' \times \vec{r}''\|}{\|\vec{r}'\|^3}$ relies strictly on the first and second derivative polynomials $\vec{r}'(t) = \langle 8t, -4, -2 \rangle$ and $\vec{r}''(t) = \langle 8, 0, 0 \rangle$ , allowing rapid numerical evaluation at $t = 1$ before taking the determinant and norms.
(c) In highway design (AASHTO and NDOT standards), dynamic vehicle stability is modeled by the lateral force balance $F_{\text{friction}} = m \frac{v^2}{\rho}$ . Civil engineers must verify that the physical radius of curvature $\rho = 1/\kappa$ in meters is greater than or equal to the minimum design turn radius ( $\rho \ge \rho_{\min}$ ) to ensure that semi-trucks and passenger cars can safely negotiate the exit ramp at the advisory speed limit without skidding outward.

Show your steps and calculations here.

Practice Problem 5

Given the 3D position vector:

\vec{r}(t) = \langle 2\cos(t), 2\sin(t), t \rangle

Solve for the unit tangent vector $\hat{T}(t)$ .
Compute the curvature $\kappa(t)$ and the radius of curvature $\rho$ .

Solution Key:

Differentiate Position Vector and Compute Speed: State the Vector Derivative Formula:
$\vec{r}'(t) = \frac{d}{dt}\langle 2\cos(t), 2\sin(t), t \rangle = \langle -2\sin(t), 2\cos(t), 1 \rangle$

State the Scalar Speed Formula:
$\|\vec{r}'(t)\| = \sqrt{(-2\sin(t))^2 + (2\cos(t))^2 + 1^2} = \sqrt{4\sin^2(t) + 4\cos^2(t) + 1} = \sqrt{4(1) + 1} = \sqrt{5}$
Formulate Unit Tangent Vector $\hat{T}(t)$ : State the Unit Tangent Vector Formula:
$\hat{T}(t) = \frac{\vec{r}'(t)}{\|\vec{r}'(t)\|} = \frac{\langle -2\sin(t), 2\cos(t), 1 \rangle}{\sqrt{5}} = \left\langle -\frac{2}{\sqrt{5}}\sin(t), \frac{2}{\sqrt{5}}\cos(t), \frac{1}{\sqrt{5}} \right\rangle$
Compute Curvature $\kappa$ via Unit Tangent Derivative Formula: State the Unit Tangent Derivative Curvature Formula:
$\kappa(t) = \frac{\|\hat{T}'(t)\|}{\|\vec{r}'(t)\|}$ (Defined in Curvature of a Space Curve.)

Differentiate $\hat{T}(t)$ :
$\hat{T}'(t) = \left\langle -\frac{2}{\sqrt{5}}\cos(t), -\frac{2}{\sqrt{5}}\sin(t), 0 \right\rangle$ Compute magnitude $\|\hat{T}'(t)\|:$ $\|\hat{T}'(t)\| = \sqrt{\left(-\frac{2}{\sqrt{5}}\cos(t)\right)^2 + \left(-\frac{2}{\sqrt{5}}\sin(t)\right)^2 + 0^2} = \sqrt{\frac{4}{5}(\cos^2(t) + \sin^2(t))} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$

Evaluate curvature:
$\kappa = \frac{\frac{2}{\sqrt{5}}}{\sqrt{5}} = \frac{2}{5} = 0.4\text{ units}^{-1}$
Calculate Radius of Curvature $\rho$ : State the Radius of Curvature Formula:
$\rho = \frac{1}{\kappa} = \frac{1}{2/5} = \frac{5}{2} = 2.5\text{ units}$ (Defined in Radius of Curvature and Osculating Circles.)

Show your steps and calculations here.

Practice Problem 6

Find the curvature $\kappa(s)$ for the curve parametrized by arc length $s$ :

\vec{r}(s) = 29\cos\left(\frac{s}{29}\right)\hat{i} + 29\sin\left(\frac{s}{29}\right)\hat{j}

Solution Key:

State the Intrinsic Arc Length Curvature Formula: When a curve is parametrized by arc length parameter $s$ , the first derivative is identically the unit tangent vector ( $\hat{T}(s) = \vec{r}'(s)$ ), and curvature is:
$\kappa(s) = \|\vec{r}''(s)\| = \|\hat{T}'(s)\|$ (Defined in Curvature of a Space Curve.)
Differentiate with Respect to $s$ (First Derivative):
$\hat{T}(s) = \vec{r}'(s) = \frac{d}{ds}\left[29\cos\left(\frac{s}{29}\right)\right]\hat{i} + \frac{d}{ds}\left[29\sin\left(\frac{s}{29}\right)\right]\hat{j}$ Applying the chain rule: $\hat{T}(s) = 29\left(-\sin\left(\frac{s}{29}\right) \cdot \frac{1}{29}\right)\hat{i} + 29\left(\cos\left(\frac{s}{29}\right) \cdot \frac{1}{29}\right)\hat{j} = -\sin\left(\frac{s}{29}\right)\hat{i} + \cos\left(\frac{s}{29}\right)\hat{j}$ (Notice $\|\hat{T}(s)\| = \sqrt{(-\sin)^2 + (\cos)^2} = 1$ , confirming intrinsic unit speed).
Differentiate to Find Second Derivative $\vec{r}''(s)$ :
$\vec{r}''(s) = \frac{d}{ds}\left[-\sin\left(\frac{s}{29}\right)\right]\hat{i} + \frac{d}{ds}\left[\cos\left(\frac{s}{29}\right)\right]\hat{j}$ $\vec{r}''(s) = -\frac{1}{29}\cos\left(\frac{s}{29}\right)\hat{i} - \frac{1}{29}\sin\left(\frac{s}{29}\right)\hat{j}$
Compute Curvature Magnitude:
$\kappa(s) = \|\vec{r}''(s)\| = \sqrt{\left(-\frac{1}{29}\cos\left(\frac{s}{29}\right)\right)^2 + \left(-\frac{1}{29}\sin\left(\frac{s}{29}\right)\right)^2} = \sqrt{\frac{1}{29^2}(\cos^2 + \sin^2)} = \sqrt{\frac{1}{29^2}} = \frac{1}{29}$ (Geometric check: $\vec{r}(s)$ is a circle of radius $R = 29$ , so $\kappa = 1/R = 1/29$ ).

Show your steps and calculations here.

Regroup 2

Compare the three curvature formulas and establish a decision tree for when to use each (trigonometric helix $\implies \hat{T}'$ , arbitrary polynomials $\implies \vec{r}' \times \vec{r}''$ , arc length $\implies \|\vec{r}''\|$ ).
Reiterate the physical meaning of the osculating circle: it shares the same location, tangent line, and normal curvature vector.

Skill Block 3

Worked Example A: Decomposing Acceleration into Tangential and Normal Components

A particle moves along a 2D path described by the position vector:

\vec{r}(t) = -7t^8\hat{i} + 7t^3\hat{j}

Decompose the acceleration vector into its tangential component

a_T

and normal component

a_N

at time

t = 1

Worked Solution:

Compute Velocity Vector $\vec{v}(t)$ and Evaluate at $t = 1$ : State the Velocity Derivative Formula:
$\vec{v}(t) = \vec{r}'(t) = \frac{d}{dt}[-7t^8]\hat{i} + \frac{d}{dt}[7t^3]\hat{j} = -56t^7\hat{i} + 21t^2\hat{j} = \langle -56t^7, 21t^2, 0 \rangle$ At $t = 1$ : $\vec{v}(1) = \langle -56, 21, 0 \rangle$
Compute Scalar Speed $\|\vec{v}(1)\|:$ State the Speed Formula:
$\|\vec{v}(1)\| = \sqrt{(-56)^2 + 21^2 + 0^2} = \sqrt{3136 + 441} = \sqrt{3577}$
Compute Acceleration Vector $\vec{a}(t)$ and Evaluate at $t = 1$ : State the Acceleration Derivative Formula:
$\vec{a}(t) = \vec{v}'(t) = \frac{d}{dt}[-56t^7]\hat{i} + \frac{d}{dt}[21t^2]\hat{j} = -392t^6\hat{i} + 42t\hat{j} = \langle -392t^6, 42t, 0 \rangle$ At $t = 1$ : $\vec{a}(1) = \langle -392, 42, 0 \rangle$
Calculate Tangential Component of Acceleration $a_T$ : State the Tangential Acceleration Formula:
$a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}$ (Defined in Tangential and Normal Components of Acceleration.)

Compute dot product $\vec{v}(1) \cdot \vec{a}(1)$ :
$\vec{v}(1) \cdot \vec{a}(1) = (-56)(-392) + (21)(42) + (0)(0) = 21952 + 882 = 22834$ Divide by speed: $a_T = \frac{22834}{\sqrt{3577}} \approx 381.82\text{ units/s}^2$
Calculate Normal Component of Acceleration $a_N$ : State the Normal Acceleration Formula:
$a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}$ (Defined in Tangential and Normal Components of Acceleration.)

Compute cross product $\vec{v}(1) \times \vec{a}(1)$ :
$\vec{v}(1) \times \vec{a}(1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -56 & 21 & 0 \\ -392 & 42 & 0 \end{vmatrix} = \hat{k}\Big((-56)(42) - (21)(-392)\Big) = \hat{k}(-2352 + 8232) = \langle 0, 0, 5880 \rangle$ Magnitude of cross product: $\|\vec{v}(1) \times \vec{a}(1)\| = 5880$ Divide by speed: $a_N = \frac{5880}{\sqrt{3577}} \approx 98.32\text{ units/s}^2$

Practice Problem 7: Concept & Reading Check

Open this Skill Block on the Memex to locate the information needed to answer the following:

(a) Explain the physical distinction between the tangential component $a_T$ and the normal component $a_N$ of acceleration.
(b) In Worked Example A, explain why a 2D planar vector in the $xy$ -plane must be treated as having a $z$ -component of $0$ to compute the cross product $\vec{v} \times \vec{a}$ .
(c) In the linked engineering application Automotive Cornering & Tire Traction Limits (Automotive & Electrical Engineering), why is the tire traction skid threshold governed by normal acceleration $a_N$ rather than tangential acceleration $a_T$ ?

Solution Key:

(a) Differentiating the velocity vector $\vec{v} = v\hat{T}$ produces $\vec{a} = v'\hat{T} + v\hat{T}' = v'\hat{T} + (\kappa v^2)\hat{N}$ . The tangential component $a_T = v'$ is the scalar time derivative of speed; it is positive when accelerating, negative when braking, and zero at constant speed. The normal component $a_N = \kappa v^2 = v^2/\rho$ is strictly non-negative; it quantifies the centripetal acceleration toward the inside of the curve.
(b) The cross product is a 3D vector operation that produces an orthogonal vector. For 2D motion in the $xy$ -plane, setting $z = 0$ yields $\vec{v} = \langle v_x, v_y, 0 \rangle$ and $\vec{a} = \langle a_x, a_y, 0 \rangle$ . The resulting cross product $\vec{v} \times \vec{a} = (v_x a_y - v_y a_x)\hat{k}$ aligns with the binormal vector $\hat{B}$ along the $z$ -axis, whose magnitude is $|v_x a_y - v_y a_x|$ .
(c) In vehicle dynamics, tire adhesion is governed by Coulomb friction: the maximum lateral centripetal force the road can supply without skidding is $F_{\text{lateral},\max} = \mu_s m g$ . Equating this to $m a_N$ gives the critical turning limit $a_N \le \mu_s g$ . If a driver enters a sharp curve with high speed, $a_N = \frac{v^2}{\rho}$ can exceed this limit even while traveling at constant speed ( $a_T = 0$ ), causing the vehicle to slide off the road.

Show your steps and calculations here.

Practice Problem 8

Given the 2D position vector:

\vec{r}(t) = -t^5\hat{i} - 7t^2\hat{j}

Decompose the acceleration vector into its tangential component

a_T

and normal component

a_N

t = 1

Solution Key:

Compute Velocity Vector $\vec{v}(t)$ and Evaluate at $t = 1$ : State the Velocity Derivative Formula:
$\vec{v}(t) = \vec{r}'(t) = \frac{d}{dt}[-t^5]\hat{i} + \frac{d}{dt}[-7t^2]\hat{j} = -5t^4\hat{i} - 14t\hat{j} = \langle -5t^4, -14t, 0 \rangle$ At $t = 1$ : $\vec{v}(1) = \langle -5, -14, 0 \rangle$
Compute Scalar Speed $\|\vec{v}(1)\|:$ State the Speed Formula:
$\|\vec{v}(1)\| = \sqrt{(-5)^2 + (-14)^2 + 0^2} = \sqrt{25 + 196} = \sqrt{221}$
Compute Acceleration Vector $\vec{a}(t)$ and Evaluate at $t = 1$ : State the Acceleration Derivative Formula:
$\vec{a}(t) = \vec{v}'(t) = \frac{d}{dt}[-5t^4]\hat{i} + \frac{d}{dt}[-14t]\hat{j} = -20t^3\hat{i} - 14\hat{j} = \langle -20t^3, -14, 0 \rangle$ At $t = 1$ : $\vec{a}(1) = \langle -20, -14, 0 \rangle$
Calculate Tangential Component of Acceleration $a_T$ : State the Tangential Acceleration Formula:
$a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}$ (Defined in Tangential and Normal Components of Acceleration.)

Compute dot product $\vec{v}(1) \cdot \vec{a}(1)$ :
$\vec{v}(1) \cdot \vec{a}(1) = (-5)(-20) + (-14)(-14) + (0)(0) = 100 + 196 = 296$ Divide by speed: $a_T = \frac{296}{\sqrt{221}} \approx 19.91\text{ units/s}^2$
Calculate Normal Component of Acceleration $a_N$ : State the Normal Acceleration Formula:
$a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}$ (Defined in Tangential and Normal Components of Acceleration.)

Compute cross product $\vec{v}(1) \times \vec{a}(1)$ :
$\vec{v}(1) \times \vec{a}(1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & -14 & 0 \\ -20 & -14 & 0 \end{vmatrix} = \hat{k}\Big((-5)(-14) - (-14)(-20)\Big) = \hat{k}(70 - 280) = \langle 0, 0, -210 \rangle$ Magnitude of cross product: $\|\vec{v}(1) \times \vec{a}(1)\| = |-210| = 210$ Divide by speed: $a_N = \frac{210}{\sqrt{221}} \approx 14.13\text{ units/s}^2$

Show your steps and calculations here.

Practice Problem 9

A reconnaissance drone moves along a 3D spatial trajectory described by the position vector:

\vec{r}(t) = \left\langle t, \frac{1}{2}t^2, \frac{1}{3}t^3 \right\rangle

Decompose the drone's acceleration vector into its tangential component $a_T$ and normal component $a_N$ at time $t = 1$ .
Verify the decomposition using the Pythagorean acceleration identity $\|\vec{a}(1)\|^2 = a_T^2 + a_N^2$ .

Solution Key:

Compute Velocity and Acceleration Vectors at $t = 1$ :
$\vec{v}(t) = \vec{r}'(t) = \langle 1, t, t^2 \rangle \implies \vec{v}(1) = \langle 1, 1, 1 \rangle$ $\vec{a}(t) = \vec{r}''(t) = \langle 0, 1, 2t \rangle \implies \vec{a}(1) = \langle 0, 1, 2 \rangle$
Compute Magnitudes: State the Speed and Acceleration Magnitude Formulas:
$\|\vec{v}(1)\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ $\|\vec{a}(1)\| = \sqrt{0^2 + 1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$
Calculate Tangential Component $a_T$ : State the Tangential Acceleration Formula:
$a_T = \frac{\vec{v} \cdot \vec{a}}{\|\vec{v}\|}$ Compute dot product: $\vec{v}(1) \cdot \vec{a}(1) = (1)(0) + (1)(1) + (1)(2) = 0 + 1 + 2 = 3$ $a_T = \frac{3}{\sqrt{3}} = \sqrt{3} \approx 1.732\text{ m/s}^2$
Calculate Normal Component $a_N$ via Cross Product: State the Normal Acceleration Formula:
$a_N = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|}$ Compute cross product: $\vec{v}(1) \times \vec{a}(1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & 2 \end{vmatrix} = \hat{i}(2 - 1) - \hat{j}(2 - 0) + \hat{k}(1 - 0) = \langle 1, -2, 1 \rangle$ Compute magnitude: $\|\vec{v}(1) \times \vec{a}(1)\| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$ Divide by speed: $a_N = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{\frac{6}{3}} = \sqrt{2} \approx 1.414\text{ m/s}^2$
Pythagorean Identity Verification: State the Pythagorean Orthogonal Decomposition Relation:
$a_T^2 + a_N^2 = (\sqrt{3})^2 + (\sqrt{2})^2 = 3 + 2 = 5$ Since $\|\vec{a}(1)\|^2 = (\sqrt{5})^2 = 5$ : $a_T^2 + a_N^2 = \|\vec{a}(1)\|^2 = 5 \quad \checkmark$ (Confirmed as established in Tangential and Normal Components of Acceleration.)

Show your steps and calculations here.

Regroup 3

Review the geometric meaning of the Pythagorean identity $\|\vec{a}\|^2 = a_T^2 + a_N^2$ .
Emphasize how $a_T$ is signed (speeding up vs. slowing down) while $a_N \ge 0$ is strictly non-negative and points inward along $\hat{N}$ .

Synthesis Wrap-up

Connecting the Triad: Arc length $s$ measures distance traversed; curvature $\kappa = \|d\hat{T}/ds\|$ measures how sharply the direction changes with distance; normal acceleration $a_N = \kappa v^2$ measures the physical force required to sustain that turning rate at speed $v$ .
Preparation for Homework & Session 07: Practice Alta homework assignments on Arc Length and Curvature and Motion in Space. In Session 07, we synthesize Weeks 1–4 for Midterm Exam 1 Review.