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:

Review core multivariable mathematical concepts and engineering calculus prerequisites.

Micro-Lecture

Engineering Context: Shaping Motion Paths

In this session, we investigate the mathematical principles of Arc Length and Curvature and explore how engineers apply these concepts to analyze real-world physical and structural systems.

Skill Block 1

Worked Example

Given the line segment given by $\vec{r}(t)=(3t^{2}-7)\hat{i}+(-4t^{2}-10)\hat{j}$ :

Find the arc length $s$ for $0\le t\le4$ .
Solve for the arc length function $s(t)$ and evaluate arc length $s$ for $0\le t\le4$ .
Find the arc length parametrization, $\vec{r}(s)$ .

Worked Solution:

Here is the complete step-by-step solution to the problem presented in the image.

Given

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

First, let's find the derivative vector $\vec{r}'(t)$ and its magnitude $||\vec{r}'(t)||$ , as they are required for all three parts:

\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}

Now, find the magnitude (speed):

||\vec{r}'(t)|| = \sqrt{(6t)^2 + (-8t)^2} = \sqrt{36t^2 + 64t^2} = \sqrt{100t^2} = 10|t|

Since our domain is $0 \le t \le 4$ , $t$ is non-negative, so:

||\vec{r}'(t)|| = 10t

1.

Find the arc length $s$ for $0 \le t \le 4$

Using the definite integral formula for arc length:

s = \int_{0}^{4} ||\vec{r}'(t)||\, dt

s = \int_{0}^{4} 10t\, dt = \left[ 5t^2 \right]_{0}^{4}

s = 5(4)^2 - 5(0)^2 = 5(16) = 80

2.

Solve for the arc length function $s(t)$ and evaluate arc length $s$ for $0 \le t \le 4$

To find the arc length function $s(t)$ starting from $a = 0$ :

s(t) = \int_{0}^{t} ||\vec{r}'(u)||\, du

s(t) = \int_{0}^{t} 10u\, du = \left[ 5u^2 \right]_{0}^{t} = 5t^2

Evaluating this function at the upper limit $t = 4$ :

s(4) = 5(4)^2 = 80

Answer: * Arc length function: $s(t) = 5t^2$

Evaluated arc length: $80$

3.

Find the arc length parametrization, $\vec{r}(s)$

Following the parametrization steps:

Invert the arc length function to solve for $t$ in terms of $s$ :

s = 5t^2 \implies t^2 = \frac{s}{5}

Since $t \ge 0$ :

t = \sqrt{\frac{s}{5}}

Substitute $t(s)$ back into the original position vector $\vec{r}(t)$ :

\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}

Answer: The arc length parametrization is $\vec{r}(s) = \left(\frac{3}{5}s - 7\right)\hat{i} + \left(-\frac{4}{5}s - 10\right)\hat{j}$ for $0 \le s \le 80$ .

Active Practice Problem

Given the position vector for a helix shape $\vec{r}(t)=3\sin(t)\hat{i}+3\cos(t)\hat{j}$ :

Find the arc length $s$ for $0\le t\le4\pi$ .
Solve for the arc length function $s(t)$ and evaluate arc length $s$ for $0\le t\le4\pi$ .
Find the arc length parametrization, $\vec{r}(s)$ .

Workspace

Solution

Given

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

First, let's find the velocity vector $\vec{r}'(t)$ and its magnitude $||\vec{r}'(t)||$ :

\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}

Now, compute the magnitude (speed):

||\vec{r}'(t)|| = \sqrt{(3\cos(t))^2 + (-3\sin(t))^2} = \sqrt{9\cos^2(t) + 9\sin^2(t)}

Using the fundamental trigonometric identity $\cos^2(t) + \sin^2(t) = 1$ :

||\vec{r}'(t)|| = \sqrt{9(1)} = 3

1. Find the arc length $s$ for $0 \le t \le 4\pi$

Using the definite integral formula for arc length:

s = \int_{0}^{4\pi} ||\vec{r}'(t)||\, dt

s = \int_{0}^{4\pi} 3\, dt = \left[ 3t \right]_{0}^{4\pi}

s = 3(4\pi) - 3(0) = 12\pi

Answer: The arc length $s$ is $12\pi$ (or approximately $37.70$ ).

2. Solve for the arc length function $s(t)$ and evaluate arc length $s$ for $0 \le t \le 4\pi$

To find the arc length function $s(t)$ starting from $a = 0$ :

s(t) = \int_{0}^{t} ||\vec{r}'(u)||\, du

s(t) = \int_{0}^{t} 3\, du = \left[ 3u \right]_{0}^{t} = 3t

Evaluating this function at the upper limit $t = 4\pi$ :

s(4\pi) = 3(4\pi) = 12\pi

Answer: * Arc length function: $s(t) = 3t$

Evaluated arc length: $12\pi$

3. Find the arc length parametrization, $\vec{r}(s)$

Following the arc length parametrization steps:

Invert the arc length function to solve for $t$ in terms of $s$ :

s = 3t \implies t = \frac{s}{3}

Substitute $t(s)$ back into the original position vector $\vec{r}(t)$ :

\vec{r}(s) = 3\sin\left(\frac{s}{3}\right)\hat{i} + 3\cos\left(\frac{s}{3}\right)\hat{j}

Answer: The arc length parametrization is $\vec{r}(s) = 3\sin\left(\frac{s}{3}\right)\hat{i} + 3\cos\left(\frac{s}{3}\right)\hat{j}$ for $0 \le s \le 12\pi$ .

Extra Practice & Extensions

Solve for the arc length from $0\le t\le\pi$ .
Explain whether or not a closed domain such as this can or cannot be used to perform an arc length parametrization of the position vector $\vec{r}(s)$ .

Regroup 1

Review common misconceptions and clarify key notations.

Skill Block 2

Worked Example

Given the position vector $\vec{r}(t)=\langle4t^{2},(-4t-3),-2t\rangle$ :

Solve for the unit tangent vector.
Solve for the curvature at $t=1$ .

Worked Solution:

Given

\vec{r}(t) = \langle 4t^2, -4t - 3, -2t \rangle \quad \text{[cite: 52]}

First, let's find the first derivative vector $\vec{r}'(t)$ and its magnitude:

\vec{r}'(t) = \langle 8t, -4, -2 \rangle

Now, compute the magnitude (speed):

||\vec{r}'(t)|| = \sqrt{(8t)^2 + (-4)^2 + (-2)^2} = \sqrt{64t^2 + 16 + 4} = \sqrt{64t^2 + 20}

We can factor out a $4$ from under the radical:

||\vec{r}'(t)|| = \sqrt{4(16t^2 + 5)} = 2\sqrt{16t^2 + 5}

1. Solve for the unit tangent vector

The formula for the unit tangent vector is $\hat{T}(t) = \frac{\vec{r}'(t)}{||\vec{r}'(t)||}$ . Dividing each component of $\vec{r}'(t)$ by its magnitude:

\hat{T}(t) = \frac{\langle 8t, -4, -2 \rangle}{2\sqrt{16t^2 + 5}}

Simplify by dividing each term by $2$ :

\hat{T}(t) = \left\langle \frac{4t}{\sqrt{16t^2 + 5}}, \frac{-2}{\sqrt{16t^2 + 5}}, \frac{-1}{\sqrt{16t^2 + 5}} \right\rangle

2. Solve for the curvature at $t = 1$

Since we already have the position vector, it is much more efficient to use the cross-product formula for curvature:

\kappa(t) = \frac{||\vec{r}'(t) \times \vec{r}''(t)||}{||\vec{r}'(t)||^3} \quad \text{[cite: 46]}

Find the second derivative vector $\vec{r}''(t)$ :

\vec{r}''(t) = \frac{d}{dt}\langle 8t, -4, -2 \rangle = \langle 8, 0, 0 \rangle

Compute the cross product $\vec{r}'(t) \times \vec{r}''(t)$ :

\vec{r}'(t) \times \vec{r}''(t) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 8t & -4 & -2 \\ 8 & 0 & 0 \end{vmatrix}

\vec{r}'(t) \times \vec{r}''(t) = \hat{i}(0 - 0) - \hat{j}(0 - (-16)) + \hat{k}(0 - (-32)) = \langle 0, -16, 32 \rangle

Find the magnitude of this cross product:

||\vec{r}'(t) \times \vec{r}''(t)|| = \sqrt{0^2 + (-16)^2 + (32)^2} = \sqrt{256 + 1024} = \sqrt{1280} = 16\sqrt{5}

Evaluate $||\vec{r}'(t)||$ at $t = 1$ :

||\vec{r}'(1)|| = 2\sqrt{16(1)^2 + 5} = 2\sqrt{21}

Calculate the curvature $\kappa(1)$ :

\kappa(1) = \frac{16\sqrt{5}}{(2\sqrt{21})^3} = \frac{16\sqrt{5}}{8 \cdot 21\sqrt{21}} = \frac{2\sqrt{5}}{21\sqrt{21}}

To rationalize the denominator:

\kappa(1) = \frac{2\sqrt{105}}{441}

Curvature at $t = 1$ : $\kappa(1) = \frac{2\sqrt{5}}{21\sqrt{21}}$ (or approximately $0.0464$ )

Active Practice Problem

Given the position vector $\vec{r}=\langle2\cos(t), 2\sin(t), t\rangle$ :

Solve for the unit tangent vector.
Solve for the radius of curvature.

Workspace

Solution

Given

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

First, let's find the first derivative vector $\vec{r}'(t)$ and its magnitude:

\vec{r}'(t) = \langle -2\sin(t), 2\cos(t), 1 \rangle

Now, compute the magnitude (speed):

||\vec{r}'(t)|| = \sqrt{(-2\sin(t))^2 + (2\cos(t))^2 + (1)^2} = \sqrt{4\sin^2(t) + 4\cos^2(t) + 1}

Using the fundamental identity $\sin^2(t) + \cos^2(t) = 1$ :

||\vec{r}'(t)|| = \sqrt{4(1) + 1} = \sqrt{5}

1. Solve for the unit tangent vector

The formula for the unit tangent vector is $\hat{T}(t) = \frac{\vec{r}'(t)}{||\vec{r}'(t)||}$ . Dividing each component of $\vec{r}'(t)$ by its constant magnitude:

\hat{T}(t) = \frac{\langle -2\sin(t), 2\cos(t), 1 \rangle}{\sqrt{5}}

\hat{T}(t) = \left\langle -\frac{2}{\sqrt{5}}\sin(t), \frac{2}{\sqrt{5}}\cos(t), \frac{1}{\sqrt{5}} \right\rangle

2. Solve for the radius of curvature

To find the radius of curvature $\rho = \frac{1}{\kappa}$ , we first need to determine the curvature $\kappa$ . Since we have a fully worked out unit tangent vector $\hat{T}(t)$ , we can use the derivative method:

\kappa(t) = \frac{||\hat{T}'(t)||}{\|\vec{r}'(t)\|}

Find $\hat{T}'(t)$ :

\hat{T}'(t) = \left\langle -\frac{2}{\sqrt{5}}\cos(t), -\frac{2}{\sqrt{5}}\sin(t), 0 \right\rangle

Compute its 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}

||\hat{T}'(t)|| = \sqrt{\frac{4}{5}\cos^2(t) + \frac{4}{5}\sin^2(t)} = \sqrt{\frac{4}{5}(\cos^2(t) + \sin^2(t))} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}

Calculate curvature $\kappa$ :

\kappa = \frac{||\hat{T}'(t)||}{\|\vec{r}'(t)\|} = \frac{\frac{2}{\sqrt{5}}}{\sqrt{5}} = \frac{2}{5}

Calculate the radius of curvature $\rho$ :

\rho = \frac{1}{\kappa} = \frac{1}{\frac{2}{5}} = \frac{5}{2}

Answer: * Unit tangent vector: $\hat{T}(t) = \left\langle -\frac{2}{\sqrt{5}}\sin(t), \frac{2}{\sqrt{5}}\cos(t), \frac{1}{\sqrt{5}} \right\rangle$

Radius of curvature: $\rho = \frac{5}{2} = 2.5$

Extra Practice & Extensions

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

Regroup 2

Reflect on the physical modeling applications and mathematical setups.

Skill Block 3

Worked Example

A particle is moving along the given path:

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

Decompose the acceleration vector into its tangential and normal components at

t=1.

Worked Solution:

Here is the step-by-step solution to the Skill 3 Demo problem shown in the image.

Given

\vec{r}(t) = -7t^8\hat{i} + 7t^3\hat{j} \quad \text{at } t = 1

To decompose the acceleration vector into its tangential component $a_T$ and normal component $a_N$ , we use the following formulas from your learning activity document:

a_T = \frac{\vec{v} \cdot \vec{a}}{||\vec{v}||}

a_N = \frac{||\vec{v} \times \vec{a}||}{||\vec{v}||}

Step 1: Find the Velocity Vector $\vec{v}(t)$ and evaluate at $t = 1$

Take the first derivative of the position vector $\vec{r}(t)$ :

\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}

Evaluating at $t = 1$ :

\vec{v}(1) = -56(1)^7\hat{i} + 21(1)^2\hat{j} = \langle -56, 21 \rangle

Step 2: Find the Speed $||\vec{v}(1)||$

Compute the magnitude of the velocity vector at $t = 1$ :

||\vec{v}(1)|| = \sqrt{(-56)^2 + (21)^2} = \sqrt{3136 + 441} = \sqrt{3577}

Step 3: Find the Acceleration Vector $\vec{a}(t)$ and evaluate at $t = 1$

Take the derivative of the velocity vector $\vec{v}(t)$ :

\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}

Evaluating at $t = 1$ :

\vec{a}(1) = -392(1)^6\hat{i} + 42(1)\hat{j} = \langle -392, 42 \rangle

Step 4: Calculate the Tangential Component of Acceleration $a_T$

Compute the dot product $\vec{v} \cdot \vec{a}$ at $t = 1$ :

\vec{v}(1) \cdot \vec{a}(1) = (-56)(-392) + (21)(42)

\vec{v}(1) \cdot \vec{a}(1) = 21952 + 882 = 22834

Now divide by the speed $||\vec{v}(1)||$ :

a_T = \frac{22834}{\sqrt{3577}}

Step 5: Calculate the Normal Component of Acceleration $a_N$

Since $\vec{v}$ and $\vec{a}$ are 2D vectors lying in the $xy$ -plane, we can treat them as 3D vectors with a $z$ -component of $0$ to find their cross product:

\vec{v}(1) \times \vec{a}(1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -56 & 21 & 0 \\ -392 & 42 & 0 \end{vmatrix}

\vec{v}(1) \times \vec{a}(1) = \hat{k} \Big( (-56)(42) - (21)(-392) \Big)

\vec{v}(1) \times \vec{a}(1) = \hat{k} (-2352 + 8232) = 5880\hat{k}

Find the magnitude of this cross product vector:

||\vec{v}(1) \times \vec{a}(1)|| = 5880

Now divide by the speed $||\vec{v}(1)||$ :

a_N = \frac{5880}{\sqrt{3577}}

Answer

The decomposition of the acceleration vector at $t = 1$ yields:

Tangential component: $a_T = \frac{22834}{\sqrt{3577}}$ (approx. $381.82$ )
Normal component: $a_N = \frac{5880}{\sqrt{3577}}$ (approx. $98.32$ )

Active Practice Problem

Given the position vector $\vec{r}(t)=-t^{5}\hat{i}-7t^{2}\hat{j}$ , decompose the acceleration vector into its tangential and normal components at $t=1.$

Solution Key:

Given

\vec{r}(t) = -t^5\hat{i} - 7t^2\hat{j} \quad \text{at } t = 1

To decompose the acceleration vector into its tangential component $a_T$ and normal component $a_N$ , we use the standard formulas:

a_T = \frac{\vec{v} \cdot \vec{a}}{||\vec{v}||}

a_N = \frac{||\vec{v} \times \vec{a}||}{||\vec{v}||}

Step 1: Find the Velocity Vector $\vec{v}(t)$ and evaluate at $t = 1$

Take the first derivative of the position vector $\vec{r}(t)$ :

\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}

Evaluating at $t = 1$ :

\vec{v}(1) = -5(1)^4\hat{i} - 14(1)\hat{j} = \langle -5, -14 \rangle

Step 2: Find the Speed $||\vec{v}(1)||$

Compute the magnitude of the velocity vector at $t = 1$ :

||\vec{v}(1)|| = \sqrt{(-5)^2 + (-14)^2} = \sqrt{25 + 196} = \sqrt{221}

Step 3: Find the Acceleration Vector $\vec{a}(t)$ and evaluate at $t = 1$

Take the derivative of the velocity vector $\vec{v}(t)$ :

\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}

Evaluating at $t = 1$ :

\vec{a}(1) = -20(1)^3\hat{i} - 14\hat{j} = \langle -20, -14 \rangle

Step 4: Calculate the Tangential Component of Acceleration $a_T$

Compute the dot product $\vec{v} \cdot \vec{a}$ at $t = 1$ :

\vec{v}(1) \cdot \vec{a}(1) = (-5)(-20) + (-14)(-14)

\vec{v}(1) \cdot \vec{a}(1) = 100 + 196 = 296

Now divide by the speed $||\vec{v}(1)||$ :

a_T = \frac{296}{\sqrt{221}}

Step 5: Calculate the Normal Component of Acceleration $a_N$

Since $\vec{v}$ and $\vec{a}$ are 2D vectors lying in the $xy$ -plane, we append a $z$ -component of $0$ to compute their cross product:

\vec{v}(1) \times \vec{a}(1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & -14 & 0 \\ -20 & -14 & 0 \end{vmatrix}

\vec{v}(1) \times \vec{a}(1) = \hat{k} \Big( (-5)(-14) - (-14)(-20) \Big)

\vec{v}(1) \times \vec{a}(1) = \hat{k} (70 - 280) = -210\hat{k}

Find the magnitude of this cross product vector:

||\vec{v}(1) \times \vec{a}(1)|| = |-210| = 210

Now divide by the speed $||\vec{v}(1)||$ :

a_N = \frac{210}{\sqrt{221}}

Answer

The acceleration vector components at $t = 1$ are:

Tangential component: $a_T = \frac{296}{\sqrt{221}}$ (approx. $19.91$ )
Normal component: $a_N = \frac{210}{\sqrt{221}}$ (approx. $14.13$ )

Show your steps and calculations here.

Regroup 3

Verify calculations and mathematical reasoning.

Synthesis Wrap-up

Core takeaways from Session 06 and overview of homework homework assignment: Arc Length and Curvature Motion in Space.

MATH-283 (Calculus III) Curriculum

Session 06 - Arc Length and Curvature

Micro-Lecture

Engineering Context: Shaping Motion Paths

Skill Block 1

Worked Example

Given

1.

2.

3.

Active Practice Problem

Workspace

Solution

Given

1. Find the arc length sss for 0≤t≤4π0 \le t \le 4\pi0≤t≤4π

2. Solve for the arc length function s(t)s(t)s(t) and evaluate arc length sss for 0≤t≤4π0 \le t \le 4\pi0≤t≤4π

3. Find the arc length parametrization, r⃗(s)\vec{r}(s)r(s)

Extra Practice & Extensions

Regroup 1

Skill Block 2

Worked Example

Given

1. Solve for the unit tangent vector

2. Solve for the curvature at t=1t = 1t=1

Active Practice Problem

Workspace

Solution

Given

1. Solve for the unit tangent vector

2. Solve for the radius of curvature

Extra Practice & Extensions

Regroup 2

Skill Block 3

Worked Example

Given

Step 1: Find the Velocity Vector v⃗(t)\vec{v}(t)v(t) and evaluate at t=1t = 1t=1

Step 2: Find the Speed ∣∣v⃗(1)∣∣||\vec{v}(1)||∣∣v(1)∣∣

Step 3: Find the Acceleration Vector a⃗(t)\vec{a}(t)a(t) and evaluate at t=1t = 1t=1

Step 4: Calculate the Tangential Component of Acceleration aTa_TaT​

Step 5: Calculate the Normal Component of Acceleration aNa_NaN​

Answer

Active Practice Problem

Given

Step 1: Find the Velocity Vector v⃗(t)\vec{v}(t)v(t) and evaluate at t=1t = 1t=1

Step 2: Find the Speed ∣∣v⃗(1)∣∣||\vec{v}(1)||∣∣v(1)∣∣

Step 3: Find the Acceleration Vector a⃗(t)\vec{a}(t)a(t) and evaluate at t=1t = 1t=1

Step 4: Calculate the Tangential Component of Acceleration aTa_TaT​

Step 5: Calculate the Normal Component of Acceleration aNa_NaN​

Answer

Regroup 3

Synthesis Wrap-up

1. Find the arc length $s$ for $0 \le t \le 4\pi$

2. Solve for the arc length function $s(t)$ and evaluate arc length $s$ for $0 \le t \le 4\pi$

3. Find the arc length parametrization, $\vec{r}(s)$

2. Solve for the curvature at $t = 1$

Step 1: Find the Velocity Vector $\vec{v}(t)$ and evaluate at $t = 1$

Step 2: Find the Speed $||\vec{v}(1)||$

Step 3: Find the Acceleration Vector $\vec{a}(t)$ and evaluate at $t = 1$

Step 4: Calculate the Tangential Component of Acceleration $a_T$

Step 5: Calculate the Normal Component of Acceleration $a_N$

Step 1: Find the Velocity Vector $\vec{v}(t)$ and evaluate at $t = 1$

Step 2: Find the Speed $||\vec{v}(1)||$

Step 3: Find the Acceleration Vector $\vec{a}(t)$ and evaluate at $t = 1$

Step 4: Calculate the Tangential Component of Acceleration $a_T$

Step 5: Calculate the Normal Component of Acceleration $a_N$