Arc Length of a Curve: Formula, Examples, and Step-by-Step Guide

The arc length of a curve is the total distance measured along a curved line between two points. Unlike straight lines, calculating the length of a curve requires integration to account for its changing direction. It is an important concept in calculus and geometry, with applications in engineering, architecture, and physics to measure curved paths or surfaces.

Get Homework Help

1. Introduction to the Arc Length of the Curve:

The arc length of a curve is a crucial concept in geometry and calculus, enabling us to calculate the total distance along a curved line. While finding the length of a straight line is straightforward, calculating the length of a curve requires more sophisticated methods, such as integration. Understanding arc length helps in various practical applications, from engineering road curves to measuring distances in physics. This guide will cover the basics, formulas, examples, and real-life applications of arc length, making it a key topic for anyone working with curves and trajectories.

2. What is Arc Length of the Curve:

The arc length of a curve refers to the total distance measured along a smooth curve between two points. Unlike straight lines, curves change direction at every point, making their length harder to determine. To find the arc length, we typically break the curve into tiny segments, measure each, and sum them up. This process becomes precise through calculus, allowing us to calculate the exact distance along various types of curves.

3. Arc Length of the Curve Formula:

The formula for the arc length of a curve varies depending on how the curve is represented:

• For a function $y = f(x)$: The formula for arc length between two points $x = a$ and $x = b$ is given by:

  $L = \int_a^b \sqrt{1 \left( \dfrac{dy}{dx} \right)^2} \, dx$

  Here, $\dfrac{dy}{dx}$ is the derivative of the function, representing the slope at each point along the curve.

• For parametric curves $x = f(t), y = g(t)$ over an interval $t_1 \leq t \leq t_2$, the formula becomes:

  $L = \int_{t_1}^{t_2} \sqrt{\left( \dfrac{dx}{dt} \right)^2 \left( \dfrac{dy}{dt} \right)^2} \, dt$

  This accounts for both the $x$- and $y$-components of the curve.

• For curves in polar coordinates, where $r = f(\theta)$, the formula is:

  $L = \int_{\theta_1}^{\theta_2} \sqrt{\left( \dfrac{dr}{d\theta} \right)^2 r^2} \, d\theta$

Each formula computes the arc length by integrating the small changes in direction and distance along the curve.

4. How to Find the Arc Length of the Curve:

To calculate the arc length of a curve, follow these steps:

1. Identify the form of the curve: Determine whether the curve is expressed as a function $y = f(x)$, parametric equations, or polar coordinates.

2. Differentiate the function or parametric equations:

   • For $y = f(x)$, find $\dfrac{dy}{dx}$.
   • For parametric equations, find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$.
   • For polar curves, compute $\dfrac{dr}{d\theta}$.

3. Substitute into the appropriate arc length formula: Use the correct formula based on how the curve is expressed (Cartesian, parametric, or polar).

4. Set the limits of integration: Determine the start and end points of the interval over which you’re calculating the arc length.

5. Evaluate the integral: Finally, compute the integral to find the total length of the curve.

5. Arc Length of the Curve Solved Examples:

Question: 1.

Arc Length of a Function with a Square Root

Find the length of the curve $y = \sqrt{x}$ from $x = 1$ to $x = 4$.

Solution:

1. Differentiate the function:

   Given $y = \sqrt{x}$, find $\dfrac{dy}{dx}$:

   $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$

2. Set up the arc length formula:

   The arc length formula is:

   $L = \int_a^b \sqrt{1 \left( \dfrac{dy}{dx} \right)^2} \, dx$

   Substituting $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$:

   $L = \int_1^4 \sqrt{1 \left( \dfrac{1}{2\sqrt{x}} \right)^2} \, dx = \int_1^4 \dfrac{\sqrt{4 \frac{1}{x}}}{2} \, dx$

3. Evaluate the integral:

   This integral can be solved using substitution. Let $u = \dfrac{{4 \frac{1}{x}}}{2}$, then integrate:

   $L \approx 3.167 \, \text{units}$

Final Answer: The length of the curve $y = \sqrt{x}$ from $x = 1$ to $x = 4$ is approximately $3.167$ units.

Question: 2.

Arc Length of a Parametric Curve

Find the length of the parametric curve $x = 3t, y = 4t$ for $t \in [0, 2]$.

Solution:

1. Differentiate the parametric equations:

   $\dfrac{dx}{dt} = 3, \space \dfrac{dy}{dt} = 4$

2. Set up the parametric arc length formula:

   The formula for the arc length of a parametric curve is:

   $L = \int_{t_1}^{t_2} \sqrt{\left( \dfrac{dx}{dt} \right)^2 \left( \dfrac{dy}{dt} \right)^2} \, dt$

   Substituting $\dfrac{dx}{dt} = 3$ and $\dfrac{dy}{dt} = 4$:

   $L = \int_0^2 \sqrt{(3)^2 (4)^2} \, dt = \int_0^2 \sqrt{9 16} \, dt = \int_0^2 5 \, dt$

3. Evaluate the integral:

   $L = 5 \times (2 - 0) = 10$

Final Answer: The length of the parametric curve is $10$ units.

Question: 3.

Arc Length of a Parametric Curve

Find the length of the parametric curve $x = t^2, y = t^3$, for $t \in [0, 2]$.

Solution:

1. Differentiate the parametric equations:

   $\dfrac{dx}{dt} = 2t, \space \dfrac{dy}{dt} = 3t^2$

2. Set up the parametric arc length formula:

   The arc length formula for parametric curves is:

   $L = \int_{t_1}^{t_2} \sqrt{\left( \dfrac{dx}{dt} \right)^2 \left( \dfrac{dy}{dt} \right)^2} \, dt$

   Substituting $\dfrac{dx}{dt} = 2t$ and $\dfrac{dy}{dt} = 3t^2$:

   $L = \int_0^2 \sqrt{(2t)^2 (3t^2)^2} \, dt = \int_0^2 \sqrt{4t^2 9t^4} \, dt$

   Factor out $t^2$:

   $L = \int_0^2 t \sqrt{4 9t^2} \, dt$

3. Solve the integral:

   Use substitution: Let $u = 4 9t^2$, then $du = 18t \, dt$. This gives:

   $L = \dfrac{1}{18} \int_4^{40} \sqrt{u} \, du = \dfrac{1}{18} \times \dfrac{2}{3} u^{3/2} \Big|_4^{40}$

   $L = \dfrac{1}{27} \left[ (40)^{3/2} - (4)^{3/2} \right]$

   Simplifying:

   $L \approx \dfrac{1}{27} \left[ 253.15 - 8 \right] = \dfrac{1}{27} \times 245.15 \approx 9.08 \, \text{units}$

Final Answer: The length of the parametric curve is approximately $9.08$ units.

Question: 4.

Arc Length of a Function

Find the arc length of the curve $y = \dfrac{x^2}{2}$ from $x = 0$ to $x = 3$.

Solution:

The formula for the arc length of a curve $y = f(x)$ is: $L = \int_a^b \sqrt{1 \left(\frac{dy}{dx}\right)^2} \, dx,$

where $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$.

Step 1: Find $\frac{dy}{dx}$ Given $y = \frac{x^2}{2}$, differentiate $y$ with respect to $x$: $\frac{dy}{dx} = x.$

Step 2: Substitute into the arc length formula The arc length formula becomes: $L = \int_0^3 \sqrt{1 \left(\frac{dy}{dx}\right)^2} \, dx.$

Substitute $\frac{dy}{dx} = x$: $L = \int_0^3 \sqrt{1 x^2} \, dx.$

Step 3: Evaluate the integral The integral: $L = \int_0^3 \sqrt{1 x^2} \, dx$ is evaluated using a standard substitution.

Let $x = \sinh u$, so $dx = \cosh u \, du$ and $1 x^2 = \cosh^2 u$.

When $x = 0$, $u = \sinh^{-1}(0) = 0$.
When $x = 3$, $u = \sinh^{-1}(3)$.

Substitute into the integral: $L = \int_0^{\sinh^{-1}(3)} \sqrt{\cosh^2 u} \cdot \cosh u \, du.$

Simplify $\sqrt{\cosh^2 u}$: $L = \int_0^{\sinh^{-1}(3)} \cosh u \cdot \cosh u \, du = \int_0^{\sinh^{-1}(3)} \cosh^2 u \, du.$

Use the identity $\cosh^2 u = \frac{1 \cosh(2u)}{2}$: $L = \int_0^{\sinh^{-1}(3)} \frac{1 \cosh(2u)}{2} \, du.$

Split the integral: $L = \frac{1}{2} \int_0^{\sinh^{-1}(3)} 1 \, du \frac{1}{2} \int_0^{\sinh^{-1}(3)} \cosh(2u) \, du.$

Evaluate each term:

1. The first term: $\frac{1}{2} \int_0^{\sinh^{-1}(3)} 1 \, du = \frac{1}{2} \left[u\right]_0^{\sinh^{-1}(3)} = \frac{1}{2} \sinh^{-1}(3).$

2. The second term: $\frac{1}{2} \int_0^{\sinh^{-1}(3)} \cosh(2u) \, du = \frac{1}{2} \left[\frac{\sinh(2u)}{2}\right]_0^{\sinh^{-1}(3)}.$

Simplify: $\frac{1}{2} \cdot \frac{\sinh(2u)}{2} = \frac{1}{4} \sinh(2u).$

Substitute the limits: $\frac{1}{4} \left[\sinh(2u)\right]_0^{\sinh^{-1}(3)} = \frac{1}{4} \left[\sinh(2 \cdot \sinh^{-1}(3)) - \sinh(0)\right].$

Using the identity $\sinh(2x) = 2\sinh(x)\cosh(x)$, and the fact that $\sinh(0) = 0$: $\sinh(2 \cdot \sinh^{-1}(3)) = 2 \cdot 3 \cdot \sqrt{1 3^2} = 6 \cdot \sqrt{10}.$

Thus: $\frac{1}{4} \sinh(2 \cdot \sinh^{-1}(3)) = \frac{6\sqrt{10}}{4} = \frac{3\sqrt{10}}{2}.$

Step 4: Combine results Combine the terms for $L$: $L = \frac{1}{2} \sinh^{-1}(3) \frac{3\sqrt{10}}{2}.$

Final Answer: The arc length of the curve is: $L = \frac{1}{2} \sinh^{-1}(3) \frac{3\sqrt{10}}{2}$

Question: 5.

Arc Length of a Function

Find the arc length of the curve $y = {x^2}$ from $x = 0$ to $x = 2$.

Solution:

The formula for the arc length of a curve $y = f(x)$ is: $L = \int_a^b \sqrt{1 \left(\frac{dy}{dx}\right)^2} \, dx,$ where $\frac{dy}{dx}$ is the derivative of $y$ with respect to $x$.

Step 1: Find $\frac{dy}{dx}$ Given $y = x^2$, differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 2x.$

Step 2: Substitute into the arc length formula The arc length formula becomes: $L = \int_0^2 \sqrt{1 \left(\frac{dy}{dx}\right)^2} \, dx.$

Substitute $\frac{dy}{dx} = 2x$: $L = \int_0^2 \sqrt{1 (2x)^2} \, dx.$

Simplify: $L = \int_0^2 \sqrt{1 4x^2} \, dx.$

Step 3: Solve the integral We solve the integral: $L = \int_0^2 \sqrt{1 4x^2} \, dx.$

Use the substitution $x = \frac{1}{2} \sinh(u)$, so $dx = \frac{1}{2} \cosh(u) \, du$ and $1 4x^2 = \cosh^2(u)$.

When $x = 0$, $u = \sinh^{-1}(0) = 0$.
When $x = 2$, $u = \sinh^{-1}(4)$.

Substitute into the integral: $L = \int_0^{\sinh^{-1}(4)} \sqrt{\cosh^2(u)} \cdot \frac{1}{2} \cosh(u) \, du.$

Simplify $\sqrt{\cosh^2(u)}$: $L = \int_0^{\sinh^{-1}(4)} \cosh(u) \cdot \frac{1}{2} \cosh(u) \, du = \frac{1}{2} \int_0^{\sinh^{-1}(4)} \cosh^2(u) \, du.$

Use the identity $\cosh^2(u) = \frac{1 \cosh(2u)}{2}$: $L = \frac{1}{2} \int_0^{\sinh^{-1}(4)} \frac{1 \cosh(2u)}{2} \, du.$

Split the integral: $L = \frac{1}{4} \int_0^{\sinh^{-1}(4)} 1 \, du \frac{1}{4} \int_0^{\sinh^{-1}(4)} \cosh(2u) \, du.$

1. The first term: $\frac{1}{4} \int_0^{\sinh^{-1}(4)} 1 \, du = \frac{1}{4} \left[u\right]_0^{\sinh^{-1}(4)} = \frac{1}{4} \sinh^{-1}(4).$

2. The second term: $\frac{1}{4} \int_0^{\sinh^{-1}(4)} \cosh(2u) \, du = \frac{1}{4} \left[\frac{\sinh(2u)}{2}\right]_0^{\sinh^{-1}(4)}.$

Simplify: $\frac{1}{4} \cdot \frac{\sinh(2u)}{2} = \frac{1}{8} \sinh(2u).$

Substitute the limits: $\frac{1}{8} \left[\sinh(2u)\right]_0^{\sinh^{-1}(4)} = \frac{1}{8} \left[\sinh(2 \cdot \sinh^{-1}(4)) - \sinh(0)\right].$

Using the identity $\sinh(2x) = 2\sinh(x)\cosh(x)$ and the fact that $\sinh(0) = 0$: $\sinh(2 \cdot \sinh^{-1}(4)) = 2 \cdot 4 \cdot \sqrt{1 4^2} = 8 \cdot \sqrt{17}.$

Thus: $\frac{1}{8} \sinh(2 \cdot \sinh^{-1}(4)) = \frac{8\sqrt{17}}{8} = \sqrt{17}.$

Step 4: Combine results Combine the terms for $L$: $L = \frac{1}{4} \sinh^{-1}(4) \sqrt{17}.$

Final Answer: The arc length of the curve is: ${L = \frac{1}{4} \sinh^{-1}(4) \sqrt{17}}$

6. Practice Questions on Arc Length of the Curve:

Q:1. Find the arc length of the curve $y = \ln(\cosh x)$ from $x = 0$ to $x = 1$.

Q:2. Calculate the arc length of the parametric curve defined by $x = t^2, y = t^3$ for $t \in [0, 2]$.

Q:3. Determine the arc length of the polar curve $r = 1 \cos \theta$ for $\theta \in [0, 2\pi]$.

Q:4. Find the length of the curve $y = x^2$ from $x = 0$ to $x = 2$.

7. FAQs on Arc Length of the Curve:

What is the arc length of a curve?

The arc length of a curve is the total distance along the curve between two points. It’s calculated using integration, which sums up the infinitesimally small line segments along the curve.

How do you find the arc length of a curve?

To find the arc length, you use the formula:

• For a function $y = f(x)$, the arc length is:

  $L = \int_a^b \sqrt{1 \left( \dfrac{dy}{dx} \right)^2} \, dx$

• For parametric curves or polar coordinates, there are specific formulas involving derivatives of the parametric equations or polar functions.

Can arc length be negative?

No, the arc length represents a physical distance, so it is always positive.

What is the difference between arc length and the straight-line distance between two points?

The straight-line distance is the shortest distance between two points, while the arc length measures the total distance along a curve, which is typically longer than the straight-line distance.

How is arc length used in real life?

Arc length is used in various fields like engineering (road and bridge design), architecture (domes and arches), and physics (measuring curved motion paths). It is also important for calculating distances along curved surfaces or trajectories.

What is the formula for arc length in polar coordinates?

For a curve in polar coordinates $r = f(\theta)$, the arc length is given by:

$L = \int_{\theta_1}^{\theta_2} \sqrt{\left( \dfrac{dr}{d\theta} \right)^2 r^2} \, d\theta$

Is there an easy way to calculate the arc length of a circle?

Yes, for a circular arc, the arc length is given by $L = r\theta$, where $r$ is the radius of the circle and $\theta$ is the central angle in radians.

What is the difference between arc length for parametric equations and functions?

For a function $y = f(x)$, the arc length depends on $\dfrac{dy}{dx}$. For parametric equations, the arc length is computed using both the $x$- and $y$-components as functions of a parameter $t$, accounting for both horizontal and vertical changes along the curve.

8. Real-Life Application of Arc Length of the Curve:

The arc length of a curve is widely used in fields such as engineering, architecture, and physics. Civil engineers use arc length to design curved roads, bridges, and tracks. Architects calculate arc lengths for designing arches and domes. In physics, arc length is crucial in understanding the motion of objects along curved paths, such as planetary orbits or the trajectory of a particle in a magnetic field. Even in everyday life, arc length plays a role in measuring distances along curved paths like racetracks or curved walkways.

9. Conclusion:

The arc length of a curve is a versatile and essential concept in both theoretical and applied mathematics. Whether working with straight lines, curves, or complex trajectories, understanding how to calculate arc length provides valuable insights into the structure and design of various systems. From solving basic geometry problems to tackling real-world engineering challenges, mastering arc length helps in fields ranging from mathematics to architecture and beyond.

If you have any suggestions regarding the improvement of the content of this page, please write to me at My Official Email Address: [email protected]

$\fcolorbox{black}{lightpink}{\color{blue}{Get Assignment Help}}$
Are you Stuck on homework, assignments, projects, quizzes, labs, midterms, or exams?
To get connected to our tutors in real-time. Sign up and get registered with us.

$\color{red} \bold{Related \space Pages:}$
Matrix Adjoint Calculator
Matrix Formula Sheet
Linear Algebra Calculators
Matrix Inverse Calculator
Matrix Scalar Multiplication Calculator