\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Parametric Equations - Arc Length

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On this page we discuss integration and arc length of parametric equations.

When a smooth curve is defined parametrically as \( x=X(t) \) and \( y = Y(t) \), the arc length between the points \( t=t_0 \) and \( t = t_1 \) can be calculated using the integral

\(\displaystyle{ s = \int_{t_0}^{t_1}{\sqrt{[X'(t)]^2 + [Y'(t)]^2} dt}}\)
Notice we use a small s here to represent the arc length. This is the standard symbol you will see in many textbooks. We reserve a capital S to represent surface area.

Here is a quick video clip going over this equation in more detail.

PatrickJMT - Arc Length [47secs]

video by PatrickJMT

Practice

Unless otherwise instructed, calculate the arc length of these parametric curves on the given intervals.

Basic

\(x = 2+6t^2\), \(y = 5+4t^3\), \( 0 \leq t \leq \sqrt{8} \)

Problem Statement

Calculate the arc length of the parametric curve \(x = 2+6t^2\), \(y = 5+4t^3\), \(0 \leq t \leq \sqrt{8}\).

Solution

3505 video

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\(x = 9t^2\), \(y = 9t-3t^3\), \( 0 \leq t \leq 2 \)

Problem Statement

Calculate the arc length of the parametric curve \(x = 9t^2\), \(y = 9t-3t^3\), \(0 \leq t \leq 2\).

Solution

3506 video

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\(x(t)=3t^2-9\), \(y(t)=t^3-3t\), \(0\leq t\leq 3\)

Problem Statement

Compute the arc length of the curve with parametric equations \(x(t)=3t^2-9\), \(y(t)=t^3-3t\), \(0\leq t\leq 3\).

Final Answer

\( 36 \)

Problem Statement

Compute the arc length of the curve with parametric equations \(x(t)=3t^2-9\), \(y(t)=t^3-3t\), \(0\leq t\leq 3\).

Solution

2001 video

video by Dr Chris Tisdell

Final Answer

\( 36 \)

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\(x(t) = t \sin(t)\) \(y(t) = t \cos(t)\); \(0 \leq t \leq1\)

Problem Statement

Calculate the arc length of the parametric curve \(x(t) = t \sin(t)\) \(y(t) = t \cos(t)\) on the interval \(0 \leq t \leq1\).

Final Answer

arc length = \(\displaystyle{ \frac{1}{2}\left[ \sqrt{2}+\ln(\sqrt{2}+1) \right] }\)

Problem Statement

Calculate the arc length of the parametric curve \(x(t) = t \sin(t)\) \(y(t) = t \cos(t)\) on the interval \(0 \leq t \leq1\).

Solution

483 video

video by Krista King Math

Final Answer

arc length = \(\displaystyle{ \frac{1}{2}\left[ \sqrt{2}+\ln(\sqrt{2}+1) \right] }\)

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\(x = 1+3t^2\), \(y = 4+2t^3\), \(0 \leq t \leq 1\)

Problem Statement

Calculate the arc length of the parametric curve \(x = 1+3t^2\), \(y = 4+2t^3\), \(0 \leq t \leq 1\).

Solution

1377 video

video by PatrickJMT

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\(x = |6-t|\), \(y = t\), \(0 \leq t \leq 3\)

Problem Statement

Calculate the arc length of the parametric curve \(x = |6-t|\), \(y = t\), \(0 \leq t \leq 3\).

Solution

1379 video

video by PatrickJMT

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Intermediate

\(x = 2t\), \(y = (2/3)t^{3/2}\); \(5 \leq t \leq 12\)

Problem Statement

Find the arc length of the parametric curve \(x = 2t\), \(y = (2/3)t^{3/2}\) for \(5 \leq t \leq 12\).

Solution

58 video

video by Krista King Math

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\(x = t^3\), \(y = t^2\) from \((0,0)\) to \((8,4)\)

Problem Statement

Calculate the arc length of the parametric curve \(x = t^3\), \(y = t^2\) from \((0,0)\) to \((8,4)\).

Solution

469 video

video by Dr Chris Tisdell

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\(x(\theta)=a\cos^3(\theta)\), \(y(\theta)=a\sin^3(\theta)\), \(0\leq\theta\leq2\pi\)

Problem Statement

Calculate the arc length of the curve \(x(\theta)=a\cos^3(\theta)\), \(y(\theta)=a\sin^3(\theta)\), \(0\leq\theta\leq2\pi\).

Solution

Upon initial inspection, this video does not seem to go with the problem statement. However, if you eliminate the parameter θ from the equation, you will get the equation that he starts with in the video.

1380 video

video by Dr Chris Tisdell

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\(x = e^t+e^{-t}\), \(y = 5-2t\), \(0 \leq t \leq 3\)

Problem Statement

Calculate the arc length of the parametric curve \(x = e^t+e^{-t}\), \(y = 5-2t\), \(0 \leq t \leq 3\).

Solution

1378 video

video by PatrickJMT

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\(x = e^t \cos(t)\), \(y = e^t \sin(t)\), \(0 \leq t \leq \pi\)

Problem Statement

Calculate the arc length of the parametric curve \(x = e^t \cos(t)\), \(y = e^t \sin(t)\), \(0 \leq t \leq \pi\).

Solution

1381 video

video by PatrickJMT

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Really UNDERSTAND Calculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Practice Instructions

Unless otherwise instructed, calculate the arc length of these parametric curves on the given intervals.

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