On this page we discuss integration and arc length of parametric equations.
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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.
video by PatrickJMT |
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Practice
Unless otherwise instructed, calculate the arc length of these parametric curves on the given intervals.
Basic |
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\(x = 2+6t^2\), \(y = 5+4t^3\), \( 0 \leq t \leq \sqrt{8} \)
Problem Statement |
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Calculate the arc length of the parametric curve \(x = 2+6t^2\), \(y = 5+4t^3\), \(0 \leq t \leq \sqrt{8}\).
Solution |
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video by The Organic Chemistry Tutor |
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\(x = 9t^2\), \(y = 9t-3t^3\), \( 0 \leq t \leq 2 \)
Problem Statement |
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Calculate the arc length of the parametric curve \(x = 9t^2\), \(y = 9t-3t^3\), \(0 \leq t \leq 2\).
Solution |
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video by The Organic Chemistry Tutor |
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\(x(t)=3t^2-9\), \(y(t)=t^3-3t\), \(0\leq t\leq 3\)
Problem Statement |
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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 |
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\( 36 \)
Problem Statement |
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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 |
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video by Dr Chris Tisdell |
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Final Answer |
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\( 36 \)
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\(x(t) = t \sin(t)\) \(y(t) = t \cos(t)\); \(0 \leq t \leq1\)
Problem Statement |
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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 |
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arc length = \(\displaystyle{ \frac{1}{2}\left[ \sqrt{2}+\ln(\sqrt{2}+1) \right] }\)
Problem Statement |
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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 |
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video by Krista King Math |
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Final Answer |
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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 |
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Calculate the arc length of the parametric curve \(x = 1+3t^2\), \(y = 4+2t^3\), \(0 \leq t \leq 1\).
Solution |
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video by PatrickJMT |
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\(x = |6-t|\), \(y = t\), \(0 \leq t \leq 3\)
Problem Statement |
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Calculate the arc length of the parametric curve \(x = |6-t|\), \(y = t\), \(0 \leq t \leq 3\).
Solution |
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video by PatrickJMT |
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Intermediate |
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\(x = 2t\), \(y = (2/3)t^{3/2}\); \(5 \leq t \leq 12\)
Problem Statement |
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Find the arc length of the parametric curve \(x = 2t\), \(y = (2/3)t^{3/2}\) for \(5 \leq t \leq 12\).
Solution |
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video by Krista King Math |
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\(x = t^3\), \(y = t^2\) from \((0,0)\) to \((8,4)\)
Problem Statement |
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Calculate the arc length of the parametric curve \(x = t^3\), \(y = t^2\) from \((0,0)\) to \((8,4)\).
Solution |
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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 |
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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 |
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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.
video by Dr Chris Tisdell |
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\(x = e^t+e^{-t}\), \(y = 5-2t\), \(0 \leq t \leq 3\)
Problem Statement |
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Calculate the arc length of the parametric curve \(x = e^t+e^{-t}\), \(y = 5-2t\), \(0 \leq t \leq 3\).
Solution |
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video by PatrickJMT |
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\(x = e^t \cos(t)\), \(y = e^t \sin(t)\), \(0 \leq t \leq \pi\)
Problem Statement |
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Calculate the arc length of the parametric curve \(x = e^t \cos(t)\), \(y = e^t \sin(t)\), \(0 \leq t \leq \pi\).
Solution |
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video by PatrickJMT |
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Here is a playlist of the videos on this page.
Really UNDERSTAND Calculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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.