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
Calculate the arc length of the parametric curve \(x = 2+6t^2\), \(y = 5+4t^3\), \(0 \leq t \leq \sqrt{8}\).
Solution
video by The Organic Chemistry Tutor |
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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
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
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
video by Dr Chris Tisdell |
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Final Answer
\( 36 \)
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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
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
video by Krista King Math |
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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
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
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
Find the arc length of the parametric curve \(x = 2t\), \(y = (2/3)t^{3/2}\) for \(5 \leq t \leq 12\).
Solution
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
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.
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
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
video by PatrickJMT |
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