17Calculus Vector Functions - Arc Length, Arc Length Function and The Arc Length Parameter
17Calculus
For the following discussion, we will consider a parameterized curve defined by the vector function \(\displaystyle{ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle }\) which is traversed once on the continuous interval \( a \leq t \leq b \).
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Arc Length
The equation for finding the arc length of a curve is \(\displaystyle{ L = \int_{a}^{b}{ \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} ~dt } }\). In shorthand notation, you will often see \(\displaystyle{ L = \int_{a}^{b}{ \| \vec{r}'(t) \| ~dt } }\) where \(\displaystyle{ \| \vec{r}'(t) \| = \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} }\). Of course, we are assuming this is a smooth, continuous curve.
Okay, time for some practice problems.
Practice
Unless otherwise instructed, find the arc length on the given interval. Give your answer in exact form.
\( \vec{r}(t) = \langle 2\sin(t), 5t, 2\cos(t) \rangle \), \( [-10,10] \)
Problem Statement |
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Find the arc length of \( \vec{r}(t) = \langle 2\sin(t), 5t, 2\cos(t) \rangle \) on the interval \( [-10,10] \).
Final Answer |
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\(20\sqrt{29}\)
Problem Statement
Find the arc length of \( \vec{r}(t) = \langle 2\sin(t), 5t, 2\cos(t) \rangle \) on the interval \( [-10,10] \).
Solution
PatrickJMT - 699 video solution
video by PatrickJMT |
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Final Answer
\(20\sqrt{29}\)
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\(\vec{r}(t) = t\sqrt{2}\vhat{i} + e^t\vhat{j} + e^{-t}\vhat{k} \), \( 0\leq t \leq 1 \)
Problem Statement |
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Find the arc length of \(\vec{r}(t) = t\sqrt{2}\vhat{i} + e^t\vhat{j} + e^{-t}\vhat{k} \) for \( 0\leq t \leq 1 \).
Final Answer |
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\( (e^2-1)/e \)
Problem Statement
Find the arc length of \(\vec{r}(t) = t\sqrt{2}\vhat{i} + e^t\vhat{j} + e^{-t}\vhat{k} \) for \( 0\leq t \leq 1 \).
Solution
Krista King Math - 2042 video solution
video by Krista King Math |
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Final Answer
\( (e^2-1)/e \)
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\( \vec{r}(t) = \langle 2\cos(t),2\sin(t),2t\rangle \), \(0 \leq t \leq \pi \)
Problem Statement |
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Find the arc length of \( \vec{r}(t) = \langle 2\cos(t),2\sin(t),2t\rangle \) for \(0 \leq t \leq \pi \)
Final Answer |
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\( 2\pi\sqrt{2} \)
Problem Statement
Find the arc length of \( \vec{r}(t) = \langle 2\cos(t),2\sin(t),2t\rangle \) for \(0 \leq t \leq \pi \)
Solution
MIP4U - 2043 video solution
video by MIP4U |
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Final Answer
\( 2\pi\sqrt{2} \)
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\( \vec{r}(t) = \langle \cos^3(t), \sin^3(t) \rangle \), \( [0, \pi] \)
Problem Statement |
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Find the arc length of \( \vec{r}(t) = \langle \cos^3(t), \sin^3(t) \rangle \) on the interval \( [0, \pi] \).
Final Answer |
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\( 3 \)
Problem Statement
Find the arc length of \( \vec{r}(t) = \langle \cos^3(t), \sin^3(t) \rangle \) on the interval \( [0, \pi] \).
Solution
MIP4U - 2044 video solution
video by MIP4U |
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Final Answer
\( 3 \)
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\( \vec{r}(t) = \langle 5\cos(4t), 5\sin(4t), 5t\rangle \), \( -5\leq t \leq 8 \)
Problem Statement |
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Find the arc length of \( \vec{r}(t) = \langle 5\cos(4t), 5\sin(4t), 5t\rangle \) for \( -5\leq t \leq 8 \).
Final Answer |
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\( 65\sqrt{17} \)
Problem Statement
Find the arc length of \( \vec{r}(t) = \langle 5\cos(4t), 5\sin(4t), 5t\rangle \) for \( -5\leq t \leq 8 \).
Solution
MIP4U - 2045 video solution
video by MIP4U |
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Final Answer
\( 65\sqrt{17} \)
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\( \vec{r}(t) = \langle 2\ln t, 2t, t^2/2 \rangle \), \( t \in [1,2] \)
Problem Statement |
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Find the arc length of \( \vec{r}(t) = \langle 2\ln t, 2t, t^2/2 \rangle \), \( t \in [1,2] \).
Final Answer |
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\( 3/2+2\ln 2 \)
Problem Statement
Find the arc length of \( \vec{r}(t) = \langle 2\ln t, 2t, t^2/2 \rangle \), \( t \in [1,2] \).
Solution
Dr Chris Tisdell - 2046 video solution
video by Dr Chris Tisdell |
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Final Answer
\( 3/2+2\ln 2 \)
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