Chapter 11: Q. 24 (page 889)

Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22–27.
$r (t) = ⟨ t - \sin t, 1 - \cos t ⟩, [0, 2 π]$

Short Answer

Expert verified

The arc length of curve $r (t) = ⟨ t - \sin t, 1 - \cos t ⟩ on [0, 2 π] is 8 .$

Step by step solution

Step 1. Given information.

The given curve is $r (t) = ⟨ t - \sin t, 1 - \cos t ⟩ on [0, 2 π .$

Step 2. Arc length.

The arc-length is given by,

$l (a, b) = \int_{a}^{b} ||r^{'} (t)|| d t ||r^{'} (t)|| = \sqrt{(1 - \cos t)^{2} + (\sin t)^{2}} = \sqrt{1 + \cos^{2} t - 2 \cos t + \sin^{2} t} = \sqrt{2} \sqrt{1 - \cos t} = \sqrt{2} \sqrt{2 \sin^{2} \frac{t}{2}} = 2 \sin \frac{t}{2} (l) = \int_{a}^{b} ||r^{'} (t)|| d t = \int_{0}^{2 π} 2 \sin \frac{t}{2} d t {= \frac{- 2 \cos \frac{t}{2}}{\frac{1}{2}}]}_{0}^{2 π} = - 4 [\cos \frac{2 π}{2} - \cos 0] = - 4 (- 1 - 1) = 8$

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Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22–27.
$r (t) = ⟨ t - \sin t, 1 - \cos t ⟩, [0, 2 π]$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Arc length.

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Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22–27.r(t)=⟨t-sint,1-cost⟩,[0,2π]

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Arc length.

One App. One Place for Learning.

Most popular questions from this chapter

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Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22–27.
$r (t) = ⟨ t - \sin t, 1 - \cos t ⟩, [0, 2 π]$