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Chapter 11: Q. 27 (page 889)

Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22鈥�27.
$r (t) = <e^{t} \sin t, e^{t} \cos t, e^{t}>, [0, 蟺]$

Short Answer

Expert verified

The length of the curve $r (t) = <e^{t} \sin t, e^{t} \cos t, e^{t}> on [0, 蟺] is \sqrt{3} (e^{蟺} - 1)$ .

Step by step solution

01

Step 1. Given information.

The given curve is $r (t) = <e^{t} \sin t, e^{t} \cos t, e^{t}> on [0, 蟺] .$

02

Step 2. Arc length.

The arc-length is given by,

$l (a, b) = {鈭�}_{a}^{b} ||r^{'} (t)|| d t ||r^{'} (t)|| = \sqrt{{(e^{t} \cos t + e^{t} \sin t)}^{2} + {(- e^{t} \sin t + e^{t} \cos t)}^{2} + {(e^{t})}^{2}} = \sqrt{{(e^{t})}^{2} (\cos^{2} t + \sin^{2} t + 2 \sin t \cos t + \sin^{2} t + \cos^{2} t - 2 \sin t \cos t + 1)} = e^{t} \sqrt{3} (l) = {鈭�}_{a}^{b} ||r^{'} (t)|| d t = {鈭�}_{0}^{蟺} ||r^{- 1} (t)|| d t {= \sqrt{3} e^{t}]}_{0}^{蟺} = \sqrt{3} (e^{n} - e^{0}) = \sqrt{3} (e^{n} - e^{0})$

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Most popular questions from this chapter

For each of the vector-valued functions, find the unit tangent vector.

$r (t) = (\sin 2 t, \cos 2 t, t)$

Given a differentiable vector-valued function r(t), what is the definition of the unit tangent vector T(t)?

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