Chapter 11: Q. 26 (page 889)

Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22–27.
$r (t) = <e^{t}, \sqrt{2} t, e^{- t}\}, [0, 1]$

Short Answer

Expert verified

The arc-length of the curve $r (t) = <e^{t}, \sqrt{2} t, e^{- t}> on [0, 1] is \frac{e^{2} - 1}{e} .$

Step by step solution

Step 1. Given information.

The given curve is $r (t) = <e^{t}, \sqrt{2} t, e^{- t}> on [0, 1] .$

Step 2. Arc-length.

The arc-length is given by,

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

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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.
$r (t) = <e^{t}, \sqrt{2} t, e^{- t}\}, [0, 1]$

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)=et,2t,e-t,[0,1]

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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Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22–27.
$r (t) = <e^{t}, \sqrt{2} t, e^{- t}\}, [0, 1]$