Chapter 11: Q. 25 (page 889)

Find the arc length of the curves defined by the vector-valued functions on the specified intervals in Exercises 22–27.
$r (t) = (4 \sin t, t^{3 / 2}, - 4 \cos t), [0, 4]$

Short Answer

Expert verified

The arc length of curve $r (t) = <4 \sin t, t^{\frac{3}{2}}, - 4 \cos t> on [0, 4] is \frac{488}{27}$ .

Step by step solution

Step 1. Given information.

The given curve is $(t) = <4 \sin t, t^{\frac{3}{2}}, - 4 \cos t> on [0, 4] .$

Step 2. Arc-Length.

The arc length is given by,

$I (a, b) = \int_{a}^{b} ||r^{'} (t)|| d t r^{'} (t) = <4 \cos t, \frac{3}{2} t^{\frac{1}{2}}, 4 \sin t> ||r^{'} (t)|| = \sqrt{(4 \cos t)^{2} + {(\frac{3}{2} \sqrt{t})}^{2} + (4 \sin t)^{2}} = \sqrt{16 (\cos^{2} t + \sin^{2} t) + \frac{9}{4} t} = \frac{1}{2} \sqrt{64 + 9 t} N o w, (l) = \int_{a}^{b} ||r^{'} (t)|| d t = \int_{0}^{4} \frac{1}{2} \sqrt{64 + 9 t} d t {= \frac{1}{2} \frac{(64 + 9 t)^{\frac{3}{2}}}{\frac{3}{2}} \cdot \frac{1}{9}]}_{0}^{4} = \frac{1}{27} [(64 + 9 (4))^{\frac{3}{2}} - (64 + 9 (0))^{\frac{3}{2}}] = \frac{1}{27} [(100)^{\frac{3}{2}} - (64)^{\frac{3}{2}}] = \frac{1}{27} [1000 - 512] = \frac{488}{27}$

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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.
$r (t) = (4 \sin t, t^{3 / 2}, - 4 \cos t), [0, 4]$

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)=4sint,t3/2,-4cost,[0,4]

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.
$r (t) = (4 \sin t, t^{3 / 2}, - 4 \cos t), [0, 4]$