Q. 29 The arc length of polar function... [FREE SOLUTION]

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Chapter 9: Q. 29 (page 775)

The arc length of polar functions: Find the arc lengths of the following polar functions.
$r = s e c 胃 for 胃鈭� [\frac{鈭� 蟺}{6}, \frac{蟺}{6}]$

Short Answer

Expert verified

The arc length is $\frac{2}{\sqrt{3}} units$

Step by step solution

Given information.

The given curve is $r = s e c 胃 for 胃鈭� [\frac{鈭� 蟺}{6}, \frac{蟺}{6}]$

Arc length

The arc length of the curve is given by,

$L = {鈭�}_{伪}^{尾} \sqrt{r^{2} + {(\frac{d r}{d 胃})}^{2}} d 胃 L = {鈭�}_{\frac{- 蟺}{6}}^{\frac{蟺}{6}} \sqrt{{(s e c 胃)}^{2} + {(s e c 胃 \tan 胃)}^{2}} d 胃 [鈭� \frac{d r}{d 胃} = s e c 胃 \tan 胃] L = {鈭�}_{\frac{- 蟺}{6}}^{\frac{蟺}{6}} \sqrt{s e c^{2} 胃 + s e c^{2} 胃 \tan^{2} 胃} d 胃 L = {鈭�}_{\frac{- 蟺}{6}}^{\frac{蟺}{6}} \sqrt{s e c^{2} 胃 (1 + \tan^{2} 胃)} d 胃 [\sin^{2} 胃 + \cos^{2} 胃 = 1] L = {鈭�}_{\frac{- 蟺}{6}}^{\frac{蟺}{6}} \sqrt{s e c^{2} 胃 (s e c^{2} 胃)} d 胃 [1 + \tan^{2} 胃 = s e c^{2} 胃] L = {鈭�}_{\frac{- 蟺}{6}}^{\frac{蟺}{6}} s e c^{2} 胃 d 胃 L = {[\tan 胃]}_{\frac{- 蟺}{6}}^{\frac{蟺}{6}} L = [\tan \frac{蟺}{6} - \tan (- \frac{蟺}{6})] L = [\tan \frac{蟺}{6} - (蟺 - \tan (\frac{蟺}{6}))] L = [2 \tan \frac{蟺}{6} - 蟺] L = \frac{2}{\sqrt{3}} units$

The arc length is $L = \frac{2}{\sqrt{3}} units$

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91影视

The arc length of polar functions: Find the arc lengths of the following polar functions.
$r = s e c 胃 for 胃鈭� [\frac{鈭� 蟺}{6}, \frac{蟺}{6}]$

Short Answer

Step by step solution

Given information.

Arc length

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The arc length of polar functions: Find the arc lengths of the following polar functions. r=sec胃for胃鈭�鈭�蟺6,蟺6

Short Answer

Step by step solution

Given information.

Arc length

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Study anywhere. Anytime. Across all devices.

The arc length of polar functions: Find the arc lengths of the following polar functions.
$r = s e c 胃 for 胃鈭� [\frac{鈭� 蟺}{6}, \frac{蟺}{6}]$