Chapter 9: Q. 29 (page 775)

The arc length of polar functions: Find the arc lengths of the following polar functions.
$r = s e c θ for θ \in [\frac{- π}{6}, \frac{π}{6}]$

Short Answer

Expert verified

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

Step by step solution

Given information.

The given curve is $r = s e c θ for θ \in [\frac{- π}{6}, \frac{π}{6}]$

Arc length

The arc length of the curve is given by,

$L = \int_{α}^{β} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ L = \int_{\frac{- π}{6}}^{\frac{π}{6}} \sqrt{{(s e c θ)}^{2} + {(s e c θ \tan θ)}^{2}} d θ [∵ \frac{d r}{d θ} = s e c θ \tan θ] L = \int_{\frac{- π}{6}}^{\frac{π}{6}} \sqrt{s e c^{2} θ + s e c^{2} θ \tan^{2} θ} d θ L = \int_{\frac{- π}{6}}^{\frac{π}{6}} \sqrt{s e c^{2} θ (1 + \tan^{2} θ)} d θ [\sin^{2} θ + \cos^{2} θ = 1] L = \int_{\frac{- π}{6}}^{\frac{π}{6}} \sqrt{s e c^{2} θ (s e c^{2} θ)} d θ [1 + \tan^{2} θ = s e c^{2} θ] L = \int_{\frac{- π}{6}}^{\frac{π}{6}} s e c^{2} θ d θ L = {[\tan θ]}_{\frac{- π}{6}}^{\frac{π}{6}} L = [\tan \frac{π}{6} - \tan (- \frac{π}{6})] L = [\tan \frac{π}{6} - (π - \tan (\frac{π}{6}))] L = [2 \tan \frac{π}{6} - π] L = \frac{2}{\sqrt{3}} units$

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

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The arc length of polar functions: Find the arc lengths of the following polar functions.
$r = s e c θ for θ \in [\frac{- π}{6}, \frac{π}{6}]$

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=secθforθ∈−π6,π6

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The arc length of polar functions: Find the arc lengths of the following polar functions.
$r = s e c θ for θ \in [\frac{- π}{6}, \frac{π}{6}]$