Chapter 9: Q. 27 (page 775)

The arc length of polar functions: Find the arc lengths of the following polar functions.
$r = 1 - \sin θ for θ \in [0, 2 π]$

Short Answer

Expert verified

The length of an arc is $\sqrt{2} units$

Step by step solution

Given information.

The curve is $r = 1 - \sin θ for θ \in [0, 2 π]$

Arc length.

The length of an arc is given by,

$L = \int_{α}^{β} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ L = \int_{0}^{2 π} \sqrt{{(1 - \sin θ)}^{2} + {(- \cos θ)}^{2}} d θ [∵ \frac{d r}{d θ} = - \cos θ] L = \int_{0}^{2 π} \sqrt{(1 + \sin^{2} θ - 2 \sin θ) + (\cos^{2} θ)} d θ L = \int_{0}^{2 π} \sqrt{(2 - 2 \sin θ)} d θ [\sin^{2} θ + \cos^{2} θ = 1] L = \int_{0}^{2 π} \sqrt{2 (1 - \sin θ)} d θ L = \int_{0}^{2 π} \sqrt{2 (1 - \cos (\frac{π}{2} - θ))} d θ L = \int_{0}^{2 π} \sqrt{2 (2 \sin^{2} (\frac{π}{4} - \frac{θ}{2}))} d θ L = 2 \int_{0}^{2 π} \sin (\frac{π}{4} - \frac{θ}{2}) d θ L = 2 {[\frac{1}{2} \cos (\frac{π}{4} - \frac{θ}{2})]}_{0}^{2 π} L = [\cos (\frac{π}{4} - \frac{2 π}{2}) - \cos (\frac{π}{4} - \frac{0}{2})] L = [\cos (\frac{- 3 π}{4}) - \cos (\frac{π}{4})] L = [- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}] L = |- \frac{2}{\sqrt{2}}| L = \sqrt{2} units$

The arc length isrole="math" localid="1663047738225" $\sqrt{2} units$

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The arc length of polar functions: Find the arc lengths of the following polar functions.
$r = 1 - \sin θ for θ \in [0, 2 π]$

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=1−sinθforθ∈[0,2π]

Short Answer

Step by step solution

Given information.

Arc length.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

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Probability and Statistics

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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.
$r = 1 - \sin θ for θ \in [0, 2 π]$