Question #d0a78

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Question #d0a78

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Answer 1 · 2017-03-29T10:43:49

$L = \int \sqrt{{d x}^{2} + {d y}^{2}}$ $L = int sqrt(dx^2 + dy^2)$

Explanation:

$L = \int \sqrt{{(\frac{d x}{d θ})}^{2} + {(\frac{d y}{d θ})}^{2}} d θ$ $L = int sqrt((dx/(d theta))^2 + (dy/(d theta))^2 )d theta$
$L = \int \sqrt{{(d \frac{r cos θ}{d θ})}^{2} + {(d \frac{r sin θ}{d θ})}^{2}} d θ$ $L = int sqrt((d(r cos theta) /(d theta))^2 + (d(r sin theta) /(d theta))^2 )d theta$
$L = \int \sqrt{{(\frac{d r}{d θ})}^{2} + r^{2}} d θ$ $L= int sqrt(((dr)/(d theta))^2 + r^2) d theta$
now
$r = 2 (1 + cos (θ))$ $r = 2(1+cos(theta))$
so
$\frac{d r}{d θ} = - 2 sin (θ)$ $(dr)/(d theta) = -2 sin(theta)$
$r^{2} = 4 (1 + 2 cos θ + {cos}^{2} θ)$ $r^2 = 4(1 + 2 cos theta + cos^2 theta)$
Adding yields
$4 {sin}^{2} θ + 4 {cos}^{2} θ + 4 + 8 cos θ = 8 (1 + cos θ) = 16 {cos}^{2} (\frac{θ}{2})$ $4 sin^2 theta + 4 cos^2 theta + 4 + 8 cos theta = 8(1+ cos theta)= 16 cos^2(theta/2)$
so
$L = \int 4 cos (\frac{θ}{2}) d θ$ $L=int 4 cos(theta/2) d theta$
$L = 8 sin (\frac{θ}{2})$ $L=8 sin(theta/2)$

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Question #d0a78

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