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Question Number 17302 by ajfour last updated on 03/Jul/17

Find the length of ρ=a(1−cos θ) . ρ=(√(x^2 +y^2 )) , θ=tan^(−1) ((y/x)) .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\: \\ $$$$\:\:\rho=\mathrm{a}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)\:. \\ $$$$\:\rho=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\:\:,\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\:. \\ $$

Commented by ajfour last updated on 03/Jul/17

it is a cardioid, Sir. easy question sir, not perhaps for you..My Answer : s=8a ds=(√(ρ^2 +((dρ/dθ))^2 )) dθ . Book′s answer: not given .

$$\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cardioid},\:\mathrm{Sir}. \\ $$$$\:\mathrm{easy}\:\mathrm{question}\:\mathrm{sir},\:\mathrm{not}\:\mathrm{perhaps} \\ $$$$\mathrm{for}\:\mathrm{you}..\mathrm{My}\:\mathrm{Answer}\::\:\:\mathrm{s}=\mathrm{8a} \\ $$$$\:\mathrm{ds}=\sqrt{\rho^{\mathrm{2}} +\left(\frac{\mathrm{d}\rho}{\mathrm{d}\theta}\right)^{\mathrm{2}} }\:\mathrm{d}\theta\:. \\ $$$$\mathrm{Book}'\mathrm{s}\:\mathrm{answer}:\:\mathrm{not}\:\mathrm{given}\:. \\ $$

Commented by mrW1 last updated on 03/Jul/17

you are right sir!

$$\mathrm{you}\:\mathrm{are}\:\mathrm{right}\:\mathrm{sir}! \\ $$

Answered by mrW1 last updated on 03/Jul/17

dL=(√((dρ)^2 +(ρdθ)^2 ))=(√(ρ^2 +((dρ/dθ))^2 )) dθ (dρ/dθ)=asin θ ρ^2 +((dρ/dθ))^2 =a^2 [(1−cos θ)^2 +sin^2 θ] =2a^2 [1−cos θ]=4a^2 sin^2 (θ/2) L=2×2a∫_0 ^π sin (θ/2) dθ L=8a∫_0 ^π sin (θ/2) d(θ/2) L=8a[−cos (θ/2)]_0 ^π =8a

$$\mathrm{dL}=\sqrt{\left(\mathrm{d}\rho\right)^{\mathrm{2}} +\left(\rho\mathrm{d}\theta\right)^{\mathrm{2}} }=\sqrt{\rho^{\mathrm{2}} +\left(\frac{\mathrm{d}\rho}{\mathrm{d}\theta}\right)^{\mathrm{2}} }\:\mathrm{d}\theta \\ $$$$\frac{\mathrm{d}\rho}{\mathrm{d}\theta}=\mathrm{asin}\:\theta \\ $$$$\rho^{\mathrm{2}} +\left(\frac{\mathrm{d}\rho}{\mathrm{d}\theta}\right)^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} \left[\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\theta\right] \\ $$$$=\mathrm{2a}^{\mathrm{2}} \left[\mathrm{1}−\mathrm{cos}\:\theta\right]=\mathrm{4a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{L}=\mathrm{2}×\mathrm{2a}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{d}\theta \\ $$$$\mathrm{L}=\mathrm{8a}\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{d}\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{L}=\mathrm{8a}\left[−\mathrm{cos}\:\frac{\theta}{\mathrm{2}}\right]_{\mathrm{0}} ^{\pi} =\mathrm{8a} \\ $$

Commented by ajfour last updated on 03/Jul/17

yes sir, thanks for confirming.

$$\mathrm{yes}\:\mathrm{sir},\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{confirming}. \\ $$

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