Polar Infinity

Calculus Level 5

If the arc length of the polar curve r 2 = π cos ⁡ ( 2 θ ) r^2=\pi\cos(2\theta) is

[ Γ ( − a b + 1 ) ] b b 1 b \frac{\left [ \Gamma(-\frac ab + 1) \right ]^{\sqrt{b}}}{{b}^\frac{1}{b}}

for positive integers a a and b b . Find a + b a+b .


The answer is 7.

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Digvijay Singh by Digvijay Singh , 21, India

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1 solution

Note that r r is unreal for θ ∈ ( π 4 , 3 π 2 ) ∪ ( 5 π 4 , 7 π 2 ) \theta \in \left(\frac \pi 4, \frac {3\pi} 2 \right) \cup \left(\frac {5\pi} 4, \frac {7\pi} 2 \right)

The arc length is given by:

L = ∫ 0 2 π r 2 + ( d r d θ ) 2 d θ As the curve is symmetrical at the origin = 4 ∫ 0 π 4 π cos ⁡ 2 θ + ( d d θ π cos ⁡ 2 θ ) 2 d θ = 4 ∫ 0 π 4 π cos ⁡ 2 θ + ( − π sin ⁡ 2 θ cos ⁡ 2 θ ) 2 d θ = 4 ∫ 0 π 4 π cos ⁡ 2 2 θ + π sin ⁡ 2 2 θ cos ⁡ 2 θ d θ = 4 π ∫ 0 π 4 1 cos ⁡ 2 θ d θ Let x = 2 θ ⟹ d x = 2 d θ = 2 π ∫ 0 π 2 1 cos ⁡ x d x = 2 π ∫ 0 π 2 sin ⁡ 0 x cos ⁡ − 1 2 x d x = π B ( 1 2 , 1 4 ) B ( m , n ) is beta function. = π Γ ( 1 2 ) Γ ( 1 4 ) Γ ( 3 4 ) Γ ( n ) is gamma function. = π ⋅ π ⋅ Γ ( 1 4 ) 2 π Γ ( 1 4 ) = ( Γ ( 1 4 ) ) 2 2 = [ Γ ( − 3 4 + 1 ) ] 4 4 1 4 \begin{aligned} L & = \int_0^{2\pi} \sqrt{r^2+\left(\frac {dr}{d \theta} \right)^2} d \theta & \small {\color{#3D99F6}\text{As the curve is symmetrical at the origin}} \\ & = 4 \int_0^{\color{#D61F06}\frac \pi 4} \sqrt{\pi\cos 2 \theta +\left(\frac d{d \theta} \sqrt{\pi\cos 2 \theta} \right)^2} d \theta \\ & = 4 \int_0^{\frac \pi 4} \sqrt{\pi\cos 2 \theta +\left(-\frac {\sqrt \pi \sin 2\theta}{\sqrt{\cos 2 \theta}} \right)^2} d \theta \\ & = 4 \int_0^{\frac \pi 4} \sqrt{\frac {\pi\cos^2 2 \theta + \pi \sin^2 2\theta}{\cos 2 \theta}} d \theta \\ & = 4\sqrt \pi \int_0^{\frac \pi 4} \frac 1{\sqrt{\cos 2 \theta}} d \theta & \small {\color{#3D99F6}\text{Let }x = 2\theta \implies dx = 2 \ d\theta} \\ & = 2 \sqrt \pi \int_0^{\color{#D61F06}\frac \pi 2} \frac 1{\sqrt{\cos x}} dx \\ & = 2\sqrt \pi \int_0^{\frac \pi 2} \sin^0 x \cos^{-\frac 12} x \ dx \\ & = \sqrt \pi B \left(\frac 12, \frac 14 \right) & \small {\color{#3D99F6} B(m,n) \text{ is beta function.}} \\ & = \frac {\sqrt \pi \Gamma \left(\frac 12 \right)\Gamma \left(\frac 14 \right)}{\Gamma \left(\frac 34 \right)} & \small {\color{#3D99F6} \Gamma (n) \text{ is gamma function.}} \\ & = \frac {\sqrt \pi \cdot \sqrt \pi \cdot \Gamma \left(\frac 14 \right)}{\frac {\sqrt 2 \pi}{\Gamma \left(\frac 14 \right)}} \\ & = \frac {\left(\Gamma \left(\frac 14 \right)\right)^2}{\sqrt 2} \\ & = \frac {\left[\Gamma \left( - \frac 34 + 1 \right)\right]^{\sqrt 4}}{4^\frac 14} \end{aligned}

⟹ a + b = 3 + 4 = 7 \implies a+b = 3+4 = \boxed{7}

5 Helpful 0 Interesting 1 Brilliant 0 Confused

Sir this problem has to be reported,gamma(n)=(n-1)! only for integer n and moreover gamma(1/4) is transcendental and cannot be expressed in any factorial notation,and yes above you have mistakenly taken cos2x=sin2x!! Regards

Chirag Shyamsundar - 4 years, 7 months ago

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Thanks for your comments. I think gamma function was created to make factorial continuous so it should take all cases. Anyway, Wolfram Alpha takes the factorial valid.

I must have been too eager to find a solution and mistaken sin ⁡ 2 θ \sin 2\theta as cos ⁡ 2 θ \cos 2 \theta . Damn it! Now, I need to find a solution.

Chew-Seong Cheong - 4 years, 7 months ago

Got it right again. Same answer.

Chew-Seong Cheong - 4 years, 7 months ago

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