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UPSC 2016 Maths Optional Paper 1 Q8d — Step-by-Step Solution 15 marks · Section B
Curvature and torsion · Vector Analysis · asked 6× in 13 yrs · Read the full method →
Question
For the cardioid r = a ( 1 + cos θ ) r=a(1+\cos\theta) r = a ( 1 + cos θ ) ( r , θ ) (r,\theta) ( r , θ ) r r r θ = 0 , π 4 , π 2 \theta=0,\dfrac{\pi}{4},\dfrac{\pi}{2} θ = 0 , 4 π , 2 π
Technique
Polar radius of curvature ρ = ( r 2 + r 1 2 ) 3 / 2 ∣ r 2 + 2 r 1 2 − r r 2 ∣ \rho=\dfrac{(r^2+r_1^2)^{3/2}}{|r^2+2r_1^2-rr_2|} ρ = ∣ r 2 + 2 r 1 2 − r r 2 ∣ ( r 2 + r 1 2 ) 3/2 ( 1 + cos θ ) = r / a (1+\cos\theta)=r/a ( 1 + cos θ ) = r / a
Solution
For a polar curve r = r ( θ ) r=r(\theta) r = r ( θ ) r 1 = d r d θ r_1=\dfrac{dr}{d\theta} r 1 = d θ d r r 2 = d 2 r d θ 2 r_2=\dfrac{d^2r}{d\theta^2} r 2 = d θ 2 d 2 r
ρ = ( r 2 + r 1 2 ) 3 / 2 ∣ r 2 + 2 r 1 2 − r r 2 ∣ . \rho=\frac{\big(r^2+r_1^2\big)^{3/2}}{\big|r^2+2r_1^2-r\,r_2\big|}. ρ = r 2 + 2 r 1 2 − r r 2 ( r 2 + r 1 2 ) 3/2 . Step 2 — Derivatives for the cardioid
r = a ( 1 + cos θ ) , r 1 = − a sin θ , r 2 = − a cos θ . r=a(1+\cos\theta),\qquad r_1=-a\sin\theta,\qquad r_2=-a\cos\theta. r = a ( 1 + cos θ ) , r 1 = − a sin θ , r 2 = − a cos θ . Numerator base:
r 2 + r 1 2 = a 2 ( 1 + cos θ ) 2 + a 2 sin 2 θ = a 2 [ 1 + 2 cos θ + cos 2 θ + sin 2 θ ] = a 2 [ 2 + 2 cos θ ] = 2 a 2 ( 1 + cos θ ) . r^2+r_1^2=a^2(1+\cos\theta)^2+a^2\sin^2\theta=a^2\big[1+2\cos\theta+\cos^2\theta+\sin^2\theta\big]=a^2\big[2+2\cos\theta\big]=2a^2(1+\cos\theta). r 2 + r 1 2 = a 2 ( 1 + cos θ ) 2 + a 2 sin 2 θ = a 2 [ 1 + 2 cos θ + cos 2 θ + sin 2 θ ] = a 2 [ 2 + 2 cos θ ] = 2 a 2 ( 1 + cos θ ) . Note r 2 + r 1 2 = 2 a ⋅ a ( 1 + cos θ ) = 2 a r r^2+r_1^2=2a\cdot a(1+\cos\theta)=2a\,r r 2 + r 1 2 = 2 a ⋅ a ( 1 + cos θ ) = 2 a r
Denominator:
r 2 + 2 r 1 2 − r r 2 = a 2 ( 1 + cos θ ) 2 + 2 a 2 sin 2 θ − a ( 1 + cos θ ) ( − a cos θ ) . r^2+2r_1^2-r r_2=a^2(1+\cos\theta)^2+2a^2\sin^2\theta-a(1+\cos\theta)(-a\cos\theta). r 2 + 2 r 1 2 − r r 2 = a 2 ( 1 + cos θ ) 2 + 2 a 2 sin 2 θ − a ( 1 + cos θ ) ( − a cos θ ) . Expand:
= a 2 [ ( 1 + 2 cos θ + cos 2 θ ) + 2 sin 2 θ + ( cos θ + cos 2 θ ) ] =a^2\big[(1+2\cos\theta+\cos^2\theta)+2\sin^2\theta+(\cos\theta+\cos^2\theta)\big] = a 2 [ ( 1 + 2 cos θ + cos 2 θ ) + 2 sin 2 θ + ( cos θ + cos 2 θ ) ] = a 2 [ 1 + 2 cos θ + cos 2 θ + 2 sin 2 θ + cos θ + cos 2 θ ] = a 2 [ 1 + 3 cos θ + 2 cos 2 θ + 2 sin 2 θ ] . =a^2\big[1+2\cos\theta+\cos^2\theta+2\sin^2\theta+\cos\theta+\cos^2\theta\big]=a^2\big[1+3\cos\theta+2\cos^2\theta+2\sin^2\theta\big]. = a 2 [ 1 + 2 cos θ + cos 2 θ + 2 sin 2 θ + cos θ + cos 2 θ ] = a 2 [ 1 + 3 cos θ + 2 cos 2 θ + 2 sin 2 θ ] . Using 2 cos 2 θ + 2 sin 2 θ = 2 2\cos^2\theta+2\sin^2\theta=2 2 cos 2 θ + 2 sin 2 θ = 2
= a 2 [ 3 + 3 cos θ ] = 3 a 2 ( 1 + cos θ ) = 3 a r . =a^2\big[3+3\cos\theta\big]=3a^2(1+\cos\theta)=3a\,r. = a 2 [ 3 + 3 cos θ ] = 3 a 2 ( 1 + cos θ ) = 3 a r . Step 3 — Radius of curvature
ρ = ( 2 a 2 ( 1 + cos θ ) ) 3 / 2 3 a 2 ( 1 + cos θ ) = 2 3 / 2 a 3 ( 1 + cos θ ) 3 / 2 3 a 2 ( 1 + cos θ ) = 2 2 3 a ( 1 + cos θ ) 1 / 2 . \rho=\frac{\big(2a^2(1+\cos\theta)\big)^{3/2}}{3a^2(1+\cos\theta)}=\frac{2^{3/2}a^3(1+\cos\theta)^{3/2}}{3a^2(1+\cos\theta)}=\frac{2\sqrt2}{3}\,a\,(1+\cos\theta)^{1/2}. ρ = 3 a 2 ( 1 + cos θ ) ( 2 a 2 ( 1 + cos θ ) ) 3/2 = 3 a 2 ( 1 + cos θ ) 2 3/2 a 3 ( 1 + cos θ ) 3/2 = 3 2 2 a ( 1 + cos θ ) 1/2 . Since r = a ( 1 + cos θ ) r=a(1+\cos\theta) r = a ( 1 + cos θ ) ( 1 + cos θ ) 1 / 2 = r / a (1+\cos\theta)^{1/2}=\sqrt{r/a} ( 1 + cos θ ) 1/2 = r / a
ρ = 2 2 3 a r a = 2 2 3 a r . \rho=\frac{2\sqrt2}{3}a\sqrt{\frac{r}{a}}=\frac{2\sqrt2}{3}\sqrt{a}\,\sqrt r. ρ = 3 2 2 a a r = 3 2 2 a r . Step 4 — Square of the radius of curvature
ρ 2 = 8 9 a 2 ( 1 + cos θ ) = 8 9 a ⋅ a ( 1 + cos θ ) = 8 a 9 r . \rho^2=\frac{8}{9}a^2(1+\cos\theta)=\frac{8}{9}\,a\cdot a(1+\cos\theta)=\frac{8a}{9}\,r. ρ 2 = 9 8 a 2 ( 1 + cos θ ) = 9 8 a ⋅ a ( 1 + cos θ ) = 9 8 a r . ρ 2 = 8 a 9 r ⟹ ρ 2 is proportional to r . \boxed{\;\rho^2=\frac{8a}{9}\,r\quad\Longrightarrow\quad \rho^2\ \text{is proportional to}\ r.\;} ρ 2 = 9 8 a r ⟹ ρ 2 is proportional to r . Step 5 — Values at θ = 0 , π 4 , π 2 \theta=0,\ \tfrac\pi4,\ \tfrac\pi2 θ = 0 , 4 π , 2 π
Using ρ = 2 2 3 a 1 + cos θ \rho=\dfrac{2\sqrt2}{3}a\sqrt{1+\cos\theta} ρ = 3 2 2 a 1 + cos θ
θ = 0 \theta=0 θ = 0 1 + cos 0 = 2 ⇒ ρ = 2 2 3 a 2 = 4 a 3 . 1+\cos0=2\Rightarrow\rho=\dfrac{2\sqrt2}{3}a\sqrt2=\dfrac{4a}{3}. 1 + cos 0 = 2 ⇒ ρ = 3 2 2 a 2 = 3 4 a . θ = π 4 \theta=\dfrac\pi4 θ = 4 π 1 + cos π 4 = 1 + 1 2 ⇒ ρ = 2 2 3 a 1 + 1 2 = 2 a 3 2 + 2 ≈ 1.232 a . 1+\cos\dfrac\pi4=1+\dfrac{1}{\sqrt2}\Rightarrow\rho=\dfrac{2\sqrt2}{3}a\sqrt{1+\tfrac1{\sqrt2}}=\dfrac{2a}{3}\sqrt{2+\sqrt2}\approx1.232\,a. 1 + cos 4 π = 1 + 2 1 ⇒ ρ = 3 2 2 a 1 + 2 1 = 3 2 a 2 + 2 ≈ 1.232 a . θ = π 2 \theta=\dfrac\pi2 θ = 2 π 1 + cos π 2 = 1 ⇒ ρ = 2 2 3 a . 1+\cos\dfrac\pi2=1\Rightarrow\rho=\dfrac{2\sqrt2}{3}a. 1 + cos 2 π = 1 ⇒ ρ = 3 2 2 a .
Answer
ρ ∣ 0 = 4 a 3 , ρ ∣ π / 4 = 2 a 3 2 + 2 , ρ ∣ π / 2 = 2 2 a 3 . \boxed{\;\rho\big|_{0}=\frac{4a}{3},\qquad \rho\big|_{\pi/4}=\frac{2a}{3}\sqrt{2+\sqrt2},\qquad \rho\big|_{\pi/2}=\frac{2\sqrt2\,a}{3}.\;} ρ 0 = 3 4 a , ρ π /4 = 3 2 a 2 + 2 , ρ π /2 = 3 2 2 a .