1.Radius of Curvature for Polar Curves: Formula,Derivation & Illustrations

Radius of Curvature for Polar Curves: Formula

In this article,”Radius of Curvature for Polar Curves,we will find out Radius of Curvature by polar formula step-by-step.
“Before diving into these solutions,make sure you understand the foundational concept of “Radius of Curvature Cartesian Form-B.sc. Math Solutions” and have reviewed the essential “Radius of Curvature for Polar Curves:B.sc. Maths“. This will make following today’s step-by-step solutions much easier.”

2.Radius of Curvature for Polar Curves

Radius of Curvature for Polar Curves: Formula

\text{Let } r = f(\theta) \text{ be the equation of a given curves and } (r, \theta) \text{ be any point on it.}
\text{Now it is clear from the figure of the previous article that } \psi = \theta + \phi
\text{Differentiating both sides with respect to } s\text{, we have}
\frac{d\psi}{ds} = \frac{d\theta}{ds} + \frac{d\phi}{ds} = \frac{d\theta}{ds} + \frac{d\phi}{d\theta} \cdot \frac{d\theta}{ds}
= \frac{d\theta}{ds} \left(1 + \frac{d\phi}{d\theta}\right) \quad \cdots(1)
\text{Also we know that}
\frac{ds}{d\theta} = \left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} \quad \cdots (2)
\text{and } \tan\phi = \frac{r}{\left(\frac{dr}{d\theta}\right)} \quad \cdots(3)
\text{Now differentiating (3) of both sides with respect to } \theta \text{, we have}
\sec^2\phi \cdot \frac{d\phi}{d\theta} = \frac{\frac{dr}{d\theta} \cdot \frac{dr}{d\theta} - r\left(\frac{d^2r}{d\theta^2}\right)}{\left(\frac{dr}{d\theta}\right)^2}
\implies \left(\frac{d\phi}{d\theta}\right) = \frac{\left(\frac{dr}{d\theta}\right)^2 - r\left(\frac{d^2r}{d\theta^2}\right)}{(1 + \tan^2\phi)\left(\frac{dr}{d\theta}\right)^2}
= \frac{\left(\frac{dr}{d\theta}\right)^2 - r\left(\frac{d^2r}{d\theta^2}\right)}{\left(\frac{dr}{d\theta}\right)^2 + \left(\frac{dr}{d\theta}\right)^2\tan^2\phi}
= \frac{\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}{\left(\frac{dr}{d\theta}\right)^2 + r^2} \quad [\text{using (3)}] \quad \cdots(4)
\text{Substituting the values of } \frac{d\theta}{ds} \text{ from (2) and } \frac{d\phi}{d\theta} \text{ from (4) in (1) we have}
\frac{d\psi}{ds} = \frac{1}{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}}} \left[1 + \frac{\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}{r^2 + \left(\frac{dr}{d\theta}\right)^2}\right]
\implies \frac{1}{\rho} = \frac{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}
\implies \rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}

Also Read This Article:- Radius of curvature

3.Radius of Curvature for Polar Curves Illustrations

Illustration: 3. Find the radius of curvature at the point (r, \theta)on the following curves:
Illustration: 3(e). r = a(1 - \cos\theta)
Solution: r = a(1 - \cos\theta)
Differentiating with respect to \theta, we have
\frac{dr}{d\theta} = a\sin\theta \quad \dots (1)
Again differentiating with respect to \theta, we have
\frac{d^2r}{d\theta^2} = a\cos\theta
Now Radius of curvature
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{3/2}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}
= \frac{\left[r^2 + a^2\sin^2\theta\right]^{3/2}}{r^2 + 2a^2\sin^2\theta - r(a\cos\theta)}
= \frac{\left[r^2 + a^2(1 - \cos^2\theta)\right]^{3/2}}{r^2 + 2a^2(1 - \cos^2\theta) - ar\cos\theta}
= \frac{\left[r^2 + a^2\left(1 - \left(1 - \frac{r}{a}\right)^2\right)\right]^{3/2}}{r^2 + 2a^2\left(1 - \left(1 - \frac{r}{a}\right)^2\right) - ar\cos\theta}
= \frac{\left[r^2 + a^2\left(1 - \left(1 - \frac{2r}{a} + \frac{r^2}{a^2}\right)\right)\right]^{3/2}}{r^2 + 2a^2\left(1 - \left(1 - \frac{2r}{a} + \frac{r^2}{a^2}\right)\right) - ar\cos\theta}
= \frac{\left[r^2 + a^2\left(1 - 1 + \frac{2r}{a} - \frac{r^2}{a^2}\right)\right]^{3/2}}{r^2 + 2a^2\left(1 - 1 + \frac{2r}{a} - \frac{r^2}{a^2}\right) - ar\cos\theta}
= \frac{(r^2 + 2ar - r^2)^{3/2}}{r^2 + 4ar - 2r^2 - ar\left(1 - \frac{r}{a}\right)}
= \frac{(2ar)^{3/2}}{3ar}
= \frac{2\sqrt{2ar}}{3}
Illustration: 3(f). r = \frac{a}{\theta}
Solution: r = \frac{a}{\theta}
= \frac{\left[r^2 + a^2\left(1 - \left(1 - \frac{r}{a}\right)^2\right)\right]^{3/2}}{r^2 + 2a^2\left(1 - \left(1 - \frac{r}{a}\right)^2\right) - ar\cos\theta}
= \frac{\left[r^2 + a^2\left(1 - \left(1 - \frac{2r}{a} + \frac{r^2}{a^2}\right)\right)\right]^{3/2}}{r^2 + 2a^2\left(1 - \left(1 - \frac{2r}{a} + \frac{r^2}{a^2}\right)\right) - ar\cos\theta}
= \frac{\left[r^2 + a^2\left(1 - 1 + \frac{2r}{a} - \frac{r^2}{a^2}\right)\right]^{3/2}}{r^2 + 2a^2\left(1 - 1 + \frac{2r}{a} - \frac{r^2}{a^2}\right) - ar\cos\theta}
= \frac{(r^2 + 2ar - r^2)^{3/2}}{r^2 + 4ar - 2r^2 - ar\left(1 - \frac{r}{a}\right)}
= \frac{(2ar)^{3/2}}{3ar}
= \frac{2\sqrt{2ar}}{3}
Illustration: 3(f). r = \frac{a}{\theta}
Solution: r = \frac{a}{\theta}
Differentiating with respect to \theta, we have
\frac{dr}{d\theta} = -\frac{a}{\theta^2}
Again differentiating with respect to \theta, we have
\frac{d^2r}{d\theta^2} = \frac{2a}{\theta^3}
Now Radius of curvature
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{3/2}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}
= \frac{\left[r^2 + \left(-\frac{a}{\theta^2}\right)^2\right]^{3/2}}{r^2 + 2\left(-\frac{a}{\theta^2}\right)^2 - r\left(\frac{2a}{\theta^3}\right)}
= \frac{\left(r^2 + \frac{a^2}{\theta^4}\right)^{3/2}}{r^2 + \frac{2a^2}{\theta^4} - \frac{2ar}{\theta^3}}
From r = \frac{a}{\theta} \implies \theta = \frac{a}{r}
= \frac{\left[r^2 + \frac{a^2}{\left(\frac{a^4}{r^4}\right)}\right]^{3/2}}{r^2 + \frac{2a^2}{\left(\frac{a^4}{r^4}\right)} - \frac{2ar}{\left(\frac{a^3}{r^3}\right)}}
= \frac{\left(r^2 + \frac{r^4}{a^2}\right)^{3/2}}{r^2 + \frac{2r^4}{a^2} - \frac{2r^4}{a^2}}
Illustration 3(g): r(1 + \cos\theta) = a
Solution:r(1 + \cos\theta) = a \implies (1 + \cos\theta) = \frac{a}{r}
Differentiating with respect to \theta, we get:
\Rightarrow -\sin\theta = -\frac{a}{r^2}\frac{dr}{d\theta}
\Rightarrow \sin\theta = \left(\frac{a}{r^2}\right)\frac{dr}{d\theta} \quad \text{--- (1)}
\Rightarrow \frac{dr}{d\theta} = \frac{r^2\sin\theta}{a}
Again differentiating with respect to \theta, we get:
\frac{d^2r}{d\theta^2} = \frac{r^2\cos\theta}{a} + \frac{2r\sin\theta}{a}\frac{dr}{d\theta}
= \frac{r^2\cos\theta}{a} + \frac{2r\sin\theta}{a} \cdot \frac{r^2\sin\theta}{a}
\Rightarrow \frac{d^2r}{d\theta^2} = \frac{r^2\cos\theta}{a} + \frac{2r^3\sin^2\theta}{a}
Now, Radius of Curvature \rho:
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}
= \frac{\left[r^2 + \left(\frac{r^2\sin\theta}{a}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{r^2\sin\theta}{a}\right)^2 - r\left(\frac{r^2\cos\theta}{a} + \frac{2r^3\sin^2\theta}{a}\right)}
= \frac{\left(r^2 + \frac{r^4\sin^2\theta}{a^2}\right)^{\frac{3}{2}}}{r^2 + \frac{2r^4\sin^2\theta}{a^2} - \frac{r^3\cos\theta}{a} - \frac{2r^4\sin^2\theta}{a^2}}
= \frac{\frac{r^3}{a^3}(a^2 + r^2\sin^2\theta)^{\frac{3}{2}}}{r^2 - \frac{r^3\cos\theta}{a}}
= \frac{r^3\left[a^2 + r^2(1 - \cos^2\theta)\right]^{\frac{3}{2}}}{a^3 r^2\left(1 - \frac{r\cos\theta}{a}\right)}
= \frac{r\left[a^2 + r^2\left(1 - \left(\frac{a}{r} - 1\right)^2\right)\right]^{\frac{3}{2}}}{a^2\left[a - r \times \left(\frac{a}{r} - 1\right)\right]}
= \frac{r\left[a^2 + r^2\left(1 - \left(\frac{a^2}{r^2} - \frac{2a}{r} + 1\right)\right)\right]^{\frac{3}{2}}}{a^2\left[a - a + r\right]}
= \frac{r\left[a^2 + r^2\left(1 - \frac{a^2}{r^2} + \frac{2a}{r} - 1\right)\right]^{\frac{3}{2}}}{a^2 \cdot r}
= \frac{1}{a^2}\left[a^2 - a^2 + 2ar\right]^{\frac{3}{2}}
= \frac{1}{a^2}(2ar)^{\frac{3}{2}}
\implies \rho = \sqrt{\frac{8r^3}{a}}
Illustration 3(h): r^2\cos 2\theta = a^2
Solution:
r^2\cos 2\theta = a^2 \implies r^2 = \frac{a^2}{\cos 2\theta}
Differentiating with respect to \theta, we get:
2r\left(\frac{dr}{d\theta}\right) = -\frac{2a^2}{\cos^2 2\theta}(-\sin 2\theta)
\Rightarrow r\frac{dr}{d\theta} = a^2\sin 2\theta
\Rightarrow \frac{dr}{d\theta} = \frac{a^2\sin 2\theta}{r\cos^2 2\theta}
Again differentiating with respect to \theta, we get:
\frac{d^2r}{d\theta^2} = -\frac{a^2}{r^2}\cdot\frac{\sin 2\theta}{\cos^2 2\theta}\frac{dr}{d\theta} + \frac{a^2}{r}\left[\frac{2\cos^2 2\theta - 2\sin 2\theta(-2\sin 2\theta)}{\cos^3 2\theta}\right]
= -\frac{a^2}{r^2}\cdot\frac{\sin 2\theta}{\cos^2 2\theta}\cdot\frac{a^2\sin 2\theta}{r\cos^2 2\theta} + \frac{a^2}{r}\left[\frac{2}{\cos^2 2\theta} + \frac{4\sin^2 2\theta}{\cos^3 2\theta}\right]
= -\frac{a^4}{r^3}\frac{\sin^2 2\theta}{\cos^4 2\theta} + \frac{2a^2}{r}\cdot\frac{1}{\cos 2\theta} + \frac{4a^2}{r}\cdot\frac{1 - \cos^2 2\theta}{\cos^3 2\theta}
= -\frac{a^4}{r^3}\left(\frac{1 - \cos^2 2\theta}{\cos^4 2\theta}\right) + \frac{2a^2}{r}\times\frac{a^2}{a^2} + \frac{4a^2}{r}\left(\frac{1}{\cos^3 2\theta} - \frac{1}{\cos 2\theta}\right)
= -\frac{a^4}{r^3}\left(\frac{1}{\cos^4 2\theta} - \frac{1}{\cos^2 2\theta}\right) + \frac{2a^4}{r^3} + \frac{4a^2}{r}\left[\frac{1}{(\frac{a^2}{r^2})^3} - \frac{1}{\frac{a^2}{r^2}}\right]
= -\frac{a^4}{r^3}\left(\frac{1}{(\frac{a^2}{r^2})^4} - \frac{1}{(\frac{a^2}{r^2})^2}\right) + \frac{2a^4}{r^3}
+ \frac{4a^2}{r}\left(\frac{r^6}{a^6} - \frac{r^2}{a^2}\right)
= -\frac{a^4}{r^3}\left(\frac{r^8}{a^8} - \frac{r^4}{a^4}\right) + \frac{2a^4}{r^3}
+ \frac{4r^5}{a^4} - 4r
= -\frac{r^5}{a^4} + r + 2r+ \frac{4r^5}{a^4} - 4r
\Rightarrow \frac{d^2r}{d\theta^2} = \frac{3r^5}{a^4} - r
Now Radius of Curvature
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}
= \frac{\left[r^2 + \left(\frac{a^2\sin 2\theta}{r\cos^2 2\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{a^2\sin 2\theta}{r\cos^2 2\theta}\right)^2 - r\left(\frac{3r^5}{a^4} - 3r\right)}
= \frac{\left[r^2 + \frac{a^4 \cdot \sin^2 2\theta}{r^2 \cos^4 2\theta}\right]^{\frac{3}{2}}}{r^2 + \frac{2a^4 \sin^2 2\theta}{r^2 \cos^4 2\theta} - \frac{3r^6}{a^4} + r^2}
= \frac{\left[r^2 + \frac{a^4}{r^2}\left(\frac{1}{\cos^4 2\theta} - \frac{1}{\cos^2 2\theta}\right)\right]^{\frac{3}{2}}}{2r^2 + \frac{2a^4}{r^2}\left(\frac{1}{\cos^4 2\theta} - \frac{1}{\cos^2 2\theta}\right) - \frac{3r^6}{a^4}}
= \frac{\left[r^2 + \frac{a^4}{r^2}\left[\frac{1}{(\frac{a^2}{r^2})^4} - \frac{1}{(\frac{a^2}{r^2})^2}\right]\right]^{\frac{3}{2}}}{2r^2 + \frac{2a^4}{r^2}\left[\frac{1}{(\frac{a^2}{r^2})^4} - \frac{1}{(\frac{a^2}{r^2})^2}\right] - \frac{3r^6}{a^4}}
= \frac{\left[r^2 + \frac{a^4}{r^2}\left(\frac{r^8}{a^8} - \frac{r^4}{a^4}\right)\right]^{\frac{3}{2}}}{2r^2 + \frac{2a^4}{r^2}\left[\frac{r^8}{a^8} - \frac{r^4}{a^4}\right] - \frac{3r^6}{a^4}}
= \frac{\left(r^2 + \frac{r^6}{a^4} - r^2\right)^{\frac{3}{2}}}{2r^2 + \frac{2r^6}{a^4} - 2r^2 - \frac{3r^6}{a^4}}
= \frac{\frac{r^9}{a^6}}{-\frac{r^6}{a^4}} = -\frac{r^3}{a^2}
\Rightarrow \rho = \frac{r^3}{a^2} \quad (\text{Numerically})

Radius of Curvature for Polar Curves: Formula

Illustration 3(i): r^2 = a^2\cos 2\theta
Solution: r^2 = a^2\cos 2\theta
Differentiating with respect to \theta, we get:
2r\frac{dr}{d\theta} = -2a^2\sin 2\theta
\Rightarrow \frac{dr}{d\theta} = -\frac{a^2\sin 2\theta}{r}
Again differentiating with respect to \theta, we get:
\frac{d^2r}{d\theta^2} = \frac{a^2}{r^2}\sin 2\theta\frac{dr}{d\theta} - \frac{2a^2\cos 2\theta}{r}
= \frac{a^2}{r^2}\left(-\frac{a^2 \sin 2\theta}{r}\right)\sin 2\theta - \frac{2a^2\cos 2\theta}{r} \times \frac{r^2}{a^2}
= -\frac{a^4}{r^3}\sin^2 2\theta - 2r
= -\frac{a^4}{r^3}(1 - \cos^2 2\theta) - 2r
= -\frac{a^4}{r^3}\left[1 - \left(\frac{r^2}{a^2}\right)^2\right] - 2r
= -\frac{a^4}{r^3}\left(1 - \frac{r^4}{a^4}\right) - 2r
= -\frac{a^4}{r^3} + r - 2r
\Rightarrow \frac{d^2r}{d\theta^2} = -\frac{a^4}{r^3} - r = -\left(\frac{a^4}{r^3} + r\right)
\left(\frac{dr}{d\theta}\right)^2 = \left(-\frac{a^2\sin 2\theta}{r}\right)^2
= \frac{a^4\sin^2 2\theta}{r^2}
= \frac{a^4}{r^2}(1 - \cos^2 2\theta)
= \frac{a^4}{r^2}\left[1 - \left(\frac{r^2}{a^2}\right)^2\right]
= \frac{a^4}{r^2}\left(1 - \frac{r^4}{a^4}\right)
\Rightarrow \left(\frac{dr}{d\theta}\right)^2 = \frac{a^4}{r^2} - r^2
Now Radius of curvature
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\left(\frac{d^2r}{d\theta^2}\right)}
= \frac{\left(r^2 + \frac{a^4}{r^2} - r^2\right)^{\frac{3}{2}}}{r^2 + 2\left(\frac{a^4}{r^2} - r^2\right) + r\left(\frac{a^4}{r^3} + r\right)}
= \frac{\frac{a^6}{r^3}}{r^2 + \frac{2a^4}{r^2} - 2r^2 + \frac{a^4}{r^2} + r^2}
= \frac{\frac{a^6}{r^3}}{\frac{3a^4}{r^2}} = \frac{a^6}{r^3} \times \frac{r^2}{3a^4}
\Rightarrow \rho = \frac{a^3}{3r}
\text{Illustration } 3(j): \frac{2a}{r} = 1 - \cos\theta
\text{Solution: } \frac{2a}{r} = 1 - \cos\theta
\text{Differentiating with respect to } \theta \text{, we get}
-\frac{2a}{r^2}\frac{dr}{d\theta} = \sin\theta
\Rightarrow \frac{dr}{d\theta} = -\frac{r^2\sin\theta}{2a}
\text{Again differentiating with respect to } \theta \text{, we obtain}
\frac{d^2r}{d\theta^2} = -\frac{2r\sin\theta}{2a}\frac{dr}{d\theta} - \frac{r^2\cos\theta}{2a}
= -\frac{r}{a}\sin\theta \times \left(-\frac{r^2\sin\theta}{2a}\right) - \frac{r^2\cos\theta}{2a}
= \frac{r^3\sin^2\theta}{2a^2} - \frac{r^2\cos\theta}{2a}
= \frac{1}{2a^2}\sin^2\theta\left(\frac{2a}{1-\cos\theta}\right)^3 - \left(\frac{2a}{1-\cos\theta}\right)^2\frac{\cos\theta}{2a}
= \frac{1}{2a^2}\left(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)^2 \times \left(\frac{2a}{2\sin^2\frac{\theta}{2}}\right)^3 - \left(\frac{2a}{2\sin^2\frac{\theta}{2}}\right)^2 \cdot \frac{\cos\theta}{2a}
= \frac{2}{a^2}\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2} \times \frac{a^3}{\sin^6\frac{\theta}{2}} - \frac{a}{2\sin^4\frac{\theta}{2}} \cdot (2\cos^2 \frac{\theta}{2}-1)
= \frac{2a\cos^2\frac{\theta}{2}}{\sin^4\frac{\theta}{2}} - \frac{a\cos^2\frac{\theta}{2}}{\sin^4\frac{\theta}{2}} + \frac{a}{2\sin^4\frac{\theta}{2}}
\implies \frac{d^2r}{d\theta^2} = \frac{a}{2\sin^4\frac{\theta}{2}} + \frac{a\cos^2\frac{\theta}{2}}{\sin^4\frac{\theta}{2}}
\left(\frac{dr}{d\theta}\right)^2 = \left(-\frac{r^2\sin\theta}{2a}\right)^2
= \frac{r^4\sin^2\theta}{4a^2}
= \left(\frac{2a}{1-\cos\theta}\right)^4 \cdot \frac{\left(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)^2}{4a^2}
= \left(\frac{2a}{2\sin^2\frac{\theta}{2}}\right)^4 \cdot \frac{4\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}}{4a^2}
= \frac{16a^4}{16\sin^8\frac{\theta}{2}} \cdot \frac{\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}}{a^2}
\implies \left(\frac{dr}{d\theta}\right)^2 = \frac{a^2\cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}}
\text{Now Radius of curvature:}
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\left(\frac{d^2r}{d\theta^2}\right)}
= \frac{\left[\left(\frac{2a}{1-\cos\theta}\right)^2 + \frac{a^2\cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}}\right]^{\frac{3}{2}}}{\left(\frac{2a}{1-\cos\theta}\right)^2 + 2\left(\frac{a^2\cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}}\right) - \left(\frac{2a}{1-\cos\theta}\right)\left(\frac{a}{2\sin^4\frac{\theta}{2}} + \frac{a\cos^2\frac{\theta}{2}}{\sin^4\frac{\theta}{2}}\right)}
= \frac{\left[\left(\frac{2a}{2\sin^2\frac{\theta}{2}}\right)^2 + \frac{a^2\cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}}\right]^{\frac{3}{2}}}{\left(\frac{2a}{2\sin^2\frac{\theta}{2}}\right)^2 + \frac{2a^2\cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}} - \left(\frac{2a}{2\sin^2\frac{\theta}{2}}\right)\left(\frac{a}{2\sin^4\frac{\theta}{2}} + \frac{a\cos^2\frac{\theta}{2}}{\sin^4\frac{\theta}{2}}\right)}
= \frac{\left(\frac{a^2}{\sin^4\frac{\theta}{2}} + \frac{a^2\cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}}\right)^{\frac{3}{2}}}{\frac{a^2}{\sin^4\frac{\theta}{2}} + \frac{2a^2\cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}} - \frac{a}{2\sin^6\frac{\theta}{2}} - \frac{a\cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}}}
= \frac{a^3 \left(\frac{\sin^2\frac{\theta}{2} + \cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}}\right)^{\frac{3}{2}}}{a^2\left(\frac{\sin^2\frac{\theta}{2} + \cos^2\frac{\theta}{2}}{\sin^6\frac{\theta}{2}}\right) - \frac{a^2}{2\sin^6\frac{\theta}{2}}}
= \frac{\frac{a^3}{\sin^9\frac{\theta}{2}}}{\left[\frac{a^2}{\sin^6\frac{\theta}{2}} - \frac{a^2}{2\sin^6\frac{\theta}{2}}\right]}
= \frac{\frac{a^3}{\sin^9\frac{\theta}{2}}}{\left(\frac{1}{2} \cdot \frac{a^2}{\sin^6\frac{\theta}{2}}\right)}
= \frac{2a}{\sin^3\frac{\theta}{2}}
\implies \rho = \frac{2a}{\sin^3\frac{\theta}{2}}
\text{Illustration } 3(k): r^3 = a^3\sin 3\theta
\text{Solution: } r^3 = a^3\sin 3\theta
\text{Differentiating with respect to } \theta \text{, we get}
3r^2\frac{dr}{d\theta} = 3a^3\cos 3\theta
\implies \frac{dr}{d\theta} = \frac{a^3\cos 3\theta}{r^2}
\text{Again differentiating with respect to } [katex]\theta \text{, we get}[/katex]
\frac{d^2r}{d\theta^2} = -\frac{2a^3\cos 3\theta}{r^3}\frac{dr}{d\theta} - \frac{3a^3\sin 3\theta}{r^2}
= -\frac{2a^3\cos 3\theta}{r^3} \times \frac{a^3\cos 3\theta}{r^2} - \frac{3a^3\sin 3\theta}{r^2}
= -\frac{2a^6\cos^2 3\theta}{r^5} - \frac{3a^3}{r^2} \times \frac{r^3}{a^3}
= -\frac{2a^6}{r^5}(1 - \sin^2 3\theta) - 3r
= -\frac{2a^6}{r^5}\left[1 - \left(\frac{r^3}{a^3}\right)^2\right] - 3r
= -\frac{2a^6}{r^5}\left(1 - \frac{r^6}{a^6}\right) - 3r
= -\frac{2a^6}{r^5} + 2r - 3r
\implies \frac{d^2r}{d\theta^2} = -\frac{2a^6}{r^5} - r
\left(\frac{dr}{d\theta}\right)^2 = \left(\frac{a^3\cos 3\theta}{r^2}\right)^2
= \frac{a^6\cos^2 3\theta}{r^4}
= \frac{a^6}{r^4}(1 - \sin^2 3\theta)
= \frac{a^6}{r^4}\left[1 - \left(\frac{r^3}{a^3}\right)^2\right] \quad \left[\because \sin 3\theta = \frac{r^3}{a^3}\right]
= \frac{a^6}{r^4}\left(1 - \frac{r^6}{a^6}\right)
\implies \left(\frac{dr}{d\theta}\right)^2 = \frac{a^6}{r^4} - r^2
\text{Now Radius of Curvature}
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}
= \frac{\left(r^2 + \frac{a^6}{r^4} - r^2\right)^{\frac{3}{2}}}{r^2 + 2\left(\frac{a^6}{r^4} - r^2\right) - r\left(-\frac{2a^6}{r^5} - r\right)}
= \frac{a^9}{r^6} \cdot \frac{1}{r^2 + \frac{2a^6}{r^4} - 2r^2 + \frac{2a^6}{r^4} + r^2}
= \frac{a^9}{r^6} \cdot \frac{1}{\frac{4a^6}{r^4}}
= \frac{a^9}{r^6} \times \frac{r^4}{4a^6} = \frac{a^3}{4r^2}
\implies \rho = \frac{a^3}{4r^2}
\text{Illustration } 4: \text{Show that the curvature of the curves } r = a\theta \text{ and } r\theta = a \text{ at their common point are in the ratio } 3:1.
\text{Solution: } r = a\theta \quad \cdots(1)
r\theta = a \quad \cdots(2)
\text{From (1) and (2), we get}
a\theta \cdot \theta = a \implies \theta^2 = 1
\implies \theta = 1 \text{ and } r = a
\text{common point is } (a, 1)
\text{Differentiating (1) with respect to } \theta \text{, we get}
\frac{dr}{d\theta} = a \implies \left(\frac{dr}{d\theta}\right)_{(a,1)} = a
\text{Again differentiating with respect to } \theta \text{, we get}
\frac{d^2r}{d\theta^2} = 0
\text{Now Radius of curvature}
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}
\rho_1 = \frac{(r^2 + a^2)^{\frac{3}{2}}}{r^2 + 2a^2 - r(0)}
\rho_1(a,1) = \frac{(a^2 + a^2)^{\frac{3}{2}}}{a^2 + 2a^2}
= \frac{(2a^2)^{\frac{3}{2}}}{3a^2} = \frac{2\sqrt{2}a^3}{3a^2}
\implies \rho_1 = \frac{2\sqrt{2}a}{3}
\text{Again differentiating (2) with respect to } \theta \text{, we get}
\theta\frac{dr}{d\theta} + r = 0 \implies \frac{dr}{d\theta} = -\frac{r}{\theta}
\left(\frac{dr}{d\theta}\right)_{(a,1)} = -a
\text{Again differentiating with respect to } \theta \text{, we get}
\frac{dr}{d\theta} + \theta\frac{d^2r}{d\theta^2} + \frac{dr}{d\theta} = 0
\implies -a + (1)\frac{d^2r}{d\theta^2} - a = 0
\implies \frac{d^2r}{d\theta^2} = 2a
\text{Now Radius of Curvature}
\rho_2 = \frac{\left[r^2 + (-a)^2\right]^{\frac{3}{2}}}{r^2 + 2(-a)^2 - r(2a)}
= \frac{(r^2 + a^2)^{\frac{3}{2}}}{r^2 + 2a^2 - 2ar}
\rho_2(a,1) = \frac{(a^2 + a^2)^{\frac{3}{2}}}{a^2 + 2a^2 - 2a^2}
= \frac{(2a^2)^{\frac{3}{2}}}{a^2}
\implies \rho_2 = \frac{2\sqrt{2}a^3}{a^2} = 2\sqrt{2}a
\rho_2 : \rho_1 = 2\sqrt{2}a : \frac{2\sqrt{2}a}{3}
\implies \rho_2 : \rho_1 = 3 : 1
\text{Illustration } 5: \text{Show that the radius of curvature at any point of the curve } r = a(1 + \cos\theta) \text{ varies as the square root of the radius vector.}
\text{Solution: } r = a(1 + \cos\theta)
\text{Differentiating with respect to } \theta \text{, we get}
\frac{dr}{d\theta} = -a\sin\theta
\text{Again differentiating with respect to } \theta \text{, we get}
\frac{d^2r}{d\theta^2} = -a\cos\theta
= -\left(\frac{r}{a} - 1\right)a \quad \left[\because r = a(1 + \cos\theta)\right]
\implies \frac{d^2r}{d\theta^2} = -r + a = a - r
\left(\frac{dr}{d\theta}\right)^2 = a^2\sin^2\theta
= a^2(1 - \cos^2\theta)
= a^2\left[1 - \left(\frac{r}{a} - 1\right)^2\right]
= a^2\left[1 - \left(\frac{r^2}{a^2} - \frac{2r}{a} + 1\right)\right]
= a^2\left[1 - \frac{r^2}{a^2} + \frac{2r}{a} - 1\right]
= a^2\left(-\frac{r^2}{a^2} + \frac{2r}{a}\right)
\implies \left(\frac{dr}{d\theta}\right)^2 = -r^2 + 2ar
\text{Now Radius of curvature}
\rho = \frac{\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{dr}{d\theta}\right)^2 - r\frac{d^2r}{d\theta^2}}
\rho = \left[r^2 + (-r^2 + 2ar)^2\right]^{\frac{3}{2}}
\rho = \frac{\left[r^2 + (-r^2 + 2ar)^2\right]^{\frac{3}{2}}}{r^2 + 2(-r^2 + 2ar) - r(a - r)}
= \frac{(2ar)^{\frac{3}{2}}}{r^2 +4ar - ar + r^2}
= \frac{2\sqrt{2} r^{\frac{3}{2}}}{3ar}
\implies \rho = \frac{2\sqrt{2}}{3a}\sqrt{r}
\rho \propto \sqrt{r}
By solving the above examples,one can understand the Radius of Curvature for Polar Curves.

4.Practice Questions of Radius of Curvature for Polar Curves for Students

Radius of Curvature for Polar Curves: Formula

(1.)r = ae^{\theta\cot\alpha} \\ (2.) \frac{2a}{r^3} = (1 + \cos 3\theta)
Answers: (1.)\rho = r\operatorname{cosec}\alpha \\ (2.) \rho = \frac{2r^{\frac{5}{2}}}{\sqrt{a}}
By solving the above questions,you can understand the Radius of Curvature for Polar Curves well because the concept is well understood when you solve it practically.
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5.Frequently Asked Questions Related to Radius of Curvature for Polar Curves

6.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students)

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*”16 रु. प्रति लीटर की लागत से मिलने वाले साबुन द्रव को पतला करके अगर 18 रु. प्रति लीटर बेचा जाना हो,ताकि उस पर 25 प्रतिशत लाभ कमाया जा सके तो उसमें किस अनुपात में पानी मिलाया जाए?
(If soap liquid obtained at a cost of Rs. 16 per litre is to be diluted and sold at Rs. 18 per litre, so as to make 25 per cent profit on it, in what proportion water should be added to it?)”*
*पिछली प्रश्नोत्तरी का हल*124
331 \text{ का } \frac{3}{8} = 331 \times \frac{3}{8} \approx 124
*Previous Quiz Solution*124
\frac{3}{8} \text{ of } 331 = 331 \times \frac{3}{8} \approx 124
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*