1.Radius of Curvature for Polar Curves :B.sc. Maths|Radius of Curvature of Pedal Curves

Radius of Curvature for Polar Curves

In this article “Radius of Curvature for Polar Curves”,we will find out radius of curvature of polar curves and pedal curves.
“Before diving into these solutions,make sure you understand the foundational concept of “Parametric and Cartesian Formula for Curvature” and have reviewed the essential “Radius of Curvature Cartesian Form“.This will make following today’s step-by-step solutions much easier.”

2.Radius of Curvature of Pedal Curves

Radius of Curvature for Polar Curves

Also Read This Article:- Radius of curvature

3.Radius of Curvature for Polar Curves Solved Illustrations

Illustration:1.Find the radius of curvature at the point (p,r) on the following curves.
\text{Illustration : } 1(a) \ p^2 = ar \text{ (Parabola)} \\ \text{Solution : } p^2 = ar \\ \text{Differentiating w.r.t. } r \text{, we get} \\ 2p \frac{dp}{dr} = a \\ \Rightarrow \frac{dp}{dr} = \frac{a}{2p} \\ \Rightarrow \frac{dp}{dr} = \frac{a}{2\sqrt{ar}} \\ \therefore \rho = r \frac{dr}{dp} \\ = \frac{r}{\frac{a}{2\sqrt{ar}}} = \frac{r \times 2\sqrt{ar}}{a} \\ \Rightarrow \rho = \frac{2r^{\frac{3}{2}}}{\sqrt{a}} \\ = \frac{2}{\sqrt{a}} \left(\frac{p^2}{a}\right)^{\frac{3}{2}} \\ \Rightarrow \rho = \frac{2p^3}{a^2} \\ \text{Illustration : } 1(b) \ pr = a^2 \text{ (Hyperbola)} \\ \text{Solution : } pr = a^2 \\ \text{Differentiating w.r.t. } r \text{, we get} \\ \frac{dp}{dr} + p = 0 \\ \Rightarrow \frac{dp}{dr} = -p\\ \rho = r \frac{dr}{dp} \\ \Rightarrow = r \left(-\frac{1}{p}\right) \\ = \frac{a^2}{p} \left(-\frac{1}{p}\right) \ [\because pr = a^2] \\ = -\frac{a^2}{p^2} \\ \Rightarrow \rho = \frac{a^2}{p^2} \text{ (numerically)} \\ \text{Illustration : } 1(c) \ r^3 = 2ap^2 \\ \text{Solution : } r^3 = 2ap^2 \text{ (cardioid)} \\ \text{Differentiating w.r.t. } r, \text{ we get} \\ 3r^2 = 2a \left(2p \frac{dp}{dr}\right) \\ \Rightarrow \frac{dr}{dp} = \frac{4ap}{3r^2} \\ \therefore \rho = r \frac{dr}{dp} \\ = r \cdot \frac{4ap}{3r^2} \\ = \frac{4ap}{3r} \\ = \frac{4a}{3r} \times \frac{r^{\frac{3}{2}}}{\sqrt{2a}} \\ = \frac{2\sqrt{2ar}}{3} \\ \text{Illustration : } 1(d) \ r^3 = a^2 p \text{ (Lemniscate)} \\ \text{Solution : } r^3 = a^2 p \\ \text{Differentiating w.r.t. } r, \text{ we get} \\ 3r^2 = a^2 \frac{dp}{dr}\\ \frac{dp}{dr} = \frac{dr}{dp} = \frac{a^2}{3r^2} \\ \rho = r \frac{dr}{dp} \\ = r \left(\frac{a^2}{3r^2}\right) = \frac{a^2}{3r} \\ \Rightarrow \rho = \frac{a^2}{3r} \\ \text{Illustration : } 1(e) \ \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} - \frac{r^2}{a^2 b^2} \text{ (ellipse)} \\ \text{Solution : } \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} - \frac{r^2}{a^2 b^2} \\ \text{Differentiating w.r.t. } r, \text{ we get} \\ -\frac{2}{p^3} \frac{dp}{dr} = -\frac{2r}{a^2 b^2} \\ \Rightarrow \frac{dr}{dp} = \frac{a^2 b^2}{r p^3} \\ \rho = r \frac{dr}{dp} \\ = r \left(\frac{a^2 b^2}{r p^3}\right) \\ \Rightarrow \rho = \frac{a^2 b^2}{p^3} \\ \text{Illustration : (f)  } p a^n = r^{n+1} \text{ (sine spiral) } \\ \text{Solution : } p a^n = r^{n+1} \\ \text{Differentiating w.r.t. } r, \text{ we get} \\ \frac{dp}{dr} a^n = (n+1)r^n \\ \Rightarrow \frac{dr}{dp} = \frac{a^n}{(n+1)r^n} \\ \rho = r \frac{dr}{dp} \\ = r \left(\frac{a^n}{(n+1)r^n}\right)\\ = \frac{a^n}{(n+1)} r^{-n+1} \\ \Rightarrow \rho = \frac{a^n r^{-n+1}}{(n+1)} \\ \text{Illustration : 2. In the curve} \\ p = \frac{r^{n+1}}{a^n}, \text{ show that the} \\ \text{radius of curvature varies} \\ \text{inversely as the } (n+1)\text{th power} \\ \text{of the radius vector.} \\ \text{Solution : Given } p = \frac{r^{n+1}}{a^n} \\ \text{Differentiating w.r.t. } r, \text{ we get} \\ \frac{dp}{dr} = \frac{(n+1)r^n}{a^n} \\ \Rightarrow \frac{dr}{dp} = \frac{a^n}{(n+1)r^n} \\ \rho = r \left(\frac{dr}{dp}\right) \\ = r \cdot \frac{a^n}{(n+1)r^n} \\ \Rightarrow \rho = \frac{a^n}{(n+1)r^{n-1}} \\ \Rightarrow \rho \propto \frac{1}{r^{n-1}}

Radius of Curvature for Polar Curves

\text{Illustration : 3. Find the radius} \\ \text{of curvature at the point } (r, \theta) \text{ on} \\ \text{the following curves :} \\ \text{Illustration : 3(a). } r = a\cos\theta \\ \text{Solution : } r = a\cos\theta\\ \text{Differentiating w.r.t. } \theta, \text{ we get} \\ \frac{dr}{d\theta} = -a\sin\theta \\ \text{Again differentiating, we have} \\ \frac{d^2 r}{d\theta^2} = -a\cos\theta \\ \therefore \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^2 r}{d\theta^2}} \\ = \frac{\left[r^2 + (-a\sin\theta)^2\right]^{\frac{3}{2}}}{r^2 + 2(-a\sin\theta)^2 - r(-a\cos\theta)} \\ = \frac{(r^2 + a^2 \sin^2\theta)^{\frac{3}{2}}}{r^2 + 2a^2 \sin^2\theta + a\cos\theta \cdot r} \\ = \frac{[r^2 + a^2(1 - \cos^2\theta)]^{\frac{3}{2}}}{r^2 + r^2 + 2a^2 \sin^2\theta} \\ = \frac{(r^2 + a^2 - a^2 \cos^2\theta)^{\frac{3}{2}}}{2r^2 + 2a^2 \sin^2\theta} \\ = \frac{(r^2 + a^2 - r^2)^{\frac{3}{2}}}{2r^2 + 2a^2(1 - \cos^2\theta)} \ [\because r = a\cos\theta] \\ = \frac{a^3}{2r^2 + 2a^2 - 2a^2 \cos^2\theta} \\ = \frac{a^3}{2r^2 + 2a^2 - 2r^2} \\ = \frac{a^3}{2a^2} \\ \Rightarrow \rho = \frac{a}{2} \\ \text{Illustration : } 3(b). \ \theta = a^{-1}(r^2 - a^2)^{\frac{1}{2}} - \cos^{-1}\left(\frac{a}{r}\right) \\ \text{Solution : } \theta = a^{-1}(r^2 - a^2)^{\frac{1}{2}} - \cos^{-1}\left(\frac{a}{r}\right) \\ \text{Differentiating w.r.t. } \theta, \text{ we get}\\ 1 = a^{-1} \frac{1}{2} (r^2 - a^2)^{-\frac{1}{2}} 2r \cdot \frac{dr}{d\theta} + \frac{1}{\sqrt{1 - \frac{a^2}{r^2}}} \cdot \frac{dr}{d\theta}  \left(-\frac{a}{r^2}\right) \\ 1 = \left[\frac{a^{-1}}{2} (r^2 - a^2)^{-\frac{1}{2}} - \frac{r}{\sqrt{r^2 - a^2}} \cdot \frac{a}{r^2}\right] \frac{dr}{d\theta} \\ \frac{d\theta}{dr} = \frac{2r}{a(r^2 - a^2)^{-\frac{1}{2}}} - \frac{a}{r\sqrt{r^2 - a^2}} \\ = \frac{r^2 - a^2}{ar\sqrt{r^2 - a^2}} \\ \Rightarrow \frac{d\theta}{dr} = \frac{\sqrt{r^2 - a^2}}{ar} \\ \frac{dr}{d\theta} = \frac{ar}{\sqrt{r^2 - a^2}} \\ \text{Again differentiating w.r.t. } \theta, \text{ we get} \\ \frac{d^2 r}{d\theta^2} = a \left[\frac{\sqrt{r^2 - a^2} \cdot 1 - 2\sqrt{r^2 - a^2} \cdot \phi}{r^2 - a^2}\right] \frac{dr}{d\theta} \\ = a \left[\frac{\sqrt{r^2 - a^2} - \frac{r^2}{\sqrt{r^2 - a^2}}}{r^2 - a^2}\right] \frac{dr}{d\theta} \\ = \frac{a \left[\frac{r^2 - a^2 - r^2}{\sqrt{r^2 - a^2}}\right] \frac{dr}{d\theta}}{r^2 - a^2} \\ = a \left[\frac{-a^2}{(r^2 - a^2)^{\frac{3}{2}}}\right] \times \frac{dr}{d\theta} \\ = -\frac{a^3}{(r^2 - a^2)^{\frac{3}{2}}} \times \frac{ar}{\sqrt{r^2 - a^2}} \\ = -\frac{a^4 r}{(r^2 - a^2)^2}\\ \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^2 r}{d\theta^2}} \\ = \frac{\left[r^2 + \left(\frac{ar}{\sqrt{r^2 - a^2}}\right)^2\right]^{\frac{3}{2}}}{r^2 + 2\left(\frac{ar}{\sqrt{r^2 - a^2}}\right)^2 - r\left(-\frac{a^4 r}{(r^2 - a^2)^2}\right)} \\ = \frac{\left[r^2 + \frac{a^2 r^2}{r^2 - a^2}\right]^{\frac{3}{2}}}{r^2 + \frac{2a^2 r^2}{r^2 - a^2} + \frac{a^4 r^2}{(r^2 - a^2)^2}} \\ = \frac{(r^4 - r^2 a^2 + a^2 r^2)^{\frac{3}{2}}}{(r^2 - a^2)^{\frac{3}{2}} \left[r^2 + \frac{2a^2 r^2}{r^2 - a^2} + \frac{a^4 r^2}{(r^2 - a^2)^2}\right]} \\ = \frac{r^6 \times (r^2 - a^2)^2}{(r^2 - a^2)^{\frac{3}{2}} [(r^4 + a^2 r^2)(r^2 - a^2) + a^4 r^2]} \\ = \frac{r^6(r^2 - a^2)^2}{(r^2 - a^2)^{\frac{3}{2}} (r^6 + a^2 r^4 - a^2 r^4 - a^4 r^2 + a^4 r^2)} \\ = \frac{r^6(r^2 - a^2)^2}{r^6 (r^2 - a^2)^{\frac{3}{2}}} \\ \Rightarrow \rho = \frac{(r^2 - a^2)^2}{(r^2 - a^2)^{\frac{3}{2}}} \\ \Rightarrow \rho = \sqrt{r^2 - a^2} \\ \text{Illustration : } 3(c). \ r^n = a^n \sin n\theta \\ \text{Solution : } r^n = a^n \sin n\theta \\ \text{Taking logarithms}\\ n\log r = n\log a + \log\sin n\theta \\ \text{Differentiating w.r.t. } \theta, \text{ we get} \\ n \cdot \frac{1}{r} \frac{dr}{d\theta} = 0 + \frac{1}{\sin n\theta} \cdot \cos n\theta \cdot n \\ \Rightarrow \frac{1}{r} \frac{dr}{d\theta} = \cot n\theta \\ \Rightarrow \frac{d\theta}{dr} = \frac{1}{r} \tan n\theta \\ \Rightarrow r \frac{d\theta}{dr} = \tan n\theta \\ \Rightarrow \tan\phi = \tan n\theta \ [r\frac{d\theta}{dr} = \tan\phi] \\ \Rightarrow \phi = n\theta \\ p = r \sin\phi \\ \Rightarrow p = r \sin n\theta \\ \Rightarrow p = r \cdot \frac{r^n}{a^n} \\ = p = \frac{r^{n+1}}{a^n} \\ \text{differentiating w.r.t. } r, \text{ we get} \\ \frac{dp}{dr} = \frac{(n+1)r^n}{a^n} \\ \rho = r \frac{dr}{dp} = r \left(\frac{a^n}{(n+1)r^n}\right) \\ \Rightarrow \rho = \frac{a^n}{(n+1)r^{n-1}} \Rightarrow \rho = \frac{a^n r^{-n+1}}{n+1} \\ \text{Illustration : } 3(d). \ r = ae^{m\theta} \\ \text{Solution : } r = ae^{m\theta} \\ \text{differentiating w.r.t. } \theta, \text{ we get} \\ \frac{dr}{d\theta} = am e^{m\theta} \\ \text{Again differentiating w.r.t. } \theta, \text{ we get} \\ \frac{d^2 r}{d\theta^2} = am^2 e^{m\theta}\\ \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^2 r}{d\theta^2}} \\ = \frac{\left[r^2 + (ame^{m\theta})^2\right]^{\frac{3}{2}}}{r^2 + 2(ame^{m\theta})^2 - r \times am^2 e^{m\theta}} \\ = \frac{(r^2 + a^2 m^2 e^{2m\theta})^{\frac{3}{2}}}{r^2 + 2a^2 m^2 e^{2m\theta} - am^2 re^{m\theta}} \\ = \frac{(r^2 + m^2 r^2)^{\frac{3}{2}}}{r^2 + 2m^2 r^2 - m^2 r \cdot r} \ [\because r = ae^{m\theta}] \\ = \frac{\left[r^2(1 + m^2)\right]^{\frac{3}{2}}}{r^2 + 2m^2 r^2 - m^2 r^2} \\ = \frac{r^3 (1 + m^2)^{\frac{3}{2}}}{r^2 + m^2 r^2} \\ = \frac{r^3 (1 + m^2)^{\frac{3}{2}}}{r^2(1 + m^2)} \\ = r\sqrt{1 + m^2} \\ = r\sqrt{1 + \cot^2\alpha} \ [\because m = \cot\alpha] \\ = r\sqrt{\csc^2\alpha} \\ \Rightarrow \rho = r\csc\alpha \text{ where } m = \cot\alpha
By solving the above examples,one can understand the Radius of Curvature for Polar Curves.

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

Radius of Curvature for Polar Curves

(1.) Find the radius of curvature at the point (p, r) on the spiral spiral p^2 = \frac{r^4}{r^2 + a^2}
(2.) Find the radius of curvature at (r, \theta) on the curve r^2 = a^2 \cos 2\theta
Ans: (1.)\frac{(r^2 + a^2)^{\frac{3}{2}}}{(r^2 +2a^2)}
(2.) \rho \propto \frac{1}{r}
By solving the above questions,one can understand the Radius of Curvature for Polar Curves well because the concept is well understood when you solve the questions 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!👇
*” \left(\frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}} + \frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}\right) को सरल कीजिए।”*
Simplify \left(\frac{1+\sqrt{2}}{\sqrt{5}+\sqrt{3}} + \frac{1-\sqrt{2}}{\sqrt{5}-\sqrt{3}}\right).
*पिछली प्रश्नोत्तरी का उत्तर*
दिया है: x = 7 - 4\sqrt{3} \\ \text{ तब } x + \frac{1}{x} = 7 - 4\sqrt{3} + \frac{1}{7 - 4\sqrt{3}} \\ = 7 - 4\sqrt{3} + \frac{1}{7 - 4\sqrt{3}} \times \frac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}} \\ = 7 - 4\sqrt{3} + \frac{7 + 4\sqrt{3}}{49 - 48} \\ \Rightarrow x + \frac{1}{x} = 7 - 4\sqrt{3} + 7 + 4\sqrt{3} = 14
*Previous Quiz Solution*
Given x = 7 - 4\sqrt{3} \\ \text{ तब } x + \frac{1}{x} = 7 - 4\sqrt{3} + \frac{1}{7 - 4\sqrt{3}} \\ = 7 - 4\sqrt{3} + \frac{1}{7 - 4\sqrt{3}} \times \frac{7 + 4\sqrt{3}}{7 + 4\sqrt{3}} \\ = 7 - 4\sqrt{3} + \frac{7 + 4\sqrt{3}}{49 - 48} \\ \Rightarrow x + \frac{1}{x} = 7 - 4\sqrt{3} + 7 + 4\sqrt{3} = 14
This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*