1.Radius of Curvature in UG Mathematics

Radius of Curvature in UG Mathematics

In this article we will read about Radius of Curvature in UG Mathematics.Knowing what is radius of curvature and how to find it for any curve.

Also Read This Article:- Euler’s Theorem of Homogeneous Function

2.Radius of Curvature for Cartesian Curve [Raj. 74,Jodhpur 80(s)]

Radius of Curvature in UG Mathematics

Let y=f(x) be the equation of a given curve.If \psi be the angle which the tangent at any point (x,y) on the curve makes with the positive direction of the x-axis,then we have
\tan\psi=\frac{dy}{dx} \ldots(1)
Also if s be the length of the arc of the curve from a fixed point to the point (x,y) then we know that
\frac{dx}{ds}=\cos\psi \cdots(2)
Now differentiating (1) with respect to s on both sides we get
\sec^2\psi\left(\frac{d\psi}{ds}\right) =\frac{d}{ds} \left(\frac{dy}{dx}\right) \\ =\frac{d}{dx}\left(\frac{dy}{dx}\right)\cdot\frac{dx}{ds} \\ =\frac{d^2y}{dx^2}\cdot\cos\psi \qquad\text{[Using (2)]} \\ \Rightarrow \frac{d\psi}{ds} = \frac{\dfrac{d^2y}{dx^2}} {\sec^3\psi} \\ \Rightarrow \rho =\frac{ds}{d\psi}=\frac{\sec^3\psi} {\dfrac{d^2y}{dx^2}} \\ =\frac{\left(1+\tan^2\psi\right)^{\frac32}}{\dfrac{d^2y}{dx^2}} \\ \Rightarrow \rho=\frac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}} \qquad\text{[Using (1)]}

Also Read This Article:- Radius of curvature

3.Radius of Curvature in UG Mathematics Illustrations

Illustration:1.Find the radius of curvature at the point on the following curves:
Illustration:1(a). s=4a \sin \psi
Solution: s=4a \sin \psi
Differentiating w.r.t. \psi ,we get
\frac{ds}{d \psi}=4a \cos \psi \\ \Rightarrow \rho=\frac{ds}{d \psi}=4a \cos \psi \\ \Rightarrow \rho=4 a\cos \psi
Illustration:1(b). s=c \log(\sec \psi)
Solution: s=c \log(\sec\psi)
Differentiating w.r.t. \psi ,we get
\frac{ds}{d\psi} = c\cdot\frac{1}{\sec\psi}\cdot(\sec\psi\tan\psi) \\ \Rightarrow \rho=c \tan\psi
Illustration:1(c). s=c \tan\psi (catanary)
Solution: s=c \tan\psi
Differentiating w.r.t. \psi ,we get
\frac{ds}{d\psi} = c \sec^{2}\psi \\ \rho=c \sec^{2}\psi
Illustration:1(d). s=8a\sin^{2}\left(\frac{\psi}{6}\right) (Cardioid)
Solution: s=8a\sin^{2}\left(\frac{\psi}{6}\right)
Differentiating w.r.t. \psi ,we get
\frac{dS}{d\psi}= 8a\cdot2\sin\left(\frac{\psi}{6}\right) \cos\left(\frac{\psi}{6}\right)\cdot\frac16\\ = 8a\sin\left(\frac{\psi}{6}\right) \cos\left(\frac{\psi}{6}\right)\\ = 4a\sin \left( 2 \cdot\frac{\psi}{6}\right) \\ \therefore\ \rho=4a\sin\left(\frac{\psi}{3}\right)
Illustration:1(e). s=a\log(\tan\psi+\sec\psi)+a\tan\psi\sec\psi \qquad\text{(Parabola)}
Solution: s=a\log(\tan\psi+\sec\psi)+a\tan\psi\sec\psi
Differentiating w.r.t., \psi we get
\frac{ds}{d\psi}=a\cdot\frac{1}{\tan\psi+\sec\psi} \left(\sec^2\psi+\sec\psi\tan\psi\right) +a\sec^3\psi +a\tan^2\psi \sec\psi\\ \therefore \rho= \frac{a\sec\psi(\sec\psi+\tan\psi)}{\tan\psi+\sec\psi}+a\sec^3\psi +a\tan^2\psi \sec\psi\\ =a\sec\psi+a\sec^3\psi+a\tan^2\psi \sec\psi\\ = a\sec\psi(1+\tan^2\psi)+a\sec^3\psi\\ =a\sec\psi\sec^2\psi +a\sec^3\psi \qquad \left[\because 1+\tan^2\psi=\sec^2\psi\right]\\ = a\sec^3\psi +a\sec^3\psi\\ \therefore \rho=2a\sec^3\psi
Illustration:2.Show that for the curve s^2=8ay , \rho=4a\sqrt{1-\frac{y}{2a}}
Solution: s^2=8ay \cdots(1)
Differentiating w.r.t. \psi ,we get
2s\frac{ds}{dy}=8a\\ \Rightarrow 2s\left(\frac{1}{\sin\psi}\right)=8a \left[\because \frac{dy}{ds}=\sin\psi \right]\\ \therefore s=4a\sin\psi \cdots(2)
Again Differentiating w.r.t. \psi ,we get
\frac{ds}{d\psi}=4a\cos\psi\\ \therefore\ \rho=4a\cos\psi\\ =4a\sqrt{1-\sin^2\psi}\\ =4a\sqrt{1-\frac{s^2}{16a^2}} \qquad\text{[From (2)]}\\ =4a\sqrt{1-\frac{8ay}{16a^2}} \qquad\text{[From (1)]}\\ \therefore\ \rho =4a\sqrt{1-\frac{y}{2a}}
Illustration:3.Show that for the curve s=ce^{\frac{x}{c}}, \rho=\frac{s\sqrt{s^2-c^2}}{c}
Solution: s=ce^{\frac{x}{c}} \ldots(1)
Differentiating w.r.t. x,we get
\frac{ds}{dx} =ce^{\frac{x}{c}}\cdot\frac{1}{c}\\ \Rightarrow \frac{1}{\cos \psi}=e^{\frac{x}{c}} \qquad \left[\because \frac{ds}{dx}=\cos \psi \right]\\ \therefore \sec\psi=e^{\frac{x}{c}}\\ \therefore \sec\psi=\frac{s}{c} \qquad\text{[From (1)]} \cdots(2)
Again differentiating w.r.t. x,we get
\sec\psi\tan\psi =\frac{1}{c}\frac{ds}{d\psi}\\ \therefore \rho=c\sec\psi\tan\psi\\ =c\left(\frac{\sin\psi}{\cos^2\psi}\right)\\ =c\left(\frac{\sqrt{1-\cos^2\psi}}{\cos^2\psi}\right)\\ =c\left(\frac{\sqrt{1-\frac{c^2}{s^2}}}{\frac{c^2}{s^2}}\right) \qquad\text{[From (2)]} \\ =c\cdot\frac{s^{2}}{c^{2}}\cdot\frac{\sqrt{s^{2}-c^{2}}}{s}\\ \therefore\ \rho=\frac{s\sqrt{s^{2}-c^{2}}}{c}

Radius of Curvature in UG Mathematics

Illustration:4.Find the radius of curvature at the point (x,y) on the following curves:
Illustration:4(a). x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}
Solution: x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}} \cdots(1)
Differentiating w.r.t. x ,we get
\frac23x^{-\frac13}+\frac23y^{-\frac13}\frac{dy}{dx}=0\\ \Rightarrow \frac23y^{-\frac13}\frac{dy}{dx} =-\frac23x^{-\frac13}\\ \Rightarrow \frac{dy}{dx} =-\frac{x^{-\frac13}}{y^{-\frac13}}\\ \therefore \frac{dy}{dx}=-\frac{y^{\frac13}}{x^{\frac13}} \cdots(2)
Again differentiating w.r.t. x,we get
\frac{d^{2}y}{dx^{2}}=-\frac{ x^{\frac13}\left(\frac{1}{3}y^{-\frac23}\frac{dy}{dx}\right)-\frac13y^{\frac13}x^{-\frac23}}{x^{\frac23}}\\ =-\frac{ \frac{x^{\frac13}}{3y^{\frac23}}\frac{dy}{dx}-\frac{y^{\frac13}}{3x^{\frac23}}}{x^{\frac23}}\\ =-\frac{ \frac{x^{\frac13}}{3y^{\frac23}}\left(-\frac{y^{\frac13}}{x^{\frac13}}\right)-\frac{y^{\frac13}}{3x^{\frac23}}}{x^{\frac23}}\\ =\frac{\frac{1}{3y^{\frac13}}+\frac{y^{\frac13}}{3x^{\frac23}}}{x^{\frac23}} \\ =\frac{1}{3}\left[\frac{\frac{x^{\frac23}+y^{\frac23}}{y^{\frac13}x^{\frac23}}}{\frac{1}{x^{\frac23}}}\right] \\ =\frac{1}{3}\cdot \frac{a^{\frac23}}{y^{\frac13}x^{\frac43}} \qquad\text{[From (1)]}\\ \therefore \frac{d^2y}{dx^2} =\frac{a^{\frac23}} {3y^{\frac13}x^{\frac43}} \ldots(3) \\ \rho= \frac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}}\\ =\frac{ \left[ 1+ \left(\dfrac{y^{\frac13}}{x^{\frac13}}\right)^2\right]^{\frac32}}{\dfrac{a^{\frac23}}{3y^{\frac13}x^{\frac43}}}\\=\left[\frac{x^{\frac23}+y^{\frac23}}{x^{\frac23}}\right]^{\frac32}\cdot \frac{3y^{\frac13}x^{\frac43}}{a^{\frac23}}\\ =\frac{\left(a^{\frac23}\right)^{\frac32}}{x} \cdot\frac{3y^{\frac13}x^{\frac43}}{a^{\frac23}}\qquad\text{[From (1)]}\\ =\frac{a}{x} \cdot \frac{3y^{\frac13}x^{\frac43}}{a^{\frac23}}\\ =3a^{\frac13}x^{\frac13}y^{\frac13}\\ \therefore \rho =3(axy)^{\frac13}
Illustration:4(b). xy=c^2
Solution: xy=c^2 \cdots(1)
Differentiating w.r.t. x ,we get
y+x\left(\frac{dy}{dx}\right)=0\\ \Rightarrow x\left(\frac{dy}{dx}\right)=-y\\ \therefore \frac{dy}{dx} = -\frac{y}{x} \ldots(2)
Again differentiating w.r.t. x,we get
\frac{d^2y}{dx^2}=-\left[\frac{x\frac{dy}{dx}-y\cdot1}{x^2}\right]\\ =-\left[\frac{x\left(-\frac{y}{x}\right)-y}{x^2}\right] \qquad\text{[From (2)]}\\ =\frac{y+y}{x^2}\\ \Rightarrow \frac{d^2y}{dx^2} = \frac{2y}{x^2} \\ \rho=\frac{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}}\\ =\frac{\left[1+\dfrac{y^2}{x^2}\right]^{\frac32}}{\dfrac{2y}{x^2}}\\ =\left[\frac{x^2+y^2}{x^2}\right]^{\frac32}\cdot\frac{x^2}{2y}\\ =\frac{(x^2+y^2)^{\frac32}}{x^3} \cdot \frac{x^2}{2y}\\ =\frac{(x^2+y^2)^{\frac32}}{2xy}\\ \Rightarrow \rho=\frac{(x^2+y^2)^{\frac32}}{2c^2}
Illustration:4(c). ay^2=x^3
Solution: ay^2=x^3 \cdots(1)
Differentiating w.r.t. x ,we get
2ay\left(\frac{dy}{dx}\right)=3x^2\\ \therefore \frac{dy}{dx}=\frac{3x^2}{2ay} \cdots(2)
Again differentiating w.r.t. x,we get
\frac{d^2y}{dx^2} = \frac{3}{2a} \left[ \frac{y\cdot2x-x^2\frac{dy}{dx}}{y^2}\right]=\frac{3}{2a} \left[ \frac{2xy-x^2\left(\dfrac{3x^2}{2ay}\right)} {y^2} \right] \qquad\text{[From (2)]} \\= \frac{3}{2a} \left[ \frac{4axy^2-3x^4} {2ay^3}\right] \\ = \frac{3}{2a} \left[ \frac{4ax\left(\dfrac{x^3}{a}\right)-3x^4} {2a\left(\dfrac{x^{\frac32}}{a^{\frac32}}\right)}\right] \qquad\text{[From (1)]} \\ =\frac{3}{2a} \left[ \frac{x^4}{2x^{\frac32}} \right]a^{\frac12} \\ =\frac{3}{4a^{\frac12}}\left( \frac{x^4}{x^{\frac32}} \right) \\= \frac{3}{4\sqrt{ax}} \\ \therefore \rho= \frac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}} \\ = \frac{\left[1+\left(\dfrac{3x^2}{2ay}\right)^2\right]^{\frac32} } {\dfrac{3}{4\sqrt{ax}}} \\ =\left[ 1+\frac{9x^4}{4a^2y^2}\right]^{\frac32}\cdot \frac{4\sqrt{ax}}{3} \\= \left[ 1+\frac{9x^4} {4a^2\left(\dfrac{x^3}{a}\right)}\right]^{\frac32} \cdot \frac{4\sqrt{ax}}{3} \\= \left[ 1+\frac{9x}{4a} \right]^{\frac32}\cdot \frac{4\sqrt{ax}}{3} \\ =\frac{(4a+9x)^{\frac32}}{(4a)^{\frac32}}\cdot \frac{4\sqrt{ax}}{3} \\ = \frac{\sqrt{x}\,(4a+9x)^{\frac32}}{6a}
Illustration:4(d). y=c\log\sec\left(\frac{x}{c}\right)
Solution: y=c\log\sec\left(\frac{x}{c}\right)
Differentiating w.r.t. x,we get
\frac{dy}{dx}=c\cdot \frac{1}{\sec\left(\frac{x}{c}\right)} \cdot \sec\left(\frac{x}{c}\right) \tan\left(\frac{x}{c}\right) \cdot \frac{1}{c}\\ \Rightarrow \frac{dy}{dx}=\tan\left(\frac{x}{c}\right)
Again differentiating w.r.t. x,we get
\frac{d^2y}{dx^2}= \sec^2\left(\frac{x}{c}\right)\cdot\frac{1}{c}\\ \Rightarrow \frac{d^2y}{dx^2}=\frac{\sec^2\left(\frac{x}{c}\right)}{c} \\ \rho= \frac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}}\\ =\frac{\left[1+\tan^2\left(\frac{x}{c}\right)\right]^{\frac32}}{\dfrac{\sec^2\left(\frac{x}{c}\right)}{c}}\\ = \left[\sec^2\left(\frac{x}{c}\right)\right]^{\frac32} \cdot \frac{c} {\sec^2\left(\frac{x}{c}\right)}\\=\frac{c\sec^3\left(\frac{x}{c}\right) }{\sec^2\left(\frac{x}{c}\right)}\\ \therefore \rho=c\sec\left(\frac{x}{c}\right)
Illustration:5.Find the radius of curvature of the following curves:
Illustration:5(a). \sqrt{x}+\sqrt{y}=1,\quad \text{at }\left(\frac14,\frac14\right)
Solution: \sqrt{x}+\sqrt{y}=1 \cdots(1)
Differentiating w.r.t. x ,we get
\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}} \frac{dy}{dx}=0 \\ \therefore \frac{dy}{dx}=-\sqrt{\frac{y}{x}} \ldots(2) \\ \left(\frac{dy}{dx}\right)_{\left(\frac14,\frac14\right)}=-\sqrt{\frac{\frac14}{\frac14}} =-1
Again differentiating w.r.t. x,we get
\frac{d^2y}{dx^2}=- \left[\frac{ \sqrt{x}\cdot \frac{1}{2\sqrt{y}}\frac{dy}{dx}- \sqrt{y}\cdot\frac{1}{2\sqrt{x}}}{x}\right] \\=-\left[ \frac{\frac{\sqrt{x}}{2\sqrt{y}}\left(-\sqrt{\frac{y}{x}}\right)-\frac{\sqrt{y}}{2\sqrt{x}}}{x}\right] \\ =\frac{\frac12+\frac{\sqrt{y}}{2\sqrt{x}}}{x} \\ \therefore \left(\frac{d^2y}{dx^2}\right)_{\left(\frac14,\frac14\right)}= \frac{\frac12+\frac{\sqrt{\frac14}}{2\sqrt{\frac14}}}{\frac14}=\frac{\frac12+\frac12}{\frac14}=\frac{1}{\frac14}=4 \\ \Rightarrow \left(\frac{d^2y}{dx^2}\right)_{\left(\frac14,\frac14\right)}=4 \\ \rho=\frac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}}\\ =\frac{\left[1+(-1)^2\right]^{\frac32}}{4} \\ =\frac{(1+1)^{\frac32}}{4} =\frac{2^{\frac32}}{4} \\ \Rightarrow \rho= \frac{1}{\sqrt2}
Illustration:5(b). y=4\sin x-\sin2x,\quad \text{at }x=\frac{\pi}{2}
Solution: y=4 \sin x-\sin 2x
Differentiating w.r.t.,we get
\frac{dy}{dx}=4\cos x-2\cos2x \cdots(2) \\ \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{2}}=4\cos\frac{\pi}{2}-2\cos\left(\frac{2\pi}{2}\right)\\ =4(0)-2\cos\pi\\ =-2(-1)\\ \Rightarrow \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{2}}=2
Again differentiating w.r.t. x,we get
\frac{d^2y}{dx^2}=-4\sin x+4\sin2x \\ \left(\frac{d^2y}{dx^2}\right)_{x=\frac{\pi}{2}}=-4\sin\frac{\pi}{2}+4\sin\left(\frac{2\pi}{2}\right)\\ =-4(1)+4\sin\pi\\ =-4+4(0)\\ \Rightarrow \left(\frac{d^2y}{dx^2}\right)_{x=\frac{\pi}{2}}=-4 \\ \rho=\frac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{\frac32}}{\dfrac{d^2y}{dx^2}} \\ =\frac{\left[1+(2)^2\right]^{\frac32}}{(-4)}\\ =\frac{(1+4)^{\frac32}}{(-4)}\\ =-\frac{5^{\frac32}}{4}\\ \Rightarrow \rho=\frac{5\sqrt5}{4} [negleting -ve sign]
With the above illustrations,one can understand the Radius of Curvature in UG Mathematics.

4.Practice Questions of Radius of Curvature for Students

Radius of Curvature in UG Mathematics

(1.)For the curve y=\frac{ax}{a+x} ,if \rho is the radius of curvature at any point (x,y),show that
\left(\frac{2\rho}{a}\right)^{\frac23}=\left(\frac{x}{y}\right)^2+\left(\frac{y}{x}\right)^2
(2.)Prove that for the curve s=a\log\tan\left(\frac{\pi}{4}+\frac{\psi}{2}\right) +a\tan\psi \sec\psi , \rho=2a\sec^{3}\psi
,and hence show that \frac{d^{2}y}{dx^{2}}=\frac{1}{2a} and also that this differential equation is satisfied by the parabola x^{2}=4ay .
By solving the above questions,you can understand the Radius of Curvature in UG Mathematics well because the concept is well understood when you solve it practically.

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5.Important Formulae of Radius of Curvature

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6.Frequently Asked Questions Related to Radius of Curvature in UG Mathematics

\begin{array}{|c|} \hline \text{**छात्र-छात्राओं से आज का प्रश्न**} \\ \text{*'पुस्तकें रखी हुई एक आलमारी (सेल्फ) का} \\ \text{ वजन 25 किग्रा था।दो तिहाई पुस्तकें बेचने} \\ \text{ के पश्चात आलमारी और पुस्तकों का वजन} \\ \text{  15 किग्रा रह गया।} \\ \text{खाली आलमारी का वजन बताओ।'*} \\ \text{दिनांक 27.06.2026 के प्रश्न का उत्तर:1,5,3}  \\ 1^3+5^3+3^3=1+125+27=153 \\  \text{**Today's Question to Students** }\\ \text{*'A cupboard containing books weighed } \\ \text{ 25 kg.After selling two-thirds of the books, } \\ \text{the shelves and books weighed 15 kg.} \\ \text{Tell me the weight of the empty cupboard.'*} \\ \text{Answer to Question Dated 27.06.2026:1,5,3} \\ 1^3+5^3+3^3=1+125+27=153 \\ \hline  \end{array}
This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*