1.Parametric and Cartesian Formula for Curvature-B.Sc. Math

Parametric and Cartesian Formula for Curvature-B.Sc. Math

In this article “Parametric and Cartesian Formula for Curvature-B.Sc. Math”,we will read about how to find radius of curvature for cartesian curves and parametric curves?
“Before moving to Part 2,read our basic introduction to Radius of Curvature in UG Mathematics.”

Also Read This Article:-Power Series in Differential Calculus

2.Illustrations Based on Parametric and Cartesian Formula for Curvature-B.Sc. Math

Illustration:5.Find the radius of curvature of the following curves:
Illustration:5(c). (x^2 + y^2)^2 = a^2(y^2 - x^2), at (0, a)
Solution: (x^2 + y^2)^2 = a^2(y^2 - x^2), \text{ at } (0, a)
Differentiating w.r.t. x,we get
2(x^2 + y^2)\left(2x + 2y\frac{dy}{dx}\right) = a^2\left(2y\frac{dy}{dx} - 2x\right) \\ \Rightarrow 2(x^2 + y^2)\left(x + y\frac{dy}{dx}\right) = a^2\left(y\frac{dy}{dx} - x\right) \quad \dots(1) \\ \text{Put } x = 0, y = a \\ 2(0^2 + a^2)\left(0 + a\frac{dy}{dx}\right) = a^2\left(a\frac{dy}{dx} - 0\right) \\ \Rightarrow 2a^3\frac{dy}{dx} = a^3\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx} = 0 \\ \text{Again differentiating w.r.t. } x, \text{ we get} \\ 2\left(2x + 2y\frac{dy}{dx}\right)\left(x + y\frac{dy}{dx}\right) + 2(x^2 + y^2)\left[1 + \left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2}\right] \\ = a^2\left[\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} - 1\right] \\ \text{Put } x = 0, y = a \text{ and } \frac{dy}{dx} = 0 \\ \Rightarrow 2(2 \times 0 + 2 \times a \times 0)(0 + a \times 0) + 2(0^2 + a^2)\left[1 + 0^2 + a\frac{d^2y}{dx^2}\right] \\ = a^2\left[0^2 + a\frac{d^2y}{dx^2} - 1\right] \\ \Rightarrow 2a^2 + 2a^3\frac{d^2y}{dx^2} = a^3\frac{d^2y}{dx^2} - a^2 \\ \Rightarrow 2a\frac{d^2y}{dx^2} - a\frac{d^2y}{dx^2} = -3 \\ \Rightarrow a\frac{d^2y}{dx^2} = -3 \\ \Rightarrow \frac{d^2y}{dx^2} = -\frac{3}{a} \\ \rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} \\ = \frac{(1 + 0^2)^{\frac{3}{2}}}{-\frac{3}{a}} \\ = -\frac{a}{3} \\ \Rightarrow \rho = \frac{a}{3} (Numerical Value)
Illustration:5(d). e^x ,at the point where it croses the y-axis
Solution: y = e^x \quad \dots(1) \\ \text{It crosses at } y\text{-axis } x = 0 \\ y = e^0 \Rightarrow y = 1 \\ \text{Point is } (0, 1) \\ \text{Differentiating } (1) \text{ w.r.t. } x, \text{ we get} \\ \frac{dy}{dx} = e^x \Rightarrow \left(\frac{dy}{dx}\right)_{(0,1)} = e^0 = 1 \\ \text{Again differentiating w.r.t. } x, \text{ we get} \\ \frac{d^2y}{dx^2} = e^x \\ \left(\frac{d^2y}{dx^2}\right)_{(0,1)} = e^0 = 1 \\ \rho = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}} \\ = \frac{\left[1 + (1)^2\right]^{3/2}}{(1)} \\ = (2)^{3/2} \\ \Rightarrow \rho = \sqrt{8}
Illustration:5(e). y^2 = \frac{4a^2(2a-x)}{x} , at the point where it meets x-axis
Solution: y^2 = \frac{4a^2(2a-x)}{x} \dots(1) \\ \text{at the point } x\text{-axis } y=0 \\ 0^2 = \frac{4a^2(2a-x)}{x} \Rightarrow x=2a \\ \text{point is } (2a,0) \\ \text{differentiating } \text{w.r.t. } x, \text{we get} \\ 2y \frac{dy}{dx} = 4a^2 \left[ \frac{x \cdot (-1) - (2a-x) \cdot 1}{x^2} \right] \\ \Rightarrow y \frac{dy}{dx} = 2a^2 \left( \frac{-x-2a+x}{x^2} \right) \\ \Rightarrow \frac{dx}{dy} = -\frac{x^2y}{4a^3} \dots(2) \\ \Rightarrow \left(\frac{dx}{dy}\right)_{(2a,0)} = -\frac{(2a)^2(0)}{4a^3} = 0 \\ \text{Again differentiating } (2) \text{ w.r.t. } y \text{ we get } \\ \frac{d^2x}{dy^2} = -\frac{1}{4a^3} \left( \frac{dx}{dy} 2xy + x^2 \cdot 1  \right) \\ \Rightarrow \left(\frac{d^2x}{dy^2}\right)_{(2a,0)} = -\frac{1}{4a^3} (2 \times 2a \times 0 + (2a)^2 \times 1) \\ \Rightarrow \left(\frac{d^2x}{dy^2}\right)_{(2a,0)} = -\frac{1}{a} \\ \rho = \frac{\left[1 + \left(\frac{dx}{dy}\right)^2\right]^{3/2}}{\frac{d^2x}{dy^2}} \\ = \frac{(1+0^2)^{\frac{3}{2}}}{-\frac{1}{a}} \\ \Rightarrow \rho = -a \\ \Rightarrow \rho = a \quad (\text{Numerical value})
Illustration:5(f). x^2y = a(x^2+y^2), at (-2a, 2a)
Solution: x^2y = a(x^2+y^2) \\ \text{Differentiating } \text{w.r.t } x, \text{we get} \\ 2xy + x^2 \frac{dy}{dx} = a \left( 2x + 2y \frac{dy}{dx} \right) \\ \Rightarrow \text{put } x = -2a, y=2a \\ \Rightarrow 2(-2a)(2a) + (-2a)^2 \frac{dy}{dx} = a(2x-2a) + 2 \times 2a \times \frac{dy}{dx} \\ \Rightarrow -8a + 4a^2 \frac{dy}{dx} = a(-4a+4a\frac{dy}{dx}) \\ \Rightarrow -8a + 4a^2 \frac{dy}{dx} = -4a^2 + 4a^2 \frac{dy}{dx} \\ \Rightarrow (-8a + 4a^2) \frac{dx}{dy} = 0 \\ \Rightarrow \frac{dx}{dy} = 0 \\ \text{Again differentiating } \text{w.r.t. } y, \text{we get } \\ \frac{dx}{dy} = \frac{2ay-x^2}{2xy-2ax} \\ \Rightarrow \frac{d^2x}{dy^2} = \frac{(2xy-2ax)(2a-2x\frac{dx}{dy})-(2ay-x^2)(2y\frac{dx}{dy}+2x-2a\frac{dx}{dy})}{(2xy-2ax)^2} \\ \Rightarrow \left(\frac{d^2x}{dy^2}\right)_{(-2a,2a)} = \frac{\begin{array}{l}(2 \times -2a \times 2a - 2a \times -2a)(2a-2 \times -2a \times 0)\\ -(2a \times 2a - (-2a)^2)(2 \times 2a \times 0 + 2 \times -2a - 2a \times 0)\end{array}}{(2 \times -2a \times 2a - 2a \times -2a)^2} \\ = \frac{(-8a^2+4a^2)(2a)-(4a^2-4a^2)(-4a)}{(-8a^2+4a^2)^2} \\ \Rightarrow \left(\frac{d^2x}{dy^2}\right)_{(-2a,2a)} = \frac{-8a^3}{16a^4} = -\frac{1}{2a} \\ \rho = \frac{[1+(\frac{dx}{dy})^2]^{\frac{3}{2}}}{\frac{d^2x}{dy^2}} \\ = \frac{(1+0^2)^{\frac{3}{2}}}{-\frac{1}{2a}} \\ \Rightarrow \rho = -2a \\ \Rightarrow \rho = 2a (\text{Numerical value})

Parametric and Cartesian Formula for Curvature-B.Sc. Math

Illustration:6(a). x = a\cos t, y = b\sin t
Solution: x = a\cos t, y = b\sin t \\ \text{Differentiating w.r.t. } t, \text{we get} \\ \frac{dx}{dt} = -a\sin t \text{ and } \frac{dy}{dt} = b\cos t \\ \text{Again differentiating w.r.t. } t, \text{we get} \\ \frac{d^2x}{dt^2} = -a\cos t \text{ and } \frac{d^2y}{dt^2} = -b\sin t \\ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{b\cos t}{-a\sin t} \\ \Rightarrow \frac{dy}{dx} = -\frac{b}{a} \frac{\cos t}{\sin t} \\ \text{Again differentiating w.r.t. } t, \text{we get} \\ \frac{d^2y}{dx^2} = -\frac{b}{a} \left[ \frac{\sin t (-\sin t) - \cos t \cdot \cos t}{\sin^2 t} \right] \frac{dt}{dx} \\ \Rightarrow = -\frac{b}{a} \frac{-\sin^2 t - \cos^2 t}{\sin^2 t} \times \frac{1}{-a\sin t} \\ = \frac{b}{a^2} \frac{-1}{\sin^3 t} \\ \Rightarrow \frac{d^2y}{dx^2} = -\frac{b}{a^2\sin^3 t} \\ \rho = \frac{[1 + (\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} \\ = \frac{[1 + (-\frac{b}{a} \cdot \frac{\cos t}{\sin t})^2]^{\frac{3}{2}}}{-\frac{b}{a^2\sin^3 t}} \\ = \frac{[\frac{a^2\sin^2 t + b^2\cos^2 t}{a^2\sin^2 t}]^{\frac{3}{2}}}{-\frac{b}{a^2\sin^3 t}} \\ = \frac{(a^2\sin^2 t + b^2\cos^2 t)^{\frac{3}{2}}}{a^3\sin^3 t} \times -\frac{a^2\sin^3 t}{b} \\ \Rightarrow \rho = \frac{(a^2\sin^2 t + b^2\cos^2 t)^{\frac{3}{2}}}{ab} Numerical value
Illustration:6(b). x = a(3\cos t - \cos 3t), y = a(3\sin t - \sin 3t)
Solution: x = a(3\cos t - \cos 3t), y = a(3\sin t - \sin 3t) \\ x = a(3\cos t - 4\cos^3 t + 3\cos t) \\ \text{differentiating w.r.t. } t, \text{we get} \\ \frac{dx}{dt} = a(-3\sin t - 12\cos^2 t \cdot (-\sin t) - 3\sin t) \\ \Rightarrow \frac{dx}{dt} = a(-6\sin t + 12\cos^2 t \sin t) \\ y = a(3\sin t - \sin 3t) \\ y = a(3\sin t - 3\sin t + 4\sin^3 t) \\ \text{differentiating w.r.t. } t, \text{we get} \\ \frac{dy}{dt} = a(12\sin^2 t \cdot \cos t) \\ \Rightarrow \frac{dy}{dt} = 12a\sin^2 t \cos t \\ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{12a\sin^2 t \cos t}{a(-6\sin t + 12\cos^2 t \sin t)} \\ \Rightarrow \frac{dy}{dx} = \frac{2\sin t \cos t}{-1 + 2\cos^2 t} \\ \text{Again differentiating w.r.t. } t, \text{we get} \\ \frac{d^2y}{dx^2} = \frac{(-1+2\cos^2 t)(2\cos^2 t - 2\sin^2 t) - (2\sin t \cos t)(4\cos t \cdot -\sin t)}{(-1+2\cos^2 t)^2} \cdot \frac{dt}{dx} \\ = \frac{-2\cos^2 t + 2\sin^2 t + 4\cos^4 t - 4\cos^2 t \sin^2 t + 8\sin^2 t \cos^2 t}{(-1+2\cos^2 t)^2} \times \frac{1}{a(-6\sin t + 12\cos^2 t \sin t)} \\ = \frac{4\sin^2 t \cos^2 t + 4\cos^4 t + 2\sin^2 t - 2\cos^2 t}{(2\cos^2 t - 1)^2} \times \frac{1}{a(-6\sin t + 12\cos^2 t \sin t)} \\ = \frac{4\cos^2 t (\sin^2 t + \cos^2 t) + 2\sin^2 t - 2\cos^2 t}{(2\cos^2 t - 1)^2} \times \frac{1}{a(-6\sin t + 12\cos^2 t \sin t)} \\ = \frac{2\cos^2 t + 2\sin^2 t}{(2\cos^2 t - 1)^2} \times \frac{1}{6a\sin t (2\cos^2 t - 1)} \\ = \frac{2}{(2\cos^2 t - 1)^3} \times \frac{1}{6a\sin t} \\ = \frac{1}{3a\sin t (2\cos^2 t - 1)^3} \\ \Rightarrow \frac{d^2y}{dx^2} = \frac{1}{3a\sin t (2\cos^2 t - 1)^3} \\ \rho = \frac{[1 + (\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} \\ = \frac{[1 + (\frac{2\sin t \cos t}{2\cos^2 t - 1})^2]^{\frac{3}{2}}}{\frac{1}{3a\sin t (2\cos^2 t - 1)^3}} \\ = \frac{[\frac{(2\cos^2 t - 1)^2 + 4\sin^2 t \cos^2 t}{(2\cos^2 t - 1)^2}]^{\frac{3}{2}}}{\frac{1}{3a\sin t (2\cos^2 t - 1)^3}} \\ = \frac{(4\cos^4 t - 4\cos^2 t + 1 + 4\sin^2 t \cos^2 t)^{\frac{3}{2}}}{(2\cos^2 t - 1)^3} \times 3a\sin t (2\cos^2 t - 1)^3 \\ = \frac{(4\cos^4 t - 4\cos^2 t + 1 + 4(1-\cos^2 t)\cos^2 t)^{\frac{3}{2}}}{1} \times 3a\sin t \\ = (4\cos^4 t - 4\cos^2 t + 1 + 4\cos^2 t - 4\cos^4 t)^{\frac{3}{2}} \times 3a\sin t \\ \Rightarrow \rho = 3a\sin t
iIllustration:6(c). x = a\sin t - b\sin\left(\frac{at}{b}\right), y = a\cos t - b\cos\left(\frac{at}{b}\right)
Solution: x = a\sin t - b\sin\left(\frac{at}{b}\right) \\ \text{Differentiating w.r.t. } t, \text{we get} \\ \frac{dx}{dt} = a\cos t - b\cos\left(\frac{at}{b}\right) \cdot \frac{a}{b} \\ \Rightarrow \frac{dx}{dt} = a\cos t - a\cos\left(\frac{at}{b}\right) \\ y = a\cos t - b\cos\left(\frac{at}{b}\right) \\ \text{Differentiating w.r.t. } t, \text{we get} \\ \frac{dy}{dt} = -a\sin t + b\sin\left(\frac{at}{b}\right) \cdot \frac{a}{b} \\ \Rightarrow \frac{dy}{dt} = -a\sin t + a\sin\left(\frac{at}{b}\right) \\ \\ \frac{dy}{dx} = \frac{-a\sin t + a\sin(\frac{at}{b})}{a\cos t - a\cos(\frac{at}{b})} \\ \Rightarrow \frac{dy}{dx} = \frac{-\sin t + \sin(\frac{at}{b})}{\cos t - \cos(\frac{at}{b})} \\ \text{Again differentiating w.r.t. } t, \text{we get} \\ \frac{d^2y}{dx^2} = \frac{[\cos t - \cos(\frac{at}{b})][-\cos t + \cos(\frac{at}{b}) \cdot \frac{a}{b}]}{[\cos t - \cos(\frac{at}{b})]^2} \\ - \frac{[-\sin t + \sin(\frac{at}{b})][-\sin t + \sin(\frac{at}{b}) \cdot \frac{a}{b}]}{[\cos t - \cos(\frac{at}{b})]^2} \cdot \frac{dt}{dx} \\ = \frac{\begin{array}{l}-\cos^2 t + \frac{a}{b}\cos t \cos(\frac{at}{b}) + \cos t \cos(\frac{at}{b}) - \frac{a}{b}\cos^2(\frac{at}{b})\\ - \sin^2 t + \frac{a}{b}\sin t \sin(\frac{at}{b}) + \sin t \sin(\frac{at}{b}) - \frac{a}{b}\sin^2(\frac{at}{b})\end{array}}{[\cos t - \cos(\frac{at}{b})]^2 \cdot a(\cos t - \cos(\frac{at}{b}))} \\ = \frac{\begin{array}{l}-(\sin^2 t + \cos^2 t) + \frac{a}{b}(\cos t \cos(\frac{at}{b}) + \sin t \sin(\frac{at}{b})) \\+ (\cos t \cos(\frac{at}{b}) + \sin t \sin(\frac{at}{b})) - \frac{a}{b}(\cos^2(\frac{at}{b}) + \sin^2(\frac{at}{b}))\end{array}}{a[\cos t - \cos(\frac{at}{b})]^3} \\ = \frac{-1 + \frac{a}{b}\cos(\frac{at}{b}-t) + \cos(\frac{at}{b}-t) - \frac{a}{b}}{a[\cos t - \cos(\frac{at}{b})]^3} \\ = \frac{-1 - \frac{a}{b} + \cos(\frac{a-b}{b}t)[\frac{a}{b} + 1]}{a[\cos t - \cos(\frac{at}{b})]^3} \\ = \frac{(\frac{a+b}{b}) [\cos(\frac{a-b}{b}t) - 1]}{a[\cos t - \cos(\frac{at}{b})]^3} \\ = \frac{(\frac{a+b}{b}) \cdot -2\sin^2(\frac{a-b}{2b}t)}{[\cos t - \cos(\frac{at}{b})]^3} \\ \Rightarrow \frac{d^2y}{dx^2} = -2(\frac{a+b}{ab}) \frac{\sin^2(\frac{a-b}{2b}t)}{[\cos t - \cos(\frac{at}{b})]^3} \\ \rho = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}} \\ \rho = \frac{[1 + [\frac{-\sin t + \sin(\frac{at}{b})}{\cos t - \cos(\frac{at}{b})}]^2]^{\frac{3}{2}}}{-2(\frac{a+b}{ab}) \frac{\sin^2(\frac{a-b}{2b}t)}{[\cos t - \cos(\frac{at}{b})]^3}} \\ = \frac{[\frac{(\cos t - \cos(\frac{at}{b}))^2 + (-\sin t + \sin(\frac{at}{b}))^2}{(\cos t - \cos(\frac{at}{b}))^2}]^{\frac{3}{2}}}{-2(\frac{a+b}{ab}) \frac{\sin^2(\frac{a-b}{2b}t)}{[\cos t - \cos(\frac{at}{b})]^3}} \\ = \frac{[\cos^2 t + \cos^2(\frac{at}{b}) + \sin^2 t + \sin^2(\frac{at}{b}) - 2\cos t \cos(\frac{at}{b}) - 2\sin t \sin(\frac{at}{b})]^{\frac{3}{2}}}{-2(\frac{a+b}{ab}) \sin^2(\frac{a-b}{2b}t)} \\ = \frac{[1+1-2\cos(\frac{at}{b}-t)]^{3/2}}{-2(\frac{a+b}{ab}) \sin^2(\frac{a-b}{2b}t)} \\ = \frac{2\sqrt{2}[1-\cos(\frac{a-b}{b}t)]^{\frac{3}{2}}}{-2(\frac{a+b}{ab}) \sin^2(\frac{a-b}{2b}t)} \\ = \frac{-\sqrt{2}[2\sin^2(\frac{a-b}{2b}t)]^{\frac{3}{2}}}{(\frac{a+b}{ab}) \sin^2(\frac{a-b}{2b}t)} \\ \Rightarrow \rho = \frac{4ab}{a+b} \sin(\frac{a-b}{2b}t) (Numerical Value)

Parametric and Cartesian Formula for Curvature-B.Sc. Math

With the above illustrations,one can understand the Parametric and Cartesian Formula for Curvature-B.Sc. Math.

3.Practice Problems of Parametric and Cartesian Formula for Curvature-B.Sc. Math for Students

Parametric and Cartesian Formula for Curvature-B.Sc. Math

Find the radius of curvature at any point t of the following curves
(1). x = a\cos^3 t, y = a\sin^3 t
(2). x = at^2, y = 2at
Answers: (1). \rho = 3p
(2). \rho = 2a(1+t^2)^{3/2}
By solving the above problems,you can understand the Parametric and Cartesian Formula for Curvature-B.Sc. Math well because the concept is well understood when you solve it practically.
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Also Read This Article:-  Curvature

4.Frequently Asked Questions Related to Parametric and Cartesian Formula for Curvature-B.Sc. Math

**छात्र-छात्राओं से आज का प्रश्न**

*”3,8,15,24,34,48,63 श्रेणी में एक संख्या बेमेल है।वह संख्या क्या है?”*
**Today’s Question to Students**
*”There is a number mismatch in the series 3,8,15,24,34,48,63. What is that number?”*
*पिछली प्रश्नोत्तरी का उत्तर*20
मान लीजिए कि क्लब में n व्यक्ति उपस्थित हैं तो प्रत्येक व्यक्ति (n-1) बार हाथ मिलाएगा,लेकिन A का B से हाथ मिलाना अथवा B का A से हाथ मिलाना एक ही बात है अतः
\frac{n(n-1)}{2} = 190 \\ \Rightarrow n^2 - n - 380 = 0 \\ \Rightarrow n^2 - 20n + 19n - 380 = 0 \\ \Rightarrow (n-20)(n+19) = 0 \\ \therefore n = 20, n = -19(अमान्य है)
*Previous Quiz Solution*
Suppose n persons are present in the club, then each person (n-1) shakes hands once, but A shaking hands with B or B shaking hands with A is the same thing.
\frac{n(n-1)}{2} = 190 \\ \Rightarrow n^2 - n - 380 = 0 \\ \Rightarrow n^2 - 20n + 19n - 380 = 0 \\ \Rightarrow (n-20)(n+19) = 0 \\ \therefore n = 20, n = -19 (impossible)
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*