Radius of Curvature at Origin

Q: Q: 3. How do you find the radius of curvature in polar coordinates?

Ans: In case of polar co-ordinates, we have At the origin \rho = \underset{x \to 0}{\lim} \frac{x^2}{2y} = \underset{x \to 0}{\lim} \frac{r^2 \cos^2 \theta}{2r \sin \theta} = \underset{\theta \to 0}{\lim} \frac{r^2 \cdot 1}{2r \cdot \theta} = \lim_{\theta \to 0} \left(\frac{r}{2\theta}\right) By answering the above questions,you can know about the primary terms of Radius of Curvature at Origin.

Sanjay Kumawat October 17, 2026 Differential Calculus(English) No Comments

Contents hide

1 1.Radius of Curvature at Origin

2 2.Radius of Curvature at Origin Illustrations

2.1 3.Practice Questions of Radius of Curvature at Origin for Students

2.2 4.Frequently Asked Questions Related to Radius of Curvature at Origin

2.3 Q:1.How do you find the radius of curvature at the origin?

2.4 Q:2.How do you find the radius of curvature at a point (h,k)?

2.5 Q: 3. How do you find the radius of curvature in polar coordinates?

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

1.Radius of Curvature at Origin

Radius of Curvature at Origin

About Radius of Curvature at Origin we will find out radius of curvature by method of expansion and Newton’s method.

Also Read This Article :- Radius of curvature

2.Radius of Curvature at Origin Illustrations

Illustration: 1. Find the radius of curvature at the origin on the following curves
Illustration: 1(a). $5x^3 + 7y^3 + 4x^2y + xy^2 + 2x^2 + 3xy + y^2 + 4x = 0$
Solution: $5x^3 + 7y^3 + 4x^2y + xy^2 + 2x^2 + 3xy + y^2 + 4x = 0$
The curve passes through (0,0) and the tangent there at $x = 0$ i.e. $y$ -axis (obtained by equating the lowest degree term in the equation to zero).
$\therefore \rho \text{ at the origin} = \underset{x \to 0}{\lim} \left(\frac{y^2}{2x}\right)$
Now the given curve is
$5x^3 + 7y^3 + 4x^2y + xy^2 + 2x^2 + 3xy + y^2 + 4x = 0$
Dividing each term by $2x$ , we get
$\frac{5}{2}x^2 + 7y\left(\frac{y^2}{2x}\right) + 2xy + \frac{1}{2}y^2 + x + \left(\frac{3}{2}y\right) + \frac{y^2}{2x} + 2 = 0$
Taking limits as $x \to 0$ and $y \to 0$ , we get
$0 + 0 + 0 + 0 + 0 + 0 + \underset{x \to 0}{\lim} \left(\frac{y^2}{2x}\right) + 2 = 0$
$\Rightarrow \rho + 2 = 0$
$\Rightarrow \rho = -2$
$\Rightarrow \rho = 2$ (numerically)
Illustration: 1(b). $2x^4 + 3y^4 + 4x^2y + xy - y^2 + 2x = 0$
Solution: $2x^4 + 3y^4 + 4x^2y + xy - y^2 + 2x = 0$
The curve passes through latex[/latex] and the tangent at latex[/latex] is $x = 0$ i.e. $y$ -axis (obtained by equating the lowest degree term in the given equation to zero).
$\therefore \rho \text{ at the origin} = \underset{x \to 0}{\lim} \left(\frac{y^2}{2x}\right)$
Now the given curve is
$2x^4 + 3y^4 + 4x^2y + xy - y^2 + 2x = 0$
Dividing each term by $2x$ , we get
$x^3 + 3y^2\left(\frac{y^2}{2x}\right) + 2xy + \frac{y}{2} - \frac{y^2}{2x} + 1 = 0$
Taking limits as $x \to 0$ and $y \to 0$ , we get
$0 + 0 + 0 + 0 - \underset{x \to 0}{\lim} \left(\frac{y^2}{2x}\right) + 1 = 0$
$\Rightarrow -\rho + 1 = 0 \Rightarrow \rho = 1$
$\Rightarrow \rho = 1$ (numerically)
Illustration: 1(c). $y^2(a - x) = x^2(a + x)$
Solution: $y^2(a - x) = x^2(a + x)$
$\Rightarrow y = \pm x\sqrt{\left(\frac{a + x}{a - x}\right)}$
$= \pm x\left(1 + \frac{x}{a}\right)^{\frac{1}{2}} \left(1 - \frac{x}{a}\right)^{-\frac{1}{2}}$
$\Rightarrow y = \pm x\left(1 + \frac{x}{2a} + \dots\right)\left(1 + \frac{x}{2a} + \dots\right)$
$y = \pm x\left(1 + \frac{x}{a} + \dots\right)$
$= \pm \left(x + \frac{x^2}{a} + \dots\right)$
which is of the form $y = px + \frac{1}{2}qx^2 + \dots$
$\Rightarrow p = 1, \ q = \frac{2}{a}$ or $p = -1, \ q = -\frac{2}{a}$
$\therefore \rho \text{ at the origin} = \frac{(1 + p^2)^{\frac{3}{2}}}{q}$
$= \frac{(1 + 1)^{\frac{3}{2}}}{\frac{2}{a}} = a\sqrt{2}, \text{ if } p = 1, q = \frac{2}{a}$
$= \frac{(1 + 1)^{\frac{3}{2}}}{-\frac{2}{a}} = -a\sqrt{2}, \text{ if } p = -1, q = -\frac{2}{a}$
$\therefore \rho \text{ at the origin} = \pm a\sqrt{2}$
Illustration: 1(d). $y = x^3 + 5x^2 + 6x$
Solution: The curve is $y = x^3 + 5x^2 + 6x$
$\frac{dy}{dx} = 3x^2 + 10x + 6$ and $\frac{d^2y}{dx^2} = 6x + 10$
At the origin i.e. at $x = 0, y = 0$
$\frac{dy}{dx} = 6$ and $\frac{d^2y}{dx^2} = 10$
$\therefore \rho \text{ at the origin} = \frac{\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}}{\frac{d^2y}{dx^2}}$
$= \frac{\left[1 + 6^2\right]^{3/2}}{10} = \frac{(1 + 36)^{3/2}}{10}$
$\Rightarrow \rho = \frac{(37)^{3/2}}{10} = \frac{37\sqrt{37}}{10}$
Illustration: 1(e). $y = x^4 - 4x^3 - 18x^2$
Solution: $y = x^4 - 4x^3 - 18x^2$
Equating to zero the lowest degree terms in $y$ , we find that the tangent at the origin is $y = 0$ i.e., the axis of $x$ .
Hence by Newton’s formula radius of curvature at (0,0) is given by
$\rho = \underset{x \to 0}{\lim} \left(\frac{x^2}{2y}\right)$
Now dividing both sides of the given equation by $2y$ , we have
$x^2\left(\frac{x^2}{2y}\right) - 4x\left(\frac{x^2}{2y}\right) - 18\left(\frac{x^2}{2y}\right) = \frac{1}{2}$
$\Rightarrow$ Taking limits of this as $x \to 0$ , we get
$\Rightarrow 0 - 4(0) - 18 \underset{x \to 0}{\lim} \frac{x^2}{2y} = \frac{1}{2}$
$\Rightarrow -18\rho = \frac{1}{2}$
$\Rightarrow \rho = -\frac{1}{36}$
$\Rightarrow \rho = \frac{1}{36}$ (Numerically)
Illustration: 1(f). $y^2 - 3xy - 4x^2 + x^3 + x^4y + y^5 = 0$
Solution: The curve passes through the origin but equating lowest degree terms in the equation to zero, we get $y^2 - 3xy - 4x^2 = 0$ which does not give any of the coordinate axes as the tangent at (0,0). Hence Newtonian Method cannot be applied to this question. Also $y$ cannot be easily expressed in terms of $x$ .
$\therefore$ Suppose that the given equation can be expressed in the form $y = px + \frac{qx^2}{2} + \dots$ and putting $px + \frac{qx^2}{2} + \dots$ for $y$ in the given equation, we get
$\left(px + \frac{qx^2}{2} + \dots\right)^2 - 3x \left(px + \frac{qx^2}{2} + \dots\right) - 4x^2 + x^3 + \dots = 0$
Equating the coefficients of $x^2$ and $x^3$ to zero, we get
$p^2 - 3p - 4 = 0 \quad \text{--- (1)}$
and $pq - \frac{3}{2}q + 1 = 0 \quad \text{--- (2)}$
$\therefore$ From (1) we get $p = -1, 4$
and from (2.) $q = \frac{2}{5}, -\frac{2}{5}$
Also at the origin
$\rho = \frac{(1 + p^2)^{\frac{3}{2}}}{q}$
When $p = -1, q = \frac{2}{5}$
$\rho = \frac{(1 + 1)^{\frac{3}{2}}}{-\frac{2}{5}} = 5\sqrt{2}$
and when $p = 4, q = \frac{2}{5}$
$\rho = \frac{(1 + 16)^{\frac{3}{2}}}{\frac{2}{5}} = \frac{85\sqrt{17}}{2}$ (numerically)

Radius of Curvature at Origin

Illustration: 2. Show that the radius of curvature at the origin of the curve $a_1x + a_2y + b_1x^2 + b_2xy + b_3y^2 + c_1x^3 \dots = 0$ is
$\frac{(a_1^2 + a_2^2)^{\frac{3}{2}}}{2(b_1a_2^2 - b_2a_1a_2 + b_3a_1^2)}$
Solution: The curve passes through the origin but equating the degree terms in the equation to zero we find that none of the co-ordinate axes is a tangent to the curve at the origin. Moreover the given equation cannot be directly expressed in the form
$y = px + \frac{qx^2}{2!} + \dots$
$\therefore$ Suppose that the given equation can be expressed as
$y = px + q\frac{x^2}{2!} + \dots$
and substituting $px + \frac{qx^2}{2!} + \dots$ for $y$ in the given equation we have the identity
$a_1x + a_2\left(px + \frac{1}{2}qx^2 + \dots\right) + b_1x^2 + b_2x\left(px + \frac{1}{2}qx^2 + \dots\right) + b_3\left(px + \frac{1}{2}qx^2 + \dots\right)^2 + c_1x^3 + \dots = 0$
$\Rightarrow x(a_1 + a_2p) + x^2\left(\frac{1}{2}a_2q + b_1 + b_2p + b_3p^2\right) + \dots = 0$
Equating the coefficients of $x$ and $x^2$ to zero, we get
$a_1 + a_2p = 0 \quad \text{--- (1)}$
and $\frac{1}{2}a_2q + b_1 + b_2p + b_3p^2 = 0 \quad \text{--- (2)}$
From (1) we get $p = -\frac{a_1}{a_2}$
From (2), $q = \left(-\frac{2}{a_2}\right)\left[b_1 + b_2\left(-\frac{a_1}{a_2}\right) + b_3\left(-\frac{a_1}{a_2}\right)^2\right]$
$= \frac{2}{a_2^3}\left(a_1b_2a_2 - b_1a_2^2 - a_1^2b_3\right)$
$\therefore \rho \text{ at the origin} = \frac{(1 + p^2)^{\frac{3}{2}}}{q}$
$= \frac{\left(1 + \frac{a_1^2}{a_2^2}\right)^{\frac{3}{2}}}{\left(\frac{2}{a_2^3}\right)\left(a_1a_2b_2 - a_2^2b_1 - a_1^2b_3\right)}$
$= \frac{(a_1^2 + a_2^2)^{\frac{3}{2}}}{2(a_1a_2b_2 - a_2^2b_1 - a_1^2b_3)}$
Illustration: 3. Find the radius of curvature at the origin for the curve
$x = t - \frac{1}{3}t^3, \ y = t^2$
Solution: At the origin $x = 0, y = 0$ and therefore from the given equations at origin $t = 0$
Also $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{1 - t^2}$
$\therefore \left(\frac{dy}{dx}\right) \text{ at the origin} = \left(\frac{2t}{1 - t^2}\right)_{\text{at } t = 0} = 0$
i.e. $x$ -axis is the tangent to the curve at (0,0).
$\therefore \rho \text{ at the origin} = \underset{x \to 0}{\lim} \left(\frac{x^2}{2y}\right)$
$= \underset{t \to 0}{\lim} \left[\frac{\left(t - \frac{1}{3}t^3\right)^2}{2t^2}\right]$
$\because t \to 0 \text{ as } x \to 0$
$\Rightarrow \rho \text{ at the origin} = \underset{t \to 0}{\lim} \left[\frac{1}{2}\left(1 - \frac{1}{3}t^2\right)^2\right]$
$\rho = \frac{1}{2}$
Illustration: 4. Find the radius of curvature at the pole for the following curves:
Illustration: 4(a). $r = a\theta$
Solution: The curve is $r = a\theta$
$\therefore \frac{dr}{d\theta} = a$
$\therefore \tan\phi = r\frac{d\theta}{dr} = \frac{r}{a}$
$= \frac{a\theta}{a} = \theta$
$\therefore$ At the pole i.e. $r = 0, \theta = 0$
we have $\tan\phi = 0$
i.e. $\phi = 0$
i.e. initial line is a tangent to the curve at the pole
$\therefore \rho \text{ at the pole} = \underset{\theta \to 0}{\lim} \left(\frac{r}{2\theta}\right)$
$= \underset{\theta \to 0}{\lim} \left(\frac{a\theta}{2\theta}\right)$
$= \underset{\theta \to 0}{\lim} \left(\frac{a}{2}\right)$
$\Rightarrow \rho = \frac{a}{2}$
Illustration: 4(b). $r = a\sin n\theta$
Solution: The curve is $r = a\sin n\theta$
$\therefore \frac{dr}{d\theta} = an\cos n\theta$
$\tan\phi = r\frac{d\theta}{dr}$
$= r \cdot \frac{1}{an\cos n\theta}$
$= \frac{a\sin n\theta}{an\cos n\theta}$
$= \frac{1}{n}\tan n\theta$
$\therefore$ At the pole i.e. $r = 0, \theta = 0$ ,
we have $\tan\phi = 0$ i.e. $\phi = 0$
i.e. initial line is a tangent to the curve at the pole
$\therefore \rho \text{ at the pole} = \underset{\theta \to 0}{\lim} \left(\frac{r}{2\theta}\right)$
$= \underset{\theta \to 0}{\lim} \left(\frac{a\sin n\theta}{2\theta}\right) \quad \left[\frac{0}{0} \text{ form}\right]$
$= \underset{\theta \to 0}{\lim} \frac{an\cos n\theta}{2} \quad [\text{By D-L'Hospital Rule}]$
$\rho = \frac{an}{2}$
Illustration: 5. Show that the radius of curvature at the origin for the curve $x^3 + y^3 = 3axy$ are each equal to $\frac{3a}{2}$ .
Solution: The curve passes through latex[/latex] and the tangents there at $x = 0$ and $y = 0$ (obtained by equating the lowest degree term in the equation to zero).
$\rho$ at the origin
$= \underset{x \to 0}{\lim} \left(\frac{y^2}{2x}\right) = \rho_1$ (say)
and $\underset{x \to 0}{\lim} \left(\frac{x^2}{2y}\right) = \rho_2$ (say)
Now the given curve is
$x^3 + y^3 = 3axy$
Dividing each term by $2xy$ , we get $\frac{x^2}{2y} + \frac{y^2}{2x} = \frac{3a}{2}$
Taking limits as $x \to 0$ and $y \to 0$ , we get
$\lim_{y \to 0} \left(\frac{x^2}{2y}\right) = \frac{3a}{2}$ or $\rho_2 = \frac{3a}{2}$
Similarly we can prove that
$\rho_1 = \frac{3a}{2}$
The required complete solution is given by
By solving the above illutrations,one can understand the Radius of Curvature at Origin.

Also Read This Article :- Radius of Curvature for Polar Curves

3.Practice Questions of Radius of Curvature at Origin for Students

Radius of Curvature at Origin

(1.) Find the radii of curvature at the origin for:
(1.) $y^2(y - 1) = 4x^2$
(2.) $x^3 + y^2 = (y - x)(y - 2x)$
By solving the above questions,you can understand the Radius of Curvature at Origin well because the concept is well understood when you solve it practically.
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Also Read This Article:- Formula for Curvature-B.Sc. Math

4.Frequently Asked Questions Related to Radius of Curvature at Origin

Q:1.How do you find the radius of curvature at the origin?

Ans:The radius of curvature at the origin for a curve y=f(x) can be found by putting x=0,y=0 in the value of $\rho$ obtained by using the formula already established.

Q:2.How do you find the radius of curvature at a point (h,k)?

Ans:The radius of curvature at any point (h,k) can be also be found by shifting the origin to the point (h,k) by writing x=x+h,y=y+k and by applying the $\rho = \underset{x \to 0}{\lim} \frac{y^2}{2x}$ result.

Q: 3. How do you find the radius of curvature in polar coordinates?

Ans: In case of polar co-ordinates, we have
At the origin $\rho = \underset{x \to 0}{\lim} \frac{x^2}{2y}$
$= \underset{x \to 0}{\lim} \frac{r^2 \cos^2 \theta}{2r \sin \theta}$
$= \underset{\theta \to 0}{\lim} \frac{r^2 \cdot 1}{2r \cdot \theta} = \lim_{\theta \to 0} \left(\frac{r}{2\theta}\right)$
By answering the above questions,you can know about the primary terms of Radius of Curvature at Origin.

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

Radius of Curvature at Origin

🎯 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!👇
किसी सांकेतिक भाषा में ‘DECEMBER’ को ‘RDEEBCME’ के तौर पर लिखा जाता है।उसी संकेत में ‘FEBRUARY’ कैसे लिखा जाएगा?
(In some sign language, ‘DECEMBER’ is written as ‘RDEEBCME’. How will ‘FEBRUARY’ be written in the same sign?):
*पिछली प्रश्नोत्तरी का हल*
$\sqrt{900} + \sqrt{0.09} - \sqrt{0.000009}$
$= 30 + \frac{3}{10} - \frac{3}{1000}$
$= 30 + 0.3 - 0.003$
$= 30.297$
*Previous Quiz Solution*
$\sqrt{900} + \sqrt{0.09} - \sqrt{0.000009}$
$= 30 + \frac{3}{10} - \frac{3}{1000}$
$= 30 + 0.3 - 0.003$
$= 30.297$
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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