Radius of Curvature
The radius of curvature measures how sharply a curve bends at a point.
1. If the curve is given as $y = f(x)$:
$$R = \frac{\left(1 + (y')^2\right)^{3/2}}{|y''|}$$
2. If the curve is given parametrically as $x(t)$ and $y(t)$:
$$R = \frac{\left(\dot{x}^2 + \dot{y}^2\right)^{3/2}}
{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}$$
3. If the curve is given in the form $x = g(y)$:
$$R = \frac{\left(1 + (x')^2\right)^{3/2}}{|x''|}$$
4. Curvature $\kappa$ (reciprocal of the radius of curvature):
$$\kappa = \frac{1}{R}$$
Radius of Curvature — CE Board Nov. 1999
Find the radius of curvature of the curve $x = y^3$ at the point $(1,\,1)$.
Show Solution
Differentiate $x = y^3$ implicitly with respect to $y$ (treat $x$ as a function of $y$):
$$\frac{dx}{dy} = 3y^2 \implies y' = \frac{dy}{dx} = \frac{1}{3y^2}$$
At $(1,1)$: $y' = \dfrac{1}{3}$.
$$y'' = \frac{d}{dx}\left(\frac{1}{3y^2}\right) = -\frac{2y \cdot y'}{3y^4} = -\frac{2y'}{3y^3}$$
At $(1,1)$: $y'' = -\dfrac{2(1/3)}{3} = -\dfrac{2}{9}$.
$$R = \frac{\left[1+(y')^2\right]^{3/2}}{|y''|} = \frac{\left[1+(1/3)^2\right]^{3/2}}{2/9} = \frac{(10/9)^{3/2}}{2/9}$$
$$R = \frac{9}{2}\left(\frac{10}{9}\right)^{3/2} \approx \mathbf{5.56\text{ units}}$$
Curvature of a Parabola
Find the curvature $K$ of the parabola $y^2 = 12x$ at the point $(3,\,6)$.
Show Solution
Differentiate $y^2 = 12x$ implicitly:
$$2y\,y' = 12 \implies y' = \frac{6}{y}$$
At $(3,6)$: $y' = 1$.
$$y'' = \frac{y(0) - 6\,y'}{y^2} = -\frac{6\,y'}{y^2}\bigg|_{(3,6)} = -\frac{6(1)}{36} = -\frac{1}{6}$$
$$K = \frac{|y''|}{\left[1+(y')^2\right]^{3/2}} = \frac{1/6}{[1+1]^{3/2}} = \frac{1/6}{2\sqrt{2}} = \frac{1}{12\sqrt{2}} = \frac{\sqrt{2}}{24}$$
Curvature: $K = \dfrac{\sqrt{2}}{24}$
Radius of Curvature at the Vertex of a Parabola
Find the radius of curvature of $y = x^2$ at its vertex $(0,\,0)$.
Show Solution
$$y' = 2x,\quad y'' = 2$$
At $(0,0)$: $y' = 0$, $y'' = 2$.
$$R = \frac{[1+(0)^2]^{3/2}}{|2|} = \frac{1}{2} = \mathbf{0.5\text{ units}}$$
Radius of Curvature of $y = \sin x$
Find the radius of curvature of $y = \sin x$ at $x = \pi/2$.
Show Solution
$$y' = \cos x,\quad y'' = -\sin x$$
At $x = \pi/2$: $y' = 0$, $y'' = -1$.
$$R = \frac{[1+0^2]^{3/2}}{|-1|} = \frac{1}{1} = \mathbf{1\text{ unit}}$$
Radius of Curvature of an Ellipse at a Vertex
Find the radius of curvature of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ at the end of the major axis $(4,\,0)$.
Show Solution
Differentiate implicitly:
$$\frac{x}{8} + \frac{2y}{9}y' = 0 \implies y' = -\frac{9x}{16y}$$
At $(4,0)$, $y' \to \infty$, so use $x = g(y)$ form. Near $(4,0)$: differentiate $\frac{x^2}{16} + \frac{y^2}{9} = 1$ with respect to $y$:
$$\frac{x}{8}\,x' + \frac{2y}{9} = 0 \implies x' = -\frac{16y}{9x},\quad x''\big|_{y=0} = -\frac{16}{9 \cdot 4} = -\frac{4}{9}$$
At $(4,0)$: $x'=0$, so:
$$R = \frac{[1+(x')^2]^{3/2}}{|x''|} = \frac{1}{4/9} = \frac{9}{4} = \mathbf{2.25\text{ units}}$$
Note: $R = b^2/a = 9/4$ is the standard formula for ellipse vertex radius of curvature.
Exam Generator Problems
Additional board-style practice items for this topic.
Question Bank: q437
MSTE - Differential Calculus / Radius of Curvature / Engr. Janclyde Espinosa (Clidez)
Find the radius of curvature of the curve y=x4 -x2 at point (0,0).
Answer:
1/2
1/3
1/4
3/4
Show Solution
For $y=x^4-x^2$:
Question Bank: q461
MSTE - Differential Calculus / Radius of Curvature / Engr. Janclyde Espinosa (Clidez)
What is the radius of curvature at point (1,2) of the curve 4x-y2 =0?
Answer:
5.66
6.67
3.33
4.33
Show Solution
From $4x-y^2=0$, write $y=2\sqrt{x}$. At $(1,2)$:
Question Bank: t792
MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6
Find the radius of curvature of the curve y = x^2 e^x at x = 1.
29.15
32.65
56.33
45.87
Show Solution
$y = x^2 e^x$: $y' = e^x(x^2 + 2x)$, $y'' = e^x(x^2 + 4x + 2)$.
Question Bank: t793
MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6
Find the radius of curvature of the curve y = x^3 - 2x^2 + 5x - 3 at x = 3.
452.32
573.57
653.32
278.86
Show Solution
$y = x^3 - 2x^2 + 5x - 3$: $y' = 3x^2 - 4x + 5$, $y'' = 6x - 4$.
Question Bank: t1476
MSTE - Differential Calculus / Radius of Curvature / BEMz
Find the curvature of the parabola $y^{2}=12x$ at (3,6).
$- \sqrt{2}/24$
$\sqrt{2}/8$
$3\sqrt{2}$
$8\sqrt{2}/3$
Solution pending in psadquestions/t1476.json.
Question Bank: t1477
MSTE - Differential Calculus / Radius of Curvature / BEMz
Locate the center of curvature of the parabola $x^{2}=4y$ at point (2,2).
(-2,6)
(-3,6)
(-2,4)
(-3,7)
Solution pending in psadquestions/t1477.json.
Question Bank: t1478
MSTE - Differential Calculus / Radius of Curvature / BEMz
Compute the radius of curvature of the parabola $x^{2}=4y$ at the point (4,4).
22.36
24.94
20.38
18.42
Solution pending in psadquestions/t1478.json.
Question Bank: t1479
MSTE - Differential Calculus / Radius of Curvature / BEMz
Find the radius of curvature of the curve $y=2x^{3}+3x^{2}$ at (1,5).
97
90
101
87
Solution pending in psadquestions/t1479.json.
Question Bank: t1480
MSTE - Differential Calculus / Radius of Curvature / BEMz
Compute the radius of curvature of the curve $x=2y^{3}-3y^{2}$ at (4,2).
-97.15
-99.38
-95.11
-84.62
Solution pending in psadquestions/t1480.json.
Question Bank: t1481
MSTE - Differential Calculus / Radius of Curvature / BEMz
Find the radius of curvature of a parabola $y^{2}-4x=0$ at point (4,4).
22.36
25.78
20.33
15.42
Solution pending in psadquestions/t1481.json.
Question Bank: t1482
MSTE - Differential Calculus / Radius of Curvature / BEMz
Find the radius of curvature of the curve $x=y^{3}$ at point (1,1).
-1.76
-1.24
2.19
2.89
Solution pending in psadquestions/t1482.json.
