On the Curvature Function: Where does a curve bend the fastest?
On the Curvature Function:
Where does a curve bend the fastest?
Loucas Chrysafi
Department of Mathematics, Farmingdale State University of New York
e-mail: chrysala@farmingdale.edu
Sheldon Gordon
Department of Mathematics, Farmingdale State University of New York
e-mail: gordonsp@farmingdale.edu
1
On the Curvature Function: Where does a curve bend the fastest?
Abstract
We examine the behavior of the curvature function associated with most common
families of functions and curves, with the focus on establishing where maximum
curvature occurs.
1
Introduction
The notion of the curvature of a function appears in all the traditional calculus texts.
However, it is typically introduced primarily as a vehicle for providing additional
manipulation practice. As a result, the books miss some lovely mathematical ideas. The
curvature of a function y  f ( x) is defined according to the formula
k (x ) =
f ¢¢(x )
3
[1 + [f ¢(x )]2 ]2
or some other form (see [3] or [7] ), but almost all examples and problems are basically
3
algebraic exercises in which the term [1 + [f ¢(x )]2 ]2 simplifies nicely. The two standard
functions used are f ( x ) 
2
3
x 3/ 2 and f ( x)  ln(cos x) . Few of the texts actually examine
how the curvature function of a particular function is related to the behavior of the
function, nor any of the other fascinating mathematical issues that can arise. As a result,
many mathematicians themselves have never thought deeply about these issues, nor have
they used them to unify some important ideas in calculus courses.
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On the Curvature Function: Where does a curve bend the fastest?
2
Curvature
Definition Let f be a twice-differentiable function. The curvature k of f at x is
defined by
k (x ) =
f ¢¢(x )
3
.
[1 + [f ¢(x )]2 ]2
Another way of thinking about curvature is to use the fact that at any point on a curve,
where the curvature is nonzero, we can construct a circle tangent to the curve having the
same curvature at that point. This is called the osculating circle and its radius is called the
radius of curvature r . Thus,
Definition The radius of curvature r at the point x = a on the curve y  f ( x) , is
3
[1 + [f ¢(a )]2 ]2
1
r (a ) =
=
.
f ¢¢(a )
k (a )
For more details see [2].
We now examine what this means for some of the standard families of functions
encountered in the first few years of the undergraduate mathematics curriculum. Many of
the results are quite surprising and run counter to what we might intuitively expect.
Linear Functions
If f (x ) = mx + b , then clearly the curvature k is zero everywhere.
Quadratic Functions
2
If f (x ) = ax + bx + c , then
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On the Curvature Function: Where does a curve bend the fastest?
2a
k (x ) =
3
[1 + (2ax + b)2 ]2
Figure 1 shows the parabola f (x ) = x 2 along with its associated curvature function
k (x ) =
2
3
. From the graphs, it is evident that the curvature function has a
[1 + 4x 2 ]2
maximum at (or very near to) the vertex of the parabola. In particular, from the
expression for k (x ) , we see that the maximum curvature occurs at x  0 and has value 2.
In fact, for an arbitrary quadratic function y = ax 2 + bx + c , we locate the maximum
curvature by finding where the derivative k ( x)  0 , thus
1
- 12a a (2ax + b)[1 + (2ax + b)2 ]2
k ¢(x ) =
[1 + (2ax + b)2 ]3
=
- 12a a (2ax + b)
5
2 2
[1 + (2ax + b) ]
4
.
On the Curvature Function: Where does a curve bend the fastest?
Therefore k ¢(x ) = 0 when 2ax + b = 0 , so that x =
-b
. Thus, by the First Derivative
2a
Test, the curvature function has a local maximum at the vertex of the parabola as
expected. The maximum value is
2a
k (- b / 2a ) =
3
= 2a .
[1 + (2a(- b / 2a ) + b)2 ]2
Therefore, the larger the leading coefficient the greater the maximum curvature.
Interestingly, the other parameters have no effect on the curvature. Moreover, for any
parabola we note that the curvature function k (x ) ® 0 as x ® ± ¥ , as one might
expect (a good student exercise). In turn, this means that the radius of curvature increases
as x increases. Figure 2 shows the parabola f (x ) = x 2 along with the osculating circles
at x = 0 and x = 1 . The locus of centers of all the osculating circles is called the evolute
to the curve. For more details, see [2].
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On the Curvature Function: Where does a curve bend the fastest?
Power Functions
If f (x ) = x n , then
k (x ) =
n (n - 1)x n - 2
2
[1 + n x
2n - 2
3
2
.
]
Example 1: n  4.
Figure 3 shows f (x ) = x 4 along with its associated curvature function. The curvature
function has two maxima at a pair of symmetrically located points. It also has a minimum
at x  0 . Let’s see why. Clearly as x   , the curve bends less and less and so
the curvature approaches zero (another good student exercise). Also, we know that higher
even degree functions are increasingly flat near the origin. Consequently, it makes sense
that the curvature becomes maximal at two points on either side of the origin. In
particular, it turns out (we do the derivation in general later) that the abscissa of these
1
 1 6
points are x     .
 56 
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On the Curvature Function: Where does a curve bend the fastest?
Example 2: n  3.
Now let’s consider the cubic function f ( x)  x3 . Its graph along with its associated
curvature function are shown in Figure 4. Maximum curvature also seems to occur at two
points that are symmetrically located about the origin, and minimum curvature seems to
occur at the origin. The latter makes sense because the curve has a point of inflection at
the origin where f ¢¢(x ) = 0 . Thus k (x ) = 0 is a minimum there.
We next consider the general case, f (x ) = x n . We have
3
k ¢(x ) =
=
1
3
n (n - 1)x n - 2[1 + n 2x 2n - 2 ]2 n 2 (2n - 2)x 2n - 3
2
[1 + n 2x 2n - 2 ]3
n (n - 1)(n - 2)x n - 3[1 + n 2x 2n - 2 ]2 -
n (n - 1)x n - 3 [n - 2 - 2n 3x 2n - 2 + n 2x 2n - 2 ]
5
[1 + n 2x 2n - 2 ]2
The critical points for k occur where
7
.
On the Curvature Function: Where does a curve bend the fastest?
n (n - 1)x n - 3[n - 2 - 2n 3x 2n - 2 + n 2x 2n - 2 ] = 0
so that
n = 0 or n = 1 or x = 0 or x 2n - 2 =
n- 2
.
2n 3 - n 2
The cases n = 0 and n = 1 are trivial- the curve is linear and so has zero curvature. If
n = 2 , then x = 0 is the only critical point for the curvature function, which was proved
earlier. If n > 2 , then x = 0 clearly represents a point where the curvature is zero, so it
is the minimum value. The maximum curvature therefore occurs at
1
2n - 2
æ n- 2 ö
÷
x = ± çç 3
÷
è2n - n 2 ø
(1)
provided that 2n 3 - n 2 ¹ 0 or n ¹ 0 , 1/ 2 , and that 2n - 2 ¹ 0 or n ¹ 1 . Notice that,
as we observed for n = 3 and n = 4 , this always happens at two points located
symmetrically about the origin. In particular, for f (x ) = x 3 , the maximum curvature
1
1
 1 4
 1 6
occurs at x     and for f (x ) = x 4 , the maximum curvature occurs at x     .
 45 
 56 
As n gets larger, the critical points get closer and closer to ± 1 . Since the radicand in (1)
is negative in the interval 1/ 2 < n < 2 , the behavior of the critical points as a function
of n in that interval will be very interesting or not defined. We leave it to the reader for
further investigation. Figure 5 shows graphically the behavior of these critical points as a
function of n .
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On the Curvature Function: Where does a curve bend the fastest?
Exponential Functions
If f (x ) = a x (a > 1) , then we have
(ln a )2 a x
k (x ) =
2x
2
3
2
.
[1 + a (ln a ) ]
To find the critical points for k (x ) , we compute
3
1
(ln a )2[a x ln a(1 + a 2x (ln a )2 ) 2 - 3(ln a )3a 3x (1 + a 2x (ln a )2 ) 2 ]
k ¢(x ) =
[1 + a 2x (ln a )2 ]3
=
(ln a )2 [a x ln a(1 + a 2x (ln a )2 ) - 3a 3x (ln a ) 3 ]
5
[1 + a 2x (ln a )2 ]2
=
a x (ln a )3 [1 + a 2x (ln a )2 - 3a 2x (ln a )2 ]
5
[1 + a 2x (ln a )2 ]2
=
a x (ln a )3 [1 - 2a 2x (ln a )2 ]
5
.
[1 + a 2x (ln a )2 ]2
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On the Curvature Function: Where does a curve bend the fastest?
Since the denominator is positive, k ¢(x ) = 0 when 1 - 2a 2x (ln a )2 = 0 , so that
a 2x = [2(ln a )2 ]- 1 . Hence, the only zero of k ¢(x ) occurs at
x=
- ln[2(ln a )2 ]
.
2 ln a
In the special case of a = e (see Figure 6), the maximum curvature occurs at x = -
ln 2
2
and is equal to
k max =
e
-
ln 2
2
[1 + e
3
- ln 2
2
]
=
2
3
.
32
Thus, while we often tend to think of an exponential function as being “centered” at its
vertical intercept, from the point of view of curvature, the function is centered at a point
to the left of the vertical axis. Figure 7 shows e x with the circle of curvature at
x= -
ln 2
.
2
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On the Curvature Function: Where does a curve bend the fastest?
Problem 1: Show that the inflection points on the curvature function associated with the
exponential function e x occur at
x=
1 éê5 ± 21 ù
ú.
ln
ú
2 êë 4
û
Natural Logarithmic Function
For f (x ) = ln x , we have
1
k (x ) =
3
é
1 ù2
x ê1 + 2 ú
êë x ú
û
2
so that
3
1
é
1 ù2 3 é
1 ù2
é
1ù 3
2x ê1 + 2 ú + x 2 ê1 + 2 ú x - 3
2x ê1 + 2 úêë x ú
êë x ú
2 êë x ú
û
û
û x .
k ¢(x ) = = 5
3
1ù
4 é
2
é
1
ù
x ê1 + 2 ú
x 4 ê1 + 2 ú
ú
ëê x û
êë x ú
û
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On the Curvature Function: Where does a curve bend the fastest?
Since the denominator is never zero and x > 0 for the logarithmic function, the only
critical point occurs where
2x +
2 3
1
2
or x =
,
= 0 or 2x =
x x
x
2
as seen in Figure 8. Therefore, the maximum curvature is
1
k ( 2 / 2) =
( 2 / 2) éëê1 + (2 /
2
3
2 2
2) ù
ú
û
=
2
3
3
2
,
which is the same value as the maximum curvature of e x . In retrospect, one might have
expected this considering the fact that the two functions are inverses of one another.
Problem 2: Show that the maximum curvature of f (x ) = loga x occurs at x =
1
.
2(ln a )2
Trigonometric Functions
Intuitively for the sine and cosine functions, we expect the maximum curvature to occur
at the turning points. Also, the minimum curvature should occur at the inflection points.
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On the Curvature Function: Where does a curve bend the fastest?
We will prove this for the sine function. (the comparable fact obviously holds for the
cosine.) We have
sin x
k (x ) =
3
[1 + cos2 x ]2
so that
3
1
3
± cos x [1 + cos2 x ]2 ± sin(x ) [1 + cos2 x ]2 2 cos x (- sin x )
2
k ¢(x ) =
[1 + cos2 x ]3
3
1
± cos x [1 + cos2 x ]2 ± 3 sin 2 x cos x [1 + cos2 x ]2
=
[1 + cos2 x ]3
1
± cos x [1 + cos2 x ]2 [1 + cos2 x + 3 sin 2 x ]
=
.
[1 + cos2 x ]3
This expression is zero when cos x = 0 (all the other terms are positive), so that
x = ± (2k + 1)
p
,
2
k  0,1, 2
Incidentally, k max = 1 . Figure 9 shows the sine curve together with its associated
curvature function from 0 to 2p .
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On the Curvature Function: Where does a curve bend the fastest?
p
p
p
Since k max = 1 and sin( ) = 1 , the circle of curvature at x =
is centered at ( , 0)
2
2
2
and has radius one. Figure 10 shows the circles of curvature of the sine function at
x =
p
p
and x = .
2
4
Normal Distributions
Let us consider the family of normal distributions with mean m = 0 and standard
deviation s ,
2
x
1
N (x , s ) =
e 2s 2 .
s 2p
The curvature of this family is given by
1
k (x , s ) =
s 3 2p
e
-
x2
2s 2
éx 2
ù
ê 2 - 1ú
êës
úû
.
3
2
x2
é
- 2 ù2
x
ê1 +
e s ú
ê 2s 6 p
ú
ë
û
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On the Curvature Function: Where does a curve bend the fastest?
Figure 11 shows N (x ,1) along with its associated curvature function. As expected, the
maximum curvature occurs at the turning point x = 0 , and is equal to
k (0, s ) =
1
s
3
2p
Therefore, as s gets larger, the curvature at x = 0 approaches zero, and as s approaches
zero, the curvature at x = 0 goes to infinity. Even though the curvature of N (x , s ) will
always have three local maximum points, the global maximum curvature occurs at x = 0
regardless of the value of s . Proving it, however easy it may sound, is a rather messy
proposition without the use of a computer algebra system (CAS) . We leave it for the
curious and interested reader.
Note that k (0,1) = N (0,1) =
centered at (0,
1
2p
1
. This implies that the circle of curvature at x = 0 is
2p
2p ) and has radius r =
15
2p , a fact that is captured in Figure 12.
On the Curvature Function: Where does a curve bend the fastest?
Problem 3: Locate the other two local maximum points of the curvature of the normal
curve N (x ,1) . Relate these two points to features of the normal distribution.
Logistic Functions
Consider the family of logistic curves given by
f (t , L , c, l ) =
L
.
1 + ce - l t
(2)
Example 3: L  1 , c  2 ,   3
Figure 13 shows f (t ,1, 2, 3) with its associated curvature function. We observe that the
maximum curvature occurs at two points, which seem to be symmetrically located about
the inflection point of the logistic curve. The minimum curvature occurs at the point of
inflection at x =
ln 2
. We suggest that the reader investigate the effect of the parameters
3
L , c , and l on the curvature of (2). This might also make a very effective student
research project.
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On the Curvature Function: Where does a curve bend the fastest?
Example 4: Hyperbolic Cosine
First consider the hyperbolic cosine function. Interestingly enough, the curvature function
associated with cosh x is k (x ) = sech 2 x =
1
, and the curvature achieves its
cosh 2 x
maximum at x = 0 , where k (0) = 1 (see Figure 14).
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On the Curvature Function: Where does a curve bend the fastest?
x
Problem 4: Show that the maximum curvature of the catenary f (x ) = a cosh( ) occurs
a
at the point (0, a ) , and that k max =
1
.
a
Example 5: Hyperbolic Sine
Consider the hyperbolic sine. Its associated curvature function is k (x ) =
sinh x
3
[1 + cosh 2 x ]2
as shown in Figure 15.
Notice that there are two symmetrically located points of maximum curvature.
Problem 5: Show that the maximum curvature of sinh x occurs at x = ln( 2 ± 1) .
Another interesting observation, based on the symmetry of the hyperbolic sine is the fact
that ln( 2 + 1) = - ln( 2 - 1) .
Problem 6: Find all x for which
ln( x + k ) = - ln( x - k ) .
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On the Curvature Function: Where does a curve bend the fastest?
Curvature and Parametric Curves
For a parametric curve x = x (t ) , y = y (t ) , the curvature function is defined as
x ¢(t )y ¢¢(t ) - x ¢¢(t )y ¢(t )
k (t ) =
3
2
.
(3)
[[x ¢(t )] + [y ¢(t )] ]
2
2
Let’s see what happens with some common parametric curves.
Example 6: Circle
For simplicity we will consider a circle with radius r centered at the origin, and therefore
having parametric equations x (t ) = r cos t and y (t ) = r sin t . Then
r 2 sin 2 t + r 2 cos2 t
k (t ) =
3
2
=
[r 2 sin 2 t + r 2 cos2 t ]
1
,
r
Naturally, the curvature along any circle is constant and is the reciprocal of the radius
which is consistent with the fact that the curvature and the radius of curvature are always
reciprocals of each other. That will make any circle its own circle of curvature at any
point on the circle.
Example 7: Ellipse
Consider the standard ellipse centered at the origin with equation
x2 y2
+
= 1
a 2 b2
a > 0, b > 0, a ¹ b
Using the parametric equations x = a cos t , y = b sin t (0 £ t £ 2p ) , we find
k (t ) =
(- a sin t ) ×(- b sin t ) + (- a cos t ) ×(- b cos t )
2
2
2
2
3
2
=
ab
3
[a 2 sin 2 t + b2 cos2 t ]2
[a sin t + b cos t ]
so that
19
,
On the Curvature Function: Where does a curve bend the fastest?
k ¢(t ) =
=
-
1
3
ab[a 2 sin 2 t + b2 cos2 t ]2 ×(2a 2 sin t cos t - 2b2 sin t cos t )
2
[a 2 sin 2 t + b2 cos2 t ]3
- 3ab[a 2 sin t cos t - b2 sin t cos t ]
2
2
2
5
2
2
[a sin t + b cos t ]
=
- 3ab(a 2 - b2 ) sin t cos t
2
2
2
2
5
2
.
[a sin t + b cos t ]
Therefore, the critical points occur at
p
3p
t = 0, , p, , 2p .
2
2
If b > a , the maximum curvature occurs at t =
p
3p
b
and t =
, and k max = 2 . The
2
2
a
minimum curvature occurs at t = 0 and t = p , and k min =
a
. Figure 16 shows the
b2
curvature function of the ellipse with a = 2 and b = 1 as a function of the parameter t .
Clearly the curvature is maximum at the vertices on the major axis. However from the
figure, there is a considerable amount of oscillation in the curvature.
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On the Curvature Function: Where does a curve bend the fastest?
Problem 6: Use Equation (3) to establish that curvature in polar coordinates is given by
dr 2
d 2r
) - r( 2 )
dq
dq .
dr 2 23
2
[r + ( ) ]
dq
r 2 + 2(
k (q) =
In an attempt to entice the reader to further experiment with curvature, we finish with
some graphs of curves with their associated curvature in polar coordinates. In Figure 17,
we show the four-leaf rose r  sin 2 (the inner curve) and the associated curvature
function. Apparently the curvature is maximum at the angles corresponding to the
maximum values of r , and is a minimum at the angles where the rose passes through the
pole.
Figure 18 shows r  sin(3 / 2) and the associated curvature function. We leave it for the
reader to determine which is the function and which is the curvature.
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On the Curvature Function: Where does a curve bend the fastest?
3 Pedagogical Considerations
Too often, the ideas and techniques of calculus I are lost to many students by the time
they reach calculus III. An expanded treatment of curvature can provide a wonderful
opportunity to jog students’ memories about these techniques. It also gives them the
chance to practice some standard differentiation rules. But most importantly, it provides
the opportunity to connect the ideas on curvature to the ideas on the behavior of functions
and of limits. Another possibility is to assign this kind of study as a group project,
particularly if students have access to a computer algebra system such as Derive,
Mathematica or Maple. Each group could be assigned a different family of functions to
explore and report on.
Curvature seems to be an ideal topic for exploration and research for undergraduate
mathematics, engineering, and computer science students in a calculus course. This is
certainly preferable to “doing” curvature problems as little more than an algebraic
exercise. For a brief history of curvature, see [1]. Some interesting applications of
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On the Curvature Function: Where does a curve bend the fastest?
curvature in physics and engineering dynamics can be found in [3]. For an application of
curvature in designing the outer panel surface of an automobile, see [4]. An application
of motion under curvature can be found in [6]. For an application of curvature in image
processing, see [5].
References
[1] J. Coolidge, The Unsatisfactory Story of Curvature, Amer. Math. Monthly, 59
pp.375-379, “ ISSN: 0002-9890 ”, 1952.
[2] A. Gray, Modern Differential Geometry of Curves and Surfaces, “ ISBN: 0-849-3
7872-9 ”, 1993, CRC Press.
[3] R Larson, R. Hostetler, B. Edwards, Calculus, 7th ed., “ ISBN: 0-618-23972-3 ”,
2002, Houghton Mifflin, MA.
[4] M. Lial, T. Hungerford, C. Miller, Mathematics and Applications, sixth ed.,
“ ISBN: 0-673-46943-3 ”, 1995, HarperCollins, NY.
[5] Malladi, R., and Sethian, J.A., Image Processing via Level Set Curvature Flow,
Proceedings of the National Academy of Science, 92, pp. 7046-7050, 1995.
[6] Sethian, J.A., A Review of Recent Numerical Algorithms for Hypersurfaces Moving
with Curvature-Dependent Speed, Journal of Differential Geometry, 31, pp. 131-161,
“ ISSN: 0022-040X ”, 1989.
[7] J. Stewart, Calculus, fifth ed., “ ISBN: 0-534-39330-6 ”, 2003, Brooks/Cole, CA.
23