The Central Idea
Curvature measures the local bending of a curve.
The essential question is:
\[ \boxed{\text{How quickly is the tangent direction changing as we move along the curve?}} \]A curve with nearly constant tangent direction bends only slightly.
A curve whose tangent direction changes rapidly bends sharply.
A Straight Line Has Zero Curvature
Along a straight line, the tangent direction never changes.
Therefore, its curvature is
\[ \boxed{\kappa=0}. \]Here the Greek letter
\[ \kappa \]is commonly used for curvature.
A Circle Has Constant Curvature
A circle bends by the same amount everywhere.
For a circle of radius \(R\),
\[ \boxed{\kappa=\frac{1}{R}}. \]A smaller circle has greater curvature because its direction changes more rapidly over a shorter distance.
Large Radius Means Gentle Bending
Compare two circles:
\[ R_1=2,\qquad R_2=10. \]Their curvatures are
\[ \kappa_1=\frac12 \]and
\[ \kappa_2=\frac1{10}. \]The circle of radius \(2\) bends more sharply.
Curvature and Tangent Direction
At a smooth point on a curve, the tangent line gives the instantaneous direction of the curve.
Let
\[ \phi \]denote the angle of that tangent direction.
Curvature measures how rapidly \(\phi\) changes as distance along the curve changes.
Arc Length as the Natural Distance Variable
Let \(s\) denote arc length measured along the curve.
Then the basic conceptual definition of curvature is
\[ \boxed{\kappa=\left|\frac{d\phi}{ds}\right|}. \]This says:
curvature is the magnitude of the rate at which tangent angle changes per unit distance traveled along the curve.
Why We Measure Change per Unit Arc Length
Suppose the same geometric path is traversed at two different speeds.
Its shape should not suddenly acquire different curvature merely because the motion is faster.
Measuring tangent change with respect to arc length rather than time makes curvature a property of the geometric curve itself.
The Unit Tangent Vector
For a regular parametric curve
\[ \mathbf r(t), \]the velocity is
\[ \mathbf r'(t). \]The corresponding unit tangent vector is
\[ \boxed{ \mathbf T(t)= \frac{\mathbf r'(t)} {\|\mathbf r'(t)\|} }. \]The unit tangent records direction while removing speed.
Curvature from the Unit Tangent
Because curvature measures change in tangent direction per unit arc length, we may write
\[ \boxed{ \kappa= \left\| \frac{d\mathbf T}{ds} \right\| }. \]This is one of the most geometric definitions of curvature.
Regular Curves
A parametric curve is called regular at a parameter value when
\[ \mathbf r'(t)\neq\mathbf 0. \]This condition ensures that the curve has a well-defined direction of motion there.
Standard curvature formulas require this nonzero-velocity condition.
Curvature of a Plane Parametric Curve
Suppose
\[ \mathbf r(t)=\langle x(t),y(t)\rangle \]is sufficiently smooth and regular.
Then its curvature is
\[ \boxed{ \kappa= \frac{|x'y''-y'x''|} {(x'^2+y'^2)^{3/2}} }. \]All derivatives are taken with respect to the parameter \(t\).
What the Parametric Formula Is Measuring
The numerator
\[ |x'y''-y'x''| \]detects how strongly the velocity direction is turning.
The denominator
\[ (x'^2+y'^2)^{3/2} \]corrects for the speed of the parametrization.
The result depends on the geometric path rather than on how quickly the parameter traverses it.
Curvature of a Graph \(y=f(x)\)
If a curve is represented explicitly as
\[ y=f(x), \]we can use the parametrization
\[ x=t,\qquad y=f(t). \]Then
\[ x'=1,\qquad x''=0. \]Substituting into the parametric formula gives
\[ \boxed{ \kappa= \frac{|f''(x)|} {\left(1+[f'(x)]^2\right)^{3/2}} }. \]Why the Second Derivative Appears
The first derivative
\[ f'(x) \]measures slope.
The second derivative
\[ f''(x) \]measures how slope changes.
Since bending involves a change in direction, it is natural that curvature depends on a second derivative.
Curvature Is Not Just the Second Derivative
For a graph,
\[ \kappa= \frac{|f''(x)|} {(1+[f'(x)]^2)^{3/2}}. \]Therefore, curvature depends on both
\[ f''(x) \]and
\[ f'(x). \]Two points can have the same second derivative but different curvature if their slopes differ.
Example: Curvature of a Parabola
Consider
\[ y=x^2. \]Then
\[ f'(x)=2x \]and
\[ f''(x)=2. \]Therefore,
\[ \boxed{ \kappa(x)= \frac{2}{(1+4x^2)^{3/2}} }. \]Where the Parabola Bends Most Sharply
For
\[ \kappa(x)= \frac{2}{(1+4x^2)^{3/2}}, \]the denominator is smallest when
\[ x=0. \]Thus the maximum curvature occurs at the vertex:
\[ \boxed{\kappa(0)=2}. \]Farther from the vertex, the parabola bends more gently.
Example: Curvature of a Line
Consider
\[ y=mx+b. \]Then
\[ f'(x)=m \]and
\[ f''(x)=0. \]Therefore,
\[ \boxed{\kappa=0}. \]This agrees with the geometric fact that a line does not bend.
Example: Curvature of a Circle from Parametric Form
Let
\[ x=R\cos t,\qquad y=R\sin t. \]Then
\[ x'=-R\sin t,\qquad y'=R\cos t, \]and
\[ x''=-R\cos t,\qquad y''=-R\sin t. \]The numerator is
\[ |x'y''-y'x''| =R^2. \]The denominator is
\[ (x'^2+y'^2)^{3/2} =(R^2)^{3/2}=R^3. \]Therefore,
\[ \boxed{\kappa=\frac1R}. \]Curvature Does Not Depend on Position
Translating a curve changes its coordinates but not its shape.
Therefore, translating a circle from the origin to another center does not change its curvature.
A circle of radius \(R\) always has
\[ \kappa=\frac1R \]regardless of where the circle is located.
Rotation Does Not Change Curvature
Rotating a curve preserves distances and angles.
Therefore, curvature is unchanged by rigid rotation.
Curvature measures intrinsic local bending, not orientation relative to the coordinate axes.
Scaling Does Change Curvature
Suppose every length in a curve is multiplied by a positive factor
\[ \lambda. \]Then radii of curvature are multiplied by \(\lambda\), so curvature is divided by \(\lambda\):
\[ \boxed{ \kappa_{\text{scaled}} = \frac{\kappa}{\lambda} }. \]Enlarging a curve makes its bending gentler.
Radius of Curvature
When
\[ \kappa\neq0, \]the radius of curvature is
\[ \boxed{\rho=\frac1\kappa}. \]Large curvature means small radius of curvature.
Small curvature means large radius of curvature.
Example: Radius of Curvature of a Parabola
For
\[ y=x^2, \]we found
\[ \kappa(x)= \frac{2}{(1+4x^2)^{3/2}}. \]Therefore,
\[ \boxed{ \rho(x)= \frac{(1+4x^2)^{3/2}}{2} }. \]At the vertex,
\[ \rho(0)=\frac12. \]The Osculating Circle
At a sufficiently smooth point with nonzero curvature, the osculating circle is the circle that best matches the curve's local bending to second order.
Its radius is the radius of curvature:
\[ \boxed{R_{\text{osc}}=\rho=\frac1\kappa}. \]It is often informally described as the circle that locally "hugs" the curve most closely.
Center of Curvature
The center of the osculating circle is called the center of curvature.
It lies along the normal direction from the point on the curve.
If \(\mathbf N\) is the appropriate unit normal and
\[ \rho=\frac1\kappa, \]then locally the center has the form
\[ \boxed{ \mathbf C=\mathbf r+\rho\,\mathbf N }. \]The sign and orientation of the chosen normal must be handled consistently.
The Unit Normal Vector
The unit tangent vector
\[ \mathbf T \]points along the curve.
Where curvature is nonzero, the principal unit normal
\[ \mathbf N \]points in the direction in which the unit tangent is turning.
The vectors \(\mathbf T\) and \(\mathbf N\) provide a moving local frame along the curve.
Tangent and Normal Directions
At a regular point:
- the tangent direction follows the curve;
- the normal direction is perpendicular to the tangent;
- the center of curvature lies along an appropriate normal direction.
This converts the visual idea of "which way the curve bends" into vector geometry.
Signed Curvature
The formulas used so far define nonnegative curvature through an absolute value:
\[ \kappa\ge0. \]In oriented plane geometry, a signed curvature may instead distinguish turning in opposite directions.
A common signed form is
\[ \boxed{ \kappa_s= \frac{x'y''-y'x''} {(x'^2+y'^2)^{3/2}} }. \]Its sign depends on the chosen orientation and convention.
Unsigned Curvature Is Usually Enough at First
For introductory geometric comparison, we usually care about the amount of bending rather than the orientation of the bend.
Therefore,
\[ \boxed{\kappa\ge0} \]is the default convention in this lesson.
Curvature and Concavity Are Related but Different
Concavity tells us whether a graph bends upward or downward.
Curvature measures how sharply the curve bends.
For a graph \(y=f(x)\), the sign of
\[ f''(x) \]contributes to concavity, while the magnitude of curvature is
\[ \frac{|f''(x)|} {(1+[f'(x)]^2)^{3/2}}. \]Thus concavity carries directional information, while ordinary curvature does not.
Inflection Points and Curvature
At a smooth inflection point of a graph, concavity changes sign.
In many familiar cases,
\[ f''(x)=0 \]there, and hence
\[ \kappa=0. \]However, identifying inflection points requires a change in concavity, not merely the equation \(f''(x)=0\).
Example: \(y=x^3\)
Let
\[ f(x)=x^3. \]Then
\[ f'(x)=3x^2 \]and
\[ f''(x)=6x. \]Therefore,
\[ \boxed{ \kappa(x)= \frac{|6x|} {(1+9x^4)^{3/2}} }. \]At the origin,
\[ \kappa(0)=0. \]Curvature of a Space Curve
For a sufficiently smooth regular space curve
\[ \mathbf r(t)= \langle x(t),y(t),z(t)\rangle, \]curvature can be computed using
\[ \boxed{ \kappa= \frac{\|\mathbf r'(t)\times\mathbf r''(t)\|} {\|\mathbf r'(t)\|^3} }. \]The cross product measures how strongly velocity and acceleration fail to remain parallel.
Why Parallel Velocity and Acceleration Give Zero Curvature
If
\[ \mathbf r'(t)\times\mathbf r''(t)=\mathbf0, \]then velocity and acceleration are parallel or one is zero.
At a regular point, this means the acceleration does not create a perpendicular turning component at that instant.
The curvature is therefore zero there.
Curvature and Tangential-Normal Acceleration
For motion along a sufficiently smooth curve, acceleration can be decomposed into tangential and normal parts:
\[ \mathbf a=a_T\mathbf T+a_N\mathbf N. \]If the speed is
\[ v=\|\mathbf r'(t)\|, \]then the normal component satisfies
\[ \boxed{a_N=\kappa v^2}. \]Thus curvature directly measures how strongly a path redirects motion.
Constant Speed Makes the Geometry Easier to See
If a particle moves with constant speed, then its tangential acceleration is zero.
Its acceleration is entirely associated with changing direction:
\[ \mathbf a=\kappa v^2\mathbf N. \]Uniform circular motion is the standard example.
Curvature of Uniform Circular Motion
For a circle of radius \(R\),
\[ \kappa=\frac1R. \]Therefore,
\[ a_N=\kappa v^2 \]becomes
\[ \boxed{a_N=\frac{v^2}{R}}. \]This is the familiar centripetal-acceleration formula.
Curvature Has Units of Reciprocal Length
Since
\[ \kappa=\frac1R \]for a circle, curvature has units
\[ \boxed{\frac{1}{\text{length}}}. \]For example, curvature might be measured in
\[ \text{m}^{-1} \]or
\[ \text{ft}^{-1}. \]Radius of Curvature Has Units of Length
Because
\[ \rho=\frac1\kappa, \]the radius of curvature has ordinary length units.
This makes the osculating-circle interpretation especially concrete.
Curvature and Reparametrization
The same geometric curve can be traced using different parameters.
A correct curvature formula produces the same geometric curvature under a regular change of parameter, even though the intermediate derivatives may differ.
This is why the denominator in the parametric formula is essential.
Example: Same Circle, Different Speed
Compare
\[ \mathbf r_1(t)= \langle R\cos t,R\sin t\rangle \]with
\[ \mathbf r_2(t)= \langle R\cos 5t,R\sin 5t\rangle. \]The second parametrization traces the circle five times as rapidly with respect to \(t\), but both have geometric curvature
\[ \boxed{\kappa=\frac1R}. \]Curvature Near a Cusp
At a cusp, a standard regular parametrization may satisfy
\[ \mathbf r'(t)=\mathbf0. \]The ordinary regular-curve curvature formula then has a zero denominator and cannot be applied directly at that parameter value.
Cusps therefore require separate local analysis.
The Cycloid Revisited
The cycloid from the previous lesson is
\[ x=a(t-\sin t), \] \[ y=a(1-\cos t). \]Its derivatives are
\[ x'=a(1-\cos t), \] \[ y'=a\sin t, \] \[ x''=a\sin t, \] \[ y''=a\cos t. \]Curvature of a Cycloid Away from the Cusps
For parameter values where the cycloid is regular,
\[ |x'y''-y'x''| = a^2|\,\cos t-1\,|. \]Also,
\[ x'^2+y'^2 = 2a^2(1-\cos t). \]Simplifying with half-angle identities gives
\[ \boxed{ \kappa= \frac{1} {4a\left|\sin\frac t2\right|} } \]away from the cusp parameters.
What the Cycloid Curvature Shows
As a cusp parameter is approached,
\[ \left|\sin\frac t2\right|\to0. \]Therefore,
\[ \kappa\to\infty. \]This matches the visual geometry: the curve becomes arbitrarily sharply bent near the cusp.
Curvature as Local Circular Approximation
A smooth curve near a point can often be approximated by:
- a tangent line at first order;
- an osculating circle when second-order bending is included.
The tangent captures direction.
Curvature captures the next layer of local shape.
First-Order and Second-Order Geometry
Position tells us where the curve is.
First derivatives tell us how the curve is directed.
Second derivatives help describe how that direction changes.
Curvature packages this second-order directional behavior into one geometric quantity.
Applications of Curvature
Curvature appears throughout mathematics, science, engineering, and design.
- road and railway alignment;
- vehicle turning paths;
- robot and tool trajectories;
- computer graphics and animation;
- surface design;
- optics and mechanical design;
- motion planning;
- differential geometry.
Curvature Constraints in Design
A path may be mathematically continuous but still bend too sharply for a physical system.
For example, a vehicle with a minimum turning radius
\[ R_{\min} \]cannot follow a path whose curvature exceeds
\[ \boxed{\kappa_{\max}=\frac1{R_{\min}}}. \]Curvature therefore translates geometric shape into an operational constraint.
Common Error: Thinking Curvature Is the Same as Slope
Slope describes direction.
Curvature describes how direction changes.
A line can have a very large slope and still have
\[ \boxed{\kappa=0}. \]Common Error: Thinking Curvature Is Just \(|f''(x)|\)
For a graph,
\[ \kappa= \frac{|f''(x)|} {(1+[f'(x)]^2)^{3/2}}. \]Ignoring the denominator makes curvature depend incorrectly on the coordinate orientation and slope.
Common Error: Forgetting the Absolute Value
Ordinary unsigned curvature is nonnegative.
For a graph,
\[ \boxed{ \kappa= \frac{|f''(x)|} {(1+[f'(x)]^2)^{3/2}} }. \]A negative second derivative does not mean negative unsigned curvature.
Common Error: Confusing Curvature with Radius of Curvature
Curvature and radius of curvature are reciprocals:
\[ \boxed{\rho=\frac1\kappa}. \]Sharp bending means:
\[ \text{large curvature} \]but
\[ \text{small radius of curvature}. \]Common Error: Applying the Parametric Formula at a Nonregular Point
The formula
\[ \kappa= \frac{|x'y''-y'x''|} {(x'^2+y'^2)^{3/2}} \]requires
\[ x'^2+y'^2\neq0. \]If both first derivatives vanish, the denominator is zero and the standard formula cannot be used directly.
Common Error: Assuming Zero Curvature Always Means a Whole Line
A curve can have curvature zero at a single point without being globally straight.
For example,
\[ y=x^3 \]has
\[ \kappa(0)=0, \]but the entire graph is not a line.
A Reliable Procedure for Curvature of \(y=f(x)\)
- Compute \(f'(x)\).
- Compute \(f''(x)\).
- Form \(1+[f'(x)]^2\).
- Raise that quantity to the \(3/2\) power.
- Take \(|f''(x)|\).
- Use \[ \kappa= \frac{|f''(x)|} {(1+[f'(x)]^2)^{3/2}}. \]
- If needed, compute \(\rho=1/\kappa\).
- Check whether the result makes geometric sense.
A Reliable Procedure for Parametric Curvature
- Compute \(x'(t)\) and \(y'(t)\).
- Compute \(x''(t)\) and \(y''(t)\).
- Check that \(x'^2+y'^2\neq0\).
- Compute \(x'y''-y'x''\).
- Take its absolute value.
- Compute \((x'^2+y'^2)^{3/2}\).
- Divide to obtain \(\kappa\).
- Simplify and interpret the result geometrically.
Example: Full Graph Curvature Workflow
Find the curvature of
\[ y=\frac12x^2 \]at
\[ x=1. \]First,
\[ f'(x)=x \]and
\[ f''(x)=1. \]Therefore,
\[ \kappa= \frac{1} {(1+x^2)^{3/2}}. \]At \(x=1\),
\[ \boxed{ \kappa= \frac{1}{(2)^{3/2}} = \frac{1}{2\sqrt2} }. \]Example: Full Parametric Curvature Workflow
Consider
\[ x=2\cos t,\qquad y=2\sin t. \]Then
\[ x'=-2\sin t,\qquad y'=2\cos t, \] \[ x''=-2\cos t,\qquad y''=-2\sin t. \]The numerator is
\[ |x'y''-y'x''|=4. \]The denominator is
\[ (x'^2+y'^2)^{3/2} =4^{3/2}=8. \]Thus
\[ \boxed{\kappa=\frac12}. \]Since the circle has radius \(2\), the result agrees with
\[ \kappa=\frac1R. \]Practice 1: Line
Find the curvature of
\[ y=4x-7. \]Practice 2: Parabola at the Vertex
Find the curvature of
\[ y=3x^2 \]at
\[ x=0. \]Practice 3: Parabola Away from the Vertex
Find the curvature of
\[ y=x^2 \]at
\[ x=1. \]Practice 4: Radius of Curvature
If
\[ \kappa=\frac14, \]find the radius of curvature.
Practice 5: Circle
Find the curvature of a circle of radius \(12\).
Practice 6: Scaling
A curve has curvature
\[ \kappa=3 \]at a particular point.
The entire curve is enlarged by a scale factor of \(6\).
What is the new curvature at the corresponding point?
Practice 7: Parametric Circle
Find the curvature of
\[ x=5\cos t,\qquad y=5\sin t. \]Practice 8: Zero Curvature Point
Find the curvature of
\[ y=x^3 \]at
\[ x=0. \]Practice 9: Units
A circular path has radius
\[ 20\text{ m}. \]Find its curvature and state the units.
Practice 10: Normal Acceleration
A particle moves at speed
\[ v=4 \]along a path whose curvature is
\[ \kappa=\frac12. \]Find the magnitude of its normal acceleration.
Practice Answers
Practice 1
For
\[ y=4x-7, \]the second derivative is
\[ 0. \]Therefore,
\[ \boxed{\kappa=0}. \]Practice 2
For
\[ y=3x^2, \]we have
\[ f'(x)=6x, \qquad f''(x)=6. \]At \(x=0\),
\[ \boxed{\kappa=6}. \]Practice 3
For
\[ y=x^2, \]we have
\[ f'(1)=2, \qquad f''(1)=2. \]Therefore,
\[ \kappa= \frac{2} {(1+4)^{3/2}} = \boxed{\frac{2}{5\sqrt5}}. \]Practice 4
Use
\[ \rho=\frac1\kappa. \]Therefore,
\[ \boxed{\rho=4}. \]Practice 5
For a circle,
\[ \kappa=\frac1R. \]Thus
\[ \boxed{\kappa=\frac1{12}}. \]Practice 6
Scaling by \(6\) divides curvature by \(6\):
\[ \kappa_{\text{new}} = \frac36 = \boxed{\frac12}. \]Practice 7
The curve is a circle of radius \(5\).
Therefore,
\[ \boxed{\kappa=\frac15}. \]Practice 8
For
\[ y=x^3, \]we have
\[ f''(0)=0. \]Therefore,
\[ \boxed{\kappa(0)=0}. \]Practice 9
For a circle of radius \(20\text{ m}\),
\[ \kappa= \frac1{20\text{ m}}. \]Thus
\[ \boxed{\kappa=0.05\text{ m}^{-1}}. \]Practice 10
Use
\[ a_N=\kappa v^2. \]Therefore,
\[ a_N= \frac12(4^2) =\frac12(16) =\boxed{8}. \]What to Remember
Curvature measures how rapidly a curve changes direction per unit arc length.
The core ideas are:
- conceptual definition: \[ \kappa= \left\| \frac{d\mathbf T}{ds} \right\|; \]
- graph formula: \[ \kappa= \frac{|f''(x)|} {(1+[f'(x)]^2)^{3/2}}; \]
- plane parametric formula: \[ \kappa= \frac{|x'y''-y'x''|} {(x'^2+y'^2)^{3/2}}; \]
- space-curve formula: \[ \kappa= \frac{\|\mathbf r'\times\mathbf r''\|} {\|\mathbf r'\|^3}; \]
- radius of curvature: \[ \rho=\frac1\kappa; \]
- circle: \[ \kappa=\frac1R. \]
The most useful intuition is simple: large curvature means sharp bending; small curvature means gentle bending.
Where We Go Next
Curvature describes the local bending of a one-dimensional curve. We now move from curves to surfaces created by rotating curves through space.
The next lesson, Surfaces of Revolution, studies how planar profiles generate three-dimensional surfaces through rotation and how their geometry can be represented parametrically and analytically.