9.3 Curvature

INTRODUCTION

Let C be a smooth curve in either 2- or 3-space traced out by a vector function r(t). In this section we are going to consider in greater detail the acceleration vector a(t) = r″(t) introduced in the last section. But before doing this, we need to examine a scalar quantity called the curvature of a curve.

A Definition

We know that r′(t) is a tangent vector to the curve C, and consequently

(1)

is a unit tangent. But recall from the end of Section 9.1 that if C is parameterized by arc length s, then a unit tangent to the curve is also given by dr/ds. The quantity r′(t) in (1) is related to arc length s by ds/dt = r′(t). Since the curve C is smooth, we know from pages 489 and 491 that ds/dt > 0. Hence by the Chain Rule,

(2)

Now suppose C is as shown in FIGURE 9.3.1. As s increases, T moves along C, changing direction but not length (it is always of unit length). Along the portion of the curve between P1 and P2 the vector T varies little in direction; along the curve between P2 and P3, where C obviously bends more sharply, the change in the direction of the tangent T is more pronounced. We use the rate at which the unit vector T changes direction with respect to arc length as an indicator of the curvature of a smooth curve C.

A smooth curve labeled C begins at a point labeled P subscript 1, passes through a point labeled P subscript 2, and ends at a point labeled P subscript 3. 9 unit tangent vectors, each labeled T, are shown on the curve.

FIGURE 9.3.1 Unit tangents

DEFINITION 9.3.1 Curvature

Let r(t) be a vector function defining a smooth curve C. If s is the arc length parameter and T = dr/ds is the unit tangent vector, then the curvature of C at a point is

(3)

The symbol κ in (3) is the Greek letter kappa. Now since curves are generally not parameterized by arc length, it is convenient to express (3) in terms of a general parameter t. Using the Chain Rule again, we can write

In other words, curvature is given by

(4)

EXAMPLE 1 Curvature of a Circle

Find the curvature of a circle of radius a.

SOLUTION

A circle can be described by the vector function r(t) = a cos ti + a sin tj. Now from r′(t) = –a sin ti + a cos tj and r′(t) = a, we get

Hence, from (4) the curvature is

(5)

The result in (5) shows that the curvature at a point on a circle is the reciprocal of the radius of the circle and indicates a fact that is in keeping with our intuition: A circle with a small radius curves more than one with a large radius. See FIGURE 9.3.2. ≡

2 circles. Circle 1. Caption. Small curvature K. A circle with a large radius is shown on the left. Circle 2. Caption. Large curvature K. A circle with a small radius is shown on the right.

FIGURE 9.3.2 Curvature of a circle in Example 1

Tangential and Normal Components of Acceleration

Suppose a particle moves in 2- or 3-space on a smooth curve C described by the vector function r(t). Then the velocity of the particle on C is v(t) = r′(t), whereas its speed is ds/dt = v = v(t). Thus, (1) implies v(t) = vT. Differentiating this last expression with respect to t gives acceleration:

(6)

Furthermore, with the help of Theorem 9.1.4(iii), it follows from the differentiation of T · T = 1 that T · dT/dt = 0. Hence, at a point P on C the vectors T and dT/dt are orthogonal. If dT/dt ≠ 0, the vector

(7)

is a unit normal to the curve C at P with direction given by dT/dt. The vector N is also called the principal normal. But since curvature is , it follows from (7) that dT/dt = κvN. Thus, (6) becomes

(8)

By writing (8) as

a(t) = aNN + aTT, (9)

we see that the acceleration vector a of the moving particle is the sum of two orthogonal vectors aNN and aTT. See FIGURE 9.3.3. The scalar functions aT = dv/dt and aN = κv2 are called the tangential and normal components of the acceleration, respectively. Note that the tangential component of the acceleration results from a change in the magnitude of the velocity v, whereas the normal component of the acceleration results from a change in the direction of v.

A graph. A curve and 5 vectors are graphed on a three dimensional x y z coordinate system. The curve is labeled C, and passes though a point labeled P. The first vector, labeled a subscript N N, and begins at the point P. The second vector is labeled N and shorter in length, begins at the point P, and goes along the direction of the first vector. The third vector, labeled a subscript T T, and begins at the point P. The fourth vector is labeled T and shorter in length, begins at the point P, and goes along the direction of the third vector. The vectors N and T are orthogonal. The fifth vector, labeled a, begins at the point P, and goes along a direction between the first 4 vectors.

FIGURE 9.3.3 Components of acceleration

The Binormal

A third unit vector defined by

B(t) = T(t) × N(t)

is called the binormal. The three unit vectors T, N, and B form a right-handed set of mutually orthogonal vectors called the moving trihedral. The plane of T and N is called the osculating plane,* the plane of N and B is said to be the normal plane, and the plane of T and B is the rectifying plane. See FIGURE 9.3.4.

A graph. A curve and 3 vectors are graphed on a three dimensional x y z coordinate system. The curve is labeled C, and passes though a point labeled P. The first vector is labeled N and begins at the point P. The second vector is labeled T, begins at the point P and is perpendicular to the vector N. The plane formed by the vectors N and T is labeled: osculating plane. The third vector, labeled B = T * N, begins at the point P, and is orthogonal to the vectors N and T.

FIGURE 9.3.4 Osculating plane

The three mutually orthogonal unit vectors T, N, and B can be thought of as a movable right-handed coordinate system since

B(t) = T(t) × N(t), N(t) = B(t) × T(t), T(t) = N(t) × B(t).

This movable coordinate system is referred to as the TNB-frame.

EXAMPLE 2 Tangent, Normal, and Binormal Vectors

The position of a moving particle is given by r(t) = 2 cos ti + 2 sin tj + 3tk. Find the vectors T, N, and B. Find the curvature.

SOLUTION

Since r′(t) = –2 sin ti + 2 cos tj + 3k, r′(t) = , and so from (1) we see that a unit tangent is

Next, we have

Hence, (7) gives the principal normal

N(t) = –cos ti – sin tj.

Now, the binormal is

Finally, using dT/dt = 2/ and r′(t) = , we obtain from (4) that the curvature at any point is the constant

≡

The fact that the curvature in Example 2 is constant is not surprising, since the curve defined by r(t) is a circular helix.

EXAMPLE 3 Osculating, Normal, Rectifying Planes

At the point corresponding to t = π/2 on the circular helix in Example 2, find an equation of (a) the osculating plane, (b) the normal plane, and (c) the rectifying plane.

SOLUTION

From the point P in question is (0, 2, 3π/2).

(a) A normal vector to the osculating plane at P is

To find an equation of a plane we do not require a unit normal, so in lieu of B(π/2) it is a bit simpler to use From (11) of Section 7.5 an equation of the osculating plane is

(b) At the point P, the vector or is normal to the plane containing N(π/2) and B(π/2). Hence an equation of the normal plane is

(c) Finally, at the point P, the vector is normal to the plane containing T(π/2) and B(π/2). An equation of the rectifying plane is

≡

With the help of Mathematica, portions of the helix and the osculating plane in Example 3 are shown in FIGURE 9.3.5. The point (0, 2, 3π/2) is indicated in the figure by the red dot.

A graph. A helix and an osculating plane are graphed on a three dimensional x y z coordinate system. The helix enters the bottom and goes up in a spiral path, passing through a point (0, 2, 3 pi over 2). The osculating plane passes though the point (0, 2, 3 pi over 2), with the value of z decreasing as x increases.

FIGURE 9.3.5 Helix and osculating plane in Example 3

Formulas for aT, aN, and Curvature

By dotting, and in turn crossing, the vector v = vT with (9), it is possible to obtain explicit formulas involving r, r′, and r″ for the tangential and normal components of the acceleration and the curvature. Observe that

yields the tangential component of acceleration

. (10)

On the other hand,

Since B = 1, it follows that the normal component of acceleration is

. (11)

Solving (11) for the curvature gives

. (12)

EXAMPLE 4 Curvature of Twisted Cubic

The curve traced by r(t) = ti + t2j + t3k is said to be a “twisted cubic.” If r(t) is the position vector of a moving particle, find the tangential and normal components of the acceleration at any t. Find the curvature.

SOLUTION

v(t) = r′(t) = i + tj + t2k, a(t) = r″(t) = j + 2tk.

Since v · a = t + 2t3 and v = , it follows from (10) that

Now,

and v × a = . Thus, (11) gives

From (12) we find that the curvature of the twisted cubic is given by

≡

Radius of Curvature

The reciprocal of the curvature, ρ = 1/κ, is called the radius of curvature. The radius of curvature at a point P on a curve C is the radius of a circle that “fits” the curve there better than any other circle. The circle at P is called the circle of curvature and its center is the center of curvature. The circle of curvature has the same tangent line at P as the curve C, and its center lies on the concave side of C. For example, a car moving on a curved track, as shown in FIGURE 9.3.6, can, at any instant, be thought to be moving on a circle of radius ρ. Hence, the normal component of its acceleration aN = κv2 must be the same as the magnitude of its centripetal acceleration a = v2/ρ. Therefore, κ = 1/ρ and ρ = 1/κ. Knowing the radius of curvature, we can determine the speed v at which a car can negotiate a banked curve without skidding. (This is essentially the idea in Problem 26 in Exercises 9.2.)

The radius of a dashed circle is labeled rho. The circle passes through a point labeled P. A line is tangent to the circle at the point P. A curve labeled C passes around the circle between the tangent line and the circle passing through the point P. The line is tangent to the curve C at the point P as well.

FIGURE 9.3.6 Radius of curvature

REMARKS

By writing (6) as

we note that the so-called scalar acceleration d2s/dt2, referred to in the last remark, is now seen to be the tangential component of the acceleration aT.

9.3 Exercises Answers to selected odd-numbered problems begin on page ANS-23.

In Problems 1 and 2, for the given position function, find the unit tangent.

  1. r(t) = (t cos t – sin t)i + (t sin t + cos t)j + t2k, t > 0
  2. r(t) = et cos ti + et sin tj + etk
  3. Use the procedure outlined in Example 2 to find T, N, B, and κ for motion on a general circular helix that is described by r(t) = a cos ti + a sin tj + ctk.
  4. Use the procedure outlined in Example 2 to show on the twisted cubic of Example 4 that at t = 1:

In Problems 5 and 6, find an equation of the osculating plane to the given space curve at the point that corresponds to the indicated value of t.

  1. The circular helix of Example 2; t = π/4
  2. The twisted cubic of Example 4; t = 1

In Problems 7–16, r(t) is the position vector of a moving particle. Find the tangential and normal components of the acceleration at any t.

  1. r(t) = i + tj + t2k
  2. r(t) = 3 cos ti + 2 sin tj + tk
  3. r(t) = t2i + (t2 – 1)j + 2t2k
  4. r(t) = t2i – t3j + t4k
  5. r(t) = 2ti + t2j
  6. r(t) = tan−1 ti + ln(1 + t2)j
  7. r(t) = 5 cos ti + 5 sin tj
  8. r(t) = cosh ti + sinh tj
  9. r(t) = e−t(i + j + k)
  10. r(t) = ti + (2t – 1)j + (4t + 2)k
  11. Find the curvature of an elliptical helix that is described by r(t) = a cos ti + b sin tj + ctk, a > 0, b > 0, c > 0.
    1. Find the curvature of an elliptical orbit that is described by r(t) = a cos ti + b sin tj + ck, a > 0, b > 0, c > 0.
    2. Show that when a = b, the curvature of a circular orbit is the constant κ = 1/a.
  13. Show that the curvature of a straight line is the constant κ = 0. [Hint: Use (2) in Section 7.5.]
  14. Find the curvature at t = π of the cycloid that is described by

    r(t) = a(t – sin t)i + a(1 – cos t)j, a > 0.

  15. Let C be a plane curve traced by r(t) = f(t)i + g(t)j, where f and g have second derivatives. Show that the curvature at a point is given by

    .

  16. Show that if y = F(x), the formula for κ in Problem 21 reduces to

    .

In Problems 23 and 24, use the result of Problem 22 to find the curvature and radius of curvature of the curve at the indicated points. Decide at which point the curve is “sharper.”

  1. y = x2; (0, 0), (1, 1)
  2. y = x3; (–1, –1), (, )

Discussion Problems

  1. Discuss the curvature near a point of inflection of y = F(x).
  2. Show that a(t)2 = + .

 

*Literally, this means the “kissing” plane.