1. 13
VECTOR FUNCTIONS

2. Arc Length and Curvature
VECTOR FUNCTIONS
13.3
Arc Length and Curvature
In this section, we will learn how to find:
The arc length of a curve and its curvature.
PLEASE HELP REWRITE/REFORMULATE OBJECTIVE.

3. PLANE CURVE LENGTH
In Section 10.2, we defined the length of a plane curve with parametric equations x = f(t), y = g(t), a ≤ t ≤ b as the limit of lengths of inscribed polygons.

4. PLANE CURVE LENGTH
Formula 1
For the case where f’ and g’ are continuous, we arrived at the following formula:

5. The length of a space curve is defined in exactly the same way.
SPACE CURVE LENGTH
The length of a space curve is defined in exactly the same way.

6. SPACE CURVE LENGTH
Suppose that the curve has the vector equation r(t) = <f(t), g(t), h(t)>, a ≤ t ≤ b
Equivalently, it could have the parametric equations x = f(t), y = g(t), z = h(t) where f’, g’ and h’ are continuous.

7. SPACE CURVE LENGTH
Formula 2
If the curve is traversed exactly once as t increases from a to b, then it can be shown that its length is:

8. ARC LENGTH
Formula 3
Notice that both the arc length formulas 1 and 2 can be put into the more compact form

9. That is because: For plane curves r(t) = f(t) i + g(t) j
ARC LENGTH
That is because:
For plane curves r(t) = f(t) i + g(t) j
For space curves r(t) = f(t) i + g(t) j + h(t) k

10. Find the length of the arc of the circular helix with vector equation
ARC LENGTH
Example 1
Find the length of the arc of the circular helix with vector equation
r(t) = cos t i + sin t j + t k
from the point (1, 0, 0) to the point (1, 0, 2π).

11. Since r’(t) = -sin t i + cos t j + k, we have:
ARC LENGTH
Example 1
Since r’(t) = -sin t i + cos t j + k, we have:

12. ARC LENGTH
Example 1
The arc from (1, 0, 0) to (1, 0, 2π) is described by the parameter interval 0 ≤ t ≤ 2π.
So, from Formula 3, we have:

13. A single curve C can be represented by more than one vector function.
ARC LENGTH
A single curve C can be represented by more than one vector function.

14. ARC LENGTH
Equations 4 & 5
For instance, the twisted cubic r1(t) = <t, t 2, t 3> ≤ t ≤ 2 could also be represented by the function r2(u) = <eu, e2u, e3u> ≤ u ≤ ln 2
The connection between the parameters t and u is given by t = eu.

15. We say that Equations 4 and 5 are parametrizations of the curve C.

16. PARAMETRIZATION
If we were to use Equation 3 to compute the length of C using Equations 4 and 5, we would get the same answer.
In general, it can be shown that, when Equation 3 is used to compute arc length, the answer is independent of the parametrization that is used.

17. r(t) = f(t) i + g(t) j + h(t) k a ≤ t ≤ b
ARC LENGTH
Now, we suppose that C is a curve given by a vector function
r(t) = f(t) i + g(t) j + h(t) k a ≤ t ≤ b
where:
r’ is continuous.
C is traversed exactly once as t increases from a to b.

18. We define its arc length function s by:
Equation 6
We define its arc length function s by:

19. Thus, s(t) is the length of the part of C between r(a) and r(t).
ARC LENGTH FUNCTION
Thus, s(t) is the length of the part of C between r(a) and r(t).

20. ARC LENGTH FUNCTION
Equation 7
If we differentiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus (FTC1), we obtain:

21. It is often useful to parametrize a curve with respect to arc length.
PARAMETRIZATION
It is often useful to parametrize a curve with respect to arc length.
This is because arc length arises naturally from the shape of the curve and does not depend on a particular coordinate system.

22. PARAMETRIZATION
If a curve r(t) is already given in terms of a parameter t and s(t) is the arc length function given by Equation 6, then we may be able to solve for t as a function of s: t = t(s)

23. REPARAMETRIZATION
Then, the curve can be reparametrized in terms of s by substituting for t: r = r(t(s))

24. REPARAMETRIZATION
Thus, if s = 3 for instance, r(t(3)) is the position vector of the point 3 units of length along the curve from its starting point.

25. REPARAMETRIZATION
Example 2
Reparametrize the helix r(t) = cos t i + sin t j + t k with respect to arc length measured from (1, 0, 0) in the direction of increasing t.

26. The initial point (1, 0, 0) corresponds to the parameter value t = 0.
REPARAMETRIZATION
Example 2
The initial point (1, 0, 0) corresponds to the parameter value t = 0.
From Example 1, we have:
So,

27. REPARAMETRIZATION
Example 2
Therefore, and the required reparametrization is obtained by substituting for t:

28. SMOOTH PARAMETRIZATION
A parametrization r(t) is called smooth on an interval I if:
r’ is continuous.
r’(t) ≠ 0 on I.

29. A curve is called smooth if it has a smooth parametrization.
SMOOTH CURVE
A curve is called smooth if it has a smooth parametrization.
A smooth curve has no sharp corners or cusps.
When the tangent vector turns, it does so continuously.

30. SMOOTH CURVES
If C is a smooth curve defined by the vector function r, recall that the unit tangent vector T(t) is given by:
This indicates the direction of the curve.

31. You can see that T(t) changes direction:
SMOOTH CURVES
You can see that T(t) changes direction:
Very slowly when C is fairly straight.
More quickly when C bends or twists more sharply.

32. CURVATURE
The curvature of C at a given point is a measure of how quickly the curve changes direction at that point.

33. CURVATURE
Specifically, we define it to be the magnitude of the rate of change of the unit tangent vector with respect to arc length.
We use arc length so that the curvature will be independent of the parametrization.

34. CURVATURE—DEFINITION
The curvature of a curve is:
where T is the unit tangent vector.

35. CURVATURE
The curvature is easier to compute if it is expressed in terms of the parameter t instead of s.

36. CURVATURE
So, we use the Chain Rule (Theorem 3 in Section 13.2, Formula 6) to write:

37. However, ds/dt = |r’(t)| from Equation 7.
CURVATURE
Equation/Formula 9
However, ds/dt = |r’(t)| from Equation 7.
So,

38. Show that the curvature of a circle of radius a is 1/a.
Example 3
Show that the curvature of a circle of radius a is 1/a.
We can take the circle to have center the origin.
Then, a parametrization is: r(t) = a cos t i + a sin t j

39. Therefore, r’(t) = –a sin t i + a cos t j and |r’(t)| = a
CURVATURE
Example 3
Therefore, r’(t) = –a sin t i + a cos t j and |r’(t)| = a
So, and

40. So, using Equation 9, we have:
CURVATURE
Example 3
This gives |T’(t)| = 1.
So, using Equation 9, we have:

41. The result of Example 3 shows—in accordance with our intuition—that:
CURVATURE
The result of Example 3 shows—in accordance with our intuition—that:
Small circles have large curvature.
Large circles have small curvature.

42. CURVATURE
We can see directly from the definition of curvature that the curvature of a straight line is always 0—because the tangent vector is constant.

43. Formula 9 can be used in all cases to compute the curvature.
Nevertheless, the formula given by the following theorem is often more convenient to apply.

44. The curvature of the curve given by the vector function r is:
Theorem 10
The curvature of the curve given by the vector function r is:

45. T = r’/|r’| and |r’| = ds/dt.
CURVATURE
Proof
T = r’/|r’| and |r’| = ds/dt.
So, we have:

46. Hence, the Product Rule (Theorem 3 in Section 13.2, Formula 3) gives:
CURVATURE
Proof
Hence, the Product Rule (Theorem 3 in Section 13.2, Formula 3) gives:

47. Using the fact that T x T = 0 (Example 2 in Section 12.4), we have:
CURVATURE
Proof
Using the fact that T x T = 0 (Example 2 in Section 12.4), we have:

48. So, T and T’ are orthogonal by Example 4 in Section 13.2
CURVATURE
Proof
Now, |T(t)| = 1 for all t.
So, T and T’ are orthogonal by Example 4 in Section 13.2

49. Hence, by Theorem 6 in Section 12.4,
CURVATURE
Proof
Hence, by Theorem 6 in Section 12.4,

50. CURVATURE
Proof
Thus,
and

51. Find the curvature of the twisted cubic r(t) = <t, t2, t3> at:
Example 4
Find the curvature of the twisted cubic r(t) = <t, t2, t3> at:
A general point
(0, 0, 0)

52. First, we compute the required ingredients:
CURVATURE
Example 4
First, we compute the required ingredients:

53. CURVATURE
Example 4

54. CURVATURE
Example 4
Then, Theorem 10 gives:
At the origin, where t = 0, the curvature is: ĸ(0) = 2

55. CURVATURE
For the special case of a plane curve with equation y = f(x), we choose x as the parameter and write:
r(x) = x i + f(x) j

56. CURVATURE
Then,
r’(x) = i + f’(x) j
and r’’(x) = f’’(x) j

57. r’(x) x r’’(x) = f’’(x) k
CURVATURE
Since i x j = k and j x j = 0, we have:
r’(x) x r’’(x) = f’’(x) k

58. CURVATURE
We also have:

59. CURVATURE
Formula 11
So, by Theorem 10,

60. CURVATURE
Example 5
Find the curvature of the parabola y = x2 at the points (0, 0), (1, 1), (2, 4)

61. Since y’ = 2x and y’’ = 2, Formula 11 gives:
CURVATURE
Example 5
Since y’ = 2x and y’’ = 2, Formula 11 gives:

62. At (0, 0), the curvature is κ(0) = 2.
Example 5
At (0, 0), the curvature is κ(0) = 2.
At (1, 1), it is κ(1) = 2/53/2 ≈ 0.18
At (2, 4), it is κ(2) = 2/173/2 ≈ 0.03

63. Observe from the expression for κ(x) or the graph of κ here that:
CURVATURE
Example 5
Observe from the expression for κ(x) or the graph of κ here that:
κ(x) → 0 as x → ±∞
This corresponds to the fact that the parabola appears to become flatter as x → ±∞

64. NORMAL AND BINORMAL VECTORS
At a given point on a smooth space curve r(t), there are many vectors that are orthogonal to the unit tangent vector T(t).

65. So, T’(t) is orthogonal to T(t).
NORMAL VECTORS
We single out one by observing that, because |T(t)| = 1 for all t, we have T(t) · T’(t) by Example 4 in Section 13.2.
So, T’(t) is orthogonal to T(t).
Note that T’(t) is itself not a unit vector.

66. NORMAL VECTOR
However, if r’ is also smooth, we can define the principal unit normal vector N(t) (simply unit normal) as:

67. NORMAL VECTORS
We can think of the normal vector as indicating the direction in which the curve is turning at each point.

68. The vector B(t) = T(t) x N(t) is called the binormal vector.

69. It is perpendicular to both T and N and is also a unit vector.
BINORMAL VECTORS
It is perpendicular to both T and N and is also a unit vector.

70. NORMAL & BINORMAL VECTORS
Example 6
Find the unit normal and binormal vectors for the circular helix r(t) = cost i + sin t j + t k

71. NORMAL & BINORMAL VECTORS
Example 6
First, we compute the ingredients needed for the unit normal vector:

72. NORMAL & BINORMAL VECTORS
Example 6

73. NORMAL & BINORMAL VECTORS
Example 6
This shows that the normal vector at a point on the helix is horizontal and points toward the z-axis.

74. NORMAL & BINORMAL VECTORS
Example 6
The binormal vector is:

75. NORMAL & BINORMAL VECTORS
The figure illustrates Example 6 by showing the vectors T, N, and B at two locations on the helix.

76. TNB FRAME
In general, the vectors T, N, and B, starting at the various points on a curve, form a set of orthogonal vectors—called the TNB frame—that moves along the curve as t varies.
SHOULD THE PREVIOUS IMAGE BE PUT HERE TOO?

77. This TNB frame plays an important role in:
The branch of mathematics known as differential geometry.
Its applications to the motion of spacecraft.

78. NORMAL PLANE
The plane determined by the normal and binormal vectors N and B at a point P on a curve C is called the normal plane of C at P.
It consists of all lines that are orthogonal to the tangent vector T.

79. OSCULATING PLANE
The plane determined by the vectors T and N is called the osculating plane of C at P.
The name comes from the Latin osculum, meaning ‘kiss.’

80. OSCULATING PLANE
It is the plane that comes closest to containing the part of the curve near P.
For a plane curve, the osculating plane is simply the plane that contains the curve.

81. OSCULATING CIRCLE
The osculating circle (the circle of curvature) of C at P is the circle that:
Lies in the osculating plane of C at P.
Has the same tangent as C at P.
Lies on the concave side of C (toward which N points).
Has radius ρ = 1/ĸ (the reciprocal of the curvature).

82. It is the circle that best describes how C behaves near P.
OSCULATING CIRCLE
It is the circle that best describes how C behaves near P.
It shares the same tangent, normal, and curvature at P.

83. NORMAL & OSCULATING PLANES
Example 7
Find the equations of the normal plane and osculating plane of the helix in Example 6 at the point P(0, 1, π/2)

84. NORMAL & OSCULATING PLANES
Example 7
The normal plane at P has normal vector r’(π/2) = <–1, 0, 1>.
So, an equation is: or

85. NORMAL & OSCULATING PLANES
Example 7
The osculating plane at P contains the vectors T and N.
So, its normal vector is: T x N = B

86. NORMAL & OSCULATING PLANES
Example 7
From Example 6, we have:

87. NORMAL & OSCULATING PLANES
Example 7
A simpler normal vector is <1, 0, 1>.
So, an equation of the osculating plane is: or

88. NORMAL & OSCULATING PLANES
The figure shows the helix and the osculating plane in Example 7.

89. OSCULATING CIRCLES
Example 8
Find and graph the osculating circle of the parabola y = x2 at the origin.
From Example 5, the curvature of the parabola at the origin is ĸ(0) = 2.
So, the radius of the osculating circle at the origin is 1/ĸ = ½ and its center is (0, ½).

90. Therefore, its equation is:
OSCULATING CIRCLES
Example 8
Therefore, its equation is:

91. OSCULATING CIRCLES
Example 8
For the graph, we use parametric equations of this circle: x = ½ cos t y = ½ + ½ sin t

92. OSCULATING CIRCLES
Example 8
The graph is displayed.

93. SUMMARY
We summarize the formulas for unit tangent, unit normal and binormal vectors, and curvature.