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Curvature and Torsion of Curves
Lesson: Curvature and Torsion of Curves Course Developer: Vivek N Sharma Department/College: Assistant Professor, Department of Mathematics, S.G.T.B. Khalsa College, University of Delhi
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves Table of Contents 1. Learning Outcomes 2. Introduction 3. Curvature of a Plane Curve 4. The Principal Unit Normal for a Plane Curve 5. Circle of Curvature of a Plane Curve 6. Curves in Space: Curvature & Normal Vectors 7. Unit Binormal Vector for a Space Curve 8. Torsion of a Space Curve 9. Functions of Several Variables: Introduction 10. Graphs & Level Curves 11. Summary and Important Formulae 12. Exercises 13. Glossary and Further Reading 14. Solutions for Exercises
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 1.
Learning Outcomes
After studying this unit, you will be able to •
state the concept of curvature of a plane curve.
•
calculate the curvature of various curves in plane and space.
•
explain the concept of torsion and binormal vectors for space curves.
•
calculate torsion & binormal vectors of various space curves.
•
describe the meaning of a function of more than one variable.
•
analyse visually a function of two or three variables.
•
explain the concept of a level curve of a function of two or more variables.
2.
Introduction:
Geometric understanding of mathematics holds high importance in scientific analysis. It enables us to dig deeply about the question at hand. In this unit, we shall be studying the geometric properties of various plane and space curves. A very important aspect of drawing any curve is the amount of bending or twisting of the curve around any point. This is a very basic question regarding any curve we aim to draw on a plane or in space. And mathematicians have answered this question quite comfortably using calculus. They have formulated the notion of curvature of a plane curve and torsion of a space curve. These are the properties that we shall be learning in this unit. Curvature of a plane curve tells us how much does the curve bend or turn around a point. Torsion of a space curve reveals the tendency of the curve to move “away” from the plane. Since a plane curve always remains in the plane (that is it never escapes the plane to enter the space, its torsion is Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves zero. Both these properties help us visualize a curve clearly and explain its various properties. Curvature and torsion, both, are the local properties of a curve. A local property of a curve is the one which explains the geometry of the curve only “around a point”. For instance, a curve may bend “too much” around a point but just “too little” around another point. Therefore, the curvature of the curve can be zero around one point, but may be very large around another point. This explains why the curvature of a curve is a local property of the curve. These geometric considerations forced mathematicians to use calculus to understand the notion of curvature. Its further development led to the fascinating branch of mathematics called “Differential Geometry”. We shall first study curvature of plane curves and shall then extend this concept to space curves. 3.
Curvature of a Plane Curve:
Intuitively, curvature explains how much does a curve bend (or turn) around a point. But whenever a curve turns, the tangent vector at that point changes its direction. And if a curve does not turn around a point, the tangent vector at that point will not change its direction around that point. In other words, as a particle moves along a smooth curve in a plane, the vector 𝑻𝑻 =
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
= Unit Tangent Vector of the curve
turns as the curve bends. This is revealed in the figure overleaf.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
𝒅𝒅𝑟𝑟
Figure 1: The unit tangent vector 𝑻𝑻 = 𝑑𝑑𝑑𝑑 at the point 𝑷𝑷 turns along the curve as the point 𝑷𝑷 proceeds along the curve.
Since 𝑻𝑻 is a unit vector, its length remains constant and only its direction changes as the particle moves along the curve. Now we are in a position to define the curvature of the smooth curve. 3.1 Definition of Curvature: The rate at which 𝑻𝑻 turns per unit of length
along the curve is called the curvature of the curve. Thus, if 𝑻𝑻 is the unit
tangent vector of a smooth curve 𝒓𝒓(𝑡𝑡), the curvature function 𝜅𝜅(𝑡𝑡) of the curve is
𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) =| 𝑑𝑑𝑑𝑑 (𝑡𝑡)|. Value Addition : Remarks 𝑑𝑑𝑻𝑻
•
| 𝑑𝑑𝑑𝑑 | ≥ 0
•
the point 𝑃𝑃 & the curvature of the curve at 𝑃𝑃 is zero.
•
•
𝑑𝑑𝑻𝑻
If | 𝑑𝑑𝑑𝑑 | = 0, the curve does not turn at all as the particle es through 𝑑𝑑𝑻𝑻
If | 𝑑𝑑𝑑𝑑 | is large, 𝑻𝑻 turns sharply as the particle es through the point 𝑃𝑃. 𝑑𝑑𝑻𝑻
If | 𝑑𝑑𝑑𝑑 | is close to zero, 𝑻𝑻 turns more slowly as the particle es through the point 𝑃𝑃 and the curvature of the curve at 𝑃𝑃 is smaller. Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
Usually, the curve 𝒓𝒓(𝑡𝑡) is given in of a parameter 𝑡𝑡 and it may not be easy to calculate the arc length 𝑠𝑠. To overcome this hurdle, mathematicians derived the following formula for calculating the curvature.
3.2 Formula for calculating curvature: If 𝒓𝒓(𝑡𝑡) is a smooth curve, then the curvature is 𝜅𝜅(𝑡𝑡) given by 𝟏𝟏
𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) = |𝒗𝒗(𝒕𝒕)| | 𝑑𝑑𝑑𝑑 | ,
where, 𝑻𝑻 =
𝒗𝒗
|𝒗𝒗|
is the unit tangent vector.
Proof: The formula is simply a consequence of chain rule for differentiation. 𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) = | 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑻𝑻
= | 𝑑𝑑𝑑𝑑 .
(Definition) 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻 1
= | 𝑑𝑑𝑑𝑑 . 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑑𝑑
1
=
(Chain Rule)
| 𝑑𝑑𝑻𝑻
. | 𝑑𝑑𝑑𝑑 |
|𝒗𝒗(𝑡𝑡)|
(Since,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=|𝒗𝒗(𝑡𝑡)|)
Let us now go through some examples Example 1: Show that the curvature of a straight line is zero. Solution: A straight line in Cartesian form is given by 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐. In vector form we can write this as: 𝒓𝒓(𝑡𝑡) = 𝑡𝑡𝒊𝒊 + (𝑚𝑚𝑚𝑚 + 𝑐𝑐)𝒋𝒋.
Hence, the unit tangent vector is 𝑻𝑻 =
= =
𝑣𝑣
|𝑣𝑣|
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 | | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
1
=
√1+𝑚𝑚 2
(1,𝑚𝑚 )
|(1,𝑚𝑚 )|
𝒊𝒊 +
𝑚𝑚
=
√1+𝑚𝑚 2
(1,𝑚𝑚 )
√1+𝑚𝑚 2
𝒋𝒋;
=(
1
√1+𝑚𝑚 2
,
𝑚𝑚
√1+𝑚𝑚 2
)
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves and, therefore, the curvature of the straight line is given by 𝜅𝜅(𝑡𝑡) = =
1
𝑑𝑑𝑻𝑻
. | 𝑑𝑑𝑑𝑑 (𝑡𝑡)|
|𝒗𝒗(𝑡𝑡)| 1
𝑑𝑑
𝑑𝑑𝒓𝒓 .|( 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑑𝑑 |
=0
1
𝑑𝑑
𝑚𝑚
(√1+𝑚𝑚 2 ), 𝑑𝑑𝑑𝑑 (√1+𝑚𝑚 2 )) (𝑡𝑡)| 𝑑𝑑𝒓𝒓
(Observe that there was no need to calculate |𝒗𝒗(𝑡𝑡)| = | 𝑑𝑑𝑑𝑑 (𝑡𝑡)| here as
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
= 𝟎𝟎.)
Hence, the curvature of a straight line is zero. Geometrically speaking, we see that the unit tangent vector 𝑻𝑻 always points in the same direction, therefore its components remain constant. (See figure 2)
Figure 2: Curvature of a straight line is zero. 1
Example 2: Show that the curvature of a circle of radius 𝑎𝑎 is 𝑎𝑎 .
Solution: We know that the equation of a circle of radius 𝑎𝑎 with centre at (0,0) in Cartesian form is given by 𝑥𝑥 2 + 𝑦𝑦 2 = 𝑎𝑎2
which can be parameterized as 𝒓𝒓(𝑡𝑡) = (𝑎𝑎 cos 𝑡𝑡, 𝑎𝑎 sin 𝑡𝑡) = 𝑎𝑎 cos 𝑡𝑡 𝒊𝒊 + 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 𝒋𝒋 .
This gives us,
𝒗𝒗(𝑡𝑡 ) =
=
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
𝑑𝑑
𝑑𝑑𝑑𝑑
[𝑎𝑎 cos 𝑡𝑡 𝒊𝒊 + 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 𝒋𝒋 ]
= (−𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 )𝒊𝒊 + (𝑎𝑎 cos 𝑡𝑡)𝒋𝒋
and, hence we may see that
|𝒗𝒗(𝑡𝑡)| = �(−𝑎𝑎 sin 𝑡𝑡) 2 + (𝑎𝑎 cos 𝑡𝑡)2 = √𝑎𝑎2 = |𝑎𝑎| = 𝑎𝑎,
since, 𝑎𝑎 = radius of the circle is positive. Therefore, the unit tangent vector is
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 𝒗𝒗(𝑡𝑡)
𝑻𝑻 =
|𝒗𝒗(𝑡𝑡)| 1
=
𝑎𝑎
[(−𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡)𝒊𝒊 + (𝑎𝑎 cos 𝑡𝑡)𝒋𝒋]
= (− sin 𝑡𝑡)𝒊𝒊 + (cos 𝑡𝑡)𝒋𝒋.
This gives |
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
|= |(− cos 𝑡𝑡)𝒊𝒊 + (− 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡)𝒋𝒋 | = 1.
Hence, we finally obtain the curvature of the circle as
𝜅𝜅(𝑡𝑡) =
1
|𝒗𝒗(𝑡𝑡)|
𝑑𝑑𝑻𝑻
. � �= 𝑑𝑑𝑑𝑑
1
𝑎𝑎
1
(1) = . 𝑎𝑎
Intuitively also, a circle bends uniformly all through. The amount of bending is the reciprocal of its radius. 4.
The Principal Unit Normal for a Plane Curve:
So far, we have seen how important is the unit tangent vector 𝑻𝑻 of a smooth
plane curve 𝒓𝒓(𝑡𝑡) is. 𝑻𝑻 points in the direction in which the curve is turning. Since 𝑻𝑻 has constant length, the derivative Therefore, the vector bends). This vector
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
remains orthogonal to 𝑻𝑻.
points in the direction in which 𝑻𝑻 turns (as the curve
is called the Principal Normal of the curve 𝒓𝒓(𝑡𝑡) and if
we make its magnitude one, it becomes the Principal Unit Normal. Thus, we led to the following definition of the principal unit normal. 4.1 Definition (Principal Unit Normal): At a point where 𝜅𝜅 (𝑡𝑡) ≠ 0, the principal unit normal vector for a smooth curve in the plane is
𝑵𝑵 =
1 𝑑𝑑𝑻𝑻
𝜅𝜅 𝑑𝑑𝑑𝑑
.
So, if we move from 𝑷𝑷𝟎𝟎 to 𝑷𝑷𝟏𝟏 , the vector 𝑻𝑻 turns clockwise and the vector
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
points to the right. And from 𝑷𝑷𝟏𝟏 to 𝑷𝑷𝟐𝟐 , the vector 𝑻𝑻 turns anti-clockwise and the vector
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
points to the left.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
Figure 3: The Principal Unit Normal Vector remains orthogonal to the tangent vector 𝑻𝑻 at any point 𝑷𝑷 on the curve. We now compute an easy formula for 𝑵𝑵. 4.2 Formula for Calculating 𝑵𝑵: If 𝒓𝒓(𝑡𝑡) is a smooth curve, then the principal unit normal is 𝑑𝑑𝑻𝑻
𝑑𝑑𝑻𝑻
𝑵𝑵= 𝑑𝑑𝑑𝑑 �| 𝑑𝑑𝑑𝑑 |
where, 𝑻𝑻 =
𝒗𝒗
|𝒗𝒗|
is the unit tangent vector.
Proof: We use chain rule. We have 1 𝑑𝑑𝑻𝑻
𝑵𝑵 =𝜅𝜅 𝑑𝑑𝑑𝑑 1
(Definition) 𝑑𝑑𝑻𝑻
= 𝑑𝑑 𝑻𝑻 . 𝑑𝑑𝑑𝑑 ,
=
�
𝑑𝑑𝑑𝑑
|
�
1
𝑑𝑑 𝑻𝑻 𝑑𝑑𝑑𝑑 . | 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
1
= 𝑑𝑑 𝑻𝑻 = =
𝑑𝑑𝑑𝑑 | || | 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑻𝑻 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑻𝑻 � � 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻
.
𝑑𝑑𝑻𝑻
since, 𝜅𝜅 = � 𝑑𝑑𝑑𝑑 �
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
( 𝑑𝑑𝑑𝑑 . 𝑑𝑑𝑑𝑑 ) (Chain Rule) 𝑑𝑑𝑻𝑻
1
1
1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 . � � 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻
1
( 𝑑𝑑𝑑𝑑 ). 𝑑𝑑𝑑𝑑
�| 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
(Because,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
≥0 and hence,
𝑑𝑑𝑑𝑑
. 𝑑𝑑𝑑𝑑
1
𝑑𝑑𝑑𝑑 | | 𝑑𝑑𝑑𝑑
= 1)
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Curvature and Torsion of Curves
Value Addition : Remark 𝑑𝑑𝑻𝑻
𝑑𝑑𝑻𝑻
𝑑𝑑𝑻𝑻
Since, 𝑵𝑵 = 𝑑𝑑𝑑𝑑 �| 𝑑𝑑𝑑𝑑 |, therefore, 𝑵𝑵. 𝑻𝑻 = [� 𝑑𝑑𝑑𝑑 � . 𝑻𝑻]
|
1
𝑑𝑑 𝑻𝑻 | 𝑑𝑑𝑑𝑑
= 0.
Example 3: Calculate 𝑵𝑵 for 𝒓𝒓(𝑡𝑡) = 𝑐𝑐𝑐𝑐𝑐𝑐 2𝑡𝑡 𝒊𝒊 + 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑡𝑡 𝒋𝒋. Solution: We first calculate 𝑻𝑻. We have 𝒓𝒓(𝑡𝑡) = cos 2𝑡𝑡 𝒊𝒊 + sin 2𝑡𝑡 𝒋𝒋,
so that,
𝒗𝒗 =
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
and hence,
= (−2 sin 2𝑡𝑡) 𝒊𝒊 + (2cos 2𝑡𝑡) 𝒋𝒋
|𝒗𝒗| = �(−2 sin 2𝑡𝑡)2 + (2 cos 2𝑡𝑡)2 = 2.
Therefore, 𝑻𝑻 =
Hence,
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝒗𝒗
|𝒗𝒗|
= (− sin 2𝑡𝑡) 𝒊𝒊 + (cos 2𝑡𝑡) 𝒋𝒋.
= (−2 cos 2𝑡𝑡) 𝒊𝒊 + (−2 sin 2𝑡𝑡) 𝒋𝒋
which gives 𝑑𝑑𝑻𝑻
| 𝑑𝑑𝑑𝑑 | = �(−2 cos 2𝑡𝑡)2 + (−2 sin 2𝑡𝑡)2 = 2.
Hence,
𝑵𝑵 = (− cos 2𝑡𝑡) 𝒊𝒊 + (− sin 2𝑡𝑡) 𝒋𝒋. 5.
Circle of Curvature of a Plane Curve:
Usually to detect the shape of a curve around a point, we study the tangent to the curve at that point. But tangent is only a first order approximation to the curve. This is because when we calculate the equation of the tangent, we need to find out the slope of the tangent which requires differentiating 𝒓𝒓 (𝑡𝑡)
only once. Simply knowing the slope of the tangent does not reveal the
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves increasing-decreasing nature of the curve. For that we require the higher order derivatives. To get a better description of the geometry of the curve around any point 𝑷𝑷,
we consider a circle ing through 𝑷𝑷 such that the circle follows three basic rules: 1)
Circle is tangent to the curve at 𝑷𝑷. This means that the tangent to the circle ing through the point 𝑷𝑷 coincides with the
2)
tangent to the curve at 𝑷𝑷.
3)
The circle lies towards inner side of the curve.
Circle has the same curvature as the curve has at the point 𝑷𝑷.
All this can be observed in the following figure.
Figure 4: The osculating circle of a given curve at a point 𝑷𝑷. Looking at the condition (2), we reason as follows. Suppose the curvature of the curve at the point 𝑷𝑷 is 𝜅𝜅. Then the curvature of the circle at 𝑷𝑷 is also 𝜅𝜅. 1
Hence, the radius of the circle is 𝜅𝜅 . But
1
𝜅𝜅
is well defined, if and only if, 𝜅𝜅 ≠ 0.
Hence, we can define the circle of curvature at any point 𝑷𝑷 on the curve
provided 𝜅𝜅 ≠ 0 where 𝜅𝜅 = curvature of the curve at the point 𝑷𝑷. The centre Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves of circle of curvature at a point P is known as Centre of Curvature of the curve at the point P. The centre lies on line along 𝑵𝑵. Example 4: Find & graph the osculating circle of the parabola 𝑦𝑦 = 𝑥𝑥 2 at the origin.
Solution: The parabola 𝑦𝑦 = 𝑥𝑥 2
can be parameterized as 𝒓𝒓(𝑡𝑡) = 𝑡𝑡 𝒊𝒊 + 𝑡𝑡 2 𝒋𝒋 .
We first find the curvature at origin, that is 𝜅𝜅(0). We have,
𝒗𝒗 = 𝑑𝑑𝒓𝒓
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
= 𝒊𝒊 + 2𝑡𝑡 𝒋𝒋
⇒| 𝑑𝑑𝑑𝑑 |= √1 + 4𝑡𝑡 2
so that,
Hence,
𝑻𝑻 = 𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
=
Therefore,
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 | | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
−4𝑡𝑡
1
2𝑡𝑡
= (√1+4𝑡𝑡 2 )𝒊𝒊 + (√1+4𝑡𝑡 2 )𝒋𝒋.
3 (1+4𝑡𝑡 2 )2
2
𝒊𝒊 + [√1+4𝑡𝑡 2 -
16𝑡𝑡 2
𝑑𝑑𝑻𝑻
2
1
3
(1+4𝑡𝑡 2 )2
| 𝑑𝑑𝑑𝑑 (𝑡𝑡)|= �1+4𝑡𝑡 2 )3 + [√1+4𝑡𝑡 2 −
Since,
8𝑡𝑡 2
(A)
]𝒋𝒋.
8𝑡𝑡 2
3
(1+4𝑡𝑡 2 )2
]2 .
𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) = |𝒗𝒗(𝑡𝑡)| | 𝑑𝑑𝑑𝑑 (𝑡𝑡)|,
therefore, the curvature at origin is 1
𝑑𝑑𝑻𝑻
𝜅𝜅(0) = |𝒗𝒗(0)| | 𝑑𝑑𝑑𝑑 (0)|=
1
√1
. √22 = 2.
Hence, the radius of curvature at origin is
1 2
unit.
Now to locate the centre, we compute 𝑵𝑵(0). Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves We know that 𝑵𝑵 (𝑡𝑡) =
⇒ 𝑵𝑵(0) =
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
By (A), we get,
(𝑡𝑡)
(0)
𝑵𝑵(0) = 0𝒊𝒊 + 2𝒋𝒋
Hence, the centre of the radius of curvature lies on 𝑥𝑥 = 0 , that is the 𝑦𝑦 − axis 1
and therefore, has the co-ordinates (0, 𝑎𝑎)for some a ∈ 𝑅𝑅. Since radius is 2, 1
1
that is, 𝑎𝑎 = 2, therefore, the centre of the radius of curvature is (0, 2). Hence, the equation of the circle of curvature at (0,0) is 1
1
(𝑥𝑥 − 0)2 + (𝑦𝑦 − 2)2 = (2)2 1
⇒ 𝑥𝑥 2 + (𝑦𝑦 − 2)2 =
1 4
Figure 4: The osculating circle of the parabola 𝑦𝑦 = 𝑥𝑥 2 at the origin. The above figure exhibits the circle of curvature at the origin. 6.
Curves in Space: Curvature & Normal Vectors
The analysis carried out for planes curves remains completely unchanged for curves in space. That is, if a smooth curve is 𝒓𝒓(𝑡𝑡), then we have the following formulae:
Unit Tangent Vector: 𝑻𝑻 =
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 | | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
=
𝒗𝒗
|𝒗𝒗|
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 𝑑𝑑𝑻𝑻
Curvature Function: 𝜅𝜅(𝑡𝑡) = | 𝑑𝑑𝑑𝑑 | = 1 𝑑𝑑𝑻𝑻
Principal Unit Normal: 𝑵𝑵 =
=
𝜅𝜅 𝑑𝑑𝑑𝑑
1
𝑑𝑑𝑻𝑻
. | 𝑑𝑑𝑑𝑑 | |𝒗𝒗| 1 𝑑𝑑𝑻𝑻
𝑑𝑑 𝑻𝑻 | 𝑑𝑑𝑑𝑑
|
𝑑𝑑𝑑𝑑
.
Example 5: Find curvature & principal unit normal of the helix 𝒓𝒓(𝑡𝑡) = 𝑎𝑎 cos 𝑡𝑡 𝒊𝒊 + 𝑎𝑎 sin 𝑡𝑡 𝒋𝒋 + 𝑏𝑏𝑏𝑏 𝒌𝒌,
where, 𝑎𝑎, 𝑏𝑏 ≥ 0, 𝑎𝑎2 + 𝑏𝑏 2 ≠ 0.
Solution: We have 𝒗𝒗 =
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
= (−𝑎𝑎 sin 𝑡𝑡) 𝒊𝒊 + (𝑎𝑎 cos 𝑡𝑡) 𝒋𝒋 + (𝑏𝑏)𝒌𝒌
𝑑𝑑𝑑𝑑
⇒|𝒗𝒗 |= | 𝑑𝑑𝑑𝑑 | = �(𝑎𝑎 sin 𝑡𝑡)2 + (𝑎𝑎 cos 𝑡𝑡)2 + 𝑏𝑏 2 = √𝑎𝑎2 + 𝑏𝑏 2 .
Hence,
𝑻𝑻 =
𝒗𝒗
|𝒗𝒗|
Therefore, 𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻
− a sin 𝑡𝑡
= (√𝑎𝑎 2
−a cos 𝑡𝑡
= (√𝑎𝑎 2
+𝑏𝑏 2
+𝑏𝑏 2
−a sin 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
−𝒂𝒂 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 ) + 𝑏𝑏 2
⇒| 𝑑𝑑𝑑𝑑 | =�(√𝑎𝑎 2
Hence,
𝜅𝜅(𝑡𝑡) =
1
a cos 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2 +𝑏𝑏 2
)𝒋𝒋
−𝒂𝒂 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 2 ) + 𝑏𝑏 2
+ (√𝑎𝑎 2
𝑑𝑑𝑻𝑻
+𝑏𝑏 2
. | 𝑑𝑑𝑑𝑑 | = |𝒗𝒗(𝑡𝑡)|
1
√𝑎𝑎 2 + 𝑏𝑏 2
.
𝑏𝑏
)𝒋𝒋 + (√𝑎𝑎 2
+𝑏𝑏 2
)𝒌𝒌.
𝑎𝑎
= √𝑎𝑎 2 𝑎𝑎
√𝑎𝑎 2 + 𝑏𝑏 2
+ 𝑏𝑏 2
=
𝑎𝑎
𝑎𝑎 2 +𝑏𝑏 2
𝑎𝑎
Therefore, the curvature of helix is the constant 𝑎𝑎 2 + 𝑏𝑏 2 .
Thus, if b increases for any fixed value of a, the curvature decreases. But increasing b means physically stretching it. This explains why stretching the string straightens it, because while stretching the string, b increases and, hence, curvature decreases, so that, the string looks more like a straight line. For principal unit normal, we have, 𝑵𝑵 = =
𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑 |𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑 | 1
𝑎𝑎 �𝑎𝑎 2 + 𝑏𝑏 2
−a cos 𝑡𝑡
[(√𝑎𝑎 2
+𝑏𝑏 2
−a sin 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
+𝑏𝑏 2
)𝒋𝒋]
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves so that 𝑵𝑵(𝑡𝑡) = (− cos 𝑡𝑡) 𝒊𝒊 + (− sin 𝑡𝑡) 𝒋𝒋. 7.
Unit Binormal Vector for a Space Curve:
Travelling along a space curve, the forward direction of motion (𝑻𝑻) and the
direction in which our path is turning (𝑵𝑵) and the direction of motion to
move away from the plane of 𝑻𝑻 and 𝑵𝑵 describe the motion even more clearly than the usual description in of unit vectors 𝒊𝒊, 𝒋𝒋, 𝒌𝒌. That is, the vectors 𝑻𝑻, 𝑵𝑵 and 𝑻𝑻 × 𝑵𝑵 describe the geometry of the space curve quite completely.
This leads us to formulate the definition of the binormal vector (𝑩𝑩) of a curve in space.
7.1 Definition (Unit Binormal Vector of a Space Curve): The unit binormal vector of a space curve is a unit vector in the plane perpendicular to the plane containing 𝑻𝑻 and 𝑵𝑵 and is, therefore, defined by 𝑩𝑩
𝑩𝑩𝟏𝟏 = |𝑩𝑩|,
where, 𝑩𝑩 = Binormal Vector = 𝑻𝑻 × 𝑵𝑵.
Sometimes, 𝑩𝑩𝟏𝟏 and 𝑩𝑩 are used interchangeably.
Value Addition: Remarks • •
Together 𝑻𝑻, 𝑵𝑵 and 𝑩𝑩 define a moving right handed vector frame 𝑻𝑻𝑻𝑻𝑻𝑻.
This 𝑻𝑻𝑻𝑻𝑻𝑻 frame is also called Frenet Frame (“fre-nay”) (named
after Jean-Frederic Frenet (1816-1900)). 8.
Torsion of a Space Curve:
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves The plane containing 𝑻𝑻 and 𝑵𝑵 is called osculating plane. Torsion describes how this plane turns about 𝑻𝑻 as the body moves along the curve. Torsion is
a scalar quantity.
Let 𝑩𝑩 = unit binormal vector of the space curve. Then,
𝑑𝑑𝑩𝑩
=
𝑑𝑑𝒔𝒔
𝑑𝑑
𝑑𝑑𝑑𝑑
(𝑻𝑻 × 𝑵𝑵) =
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
× 𝑵𝑵 + 𝑻𝑻 ×
𝑑𝑑𝑻𝑻
𝑑𝑑𝑵𝑵 𝑑𝑑𝑑𝑑
Since, 𝑵𝑵 is the unit vector along 𝑑𝑑𝑑𝑑 , therefore, 𝑑𝑑𝑻𝑻
x 𝑵𝑵 = 𝟎𝟎.
𝑑𝑑𝑑𝑑
Hence,
𝑑𝑑𝑩𝑩
Thus,
𝑑𝑑𝑑𝑑
𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
= 𝑻𝑻 ×
𝑑𝑑𝑵𝑵 𝑑𝑑𝑑𝑑
is orthogonal to 𝑻𝑻 as well as to 𝑩𝑩.
(This is because |𝑩𝑩 | = 1, that is, 𝑩𝑩 has constant length). Hence,
𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
is orthogonal to the plane of 𝑩𝑩 and 𝑻𝑻. 𝑑𝑑𝑩𝑩
In other words
𝑑𝑑𝑑𝑑
That is, 𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
is parallel to 𝑵𝑵 and is, therefore, a scalar multiple of 𝑵𝑵.
= −𝜏𝜏𝑵𝑵.
The scalar 𝜏𝜏 is called torsion along the curve. The negative sign in this
equation is simply for convention. (It simplifies matters in differential geometry.) We may see that 𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
Hence,
. 𝑵𝑵 = −(𝜏𝜏𝑵𝑵). 𝑵𝑵 = −𝜏𝜏(𝑵𝑵. 𝑵𝑵) = −𝜏𝜏(1) = −𝜏𝜏 𝑑𝑑𝑩𝑩
𝜏𝜏 = − 𝑑𝑑𝑑𝑑 𝑵𝑵.
Hence, we may now give the definition of torsion of a space curve.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 8.1 Definition (Torsion of a smooth curve): Let 𝒓𝒓(𝑡𝑡) be a smooth curve 𝑑𝑑𝑩𝑩
in space & let 𝑩𝑩 = 𝑻𝑻 × 𝑵𝑵.The torsion of the curve is 𝜏𝜏 = − 𝑑𝑑𝑑𝑑 𝑵𝑵. Value Addition: Remarks • •
𝜏𝜏 can be positive, negative or zero. If 𝜏𝜏 = 0, then
𝑑𝑑𝑩𝑩
= 𝟎𝟎 because 𝑵𝑵 ≠ 𝟎𝟎. Hence, there is no change in 𝑩𝑩
𝑑𝑑𝑑𝑑
along the arc-length and therefore, for plane curves, 𝜏𝜏 = 0.
Example 6: Find 𝑩𝑩 & 𝜏𝜏 for the helix 𝒓𝒓(𝑡𝑡) = (𝑎𝑎 cos 𝑡𝑡) 𝒊𝒊 + (𝑎𝑎 sin 𝑡𝑡) 𝒋𝒋 + (𝑏𝑏𝑏𝑏)𝒌𝒌. Solution: We have, from example 5,
and
− a sin 𝑡𝑡
𝑻𝑻 = (√𝑎𝑎 2
+𝑏𝑏 2
a cos 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
+𝑏𝑏 2
𝑏𝑏
)𝒋𝒋 + (√𝑎𝑎 2
+𝑏𝑏 2
)𝒌𝒌
𝑵𝑵 = (− cos 𝑡𝑡) 𝒊𝒊 + (− sin 𝑡𝑡) 𝒋𝒋.
Hence,
𝑩𝑩 = 𝑻𝑻 × 𝑵𝑵 =
i −a sin t
j a cos t
k b
a 2 + b2 − cos t
a 2 + b2 − sin t
a 2 + b2 0
which gives
Now,
and
𝑏𝑏 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
𝑩𝑩 = (√𝑎𝑎 2 𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
+𝑏𝑏 2
= (√𝑎𝑎 2
+𝑏𝑏 2
−b cos 𝑡𝑡 +𝑏𝑏 2
b sin 𝑡𝑡
𝑎𝑎
)𝒋𝒋 + (√𝑎𝑎 2
+𝑏𝑏 2
+𝑏𝑏 2
)𝒌𝒌
)𝒋𝒋
|𝒗𝒗| = √𝑎𝑎2 + 𝑏𝑏 2 , by example (5).
Hence,
𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
= =
1 𝑑𝑑𝑩𝑩
|𝒗𝒗| 𝑑𝑑𝑑𝑑 1
[
𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡
√𝑎𝑎 2 +𝑏𝑏 2 √𝑎𝑎 2 +𝑏𝑏 2
𝒊𝒊 +
b sin 𝑡𝑡
√𝑎𝑎 2 +𝑏𝑏 2
𝒋𝒋]
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves =
𝑏𝑏 cos 𝑡𝑡
𝑎𝑎 2 + 𝑏𝑏 2
Therefore,
𝒊𝒊 +
𝑏𝑏 sin 𝑡𝑡
𝑎𝑎 2 + 𝑏𝑏 2
𝑑𝑑𝑩𝑩
𝜏𝜏 = − 𝑑𝑑𝑑𝑑 . 𝑵𝑵
𝑏𝑏 cos 𝑡𝑡
= −[ 𝑎𝑎 2 + 𝑏𝑏 2 𝒊𝒊 + =
𝑏𝑏
𝑎𝑎 2 + 𝑏𝑏 2
𝒋𝒋
𝑏𝑏 sin 𝑡𝑡
𝑎𝑎 2 + 𝑏𝑏 2
𝒋𝒋]. [(−𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡) 𝒊𝒊 + (−𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡) 𝒋𝒋]
The given helix is shown below.
Figure 5: The helix for example 6.
In the general setting of space curves, often given in parameterized form, we have some formulae for calculation of curvature and torsion. The derivations are cumbersome and therefore, omitted. 8.2 Formulae for calculating the Torsion and Curvature: We state the formulae for calculating the torsion and curvature of the space curve 𝒓𝒓(𝑡𝑡) = (𝑥𝑥(𝑡𝑡), 𝑦𝑦(𝑡𝑡), 𝑧𝑧(𝑡𝑡)). For curvature, we have the following formula:
𝜅𝜅(𝑡𝑡) =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
For torsion, we have the following formula:
𝜏𝜏 =
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
.
Of course, we assume that 𝒗𝒗 ≠ 𝟎𝟎 and 𝒗𝒗 × 𝒂𝒂 ≠ 𝟎𝟎. Further, 𝑥𝑥 ′ (𝑡𝑡) = Value Addition: Remarks
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
, and so on.
The tuple (𝑻𝑻, 𝜅𝜅, 𝑵𝑵, 𝑩𝑩, 𝜏𝜏) is known as Frenet Apparatus. 9.
Functions of Several Variables: Introduction
In practical life, we encounter various problems whose solutions are influenced by various factors. For instance, in transportation, stockmarkets, economics, finance, psychology, statistics, physics, geometry one is very naturally led to consider functions of more than one variable. As an elementary instance, the volume 𝑉𝑉 of a cylinder 𝑉𝑉 = 𝜋𝜋𝑟𝑟 2 ℎ is a function of the
radius & height of the cylinder. In atmospheric sciences, the speed of the wind at any place on earth is a function of the latitude & longitude of that place. As a first introduction, we shall learn how to graph these functions. We begin with the following definitions. 9.1 Definition (Function of n independent variables): Suppose 𝐷𝐷 is a
set of 𝑛𝑛-tuples of real numbers (𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ). A real-valued function 𝑓𝑓 on 𝐷𝐷 is
a rule that assigns a unique (single) real number 𝑤𝑤 = 𝑓𝑓(𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 )
to each element in 𝐷𝐷. The set 𝐷𝐷 is the function’s domain. The set of 𝑤𝑤-
values taken on by 𝑓𝑓 is the function’s range. The symbol 𝑤𝑤 is the dependent variable of 𝑓𝑓, and 𝑓𝑓 is said to be a function of the 𝑛𝑛
independent variables 𝑥𝑥1 to 𝑥𝑥𝑛𝑛 . We also call the 𝑥𝑥𝑗𝑗 ′s the function’s input variables and call 𝑤𝑤 the function’s output variable.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
10.
Graphs & Level Curves:
We shall study the graphing of only two and three variable functions. We begin with the two-variable case. There are two standard ways to view the values of a function 𝑓𝑓(𝑥𝑥, 𝑦𝑦). One way is to draw the curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝐶𝐶 in the 𝑥𝑥 − 𝑦𝑦 plane where 𝐶𝐶 is any point in the range of 𝑓𝑓. This is called the levelcurve technique. The other way is to draw the surface 𝑧𝑧 = 𝑓𝑓(𝑥𝑥, 𝑦𝑦) in space. This surface 𝑧𝑧 = 𝑓𝑓(𝑥𝑥, 𝑦𝑦) is called the graph of the function 𝑓𝑓(𝑥𝑥, 𝑦𝑦). Example 7: Graph 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 100 − 𝑥𝑥 2 − 𝑦𝑦 2 ,
and plot the level curves 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0, 51, 75
in the domain of 𝑓𝑓 in the plane.
Solution: The function 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 100 − 𝑥𝑥 2 − 𝑦𝑦 2 is continuous ∀ (𝑥𝑥, 𝑦𝑦)𝜀𝜀 ℝ2 .
Therefore domain of 𝑓𝑓 is the entire plane. We first analyse the level curves. For the level curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0,
we see that,
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0
⇒ 100 = 𝑥𝑥 2 + 𝑦𝑦 2 .
Hence, the level curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0 is the circle of radius 10 centred at (0,0). Similarly, the level curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 51
is the circle
𝑥𝑥 2 + 𝑦𝑦 2 = 49;
and the level curve
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 75
is the circle
𝑥𝑥 2 + 𝑦𝑦 2 = 25.
The graph and the level curves are shown below.
Figure 6: The graph and level curves for example 7. Let us now take up the 3-variable case. Example 8: Describe the level surfaces of the function 𝑔𝑔(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 .
Solution: The value of 𝑔𝑔 at any point (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) is the distance of the point
(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) from the origin (0,0,0). A level surface is given by �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 𝑐𝑐, where 𝑐𝑐 is a real number. Clearly, 𝑐𝑐 cannot be negative. For positive values
of 𝑐𝑐, each level surface �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 𝑐𝑐, is a sphere of radius 𝑐𝑐 centered at
origin. For 𝑐𝑐 = 0 the level surface �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 0 reduces to the point (0,0,0).
The figure is given below.
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Curvature and Torsion of Curves
Figure 7: The graph and level curves for example 8.
11. Summary and Important Formulae for curves in plane and space
•
Unit Tangent Vector:
•
Curvature Function:
•
Principal Unit Normal Vector:
•
Binormal Vector:
•
Torsion of a Space Curve:
•
Formula for the Curvature:
•
Formula for the Torsion:
𝑻𝑻 =
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 𝑑𝑑𝑑𝑑
| | 𝑑𝑑𝑑𝑑
=
𝟏𝟏
𝒗𝒗
|𝒗𝒗|
𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) = |𝒗𝒗(𝒕𝒕)| | 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑
1 𝑑𝑑𝑻𝑻
𝑵𝑵 = |𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑 | = 𝑵𝑵 = 𝜅𝜅 𝑑𝑑𝑑𝑑 𝑩𝑩 = 𝑻𝑻 × 𝑵𝑵 𝑑𝑑𝑩𝑩
𝜏𝜏 = − 𝑑𝑑𝑑𝑑 𝑵𝑵 𝜅𝜅(𝑡𝑡) =
𝜏𝜏 =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
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Curvature and Torsion of Curves
Exercises Find 𝑻𝑻, 𝑵𝑵 and 𝜿𝜿 for the Plane Curves in Exercises 1-3 and Space Curves
in Exercises 4 to 6. Find 𝑩𝑩 and 𝝉𝝉 for the Space Curves in Exercises 4 to
6.
1. 2. 3. 4. 5. 6.
𝒓𝒓(𝑡𝑡) = 𝑡𝑡 𝒊𝒊 + 𝑙𝑙𝑙𝑙(𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡)𝒋𝒋
𝒓𝒓(𝑡𝑡) = (2𝑡𝑡 + 3)𝒊𝒊 + (5 − 𝑡𝑡 2 )𝒋𝒋 𝒓𝒓(𝑡𝑡) = (𝑐𝑐𝑐𝑐𝑐𝑐 3 𝑡𝑡)𝒊𝒊 + (𝑠𝑠𝑠𝑠𝑠𝑠3 𝑡𝑡)𝒋𝒋
𝒓𝒓(𝑡𝑡) = (3 sin 𝑡𝑡) 𝒊𝒊 + (3 cos 𝑡𝑡)𝒋𝒋 + 4𝑡𝑡 𝒌𝒌
𝒓𝒓(𝑡𝑡) = (cos 𝑡𝑡 + 𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)𝒊𝒊 + (sin 𝑡𝑡 − 𝑡𝑡 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)𝒋𝒋 + 3𝒌𝒌 𝒓𝒓(𝑡𝑡) = (𝑒𝑒 𝑡𝑡 cos 𝑡𝑡)𝒊𝒊 + (𝑒𝑒 𝑡𝑡 sin 𝑡𝑡)𝒋𝒋 + 2𝒌𝒌
Plot the typical (a) level curves and (b) graphs for the functions in exercises 7 to 9.
7. 8. 9.
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑦𝑦 2
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 4 − 𝑦𝑦 2
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥 2 + 𝑦𝑦 2
Plot a typical level surface for the functions in exercises 10 to 12.
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Curvature and Torsion of Curves 10. 11.
𝑓𝑓(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2
12.
𝑓𝑓(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑧𝑧
13.
Determine the curvature and torsion along the curve
𝑓𝑓(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑦𝑦 2 + 𝑧𝑧 2
𝒓𝒓(𝑡𝑡) = (𝑡𝑡 − 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡, 1 − cos 𝑡𝑡 , 𝑡𝑡). 14. 15.
Show that the curve 𝒓𝒓(𝑡𝑡) = (𝑡𝑡,
1+𝑡𝑡 1−𝑡𝑡 2 𝑡𝑡
,
𝑡𝑡
) lies in a plane.
Find the principle unit normal and unit binormal along the space curve 𝒓𝒓(𝑡𝑡) = (3𝑡𝑡 − 𝑡𝑡 3 , 3𝑡𝑡 2 , 3𝑡𝑡 + 𝑡𝑡 3 ). Further, prove that the curvature and the torsion along the curve are same.
Glossary Curvature, unit tangent, normal, unit binormal, torsion Further Reading It is always welcome to practice more exercises from various books available in libraries and elsewhere. 1. Calculus and Analytic Geometry (11th Edition) by George B. Thomas and Ross L. Finney; ISBN: 13: 978-0201531749 2. Calculus by Howard Anton and Irl C. Bivens (10th Edition); ISBN: 13: 978-0470647691
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Curvature and Torsion of Curves Solutions For Exercises 𝑻𝑻, 𝑵𝑵 and 𝜿𝜿 for Exercises 1-3 1.
2.
3.
𝑩𝑩 and 𝜏𝜏 for Exercises 4-6 4.
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Curvature and Torsion of Curves
5.
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Curvature and Torsion of Curves 6.
Level Curves and Graphs for Exercises 7 to 9 7.
8.
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Curvature and Torsion of Curves
9.
Level Surfaces for Exercises 10 to 12 10.
11.
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Curvature and Torsion of Curves
12.
13.
The given curve 𝒓𝒓(𝑡𝑡) = (𝑡𝑡 − 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡, 1 − cos 𝑡𝑡 , 𝑡𝑡) is a space curve and hence we use the formula:
𝜅𝜅(𝑡𝑡) =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
Now, we see that,
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Curvature and Torsion of Curves 𝑑𝑑𝒓𝒓
𝒗𝒗(𝑡𝑡) = 𝑑𝑑𝑑𝑑 = (1 − cos 𝑡𝑡 , sin 𝑡𝑡 , 1) and hence,
𝒂𝒂(𝑡𝑡) =
𝑑𝑑𝒗𝒗 𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝒓𝒓
= 𝑑𝑑𝑡𝑡 2 = (sin 𝑡𝑡, cos 𝑡𝑡, 0).
Curvature:
i j k We have, 𝒗𝒗 × 𝒂𝒂 = 1 − cos t sin t 1 sin t cos t 0
= (− cos 𝑡𝑡)𝒊𝒊 + (sin 𝑡𝑡)𝒋𝒋 + (cos 𝑡𝑡 − 1)𝒌𝒌.
Hence,
|𝒗𝒗 × 𝒂𝒂| = �(−𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡)𝟐𝟐 + (sin 𝑡𝑡) 𝟐𝟐 + (cos 𝑡𝑡 − 1) 𝟐𝟐 𝑡𝑡
= �1 + 4𝑠𝑠𝑠𝑠𝑠𝑠4 (2);
and |𝒗𝒗| = �(1 − cos 𝑡𝑡)2 + (sin 𝑡𝑡) 2 + 1 𝑡𝑡
= �1 + 4𝑠𝑠𝑠𝑠𝑠𝑠2 (2).
Thus, the curvature along the given curve is
𝜅𝜅(𝑡𝑡) =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
=
𝑡𝑡 2
�1+4𝑠𝑠𝑠𝑠𝑠𝑠 4 ( ) 𝑡𝑡 2
3
(1+4𝑠𝑠𝑠𝑠𝑠𝑠 2 ( ))2
.
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Curvature and Torsion of Curves
Torsion: We have, x ' x '' x ''' 1 − cos t sin t y ' y '' y ''' = sin t cos t z ' z '' z ''' 1 0
cos t − sin t 0
= −1. Hence, torsion along the given curve is
𝜏𝜏 =
14.
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
=
−1
𝑡𝑡 2
.
1+4𝑠𝑠𝑠𝑠𝑠𝑠 4 ( )
To show that the given curve
𝒓𝒓(𝑡𝑡) = (𝑡𝑡,
1+𝑡𝑡 1−𝑡𝑡 2 𝑡𝑡
,
𝑡𝑡
)
lies in a plane, it is enough to show that the torsion of the curve is zero. So, we calculate the torsion of the curve. We have,
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Curvature and Torsion of Curves
1 x ' x '' x ''' 1 y ' y '' y ''' = 1 − 2 t z ' z '' z ''' 1 −1 − 2 t
0 2 t3 2 t3
0 6 − 4 = 0. t 6 − 4 t
Therefore,
𝜏𝜏 =
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
= 0.
Hence, the given curve is a plane curve. 15. The space curve given to us is 𝒓𝒓(𝑡𝑡) = (3𝑡𝑡 − 𝑡𝑡 3 , 3𝑡𝑡 2 , 3𝑡𝑡 + 𝑡𝑡 3 ). Hence, the velocity vector 𝒗𝒗(𝑡𝑡) along the curve is 𝑑𝑑𝒓𝒓
𝒗𝒗(𝑡𝑡) = 𝑑𝑑𝑑𝑑 = (3 − 3𝑡𝑡 2 , 6𝑡𝑡, 3 + 3𝑡𝑡 2 ), and hence, the speed |𝒗𝒗(𝑡𝑡)| along the curve is |𝒗𝒗(𝑡𝑡)| = �(3 − 3𝑡𝑡 2 )2 + 36𝑡𝑡 2 + (3 + 3𝑡𝑡 2 )2 = 3√2(1 + 𝑡𝑡 2 ). Hence, the unit tangent vector 𝑻𝑻 along the curve is
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Curvature and Torsion of Curves 𝑻𝑻 =
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 | | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝒗𝒗
1
= |𝒗𝒗| = 3√2(1+𝑡𝑡 2 ) (3 − 3𝑡𝑡 2 , 6𝑡𝑡, 3 + 3𝑡𝑡 2 )
which simplifies to
𝑻𝑻 =
1
√2(1+𝑡𝑡 2 )
(1 − 𝑡𝑡 2 , 2𝑡𝑡, 1 + 𝑡𝑡 2 ).
Principal Unit Normal: We have, 𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
�1−𝑡𝑡 2 �
2√2𝑡𝑡
= (− (1+𝑡𝑡 2 )2 , √2 (1+𝑡𝑡 2 )2 , 0)
and therefore,
𝑑𝑑𝑻𝑻
(1−𝑡𝑡 2 )
2√2𝑡𝑡
2
√2
� 𝑑𝑑𝑑𝑑 � = �(− (1+𝑡𝑡 2 )2 )2 + (√2 (1+𝑡𝑡 2 )2 )2 = �(1+𝑡𝑡 2 )2 = (1+𝑡𝑡 2 ). Hence, the principal unit normal 𝑵𝑵 along the curve is 𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑
𝑵𝑵 = |𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑 | = √𝟐𝟐
1
� (1+𝑡𝑡 2 )
2√2𝑡𝑡
�1−𝑡𝑡 2 �
(− (1+𝑡𝑡 2 )2 , √2 (1+𝑡𝑡 2 )2 , 0)
and this simplifies to 2𝑡𝑡
1−𝑡𝑡 2
𝑵𝑵 = (− 1+𝑡𝑡 2 , 1+𝑡𝑡 2 , 0). Principal Unit Binormal:
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Curvature and Torsion of Curves The principal unit binormal 𝑩𝑩 along the curve is the crossproduct of the vectors 𝑻𝑻 and 𝑵𝑵 and is given by
𝑩𝑩 = 𝑻𝑻 × 𝑵𝑵 =
i
j
k
(1 − t 2 ) 2 1+ t2
2t 2 1+ t2
(1 + t 2 ) 2 1+ t2
(
−
)
2t 1+ t2
(
1− t2 1+ t2
)
(
)
0
and this simplifies to
𝑩𝑩 =
1 2 2 ) (𝑡𝑡 (1+𝑡𝑡 √2
− 1, −2𝑡𝑡, 𝑡𝑡 2 + 1).
Curvature: The acceleration vector 𝒂𝒂(𝑡𝑡) of the given curve is 𝒂𝒂(𝑡𝑡) =
𝑑𝑑𝒗𝒗 𝑑𝑑𝑑𝑑
= (−6𝑡𝑡, 6, 6𝑡𝑡).
Hence, we have
i 𝒗𝒗 × 𝒂𝒂 = 3 − 3t 2 −6t
j k 6t 3 + 3t 2 = 18(𝑡𝑡 2 − 1)𝒊𝒊 − 36𝑡𝑡 𝒋𝒋 + 18(𝑡𝑡 2 + 1)𝒌𝒌 6 6t
Hence, the curvature 𝜅𝜅(𝑡𝑡) of the given curve is
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Curvature and Torsion of Curves 𝜅𝜅(𝑡𝑡) =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
=
|18�𝑡𝑡 2 −1�𝒊𝒊−36𝑡𝑡 𝒋𝒋+18�𝑡𝑡 2 +1�𝒌𝒌| |3−3𝑡𝑡 2 𝒊𝒊+ 6𝑡𝑡𝒋𝒋+ 3+3𝑡𝑡 2 𝒌𝒌|3
2
= 3(1+𝑡𝑡 2 )2 .
Torsion: We now see that
3 − 3t 2 x ' x '' x ''' y ' y '' y ''' = 6t z ' z '' z ''' 3 + 3t 2
−6t 6 6t
−6 0 = 216, 6
and i 𝒗𝒗 × 𝒂𝒂 = 3 − 3t 2 −6t
j k 6t 3 + 3t 2 = 18(𝑡𝑡 2 − 1)𝒊𝒊 − 36𝑡𝑡 𝒋𝒋 + 18(𝑡𝑡 2 + 1)𝒌𝒌 6 6t
and hence, the torsion 𝜏𝜏 of the given curve is
𝜏𝜏 =
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
216
2
= |18(𝑡𝑡 2 −1)𝒊𝒊−36𝑡𝑡 𝒋𝒋+18(𝑡𝑡 2 +1)𝒌𝒌|2 = 3(1+𝑡𝑡 2 )2
and therefore, 𝜅𝜅 = 𝜏𝜏.
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Lesson: Curvature and Torsion of Curves Course Developer: Vivek N Sharma Department/College: Assistant Professor, Department of Mathematics, S.G.T.B. Khalsa College, University of Delhi
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Curvature and Torsion of Curves Table of Contents 1. Learning Outcomes 2. Introduction 3. Curvature of a Plane Curve 4. The Principal Unit Normal for a Plane Curve 5. Circle of Curvature of a Plane Curve 6. Curves in Space: Curvature & Normal Vectors 7. Unit Binormal Vector for a Space Curve 8. Torsion of a Space Curve 9. Functions of Several Variables: Introduction 10. Graphs & Level Curves 11. Summary and Important Formulae 12. Exercises 13. Glossary and Further Reading 14. Solutions for Exercises
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Curvature and Torsion of Curves 1.
Learning Outcomes
After studying this unit, you will be able to •
state the concept of curvature of a plane curve.
•
calculate the curvature of various curves in plane and space.
•
explain the concept of torsion and binormal vectors for space curves.
•
calculate torsion & binormal vectors of various space curves.
•
describe the meaning of a function of more than one variable.
•
analyse visually a function of two or three variables.
•
explain the concept of a level curve of a function of two or more variables.
2.
Introduction:
Geometric understanding of mathematics holds high importance in scientific analysis. It enables us to dig deeply about the question at hand. In this unit, we shall be studying the geometric properties of various plane and space curves. A very important aspect of drawing any curve is the amount of bending or twisting of the curve around any point. This is a very basic question regarding any curve we aim to draw on a plane or in space. And mathematicians have answered this question quite comfortably using calculus. They have formulated the notion of curvature of a plane curve and torsion of a space curve. These are the properties that we shall be learning in this unit. Curvature of a plane curve tells us how much does the curve bend or turn around a point. Torsion of a space curve reveals the tendency of the curve to move “away” from the plane. Since a plane curve always remains in the plane (that is it never escapes the plane to enter the space, its torsion is Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves zero. Both these properties help us visualize a curve clearly and explain its various properties. Curvature and torsion, both, are the local properties of a curve. A local property of a curve is the one which explains the geometry of the curve only “around a point”. For instance, a curve may bend “too much” around a point but just “too little” around another point. Therefore, the curvature of the curve can be zero around one point, but may be very large around another point. This explains why the curvature of a curve is a local property of the curve. These geometric considerations forced mathematicians to use calculus to understand the notion of curvature. Its further development led to the fascinating branch of mathematics called “Differential Geometry”. We shall first study curvature of plane curves and shall then extend this concept to space curves. 3.
Curvature of a Plane Curve:
Intuitively, curvature explains how much does a curve bend (or turn) around a point. But whenever a curve turns, the tangent vector at that point changes its direction. And if a curve does not turn around a point, the tangent vector at that point will not change its direction around that point. In other words, as a particle moves along a smooth curve in a plane, the vector 𝑻𝑻 =
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
= Unit Tangent Vector of the curve
turns as the curve bends. This is revealed in the figure overleaf.
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Curvature and Torsion of Curves
𝒅𝒅𝑟𝑟
Figure 1: The unit tangent vector 𝑻𝑻 = 𝑑𝑑𝑑𝑑 at the point 𝑷𝑷 turns along the curve as the point 𝑷𝑷 proceeds along the curve.
Since 𝑻𝑻 is a unit vector, its length remains constant and only its direction changes as the particle moves along the curve. Now we are in a position to define the curvature of the smooth curve. 3.1 Definition of Curvature: The rate at which 𝑻𝑻 turns per unit of length
along the curve is called the curvature of the curve. Thus, if 𝑻𝑻 is the unit
tangent vector of a smooth curve 𝒓𝒓(𝑡𝑡), the curvature function 𝜅𝜅(𝑡𝑡) of the curve is
𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) =| 𝑑𝑑𝑑𝑑 (𝑡𝑡)|. Value Addition : Remarks 𝑑𝑑𝑻𝑻
•
| 𝑑𝑑𝑑𝑑 | ≥ 0
•
the point 𝑃𝑃 & the curvature of the curve at 𝑃𝑃 is zero.
•
•
𝑑𝑑𝑻𝑻
If | 𝑑𝑑𝑑𝑑 | = 0, the curve does not turn at all as the particle es through 𝑑𝑑𝑻𝑻
If | 𝑑𝑑𝑑𝑑 | is large, 𝑻𝑻 turns sharply as the particle es through the point 𝑃𝑃. 𝑑𝑑𝑻𝑻
If | 𝑑𝑑𝑑𝑑 | is close to zero, 𝑻𝑻 turns more slowly as the particle es through the point 𝑃𝑃 and the curvature of the curve at 𝑃𝑃 is smaller. Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
Usually, the curve 𝒓𝒓(𝑡𝑡) is given in of a parameter 𝑡𝑡 and it may not be easy to calculate the arc length 𝑠𝑠. To overcome this hurdle, mathematicians derived the following formula for calculating the curvature.
3.2 Formula for calculating curvature: If 𝒓𝒓(𝑡𝑡) is a smooth curve, then the curvature is 𝜅𝜅(𝑡𝑡) given by 𝟏𝟏
𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) = |𝒗𝒗(𝒕𝒕)| | 𝑑𝑑𝑑𝑑 | ,
where, 𝑻𝑻 =
𝒗𝒗
|𝒗𝒗|
is the unit tangent vector.
Proof: The formula is simply a consequence of chain rule for differentiation. 𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) = | 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑻𝑻
= | 𝑑𝑑𝑑𝑑 .
(Definition) 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻 1
= | 𝑑𝑑𝑑𝑑 . 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑑𝑑
1
=
(Chain Rule)
| 𝑑𝑑𝑻𝑻
. | 𝑑𝑑𝑑𝑑 |
|𝒗𝒗(𝑡𝑡)|
(Since,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=|𝒗𝒗(𝑡𝑡)|)
Let us now go through some examples Example 1: Show that the curvature of a straight line is zero. Solution: A straight line in Cartesian form is given by 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐. In vector form we can write this as: 𝒓𝒓(𝑡𝑡) = 𝑡𝑡𝒊𝒊 + (𝑚𝑚𝑚𝑚 + 𝑐𝑐)𝒋𝒋.
Hence, the unit tangent vector is 𝑻𝑻 =
= =
𝑣𝑣
|𝑣𝑣|
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 | | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
1
=
√1+𝑚𝑚 2
(1,𝑚𝑚 )
|(1,𝑚𝑚 )|
𝒊𝒊 +
𝑚𝑚
=
√1+𝑚𝑚 2
(1,𝑚𝑚 )
√1+𝑚𝑚 2
𝒋𝒋;
=(
1
√1+𝑚𝑚 2
,
𝑚𝑚
√1+𝑚𝑚 2
)
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves and, therefore, the curvature of the straight line is given by 𝜅𝜅(𝑡𝑡) = =
1
𝑑𝑑𝑻𝑻
. | 𝑑𝑑𝑑𝑑 (𝑡𝑡)|
|𝒗𝒗(𝑡𝑡)| 1
𝑑𝑑
𝑑𝑑𝒓𝒓 .|( 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑑𝑑 |
=0
1
𝑑𝑑
𝑚𝑚
(√1+𝑚𝑚 2 ), 𝑑𝑑𝑑𝑑 (√1+𝑚𝑚 2 )) (𝑡𝑡)| 𝑑𝑑𝒓𝒓
(Observe that there was no need to calculate |𝒗𝒗(𝑡𝑡)| = | 𝑑𝑑𝑑𝑑 (𝑡𝑡)| here as
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
= 𝟎𝟎.)
Hence, the curvature of a straight line is zero. Geometrically speaking, we see that the unit tangent vector 𝑻𝑻 always points in the same direction, therefore its components remain constant. (See figure 2)
Figure 2: Curvature of a straight line is zero. 1
Example 2: Show that the curvature of a circle of radius 𝑎𝑎 is 𝑎𝑎 .
Solution: We know that the equation of a circle of radius 𝑎𝑎 with centre at (0,0) in Cartesian form is given by 𝑥𝑥 2 + 𝑦𝑦 2 = 𝑎𝑎2
which can be parameterized as 𝒓𝒓(𝑡𝑡) = (𝑎𝑎 cos 𝑡𝑡, 𝑎𝑎 sin 𝑡𝑡) = 𝑎𝑎 cos 𝑡𝑡 𝒊𝒊 + 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 𝒋𝒋 .
This gives us,
𝒗𝒗(𝑡𝑡 ) =
=
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
𝑑𝑑
𝑑𝑑𝑑𝑑
[𝑎𝑎 cos 𝑡𝑡 𝒊𝒊 + 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 𝒋𝒋 ]
= (−𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡 )𝒊𝒊 + (𝑎𝑎 cos 𝑡𝑡)𝒋𝒋
and, hence we may see that
|𝒗𝒗(𝑡𝑡)| = �(−𝑎𝑎 sin 𝑡𝑡) 2 + (𝑎𝑎 cos 𝑡𝑡)2 = √𝑎𝑎2 = |𝑎𝑎| = 𝑎𝑎,
since, 𝑎𝑎 = radius of the circle is positive. Therefore, the unit tangent vector is
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 𝒗𝒗(𝑡𝑡)
𝑻𝑻 =
|𝒗𝒗(𝑡𝑡)| 1
=
𝑎𝑎
[(−𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡)𝒊𝒊 + (𝑎𝑎 cos 𝑡𝑡)𝒋𝒋]
= (− sin 𝑡𝑡)𝒊𝒊 + (cos 𝑡𝑡)𝒋𝒋.
This gives |
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
|= |(− cos 𝑡𝑡)𝒊𝒊 + (− 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡)𝒋𝒋 | = 1.
Hence, we finally obtain the curvature of the circle as
𝜅𝜅(𝑡𝑡) =
1
|𝒗𝒗(𝑡𝑡)|
𝑑𝑑𝑻𝑻
. � �= 𝑑𝑑𝑑𝑑
1
𝑎𝑎
1
(1) = . 𝑎𝑎
Intuitively also, a circle bends uniformly all through. The amount of bending is the reciprocal of its radius. 4.
The Principal Unit Normal for a Plane Curve:
So far, we have seen how important is the unit tangent vector 𝑻𝑻 of a smooth
plane curve 𝒓𝒓(𝑡𝑡) is. 𝑻𝑻 points in the direction in which the curve is turning. Since 𝑻𝑻 has constant length, the derivative Therefore, the vector bends). This vector
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
remains orthogonal to 𝑻𝑻.
points in the direction in which 𝑻𝑻 turns (as the curve
is called the Principal Normal of the curve 𝒓𝒓(𝑡𝑡) and if
we make its magnitude one, it becomes the Principal Unit Normal. Thus, we led to the following definition of the principal unit normal. 4.1 Definition (Principal Unit Normal): At a point where 𝜅𝜅 (𝑡𝑡) ≠ 0, the principal unit normal vector for a smooth curve in the plane is
𝑵𝑵 =
1 𝑑𝑑𝑻𝑻
𝜅𝜅 𝑑𝑑𝑑𝑑
.
So, if we move from 𝑷𝑷𝟎𝟎 to 𝑷𝑷𝟏𝟏 , the vector 𝑻𝑻 turns clockwise and the vector
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
points to the right. And from 𝑷𝑷𝟏𝟏 to 𝑷𝑷𝟐𝟐 , the vector 𝑻𝑻 turns anti-clockwise and the vector
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
points to the left.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
Figure 3: The Principal Unit Normal Vector remains orthogonal to the tangent vector 𝑻𝑻 at any point 𝑷𝑷 on the curve. We now compute an easy formula for 𝑵𝑵. 4.2 Formula for Calculating 𝑵𝑵: If 𝒓𝒓(𝑡𝑡) is a smooth curve, then the principal unit normal is 𝑑𝑑𝑻𝑻
𝑑𝑑𝑻𝑻
𝑵𝑵= 𝑑𝑑𝑑𝑑 �| 𝑑𝑑𝑑𝑑 |
where, 𝑻𝑻 =
𝒗𝒗
|𝒗𝒗|
is the unit tangent vector.
Proof: We use chain rule. We have 1 𝑑𝑑𝑻𝑻
𝑵𝑵 =𝜅𝜅 𝑑𝑑𝑑𝑑 1
(Definition) 𝑑𝑑𝑻𝑻
= 𝑑𝑑 𝑻𝑻 . 𝑑𝑑𝑑𝑑 ,
=
�
𝑑𝑑𝑑𝑑
|
�
1
𝑑𝑑 𝑻𝑻 𝑑𝑑𝑑𝑑 . | 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
1
= 𝑑𝑑 𝑻𝑻 = =
𝑑𝑑𝑑𝑑 | || | 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑻𝑻 𝑑𝑑𝑑𝑑 𝑑𝑑 𝑻𝑻 � � 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻
.
𝑑𝑑𝑻𝑻
since, 𝜅𝜅 = � 𝑑𝑑𝑑𝑑 �
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
( 𝑑𝑑𝑑𝑑 . 𝑑𝑑𝑑𝑑 ) (Chain Rule) 𝑑𝑑𝑻𝑻
1
1
1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 . � � 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻
1
( 𝑑𝑑𝑑𝑑 ). 𝑑𝑑𝑑𝑑
�| 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
(Because,
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
≥0 and hence,
𝑑𝑑𝑑𝑑
. 𝑑𝑑𝑑𝑑
1
𝑑𝑑𝑑𝑑 | | 𝑑𝑑𝑑𝑑
= 1)
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
Value Addition : Remark 𝑑𝑑𝑻𝑻
𝑑𝑑𝑻𝑻
𝑑𝑑𝑻𝑻
Since, 𝑵𝑵 = 𝑑𝑑𝑑𝑑 �| 𝑑𝑑𝑑𝑑 |, therefore, 𝑵𝑵. 𝑻𝑻 = [� 𝑑𝑑𝑑𝑑 � . 𝑻𝑻]
|
1
𝑑𝑑 𝑻𝑻 | 𝑑𝑑𝑑𝑑
= 0.
Example 3: Calculate 𝑵𝑵 for 𝒓𝒓(𝑡𝑡) = 𝑐𝑐𝑐𝑐𝑐𝑐 2𝑡𝑡 𝒊𝒊 + 𝑠𝑠𝑠𝑠𝑠𝑠 2𝑡𝑡 𝒋𝒋. Solution: We first calculate 𝑻𝑻. We have 𝒓𝒓(𝑡𝑡) = cos 2𝑡𝑡 𝒊𝒊 + sin 2𝑡𝑡 𝒋𝒋,
so that,
𝒗𝒗 =
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
and hence,
= (−2 sin 2𝑡𝑡) 𝒊𝒊 + (2cos 2𝑡𝑡) 𝒋𝒋
|𝒗𝒗| = �(−2 sin 2𝑡𝑡)2 + (2 cos 2𝑡𝑡)2 = 2.
Therefore, 𝑻𝑻 =
Hence,
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝒗𝒗
|𝒗𝒗|
= (− sin 2𝑡𝑡) 𝒊𝒊 + (cos 2𝑡𝑡) 𝒋𝒋.
= (−2 cos 2𝑡𝑡) 𝒊𝒊 + (−2 sin 2𝑡𝑡) 𝒋𝒋
which gives 𝑑𝑑𝑻𝑻
| 𝑑𝑑𝑑𝑑 | = �(−2 cos 2𝑡𝑡)2 + (−2 sin 2𝑡𝑡)2 = 2.
Hence,
𝑵𝑵 = (− cos 2𝑡𝑡) 𝒊𝒊 + (− sin 2𝑡𝑡) 𝒋𝒋. 5.
Circle of Curvature of a Plane Curve:
Usually to detect the shape of a curve around a point, we study the tangent to the curve at that point. But tangent is only a first order approximation to the curve. This is because when we calculate the equation of the tangent, we need to find out the slope of the tangent which requires differentiating 𝒓𝒓 (𝑡𝑡)
only once. Simply knowing the slope of the tangent does not reveal the
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves increasing-decreasing nature of the curve. For that we require the higher order derivatives. To get a better description of the geometry of the curve around any point 𝑷𝑷,
we consider a circle ing through 𝑷𝑷 such that the circle follows three basic rules: 1)
Circle is tangent to the curve at 𝑷𝑷. This means that the tangent to the circle ing through the point 𝑷𝑷 coincides with the
2)
tangent to the curve at 𝑷𝑷.
3)
The circle lies towards inner side of the curve.
Circle has the same curvature as the curve has at the point 𝑷𝑷.
All this can be observed in the following figure.
Figure 4: The osculating circle of a given curve at a point 𝑷𝑷. Looking at the condition (2), we reason as follows. Suppose the curvature of the curve at the point 𝑷𝑷 is 𝜅𝜅. Then the curvature of the circle at 𝑷𝑷 is also 𝜅𝜅. 1
Hence, the radius of the circle is 𝜅𝜅 . But
1
𝜅𝜅
is well defined, if and only if, 𝜅𝜅 ≠ 0.
Hence, we can define the circle of curvature at any point 𝑷𝑷 on the curve
provided 𝜅𝜅 ≠ 0 where 𝜅𝜅 = curvature of the curve at the point 𝑷𝑷. The centre Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves of circle of curvature at a point P is known as Centre of Curvature of the curve at the point P. The centre lies on line along 𝑵𝑵. Example 4: Find & graph the osculating circle of the parabola 𝑦𝑦 = 𝑥𝑥 2 at the origin.
Solution: The parabola 𝑦𝑦 = 𝑥𝑥 2
can be parameterized as 𝒓𝒓(𝑡𝑡) = 𝑡𝑡 𝒊𝒊 + 𝑡𝑡 2 𝒋𝒋 .
We first find the curvature at origin, that is 𝜅𝜅(0). We have,
𝒗𝒗 = 𝑑𝑑𝒓𝒓
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
= 𝒊𝒊 + 2𝑡𝑡 𝒋𝒋
⇒| 𝑑𝑑𝑑𝑑 |= √1 + 4𝑡𝑡 2
so that,
Hence,
𝑻𝑻 = 𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
=
Therefore,
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 | | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
−4𝑡𝑡
1
2𝑡𝑡
= (√1+4𝑡𝑡 2 )𝒊𝒊 + (√1+4𝑡𝑡 2 )𝒋𝒋.
3 (1+4𝑡𝑡 2 )2
2
𝒊𝒊 + [√1+4𝑡𝑡 2 -
16𝑡𝑡 2
𝑑𝑑𝑻𝑻
2
1
3
(1+4𝑡𝑡 2 )2
| 𝑑𝑑𝑑𝑑 (𝑡𝑡)|= �1+4𝑡𝑡 2 )3 + [√1+4𝑡𝑡 2 −
Since,
8𝑡𝑡 2
(A)
]𝒋𝒋.
8𝑡𝑡 2
3
(1+4𝑡𝑡 2 )2
]2 .
𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) = |𝒗𝒗(𝑡𝑡)| | 𝑑𝑑𝑑𝑑 (𝑡𝑡)|,
therefore, the curvature at origin is 1
𝑑𝑑𝑻𝑻
𝜅𝜅(0) = |𝒗𝒗(0)| | 𝑑𝑑𝑑𝑑 (0)|=
1
√1
. √22 = 2.
Hence, the radius of curvature at origin is
1 2
unit.
Now to locate the centre, we compute 𝑵𝑵(0). Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves We know that 𝑵𝑵 (𝑡𝑡) =
⇒ 𝑵𝑵(0) =
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
By (A), we get,
(𝑡𝑡)
(0)
𝑵𝑵(0) = 0𝒊𝒊 + 2𝒋𝒋
Hence, the centre of the radius of curvature lies on 𝑥𝑥 = 0 , that is the 𝑦𝑦 − axis 1
and therefore, has the co-ordinates (0, 𝑎𝑎)for some a ∈ 𝑅𝑅. Since radius is 2, 1
1
that is, 𝑎𝑎 = 2, therefore, the centre of the radius of curvature is (0, 2). Hence, the equation of the circle of curvature at (0,0) is 1
1
(𝑥𝑥 − 0)2 + (𝑦𝑦 − 2)2 = (2)2 1
⇒ 𝑥𝑥 2 + (𝑦𝑦 − 2)2 =
1 4
Figure 4: The osculating circle of the parabola 𝑦𝑦 = 𝑥𝑥 2 at the origin. The above figure exhibits the circle of curvature at the origin. 6.
Curves in Space: Curvature & Normal Vectors
The analysis carried out for planes curves remains completely unchanged for curves in space. That is, if a smooth curve is 𝒓𝒓(𝑡𝑡), then we have the following formulae:
Unit Tangent Vector: 𝑻𝑻 =
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 | | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
=
𝒗𝒗
|𝒗𝒗|
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 𝑑𝑑𝑻𝑻
Curvature Function: 𝜅𝜅(𝑡𝑡) = | 𝑑𝑑𝑑𝑑 | = 1 𝑑𝑑𝑻𝑻
Principal Unit Normal: 𝑵𝑵 =
=
𝜅𝜅 𝑑𝑑𝑑𝑑
1
𝑑𝑑𝑻𝑻
. | 𝑑𝑑𝑑𝑑 | |𝒗𝒗| 1 𝑑𝑑𝑻𝑻
𝑑𝑑 𝑻𝑻 | 𝑑𝑑𝑑𝑑
|
𝑑𝑑𝑑𝑑
.
Example 5: Find curvature & principal unit normal of the helix 𝒓𝒓(𝑡𝑡) = 𝑎𝑎 cos 𝑡𝑡 𝒊𝒊 + 𝑎𝑎 sin 𝑡𝑡 𝒋𝒋 + 𝑏𝑏𝑏𝑏 𝒌𝒌,
where, 𝑎𝑎, 𝑏𝑏 ≥ 0, 𝑎𝑎2 + 𝑏𝑏 2 ≠ 0.
Solution: We have 𝒗𝒗 =
𝑑𝑑𝒓𝒓 𝑑𝑑𝑑𝑑
= (−𝑎𝑎 sin 𝑡𝑡) 𝒊𝒊 + (𝑎𝑎 cos 𝑡𝑡) 𝒋𝒋 + (𝑏𝑏)𝒌𝒌
𝑑𝑑𝑑𝑑
⇒|𝒗𝒗 |= | 𝑑𝑑𝑑𝑑 | = �(𝑎𝑎 sin 𝑡𝑡)2 + (𝑎𝑎 cos 𝑡𝑡)2 + 𝑏𝑏 2 = √𝑎𝑎2 + 𝑏𝑏 2 .
Hence,
𝑻𝑻 =
𝒗𝒗
|𝒗𝒗|
Therefore, 𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
𝑑𝑑𝑻𝑻
− a sin 𝑡𝑡
= (√𝑎𝑎 2
−a cos 𝑡𝑡
= (√𝑎𝑎 2
+𝑏𝑏 2
+𝑏𝑏 2
−a sin 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
−𝒂𝒂 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 ) + 𝑏𝑏 2
⇒| 𝑑𝑑𝑑𝑑 | =�(√𝑎𝑎 2
Hence,
𝜅𝜅(𝑡𝑡) =
1
a cos 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2 +𝑏𝑏 2
)𝒋𝒋
−𝒂𝒂 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 2 ) + 𝑏𝑏 2
+ (√𝑎𝑎 2
𝑑𝑑𝑻𝑻
+𝑏𝑏 2
. | 𝑑𝑑𝑑𝑑 | = |𝒗𝒗(𝑡𝑡)|
1
√𝑎𝑎 2 + 𝑏𝑏 2
.
𝑏𝑏
)𝒋𝒋 + (√𝑎𝑎 2
+𝑏𝑏 2
)𝒌𝒌.
𝑎𝑎
= √𝑎𝑎 2 𝑎𝑎
√𝑎𝑎 2 + 𝑏𝑏 2
+ 𝑏𝑏 2
=
𝑎𝑎
𝑎𝑎 2 +𝑏𝑏 2
𝑎𝑎
Therefore, the curvature of helix is the constant 𝑎𝑎 2 + 𝑏𝑏 2 .
Thus, if b increases for any fixed value of a, the curvature decreases. But increasing b means physically stretching it. This explains why stretching the string straightens it, because while stretching the string, b increases and, hence, curvature decreases, so that, the string looks more like a straight line. For principal unit normal, we have, 𝑵𝑵 = =
𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑 |𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑 | 1
𝑎𝑎 �𝑎𝑎 2 + 𝑏𝑏 2
−a cos 𝑡𝑡
[(√𝑎𝑎 2
+𝑏𝑏 2
−a sin 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
+𝑏𝑏 2
)𝒋𝒋]
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Curvature and Torsion of Curves so that 𝑵𝑵(𝑡𝑡) = (− cos 𝑡𝑡) 𝒊𝒊 + (− sin 𝑡𝑡) 𝒋𝒋. 7.
Unit Binormal Vector for a Space Curve:
Travelling along a space curve, the forward direction of motion (𝑻𝑻) and the
direction in which our path is turning (𝑵𝑵) and the direction of motion to
move away from the plane of 𝑻𝑻 and 𝑵𝑵 describe the motion even more clearly than the usual description in of unit vectors 𝒊𝒊, 𝒋𝒋, 𝒌𝒌. That is, the vectors 𝑻𝑻, 𝑵𝑵 and 𝑻𝑻 × 𝑵𝑵 describe the geometry of the space curve quite completely.
This leads us to formulate the definition of the binormal vector (𝑩𝑩) of a curve in space.
7.1 Definition (Unit Binormal Vector of a Space Curve): The unit binormal vector of a space curve is a unit vector in the plane perpendicular to the plane containing 𝑻𝑻 and 𝑵𝑵 and is, therefore, defined by 𝑩𝑩
𝑩𝑩𝟏𝟏 = |𝑩𝑩|,
where, 𝑩𝑩 = Binormal Vector = 𝑻𝑻 × 𝑵𝑵.
Sometimes, 𝑩𝑩𝟏𝟏 and 𝑩𝑩 are used interchangeably.
Value Addition: Remarks • •
Together 𝑻𝑻, 𝑵𝑵 and 𝑩𝑩 define a moving right handed vector frame 𝑻𝑻𝑻𝑻𝑻𝑻.
This 𝑻𝑻𝑻𝑻𝑻𝑻 frame is also called Frenet Frame (“fre-nay”) (named
after Jean-Frederic Frenet (1816-1900)). 8.
Torsion of a Space Curve:
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Curvature and Torsion of Curves The plane containing 𝑻𝑻 and 𝑵𝑵 is called osculating plane. Torsion describes how this plane turns about 𝑻𝑻 as the body moves along the curve. Torsion is
a scalar quantity.
Let 𝑩𝑩 = unit binormal vector of the space curve. Then,
𝑑𝑑𝑩𝑩
=
𝑑𝑑𝒔𝒔
𝑑𝑑
𝑑𝑑𝑑𝑑
(𝑻𝑻 × 𝑵𝑵) =
𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
× 𝑵𝑵 + 𝑻𝑻 ×
𝑑𝑑𝑻𝑻
𝑑𝑑𝑵𝑵 𝑑𝑑𝑑𝑑
Since, 𝑵𝑵 is the unit vector along 𝑑𝑑𝑑𝑑 , therefore, 𝑑𝑑𝑻𝑻
x 𝑵𝑵 = 𝟎𝟎.
𝑑𝑑𝑑𝑑
Hence,
𝑑𝑑𝑩𝑩
Thus,
𝑑𝑑𝑑𝑑
𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
= 𝑻𝑻 ×
𝑑𝑑𝑵𝑵 𝑑𝑑𝑑𝑑
is orthogonal to 𝑻𝑻 as well as to 𝑩𝑩.
(This is because |𝑩𝑩 | = 1, that is, 𝑩𝑩 has constant length). Hence,
𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
is orthogonal to the plane of 𝑩𝑩 and 𝑻𝑻. 𝑑𝑑𝑩𝑩
In other words
𝑑𝑑𝑑𝑑
That is, 𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
is parallel to 𝑵𝑵 and is, therefore, a scalar multiple of 𝑵𝑵.
= −𝜏𝜏𝑵𝑵.
The scalar 𝜏𝜏 is called torsion along the curve. The negative sign in this
equation is simply for convention. (It simplifies matters in differential geometry.) We may see that 𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
Hence,
. 𝑵𝑵 = −(𝜏𝜏𝑵𝑵). 𝑵𝑵 = −𝜏𝜏(𝑵𝑵. 𝑵𝑵) = −𝜏𝜏(1) = −𝜏𝜏 𝑑𝑑𝑩𝑩
𝜏𝜏 = − 𝑑𝑑𝑑𝑑 𝑵𝑵.
Hence, we may now give the definition of torsion of a space curve.
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Curvature and Torsion of Curves 8.1 Definition (Torsion of a smooth curve): Let 𝒓𝒓(𝑡𝑡) be a smooth curve 𝑑𝑑𝑩𝑩
in space & let 𝑩𝑩 = 𝑻𝑻 × 𝑵𝑵.The torsion of the curve is 𝜏𝜏 = − 𝑑𝑑𝑑𝑑 𝑵𝑵. Value Addition: Remarks • •
𝜏𝜏 can be positive, negative or zero. If 𝜏𝜏 = 0, then
𝑑𝑑𝑩𝑩
= 𝟎𝟎 because 𝑵𝑵 ≠ 𝟎𝟎. Hence, there is no change in 𝑩𝑩
𝑑𝑑𝑑𝑑
along the arc-length and therefore, for plane curves, 𝜏𝜏 = 0.
Example 6: Find 𝑩𝑩 & 𝜏𝜏 for the helix 𝒓𝒓(𝑡𝑡) = (𝑎𝑎 cos 𝑡𝑡) 𝒊𝒊 + (𝑎𝑎 sin 𝑡𝑡) 𝒋𝒋 + (𝑏𝑏𝑏𝑏)𝒌𝒌. Solution: We have, from example 5,
and
− a sin 𝑡𝑡
𝑻𝑻 = (√𝑎𝑎 2
+𝑏𝑏 2
a cos 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
+𝑏𝑏 2
𝑏𝑏
)𝒋𝒋 + (√𝑎𝑎 2
+𝑏𝑏 2
)𝒌𝒌
𝑵𝑵 = (− cos 𝑡𝑡) 𝒊𝒊 + (− sin 𝑡𝑡) 𝒋𝒋.
Hence,
𝑩𝑩 = 𝑻𝑻 × 𝑵𝑵 =
i −a sin t
j a cos t
k b
a 2 + b2 − cos t
a 2 + b2 − sin t
a 2 + b2 0
which gives
Now,
and
𝑏𝑏 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡
)𝒊𝒊 + (√𝑎𝑎 2
𝑩𝑩 = (√𝑎𝑎 2 𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
+𝑏𝑏 2
= (√𝑎𝑎 2
+𝑏𝑏 2
−b cos 𝑡𝑡 +𝑏𝑏 2
b sin 𝑡𝑡
𝑎𝑎
)𝒋𝒋 + (√𝑎𝑎 2
+𝑏𝑏 2
+𝑏𝑏 2
)𝒌𝒌
)𝒋𝒋
|𝒗𝒗| = √𝑎𝑎2 + 𝑏𝑏 2 , by example (5).
Hence,
𝑑𝑑𝑩𝑩 𝑑𝑑𝑑𝑑
= =
1 𝑑𝑑𝑩𝑩
|𝒗𝒗| 𝑑𝑑𝑑𝑑 1
[
𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡
√𝑎𝑎 2 +𝑏𝑏 2 √𝑎𝑎 2 +𝑏𝑏 2
𝒊𝒊 +
b sin 𝑡𝑡
√𝑎𝑎 2 +𝑏𝑏 2
𝒋𝒋]
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Curvature and Torsion of Curves =
𝑏𝑏 cos 𝑡𝑡
𝑎𝑎 2 + 𝑏𝑏 2
Therefore,
𝒊𝒊 +
𝑏𝑏 sin 𝑡𝑡
𝑎𝑎 2 + 𝑏𝑏 2
𝑑𝑑𝑩𝑩
𝜏𝜏 = − 𝑑𝑑𝑑𝑑 . 𝑵𝑵
𝑏𝑏 cos 𝑡𝑡
= −[ 𝑎𝑎 2 + 𝑏𝑏 2 𝒊𝒊 + =
𝑏𝑏
𝑎𝑎 2 + 𝑏𝑏 2
𝒋𝒋
𝑏𝑏 sin 𝑡𝑡
𝑎𝑎 2 + 𝑏𝑏 2
𝒋𝒋]. [(−𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡) 𝒊𝒊 + (−𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡) 𝒋𝒋]
The given helix is shown below.
Figure 5: The helix for example 6.
In the general setting of space curves, often given in parameterized form, we have some formulae for calculation of curvature and torsion. The derivations are cumbersome and therefore, omitted. 8.2 Formulae for calculating the Torsion and Curvature: We state the formulae for calculating the torsion and curvature of the space curve 𝒓𝒓(𝑡𝑡) = (𝑥𝑥(𝑡𝑡), 𝑦𝑦(𝑡𝑡), 𝑧𝑧(𝑡𝑡)). For curvature, we have the following formula:
𝜅𝜅(𝑡𝑡) =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
.
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Curvature and Torsion of Curves
For torsion, we have the following formula:
𝜏𝜏 =
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
.
Of course, we assume that 𝒗𝒗 ≠ 𝟎𝟎 and 𝒗𝒗 × 𝒂𝒂 ≠ 𝟎𝟎. Further, 𝑥𝑥 ′ (𝑡𝑡) = Value Addition: Remarks
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
, and so on.
The tuple (𝑻𝑻, 𝜅𝜅, 𝑵𝑵, 𝑩𝑩, 𝜏𝜏) is known as Frenet Apparatus. 9.
Functions of Several Variables: Introduction
In practical life, we encounter various problems whose solutions are influenced by various factors. For instance, in transportation, stockmarkets, economics, finance, psychology, statistics, physics, geometry one is very naturally led to consider functions of more than one variable. As an elementary instance, the volume 𝑉𝑉 of a cylinder 𝑉𝑉 = 𝜋𝜋𝑟𝑟 2 ℎ is a function of the
radius & height of the cylinder. In atmospheric sciences, the speed of the wind at any place on earth is a function of the latitude & longitude of that place. As a first introduction, we shall learn how to graph these functions. We begin with the following definitions. 9.1 Definition (Function of n independent variables): Suppose 𝐷𝐷 is a
set of 𝑛𝑛-tuples of real numbers (𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 ). A real-valued function 𝑓𝑓 on 𝐷𝐷 is
a rule that assigns a unique (single) real number 𝑤𝑤 = 𝑓𝑓(𝑥𝑥1 , 𝑥𝑥2 , … , 𝑥𝑥𝑛𝑛 )
to each element in 𝐷𝐷. The set 𝐷𝐷 is the function’s domain. The set of 𝑤𝑤-
values taken on by 𝑓𝑓 is the function’s range. The symbol 𝑤𝑤 is the dependent variable of 𝑓𝑓, and 𝑓𝑓 is said to be a function of the 𝑛𝑛
independent variables 𝑥𝑥1 to 𝑥𝑥𝑛𝑛 . We also call the 𝑥𝑥𝑗𝑗 ′s the function’s input variables and call 𝑤𝑤 the function’s output variable.
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Curvature and Torsion of Curves
10.
Graphs & Level Curves:
We shall study the graphing of only two and three variable functions. We begin with the two-variable case. There are two standard ways to view the values of a function 𝑓𝑓(𝑥𝑥, 𝑦𝑦). One way is to draw the curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝐶𝐶 in the 𝑥𝑥 − 𝑦𝑦 plane where 𝐶𝐶 is any point in the range of 𝑓𝑓. This is called the levelcurve technique. The other way is to draw the surface 𝑧𝑧 = 𝑓𝑓(𝑥𝑥, 𝑦𝑦) in space. This surface 𝑧𝑧 = 𝑓𝑓(𝑥𝑥, 𝑦𝑦) is called the graph of the function 𝑓𝑓(𝑥𝑥, 𝑦𝑦). Example 7: Graph 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 100 − 𝑥𝑥 2 − 𝑦𝑦 2 ,
and plot the level curves 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0, 51, 75
in the domain of 𝑓𝑓 in the plane.
Solution: The function 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 100 − 𝑥𝑥 2 − 𝑦𝑦 2 is continuous ∀ (𝑥𝑥, 𝑦𝑦)𝜀𝜀 ℝ2 .
Therefore domain of 𝑓𝑓 is the entire plane. We first analyse the level curves. For the level curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0,
we see that,
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0
⇒ 100 = 𝑥𝑥 2 + 𝑦𝑦 2 .
Hence, the level curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 0 is the circle of radius 10 centred at (0,0). Similarly, the level curve 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 51
is the circle
𝑥𝑥 2 + 𝑦𝑦 2 = 49;
and the level curve
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Curvature and Torsion of Curves 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 75
is the circle
𝑥𝑥 2 + 𝑦𝑦 2 = 25.
The graph and the level curves are shown below.
Figure 6: The graph and level curves for example 7. Let us now take up the 3-variable case. Example 8: Describe the level surfaces of the function 𝑔𝑔(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 .
Solution: The value of 𝑔𝑔 at any point (𝑥𝑥, 𝑦𝑦, 𝑧𝑧) is the distance of the point
(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) from the origin (0,0,0). A level surface is given by �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 𝑐𝑐, where 𝑐𝑐 is a real number. Clearly, 𝑐𝑐 cannot be negative. For positive values
of 𝑐𝑐, each level surface �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 𝑐𝑐, is a sphere of radius 𝑐𝑐 centered at
origin. For 𝑐𝑐 = 0 the level surface �𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 0 reduces to the point (0,0,0).
The figure is given below.
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Curvature and Torsion of Curves
Figure 7: The graph and level curves for example 8.
11. Summary and Important Formulae for curves in plane and space
•
Unit Tangent Vector:
•
Curvature Function:
•
Principal Unit Normal Vector:
•
Binormal Vector:
•
Torsion of a Space Curve:
•
Formula for the Curvature:
•
Formula for the Torsion:
𝑻𝑻 =
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 𝑑𝑑𝑑𝑑
| | 𝑑𝑑𝑑𝑑
=
𝟏𝟏
𝒗𝒗
|𝒗𝒗|
𝑑𝑑𝑻𝑻
𝜅𝜅(𝑡𝑡) = |𝒗𝒗(𝒕𝒕)| | 𝑑𝑑𝑑𝑑 | 𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑
1 𝑑𝑑𝑻𝑻
𝑵𝑵 = |𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑 | = 𝑵𝑵 = 𝜅𝜅 𝑑𝑑𝑑𝑑 𝑩𝑩 = 𝑻𝑻 × 𝑵𝑵 𝑑𝑑𝑩𝑩
𝜏𝜏 = − 𝑑𝑑𝑑𝑑 𝑵𝑵 𝜅𝜅(𝑡𝑡) =
𝜏𝜏 =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
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Curvature and Torsion of Curves
Exercises Find 𝑻𝑻, 𝑵𝑵 and 𝜿𝜿 for the Plane Curves in Exercises 1-3 and Space Curves
in Exercises 4 to 6. Find 𝑩𝑩 and 𝝉𝝉 for the Space Curves in Exercises 4 to
6.
1. 2. 3. 4. 5. 6.
𝒓𝒓(𝑡𝑡) = 𝑡𝑡 𝒊𝒊 + 𝑙𝑙𝑙𝑙(𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡)𝒋𝒋
𝒓𝒓(𝑡𝑡) = (2𝑡𝑡 + 3)𝒊𝒊 + (5 − 𝑡𝑡 2 )𝒋𝒋 𝒓𝒓(𝑡𝑡) = (𝑐𝑐𝑐𝑐𝑐𝑐 3 𝑡𝑡)𝒊𝒊 + (𝑠𝑠𝑠𝑠𝑠𝑠3 𝑡𝑡)𝒋𝒋
𝒓𝒓(𝑡𝑡) = (3 sin 𝑡𝑡) 𝒊𝒊 + (3 cos 𝑡𝑡)𝒋𝒋 + 4𝑡𝑡 𝒌𝒌
𝒓𝒓(𝑡𝑡) = (cos 𝑡𝑡 + 𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)𝒊𝒊 + (sin 𝑡𝑡 − 𝑡𝑡 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)𝒋𝒋 + 3𝒌𝒌 𝒓𝒓(𝑡𝑡) = (𝑒𝑒 𝑡𝑡 cos 𝑡𝑡)𝒊𝒊 + (𝑒𝑒 𝑡𝑡 sin 𝑡𝑡)𝒋𝒋 + 2𝒌𝒌
Plot the typical (a) level curves and (b) graphs for the functions in exercises 7 to 9.
7. 8. 9.
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑦𝑦 2
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 4 − 𝑦𝑦 2
𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥 2 + 𝑦𝑦 2
Plot a typical level surface for the functions in exercises 10 to 12.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 10. 11.
𝑓𝑓(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2
12.
𝑓𝑓(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑧𝑧
13.
Determine the curvature and torsion along the curve
𝑓𝑓(𝑥𝑥, 𝑦𝑦, 𝑧𝑧) = 𝑦𝑦 2 + 𝑧𝑧 2
𝒓𝒓(𝑡𝑡) = (𝑡𝑡 − 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡, 1 − cos 𝑡𝑡 , 𝑡𝑡). 14. 15.
Show that the curve 𝒓𝒓(𝑡𝑡) = (𝑡𝑡,
1+𝑡𝑡 1−𝑡𝑡 2 𝑡𝑡
,
𝑡𝑡
) lies in a plane.
Find the principle unit normal and unit binormal along the space curve 𝒓𝒓(𝑡𝑡) = (3𝑡𝑡 − 𝑡𝑡 3 , 3𝑡𝑡 2 , 3𝑡𝑡 + 𝑡𝑡 3 ). Further, prove that the curvature and the torsion along the curve are same.
Glossary Curvature, unit tangent, normal, unit binormal, torsion Further Reading It is always welcome to practice more exercises from various books available in libraries and elsewhere. 1. Calculus and Analytic Geometry (11th Edition) by George B. Thomas and Ross L. Finney; ISBN: 13: 978-0201531749 2. Calculus by Howard Anton and Irl C. Bivens (10th Edition); ISBN: 13: 978-0470647691
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves Solutions For Exercises 𝑻𝑻, 𝑵𝑵 and 𝜿𝜿 for Exercises 1-3 1.
2.
3.
𝑩𝑩 and 𝜏𝜏 for Exercises 4-6 4.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
5.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 6.
Level Curves and Graphs for Exercises 7 to 9 7.
8.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
9.
Level Surfaces for Exercises 10 to 12 10.
11.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
12.
13.
The given curve 𝒓𝒓(𝑡𝑡) = (𝑡𝑡 − 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡, 1 − cos 𝑡𝑡 , 𝑡𝑡) is a space curve and hence we use the formula:
𝜅𝜅(𝑡𝑡) =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
Now, we see that,
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 𝑑𝑑𝒓𝒓
𝒗𝒗(𝑡𝑡) = 𝑑𝑑𝑑𝑑 = (1 − cos 𝑡𝑡 , sin 𝑡𝑡 , 1) and hence,
𝒂𝒂(𝑡𝑡) =
𝑑𝑑𝒗𝒗 𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝒓𝒓
= 𝑑𝑑𝑡𝑡 2 = (sin 𝑡𝑡, cos 𝑡𝑡, 0).
Curvature:
i j k We have, 𝒗𝒗 × 𝒂𝒂 = 1 − cos t sin t 1 sin t cos t 0
= (− cos 𝑡𝑡)𝒊𝒊 + (sin 𝑡𝑡)𝒋𝒋 + (cos 𝑡𝑡 − 1)𝒌𝒌.
Hence,
|𝒗𝒗 × 𝒂𝒂| = �(−𝑐𝑐𝑐𝑐𝑐𝑐 𝑡𝑡)𝟐𝟐 + (sin 𝑡𝑡) 𝟐𝟐 + (cos 𝑡𝑡 − 1) 𝟐𝟐 𝑡𝑡
= �1 + 4𝑠𝑠𝑠𝑠𝑠𝑠4 (2);
and |𝒗𝒗| = �(1 − cos 𝑡𝑡)2 + (sin 𝑡𝑡) 2 + 1 𝑡𝑡
= �1 + 4𝑠𝑠𝑠𝑠𝑠𝑠2 (2).
Thus, the curvature along the given curve is
𝜅𝜅(𝑡𝑡) =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
=
𝑡𝑡 2
�1+4𝑠𝑠𝑠𝑠𝑠𝑠 4 ( ) 𝑡𝑡 2
3
(1+4𝑠𝑠𝑠𝑠𝑠𝑠 2 ( ))2
.
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
Torsion: We have, x ' x '' x ''' 1 − cos t sin t y ' y '' y ''' = sin t cos t z ' z '' z ''' 1 0
cos t − sin t 0
= −1. Hence, torsion along the given curve is
𝜏𝜏 =
14.
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
=
−1
𝑡𝑡 2
.
1+4𝑠𝑠𝑠𝑠𝑠𝑠 4 ( )
To show that the given curve
𝒓𝒓(𝑡𝑡) = (𝑡𝑡,
1+𝑡𝑡 1−𝑡𝑡 2 𝑡𝑡
,
𝑡𝑡
)
lies in a plane, it is enough to show that the torsion of the curve is zero. So, we calculate the torsion of the curve. We have,
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves
1 x ' x '' x ''' 1 y ' y '' y ''' = 1 − 2 t z ' z '' z ''' 1 −1 − 2 t
0 2 t3 2 t3
0 6 − 4 = 0. t 6 − 4 t
Therefore,
𝜏𝜏 =
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
= 0.
Hence, the given curve is a plane curve. 15. The space curve given to us is 𝒓𝒓(𝑡𝑡) = (3𝑡𝑡 − 𝑡𝑡 3 , 3𝑡𝑡 2 , 3𝑡𝑡 + 𝑡𝑡 3 ). Hence, the velocity vector 𝒗𝒗(𝑡𝑡) along the curve is 𝑑𝑑𝒓𝒓
𝒗𝒗(𝑡𝑡) = 𝑑𝑑𝑑𝑑 = (3 − 3𝑡𝑡 2 , 6𝑡𝑡, 3 + 3𝑡𝑡 2 ), and hence, the speed |𝒗𝒗(𝑡𝑡)| along the curve is |𝒗𝒗(𝑡𝑡)| = �(3 − 3𝑡𝑡 2 )2 + 36𝑡𝑡 2 + (3 + 3𝑡𝑡 2 )2 = 3√2(1 + 𝑡𝑡 2 ). Hence, the unit tangent vector 𝑻𝑻 along the curve is
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 𝑻𝑻 =
1 𝑑𝑑𝒓𝒓
𝑑𝑑 𝒓𝒓 | | 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝒗𝒗
1
= |𝒗𝒗| = 3√2(1+𝑡𝑡 2 ) (3 − 3𝑡𝑡 2 , 6𝑡𝑡, 3 + 3𝑡𝑡 2 )
which simplifies to
𝑻𝑻 =
1
√2(1+𝑡𝑡 2 )
(1 − 𝑡𝑡 2 , 2𝑡𝑡, 1 + 𝑡𝑡 2 ).
Principal Unit Normal: We have, 𝑑𝑑𝑻𝑻 𝑑𝑑𝑑𝑑
�1−𝑡𝑡 2 �
2√2𝑡𝑡
= (− (1+𝑡𝑡 2 )2 , √2 (1+𝑡𝑡 2 )2 , 0)
and therefore,
𝑑𝑑𝑻𝑻
(1−𝑡𝑡 2 )
2√2𝑡𝑡
2
√2
� 𝑑𝑑𝑑𝑑 � = �(− (1+𝑡𝑡 2 )2 )2 + (√2 (1+𝑡𝑡 2 )2 )2 = �(1+𝑡𝑡 2 )2 = (1+𝑡𝑡 2 ). Hence, the principal unit normal 𝑵𝑵 along the curve is 𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑
𝑵𝑵 = |𝑑𝑑𝑻𝑻⁄𝑑𝑑𝑑𝑑 | = √𝟐𝟐
1
� (1+𝑡𝑡 2 )
2√2𝑡𝑡
�1−𝑡𝑡 2 �
(− (1+𝑡𝑡 2 )2 , √2 (1+𝑡𝑡 2 )2 , 0)
and this simplifies to 2𝑡𝑡
1−𝑡𝑡 2
𝑵𝑵 = (− 1+𝑡𝑡 2 , 1+𝑡𝑡 2 , 0). Principal Unit Binormal:
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves The principal unit binormal 𝑩𝑩 along the curve is the crossproduct of the vectors 𝑻𝑻 and 𝑵𝑵 and is given by
𝑩𝑩 = 𝑻𝑻 × 𝑵𝑵 =
i
j
k
(1 − t 2 ) 2 1+ t2
2t 2 1+ t2
(1 + t 2 ) 2 1+ t2
(
−
)
2t 1+ t2
(
1− t2 1+ t2
)
(
)
0
and this simplifies to
𝑩𝑩 =
1 2 2 ) (𝑡𝑡 (1+𝑡𝑡 √2
− 1, −2𝑡𝑡, 𝑡𝑡 2 + 1).
Curvature: The acceleration vector 𝒂𝒂(𝑡𝑡) of the given curve is 𝒂𝒂(𝑡𝑡) =
𝑑𝑑𝒗𝒗 𝑑𝑑𝑑𝑑
= (−6𝑡𝑡, 6, 6𝑡𝑡).
Hence, we have
i 𝒗𝒗 × 𝒂𝒂 = 3 − 3t 2 −6t
j k 6t 3 + 3t 2 = 18(𝑡𝑡 2 − 1)𝒊𝒊 − 36𝑡𝑡 𝒋𝒋 + 18(𝑡𝑡 2 + 1)𝒌𝒌 6 6t
Hence, the curvature 𝜅𝜅(𝑡𝑡) of the given curve is
Institute of Lifelong Learning, University of Delhi
Curvature and Torsion of Curves 𝜅𝜅(𝑡𝑡) =
|𝒗𝒗×𝒂𝒂| |𝒗𝒗|3
=
|18�𝑡𝑡 2 −1�𝒊𝒊−36𝑡𝑡 𝒋𝒋+18�𝑡𝑡 2 +1�𝒌𝒌| |3−3𝑡𝑡 2 𝒊𝒊+ 6𝑡𝑡𝒋𝒋+ 3+3𝑡𝑡 2 𝒌𝒌|3
2
= 3(1+𝑡𝑡 2 )2 .
Torsion: We now see that
3 − 3t 2 x ' x '' x ''' y ' y '' y ''' = 6t z ' z '' z ''' 3 + 3t 2
−6t 6 6t
−6 0 = 216, 6
and i 𝒗𝒗 × 𝒂𝒂 = 3 − 3t 2 −6t
j k 6t 3 + 3t 2 = 18(𝑡𝑡 2 − 1)𝒊𝒊 − 36𝑡𝑡 𝒋𝒋 + 18(𝑡𝑡 2 + 1)𝒌𝒌 6 6t
and hence, the torsion 𝜏𝜏 of the given curve is
𝜏𝜏 =
x ' x '' x ''' y ' y '' y ''' z ' z '' z ''' |𝒗𝒗×𝒂𝒂|2
216
2
= |18(𝑡𝑡 2 −1)𝒊𝒊−36𝑡𝑡 𝒋𝒋+18(𝑡𝑡 2 +1)𝒌𝒌|2 = 3(1+𝑡𝑡 2 )2
and therefore, 𝜅𝜅 = 𝜏𝜏.
Institute of Lifelong Learning, University of Delhi
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