Vector-valued functions

Transcription

1 Vector-valued functions MATH Victoria University of Wellington Wellington, New Zealand For a complimentary reading, this set of slides corresponds roughly to Chapter of the same title from Anton, Bivens, Davis: Calculus - Early Transcendental

2 This chapter s content 1 Introduction to vector-valued functions 3 2 Calculus of vector-valued functions 14 3 Change of parameter; Arc length 48 4 Unit tangents, normals and binormals 63 5 Curvature 71 6 Motion along a curve 86 7 Kepler s Laws of planetary motion 90 2

3 1 Introduction to vector-valued functions ABD section: Introduction to vector-valued functions 3

4 Parametric curves in 3D Warm up: Parametric curves in 2D Given two continuous functions f, g of a single variable t, i.e. f : R 1 t R 1 x, g : R 1 t R 1 y x = f (t), and thus a point C(t) in 2D-space R 2 = R 3 x,y : y = g(t). C(t) = (x, y) = (f (t), g(t)) R 2 x,y. C : R 1 R 2 C(t) = (f (t), g(t)) Increasing t, this point moves continuously in 2D space, tracing out a parametrised curve C(t): each point on the curve inherits a label the parameter t which determines a direction ( orientation ) of the curve. ABD section: Introduction to vector-valued functions 4

5 Examples: familiar parametrised curves in 2D Ordinary graphs: y = g(x) in R 2 = R 2 x,y t as x x = f (t) = t, y = g(t). (x, y) = (t, g(t)) = (x, g(x)) since t = x. Curve C : the usual graph of y = g(x) in the plane, parametrized by x. Polar curves : r = r(θ) such as r = cos(2θ) (in R 2 = R 2 x,y) t as θ x = f (t) = r(t) cos t, y = g(t) = r(t) sin t. (x, y) = (r(t) cos t, r(t) sin t) = (r(θ) cos θ, r(θ) sin θ) if t = θ. Curve C : the polar curve of r = r(θ) in the plane, parametrized by θ. ABD section: Introduction to vector-valued functions 5

6 Parametric curves in 3D Given three continuous functions f, g, h of a single variable t, i.e. f : R 1 t R 1 x, g : R 1 t R 1 y, h : R 1 t R 1 z, x = f (t), y = g(t) z = h(t). and thus a point C(t) in 3D-space R 3 = R 3 x,y,z : C(t) = (x, y, z) = (f (t), g(t), h(t)) R 3 x,y,z. C : R 1 R 3 C(t) = (f (t), g(t), h(t)) Increasing t, this point moves continuously in 3D space, tracing out a parametrised curve C(t): each point on the curve inherits a label the parameter t which determines a direction ( orientation ) of the curve. ABD section: Introduction to vector-valued functions 6

7 Parametric curves in 3D Helix in R 3. x = a cos t, y = a sin t, z = ct x = cos t, y = sin t, z = t Fig.: RobHar, wikipedia helix ABD section: Introduction to vector-valued functions 7

8 Example: an intersection of two surfaces: z = x 3 and y = x 2!Twisted cubic" ABD section: Introduction to vector-valued functions 8

9 Fig.: Adapted from Povray code, Claudio Rocchini, wikipedia twisted cubic ABD section: Introduction to vector-valued functions 9 Example: How to parametrize the twisted cubic curve {(x, y, z) y = x 2 } {(x, y, z) z = x 3 } = {(x, y, z) y = x 2 and z = x 3 } Notice that both y and z are functions of x: Set x = t, then y = t 2, z = t 3 t (t, t 2, t 3 )

10 Vector-valued functions Write x = t, y = t 2, z = t 3 in vector form : r(t) = x(t), y(t), z(t) = t, t 2, t 3 = ti + t 2 j + t 3 k This makes r(t) an example of a vector-valued function of the real parameter t. The component functions are x(t), y(t) and z(t). The domain of r(t) is the set of allowable values of t. The natural domain of r(t): every component is defined and real-valued for these values of t. This is the intersection of the natural domains of x(t), y(t) and z(t). ABD section: Introduction to vector-valued functions 10

11 Vector-valued functions as parametric curves Any vector-valued function r(t) = x(t), y(t), z(t) describes a parametric curve C : R 1 R 3, specified by the component functions x(t), y(t) and z(t). A path or curve traced out by a moving point (in 2D, 3D,... ) as t changes; or a path traced out by the tip of the vector r(t). We call r(t) the position vector, or radius vector, of C. ABD section: Introduction to vector-valued functions 11

12 Convention on positional vectors vs. directional vectors We will predominantly think of a vector as a positional vector i.e. to indicate the position of its tip if its base is put at the origin. When a vector is used as a directional vector, it will be clear from the context, and on picture, it will be draw with an arrow. Eg. r(t 0 ) is (used as) the positional vector (as we only illustrate the position of its tip), whereas r (t 0 ) is a directional vector. ABD section: Introduction to vector-valued functions 12

13 Vector form of a line segment r 0 Vector form of a line r r 1 The line through the tip of r 0 and parallel to a vector v is, in vector form, r = r 0 + tv O v The 2!point vector form of a line: The line through the tips of the vectors r 0 and r 1 is r = r 0 + t(r 1 r 0 ) or r = (1 t)r 0 + tr 1 ABD section: Introduction to vector-valued functions 13 21

14 2 Calculus of vector-valued functions ABD section: Calculus of vector-valued functions 14

15 Limits and continuity of vector-valued functions Geometric idea of a limit lim r(t) = L t a iff r(t) approaches L in length and in direction as t a. r(t) is close to L if the length r(t) L is small. For positional vectors: In 2D, r(t) L < ɛ is true for all r(t) with tips that lie inside a disc of radius ɛ and centre L. In 3D, the disc becomes a ball. ABD section: Calculus of vector-valued functions 15

16 A formal definition of limit for vector-valued functions Let r(t) be a vector-valued function defined for all t in an open interval containing t = a, with r perhaps undefined at t = a. We write lim r(t) = L t a if given any positive real number ɛ > 0 we can find a number δ > 0 such that r(t) L < ɛ is true for all t close enough to a so that 0 < t a < δ. ABD section: Calculus of vector-valued functions 16

17 Example (formal definition): twisted cubic r(t) = x(t), y(t), z(t) = t, t 2, t 3 lim r(t) = 1, 1, 1 t 1 Given any positive real number ɛ > 0 we must find a number δ > 0 so that Now 0 < t 1 < δ t, t 2, t 3 1, 1, 1 < ɛ. t, t 2, t 3 1, 1, 1 2 = (t 1) 2 + (t 2 1) 2 + (t 3 1) 2 = (t 1) 2 [1 + (t + 1) 2 + (t 2 + t + 1) 2 ] < 59(t 1) 2 so long as 0 < t 1 < 1 because then 0 < t < 2. ABD section: Calculus of vector-valued functions 17

18 Example twisted cubic (continued) So, we can choose any δ ɛ 59 for example, δ = 0 < t 1 < δ, ɛ 59 or δ = ɛ 2. As long as δ 1 as well, then for 59 59(t 1) 2 < 59δ 2 ɛ 59( ) 2 = ɛ 2 59 and so the segment of the twisted cubic for 0 < t 1 < δ lies inside the ball of radius ɛ. Here is the illustration when ε = 1/2 and we take δ = 1/2 59. ABD section: Calculus of vector-valued functions 18

19 Limits of vector-valued functions: an equivalent formulation Theorem lim r(t) = L each component of r(t) has limit the corresponding t a component of L. (Limits in higher dimension are just 1-dimension limits, applied to each component.) ABD section: Calculus of vector-valued functions 19

20 Example (componentwise reduction): twisted cubic Consider again r(t) = x(t), y(t), z(t) = t, t 2, t 3 Then (Equivalently lim r(t) = lim ti + lim t 1 t 1 t 1 (t2 )j + lim(t 3 )k. t 1 lim t, t 2, t 3 = lim t, lim t 1 t 1 t 1 t2, lim t 3 ) t 1 As each of the functions t, t 2, and t 3 are continuous, it follows that lim r(t) = lim ti + lim t 1 t 1 t 1 (t2 )j + lim(t 3 )k t 1 = 1i j k = 1, 1, 1. ABD section: Calculus of vector-valued functions 20

21 Limits and continuity Definition : A vector-valued function r(t) is continuous at t = c if the following conditions are satisfied: 1 r(c) is defined 2 lim t c r(t) exists 3 lim t c r(t) = r(c). A vector valued function r(t) is continuous if it is continuous at every t in its domain. ABD section: Calculus of vector-valued functions 21

22 Example: Consider r(t) = t 3, e t, 2 sin t = t 3 i + e t j + ( 2 sin t)k Then lim r(t) = lim t 0 t 0 (t3 )i + lim(e t )j + lim( 2 sin t)k. t 0 t 0 As each of the function t 3, e t, and 2 sin t are continuous, it follows that lim r(t) = lim t 0 t 0 (t3 )i + lim(e t )j + lim( 2 sin t)k t 0 t 0 = 0i + e 0 j + ( 2 sin 0)k = j = r(0). By definition, r(t) is continuous at 0. ABD section: Calculus of vector-valued functions 22

23 Continuity: componentwise reduction For the vector-valued function r(t) = x(t), y(t), z(t), it can be seen that Theorem r(t) is continuous at t = c each of the 1-dimension component functions x(t), y(t), z(t) are continuous at t = c. ABD section: Calculus of vector-valued functions 23

24 Example: (the twisted cubic revisited) Consider the twisted cubic r(t) = x(t), y(t), z(t) = t, t 2, t 3. Find lim r(t) t 1 Solution: As the components t, t 2, t 3 of r(t) are continuous functions, we see that r(t) is also continuous, and so lim r(t) = r(1) = 1, 1, 1. t 1 ABD section: Calculus of vector-valued functions 24

25 Derivatives of vector-valued functions Definition : If r(t) is a vector-valued function, we define the derivative of r(t) to be the vector-valued function r (t) given by r(w) r(t) lim w t w t which is a vector: we also denote this by dr(t) dt r(w) r(t) = lim. w t w t If r (t) exists, we says that r(t) is differentiable. The domain of r (t) is the set of values of t in the domain of r(t) for which this limit exists. ABD section: Calculus of vector-valued functions 25

26 Differentiation: componentwise reduction Theorem The vector r (t) exists each of the component functions for r(t) is differentiable at t. In that case, the component functions for r (t) are the derivatives of the component functions of r (t). Proof for 2D: (higher dimensions are similar). Let r(t) = x(t)i + y(t)j. Then r (t) = lim w t r(w) r(t) w t = lim w t x(w)i+y(w)j x(t)i y(t)j w t = lim w t ( x(w) x(t) w t = x (t)i + y (t)j ) ( i + y(w) y(t) ) limw t w t j ABD section: Calculus of vector-valued functions 26

27 Example: Differentiation of components Let Then and r(t) = t 3 i + e t j 2 sin t k. r (t) = 3t 2 i + e t j 2 cos t k, r (1) = 3 i + e j 2 cos(1) k. ABD section: Calculus of vector-valued functions 27

28 Geometric interpretation of derivatives: Tangent lines to curves Definition Let P = r(t 0 ) be a point on the parametric curve r(t). If r (t 0 ) exists and r (t 0 ) 0, then r (t 0 ) is called the tangent vector to the curve at P. The line through P that is parallel to r (t 0 ) is called the tangent line at r(t 0 ). ABD section: Calculus of vector-valued functions 28

29 Explanation of the derivative formula for the tangent vector: The explanation for this is similar to the case of tangent lines to graphs y = f (x) of real-valued functions f as seen in MATH 141/142. The vector r(t) r(t 0 ) gives the direction of the secant line joining r(t 0 ) and r(t). So does the difference quotient r(t) r(t 0) t t 0. As t t 0, the secant line could be thought of as approaching the tangent line (if the latter exists). So, as t t 0, if the above difference quotient approaches some vector, i.e. if r (t 0 ) exists, then r (t 0 ) should be parallel to the imaginative tangent line. Hence, if r (t 0 ) exists and 0, we could think of it as indicating the direction of the tangent line. What if r (t 0 ) = 0, or if r (t 0 ) doesn t exist at all? ABD section: Calculus of vector-valued functions 29

30 When r (t 0 ) doesn t exist: It could be that tangent to the parametric curve r(t) at r(t 0 ) does not exists. Example Consider the 2D parametric curve r(t) = t, t, which is the graph of y = x, and so the tangent line at r(0) = 0 does not exist. Of course, r (0) does not exist either. But it could also means that our parametrisation is not good enough to describe all the feature of the geometric curve described by r(t) near t 0. Example Consider another parametric curve r(t) = 3 t, 3 t, which is the line y = x, and so the tangent line at r(0) = 0 should be the same line. But r (0) does not exist. ABD section: Calculus of vector-valued functions 30

31 When r (t 0 ) exists but = 0: Again, it could be that tangent to the parametric curve r(t) at r(t 0 ) does not exists. Example Consider the parametric curve r(t) = t 3, t 3, which is the graph of y = x, and so the tangent line at r(0) = 0 does not exist. But, you could see that r (0) = 0, 0 by consider the two branches: when t > 0 and when t < 0. But again it could also means that our parametrisation is not good enough to describe all the feature of the geometric curve described by r(t) near t 0. Example Consider the parametric curve r(t) = t 3, t 3, which is the line y = x, and so the tangent line at r(0) = 0 should be the same line. Again, r (0) = 0. ABD section: Calculus of vector-valued functions 31

32 So what is the difference between parametric curves and graphs of functions in the relation of tangents vs. derivatives? Why, for the graphs of functions of one variable, the relation is simply that: a function has a derivative at a point if and only if its graph has a tangent line at the corresponding point? But for parametric curves, the relation is more complicated? The short answer is that the variable of a function is encoded in its graph whereas the parameter of a parametric curve is not! ABD section: Calculus of vector-valued functions 32

33 Graphs of functions as parametric curves Let f : R 1 R 1. Recall that the graph of y = f (x) is the same as the parametric curve r(t) = t, f (t). Thus r (t) = 1, f (t) whenever f is differentiable (i.e. f (t) exists). Moreover, r (t) is never zero in this case. ABD section: Calculus of vector-valued functions 33

34 Warning! What we call the curve (traced/parametrised by a v.v.f.) r(t) is called the graph of a v.v.f. r(t) in ABD s Calculus. The latter terminology could lead to the confusion since the graph of any function r(t) is usually understood as the curve traced by t, r(t). So be warned when reading the literature. ABD section: Calculus of vector-valued functions 34

35 Example: The helix in 3D Parametric curve: r(t) = cos ti + sin tj + tk t 0 = 3π 4 = r(t 0 ) = ( 2 2, 2 2, 3π 4 ) Tangent vector: r (t) = sin ti + cos tj + k (the general one) t 0 = 3π 4 = r (t 0 ) = ( 2 2, 2 2, 1) Tangent line equation: r = r(t 0 ) + tr (t 0 ) = ( 2 2, 2 2, 3π 4 ) + t( 2 2, 2 2, 1) Warning: r here is used as a vector variable. To avoid confusion with r as the name of the function r(t) used for the Helix above, we could give the equation of the tangent line a new name, says: l(t) = r(t 0) + tr (t 0) so that now the tangent line is traced by this function l(t). ABD section: Calculus of vector-valued functions 35

36 Derivative rules for vector-valued functions We have the following differentiation rules Given vector-valued functions r(t), r 1 (t) and r 2 (t) in 2D or 3D, and any real-valued ( scalar ) function f (t), constant scalar k R, and constant vector c: dc dt = 0 d[kr(t)] = k dr(t) dt dt d[r 1 (t) + r 2 (t)] dt d[f (t)r(t)] dt = d[r 1(t)] dt + d[r 2(t)] dt = f (t)r(t) + f (t) dr(t) dt ABD section: Calculus of vector-valued functions 36

37 Example d [ (t 2 + t) sin t, e 2t ] = d(t2 + t) sin t, e 2t + (t 2 + t) d sin t, e 2t dt dt dt =(2t + 1) sin t, e 2t + (t 2 + t) cos t, 2e 2t = (2t + 1) sin t + (t 2 + t) cos t, (2t 2 + 4t + 1)e 2t Of course, if we don t use the product rule for vector-valued differentiation, then it could be proceeded as follows: d [ (t 2 + t) sin t, e 2t ] = d (t 2 + t) sin t, (t 2 + t)e 2t dt dt = [(t 2 + t) sin t], [(t 2 + t)e 2t ] = (2t + 1) sin t + (t 2 + t) cos t, (2t + 1)e 2t + (t 2 + t)2e 2t = (2t + 1) sin t + (t 2 + t) cos t, (2t 2 + 4t + 1)e 2t The 2nd calculation above essentially explains the new product rule from the product rule for real-valued functions. ABD section: Calculus of vector-valued functions 37

38 Derivatives of dot and cross products (r 1 r 2 ) = r 1 r 2 + r 1 r 2 d(r 1 r 2 ) dt = d(r 1) dt r 2 + r 1 d(r 2) dt (r 1 r 2 ) = r 1 r 2 + r 1 r 2 d(r 1 r 2 ) dt = d(r 1) dt r 2 + r 1 d(r 2) dt ABD section: Calculus of vector-valued functions 38

39 Example d [ t, t 2 ] tan t, cos t = d dt dt t, t 2 tan t, cos t + t, t 2 t, t 2 = 1, 2t tan t, cos t + = tan t + 2t cos t + t sec 2 t t 2 sin t d dt tan t, cos t sec 2 t, sin t In the above, we use the new dot-product rule, but if we don t the calculation goes as d [ t, t 2 ] tan t, cos t = d [ ] t tan t + t 2 cos t dt dt =(1 tan t + t sec 2 t) + (2t cos t + t 2 ( sin t)) = tan t + 2t cos t + t sec 2 t t 2 sin t This again illustrates the explanation of the new rule from the old product rule for R 1 -functions of one real variable. ABD section: Calculus of vector-valued functions 39

40 Example d [ ] (t 2 i + 3j + 4k) (i j + t 3 k) dt = d dt (t2 i + 3j + 4k) (i j + t 3 k) + (t 2 i + 3j + 4k) d dt (i j + t3 k) =(2ti) (i j + t 3 k) + (t 2 i + 3j + 4k) (3t 2 k) =( 2ti j 2t 4 i k) + (3t 4 i k + 9t 2 j k) = 9t 2 i t 4 j 2tk Of course, we can again calculate the cross product first, before differentiating. This would use the product rule for real-valued functions, again showing that these new product rules are immediate consequences of the usual product rule for differentiation of R 1 -valued functions: ( f (t)g(t) ) = f (t)g(t) + f (t)g (t) ABD section: Calculus of vector-valued functions 40

41 Radius and tangent vectors Flaming Poi Theorem If r(t) is constant for all t, then r (t) r(t) = 0 t. Proof. If r(t) = c is constant, then also r(t) r(t) = r(t) 2 = c 2 = constant. Thus d dt (r(t) r(t)) = d dt (c2 ) = 0, and we find 0 = d dt (r(t) r(t)) = r (t) r(t) + r(t) r (t) = 2r(t) r (t), giving r(t) r (t) = 0. ABD section: Calculus of vector-valued functions 41

42 Example Consider r(t) = cos ti + sin tj. Then r(t) = 1 for every t. We have r (t) = sin ti + cos tj So r (t) r(t) = cos t( sin t) + sin t cos t = 0 This agrees with the Flaming Poi theorem (as should be expected!). ABD section: Calculus of vector-valued functions 42

43 Integrals of vector-valued functions Definition The definite integral of a vector-valued function a a r(t) = x(t)i + y(t)j + z(t)k from t = a to t = b is the vector-valued function b ( b ) ( b ) ( b r(t) dt = x(t) dt i + y(t) dt j + a a ) z(t) dt k ABD section: Calculus of vector-valued functions 43

44 Antiderivatives of vector-valued functions An antiderivative of r(t) is a function R(t) such that: R (t) = r(t). The indefinite integral of r is defined as r(t) dt = R(t) + C where C is some constant vector. Basic rules of antiderivative For a constant scalar k, vector-valued functions r(t), r 1 (t), r 2 (t), and antiderivatives R(t), R 1 (t), R 2 (t) respectively: kr(t) dt = kr(t) + C [r 1 (t) + r 2 (t)] dt = R 1 (t) + R 2 (t) + C ABD section: Calculus of vector-valued functions 44

45 Fundamental theorem of vector integration Antiderivative is also called indefinite integral is because of the following Theorem Given a continuous vector-valued function r(t) with an antiderivative R(t): d [ t r(u) du ] = r(t) dt b a a r(t) dt = R(t) ] b = R(b) R(a) a ABD section: Calculus of vector-valued functions 45

46 Basic rules of integration For a constant scalar k and vector-valued functions r(t), r 1 (t), r 2 (t): b a b a kr(t) dt = k [r 1 (t) + r 2 (t)] dt = b a b a r(t) dt r 1 (t) dt + b a r 2 (t) dt But in practice, one often calculates the integral of a vector-valued function by reducing to the integrals of its components: Definition (recalled): b ( b r(t) dt = a a ) ( b ) ( b ) x(t) dt i + y(t) dt j + z(t) dt k a a ABD section: Calculus of vector-valued functions 46

47 Example 1 1, t, t2 dt = t t dt, t dt, t 2 dt = ln t, t2 2, t3 + C 3 Or 5 2 1, t, t2 dt = ln t, t2 t = ln 5, 52 2, 53 3 ] 5 2, t3 3 t=2 ln 2, 22 2, 23 3 = ln 5 2, 21 2, , t, t2 dt = t 2 t dt, = ln t ] 5 t=2 5, t2 2 2 ] 5 t dt, t=2, t ] 5 t 2 dt t=2 = ln 5 2, 21 2, ABD section: Calculus of vector-valued functions 47

48 3 Change of parameter; Arc length ABD section: Change of parameter; Arc length 48

49 Substitution of variables Twisted cubic curve: r(t) = t, t 2, t 3 : variable name t. Substitute s for t. r(s) = s, s 2, s 3 : variable name s, same formula, image set, orientation. 1 Substitute s = 2t same orientation as r(t) r 1 (t) = r(2t) = 2t, 4t 2, 8t 3 2 Substitute s = t 3 same orientation as r(t) r 2 (t) = r(t 3 ) = t 3, t 6, t 9 3 Substitute s = t opposite orientation as r(t) r 3 (t) = r( t) = t, t 2, t 3 The set of image points is exactly the same for each r 1 (t), r 2 (t), r 3 (t). Fix t 0 : corresponding values for r(t 0 ), r 1 (t 0 ), r 2 (t 0 ), r 3 (t 0 ) generally differ. Orientations may differ. ABD section: Change of parameter; Arc length 49

50 Change of parameter Definition Given a vector-valued function r(t), a change of parameter is a substitution t = g(τ) a where g is a bijective (i.e. 1:1 and onto) function (preferably continuous!). a τ is a Greek letter, pronounces tau. All 3 above substitutions are changes of parameter for the twisted cubic. Twisted cubic curve: r(t) = t, t 2, t 3 or, using variable s instead of t: r(s) = s, s 2, s 3 4 Substitute s = t 2 r 4 (t) = r(t 2 ) = t 2, t 4, t 6 This is not a change of parameter for the twisted cubic. (Is it a change of parameter for the restriction of the twisted cubic on R 3 x>0 = {(x, y, z) R3 : x > 0}?) ABD section: Change of parameter; Arc length 50

51 Smooth parametrisation So as a geometric curve could have many parametrisations, which one(s) should we work with? We prefer to use well-behaved parametrizations, which correspond to moving along a curve without stopping: Definition A vector valued function r(t) is smoothly parametrized if r (t) is continuous and r (t) 0 for all t in the domain of r. Remark: many authors use smooth function to mean a function which can be differentiated as many times as you like ABD section: Change of parameter; Arc length 51

52 Examples The function: r(t) = t 1/3, t 2/3, t (on its natural domain) is not a smoothly parametrised as r (0) does not exists. The function: r(t) = t 3, t 6, t 9 is not a smoothly parametrised as r (0) = 0. But both these functions are parametrisation of the twisted cubic, which has a smooth parametrisation as: r(t) = t, t 2, t 3 ABD section: Change of parameter; Arc length 52

53 ABD section: Change In of fact, parameter; it can Arcbe length seen that there is no change of 53 The cusp : A curve cannot be smoothly parametrized r(t) = t 2 i + t 3 j r (t) = 2ti + 3t 2 j. y 2 = x 3 (t 3 ) 2 = (t 2 ) 3 So r has a continuous derivative. But r (0) = 0, so r is not smooth at 0. The limit of its direction, as t 0, is not defined: the tangent vector has a reversal of direction when t = 0: If θ is the angle between the tangent vector and the x-axis, then cos θ = r (t) i r i = 2t 4t 2 + 9t 4 = 2t t 4 + 9t = sgn(t) t 2 Thus lim t 0 + cos θ = +1 lim t 0 + θ = 0; but lim t 0 cos θ = 1 lim t 0 θ = π.

54 Smooth change of parameter A change of parameter t = g(τ) for which g (τ) is continuous and g (τ) 0 for all τ is called a smooth change of parameter. g (τ) > 0 positive c.o.p. preserves orientation g (τ) < 0 negative c.o.p. reverses orientation Previous example: the twisted cubic r(t) = t, t 2, t 3 : 2τ, 4τ 2, 8τ 3 = r(2τ) smooth positive c.o.p. τ 3, τ 6, τ 9 = r(τ 3 ) non-smooth positive c.o.p. τ, τ 2, τ 3 = r( τ) smooth, negative c.o.p. ABD section: Change of parameter; Arc length 54

55 Chain rule for smooth changes of parameter We have the following result that holds in a slightly more general situation. Theorem If r(t) is a differentiable v.v.f. and t = g(τ) is a differentiable substitution, then r(g(τ)) is differentiable and Proof d dτ [r(g(τ))] = g (τ)r (g(τ)) t = g(τ), r(t) = x(t), y(t), z(t) r(g(τ)) = x(g(τ)), y(g(τ)), z(g(τ)) d dτ [r(g(τ))] = = d x(g(τ)), y(g(τ)), z(g(τ)) dτ d dτ z(g(τ)) d dy x(g(τ)), dτ dτ y(g(τ)), = x (g(τ))g (τ), y (g(τ))g (τ), z (g(τ))g (τ) = g (τ)r (g(τ)) ABD section: Change of parameter; Arc length 55

56 Example Consider the twisted cubic r(t) = ti + t 2 j + t 3 k. Then r (t) = i + 2tj + 3t 2 k. Consider a substitution t = τ 3. Then dt dτ = 3τ 2. Hence, by the new chain rule above d dτ r(τ 3 ) = dt dτ r (τ 3 ) = 3τ 2 (i + 2τ 3 j + 3τ 6 k) = 3τ 2 i + 6τ 5 j + 9τ 8 k What if one calculates d dτ r(τ 3 ) directly without using the above theorem? ABD section: Change of parameter; Arc length 56

57 Reparametrising smooth parametrisations Corollary If r(t) is a smooth v.v.f. and t = g(τ) is a smooth change of parameter, then q(τ) = r(g(τ)) is a smooth v.v.f with ( ) q (τ) = g (τ)r (g(τ)). In particular, r and q are both smooth parametrisations of the same geometric curve. Conversely, it can be shown that If r(t) and q(τ) are smooth parametrisations of a given geometric curve that trace any part of the curve the same number of times, then there exists a smooth change of parameter t = g(τ) such that q(τ) = r(g(τ)). ABD section: Change of parameter; Arc length 57

58 Properties of a parametrisation v.s. those of a curve The previous results allow us to discuss among properties that are defined using a parametrisation to a given geometric curve, which one are dependent on the chosen parametrisation and which ones are intrinsic to the curve. For examples, from the equation q (τ) = g (τ)r (g(τ)) between two smooth parametrisations r(t) and q(τ) = r(g(τ)) of the same geometric curve we see that 1 tangent vector at a point on a curve is dependent on a particular smooth parametrisation, 2 whereas the tangent line at that point is independent of any chosen smooth parametrisation, and hence is intrinsic to the curve itself. ABD section: Change of parameter; Arc length 58

59 Arc-length: a natural choice for curve parameter Arc Length Given a v.v.f. r!t", we would like to measure its length between t = a and t = b. Suppose r is smooth and r(t) = x(t),y(t),z(t) Divide b! a into intervals of length!t and approximate the arc from r(t) to r(t +!t) by the secant of length!l r(t +!t)!l r(t +!t) r(t) r(t)!l = [x(t +!t) x(t)] 2 + [y(t +!t) y(t)] 2 + [z(t +!t) z(t)] 2 [x(t ] 2 [ ] 2 [ ] 2 +!t) x(t) y(t +!t) y(t) z(t +!t) z(t) = + +!t!t!t!t Summing over all intervals and taking limit as!t 0 gives: Z b (dx L = t=a dt ) 2 + ( ) 2 ( ) 2 dy dz + dt dt dt ABD section: Change of parameter; Arc length 59 43

60 Theorem. traced The arc length of the curve graph by of a smooth v.v.f. r!t" from t = a to t = b is Z b L = r (t) dt a Example. Find the arc length of helix r(t) = cost i + sint j +t k from t = 0 to t =! so: r (t) = ( sint) 2 + (cost) = 2 Z! L = r (t) dt = 0 Z! 0 2dt = 2! ABD section: Change of parameter; Arc length 60

61 Arc Length Parametrization We can use arc length of the graph of a smooth v.v.f. r!t" as a new parameter. Fix a point r(t 0 ) on the graph curveof r. Then let Z t curve traced by a smooth s = s(t) = r (u) du t 0 Important Facts! signed 1" s(t 1 ) measures arc length from t = t 0 to t = t 1 2" s(t) is a smooth change of parameter. By the Fundamental Theorem of Calculus: ds = r (t) > 0 dt r smooth ABD section: Change of parameter; Arc length 61

62 Arc-length parametrization (cont.) 3" Hence dr ds = dt dr ds dt dr ds = 1 r (t) r (t) = 1 Example. Find the arc length parametrization of the helix r(t) = cost i + sint j +t k As above so and s(t) = Z t 0 r (u) du = 2t t = s/ 2 s r(s) = cos i + sin arc 2 s j + s 2 2 k ABD section: Change of parameter; Arc length 62

63 4 Unit tangents, normals and binormals ABD section: Unit tangents, normals and binormals 63

64 Unit Tangent Vectors Consider a smooth v.v.f. r!t". Then r (t) points in the tangent direction at any point on the curve. graph. De#ne the unit tangent vector to be Notes: 1. T(t) = r (t) r (t) T(t) = 1 well de#ned as r smooth 2. If r!s" is parametrized by arc length then T(s) = r (s) ABD section: Unit tangents, normals and binormals 64

65 Unit ( principal ) normal vectors Principal Normal Since T(t) = 1, the Flaming Poi Theorem tells us that T(t) T (t) = 0 Either T (t) = 0 or T (t) is orthogonal to T(t). In the latter case de#ne the principal!unit" normal vector to be N(t) = T (t) T (t) In particular: N(t) T(t) = 0 Note: If r!s" is parametrized by arc length then: N(s) = r (s) r (s) ABD section: Unit tangents, normals and binormals 65 48

66 Principal normals for the helix in 3D Example. Find the principal normal to the graph helix: of the helix: r(t) = acost i + asint j + ct k, a,c = 0, a > 0 Di!erentiating: r (t) = asint i + acost j + ck r (t) = ( asint) 2 + (acost) 2 + c 2 = a 2 + c 2 T(t) = T (t) = asint a2 + c i + acost 2 a2 + c j + c2 2 a2 + c k 2 This is the unit tangent vector. To "nd the principal normal: acost a2 + c i asint 2 a2 + c j 2 T (t) = acost a2 + c asint a2 + c 2 2 = a a 2 + c 2 ABD section: Unit tangents, normals and binormals 66

67 Principal normals for the helix in 3D Hence: N(t) = T (t) T = cost i sint j (t) So the principal normal is always parallel to the xy plane. In fact it points towards the z axis. ABD section: Unit tangents, normals and binormals 67 50

68 Binormal vectors in 3-space Normals in 3D In 3D, there is a plane normal to the graph curve C of traced a by a smooth v.v.f. r!t" at any point r 0 = r(t 0 ). Its equation is. T (r r 0 ) = 0 B N r 0 T C If T = 0 then N is a vector in this normal plane. T and N are orthogonal so span a plane # the osculating plan! A normal vector to the osculating plane is B(t) = T(t) N(t) ABD section: Unit tangents, normals and binormals 68

69 Binormal vectors in 3-space The Binormal B(t) = T(t) N(t) is called the binormal vector to C. Since T and N are orthogonal B(t) = T(t) N(t) sin! 2 = 1 Hence {T(t),N(t),B(t)} forms a right#handed system of orthonormal vectors!i.e. mutually orthogonal unit vectors". ABD section: Unit tangents, normals and binormals 69 52

70 Example: binormals for the helix in 3D Consider again the helix Then we have computed Hence, r(t) = a cos ti + a sin tj + ctk, c 0, a > 0 T(t) = a sin t a 2 + c i + 2 N(t) = cos ti sin tj B(t) = T(t) N(t) = a cos t a 2 + c 2 j + c sin t a 2 + c 2 i c a 2 + c 2 k c cos t a 2 + c 2 j + a a 2 + c 2 k ABD section: Unit tangents, normals and binormals 70

71 5 Curvature ABD section: Curvature 71

72 Frenet frames: tangent, normal, binormal Summary of tangents and normals Unit tangent: T(t) = r (t) r (t) If T (t) = 0 then: Principal N(t) = T (t) normal: T (t) In 3!space: Binormal: B(t) = T(t) N(t) N B r 0 T C Arc length param. T(s) = r (s) N(s) = r (s) r (s) B(s) = r (s) r (s) r (s) ABD section: Curvature 72

73 Intrinsic to the curve: independence of parametrization Important note: T, N and B depend only on the graph curve C, not on the particular parametrization. We say they are invariant up to a!! positive, smooth change of parameter. E.g. suppose t = g(!) and r 1 (!) = r(g(!)) then dr 1 = r (g(!)).g (!) = r (t).g (!) d! so dr 1 d! and T 1 (!) = dr 1 d! = r (t). g (!) = r (t).g (!) dr1 d! = " g positive# c.o.p.) r (t).g (!) r (t).g (!) = r (t) r (t) = T(t) ABD section: Curvature 73

74 Smooth curves in 3D and arc length Call a set C in 2#space or 3#space a smooth curve if it is the parametrised graph of by a smooth vector#valued function. Since properties of C related to tangents and normals are independent of the!positive, smooth" parametrization, we can use arc length if it makes life easier. ABD section: Curvature 74

75 Definition of curvature Curvature We consider how T changes along a smooth curve. It$s always a unit vector so only the direction changes. T = 0 T small T large De!nition. If C is a smooth curve parametrized by arc length s then its curvature, denoted! =!(s), is the scalar kappa!(s) = dt ds = r (s) 56 ABD section: Curvature 75

76 Example: curvature of the circle of radius a Examples 1. The circle radius a, centre O, in 2!space has " " arc length parametrization: ( s ( s r(s) = acos i + asin j a) a) ( s ( s Then r (s) = sin i + cos j a) a) r (s) = 1 ( s ) a cos i 1 ( s ) a a sin j a so [ r (s) = 1 ( s ) ] 2 [ a cos + 1 ( s ) ] 2!(s) = a a sin a = 1 a The smaller the circle, the greater its curvature 57 ABD section: Curvature 76

77 Example: curvature of a straight line 2. The straight line through r 0 with direction u has arc length parametrization: r(s) = r 0 + su Then r (s) = u and r (s) = 0 so curvature!(s) = 0 What we expect for a line? But... the arc length parametrization is hard to calculate in general. Can we #nd! from another parametrization? 58 ABD section: Curvature 77

78 Formulae for curvature Theorem.!13.5.2" If C is a smooth curve in 3#space which is parametrised the graph of by the smooth v.v.f. r(t) and if r (t) exists for all t then the curvature of C is given by Proof.!(t) = r (t) r (t) r (t) 3 Consider the change of parameter t = t(s) dt Recall that!1" ds = 1 ds/dt = 1!from def n of s" r (t) Hence:!2"!(t) = dt ds defn of curvature = dt dt dt ds = chain rule by!1" T (t) r (t) ABD section: Curvature 78

79 Now!suppressing the variable t", from de$nition of the unit tangent vector:!3" r = r T Di%erentiating w.r.t. t: r = r T + r T by the v.v.f. Product Rule = r T + r T N defn. of principal normal = r T + r 2! N by equation!2" Take cross product with equation!3": r r = r r (T T) + r 3! (T N) but T T = 0 and T N = B. Since B = 1, r r = r 3! and rearranging gives the Theorem. 60 ABD section: Curvature 79

80 Interpretation of curvature in 2-space The Theorem holds in 2!space if we embed the plane curve r(t) = x(t)i + y(t)j in 3!space as r(t) = x(t)i + y(t)j + 0k Then r r = 0i + 0j + (x y x y )k so r r = x y x y and! = x y x y [(x ) 2 + (y ) 2 ] 3/2 61 ABD section: Curvature 80

81 Example: the helix Find the curvature of the circular helix parametrized by: As calculated earlier r(t) = a cos ti + a sin tj + ctk, c 0, a > 0 r (t) = a sin ti + a cos tj + ck r (t) = a cos ti a sin tj so r (t) = a 2 + c 2 and r (t) r i j k (t) = a sin t a cos t c a cos t a sin t 0 = ac sin ti ac cos tj + a2 k so r r = (ac sin t) 2 + ( ac cos t) 2 + (a 2 ) 2 = a a 2 + c 2 as a > 0. Finally: κ = a a 2 + c 2 a 2 + c 2 3 = a a 2 + c 2. ABD section: Curvature 81

82 Example: the ellipse Find the curvature of the ellipse parametrized by: r = 3cost i + 2sint j Easily get: r = 3sint i + 2cost j r = 3cost i 2sint j then!as on p10": r r = 6sin 2 t + 6cos 2 t = 6 r = 9sin 2 t + 4cos 2 t hence: 6! = (9sin 2 t + 4cos 2 t) 3/2 ABD section: Curvature 82

83 Curvature at different points of the ellipse Curvature of the ellipse as a function of parameter t:! t 0 t 2! 64 ABD section: Curvature 83

84 Radius of curvature Radius of Curvature The tangent line can be interpreted as the best linear approximation to a curve. Curvature tells us about the best circular approximation. A circle of radius a has constant curvature 1/a. Hence, a point of a smooth curve of curvature!, can be approximated by a circle of radius 1/!. rho De!nition. The radius of curvature! of a smooth curve C is the reciprocal 1/! of its curvature! so long as! ABD section: Curvature 84

85 Circles of curvature or osculating circl! The circle of curvature at a point P on C is the unique circle of radius! having a common tangent with C and centre on the concave side of C. (on the osculating plane). If C is parametrised the graph of by a smooth v.v.f. r, then the centre of the circle of curvature is r(t) +!(t)n(t) centre of curvatur! In 3!space the centre of curvature lies in the osculating plane.! C ABD section: Curvature 85

86 6 Motion along a curve ABD section: Motion along a curve 86

87 The origins of calculus Isaac Newton Isaac Newton developed the ideas of calculus!!uxions" to solve problems of #nding tangents and calculating areas. Also, in his famous book Philosophia" Naturalis Principia Mathematica, he set out Laws of Motion and of gravity that provided a theoretical foundation for the three laws of planetary motion stated by Johannes Kepler. While the methods of Principia were essentially those of Euclidean geometry, the calculus of vector$valued functions is an e%ective language for expressing this theory. ABD section: Motion along a curve 87

88 Motion of a particle along a curve Start by considering a particle moving along a curve in 2$space or 3$space. The curve is the traced graph byof a v.v.f r!t", where t denotes time. Since t now represents a fundamental physical variable, we do not want to make a change of parameter. We are interested in the #elocity and acceleration of the particle. ABD section: Motion along a curve 88

89 Velocity and Acceleration De!nition If r!t" represents the position of a particle in 2# or 3# space then its instantaneous velocity, speed and acceleration at time t are: v(t)=r (t) r(t) v(t) v(t) = r (t) a(t)=v (t) = r (t) Note: If s represents arc length then by the Chain Rule: dr dt = dr ds ds dt = ds dt T(t) so the speed is given by: v(t) = ds dt 69 ABD section: Motion along a curve 89

90 7 Kepler s Laws of planetary motion You are not required to remember the content of this section for your test or exam. ABD section: Kepler s Laws of planetary motion 90

91 Kepler s Three Laws Analysing the observational data collected by Tycho Brahe, Johannes Kepler was able to formulate his three laws of planetary motion: 1 The orbits of the planets are ellipses with the Sun at a focus. 2 The line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. 3 The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. ABD section: Kepler s Laws of planetary motion 91

92 Newton s physics Gravitation The formula for the force of gravity on a projectile was derived from: Newton!s Law of Universal Gravitatio" Any two particles of masses M and m and distance r apart exert equal and opposite forces F and!f on one another of magnitude F = GMm r 2 universal gravitational constan! "inverse square law# If the vector from mass M to mass m is r then F is in the direction r so the gravitational force and corresponding acceleration are: r F = GMm r 3 r and a = GM r 3 r M F F m 72 ABD section: Kepler s Laws of planetary motion 92

93 Relative movement of m and M The above a = GM r is the absolute acceleration of m. r 3 In fact, m also exerts a force on M that is equal but opposite to F. That is, M is also being pulled by Thus M is accelerated by F = GMm r 3 r Gm r (i.e. towards m). r 3 So the acceleration of m relative to M is a = GM r 3 r Gm + m) r = G(M r 3 r 3 r For the rest of this section, we will take a as this relative acceleration of m. Of course, if M is significantly larger than m (as the case of the sun vs. a planet), then this is approximately GM r 3 r i.e. M could be thought of as stationary in this system. ABD section: Kepler s Laws of planetary motion 93

94 Mathematical setting Now, choose a coordinates so that the mass M is at 0. Also, for simplicity, set k := G(M + m) a positive constant. The equation for the movement of m (relative to M) becomes a = k r 3 r where recall that r is the vector indicating the position of m, r is the length of r (i.e. distance from m to M), and a = d2 r is the acceleration of m. dt2 ABD section: Kepler s Laws of planetary motion 94

95 A constant vector from angular momentum Since a = k r, we have r a = 0, and hence r 3 This implies that d dv [r v] = r dt dt + dr dt v = r a + v v = 0. r v = b is a constant vector. Remark that mr v is nothing but the angular momentum of m. So, physically, this means that the angular momentum is preserved. ABD section: Kepler s Laws of planetary motion 95

96 Energy is also preserved Notice that ( ) 1 = r r r 2 = 2rr 2r 3 So ( ) k = a v = r ( v v 2 d = dt [r 2 ] 2r 3 = ) ( ) v 2 ( v 2 =. Hence v 2 2 k r 2 d dt [r r] 2r 3 = r r + r r 2r 3 = r v r 3 = α is a constant. 2 k r Remark that, physically, mα is nothing but the total energy of the mass m: ) = 0, and so Thus mv 2 2 mk r is the kinetic energy of m, is the potential of the gravitational force exerted on m by M. α < 0 the kinetic energy < the gravitational potential. ABD section: Kepler s Laws of planetary motion 96

97 Another constant vector Consider w = r r v b. k We see, using some earlier calculations, that ( ) d(w) 1 = r + 1 dt r r r v b = r v k r 3 r + 1 r v a b k = r v r 3 r + 1 r v [ r (r v) = r 3 r 2 r 3 v (r v)r + r (r v) ] Denote the formula in the last square bracket by q. It can be seen that q = 0 either geometrically, or by algebraic calculation as follows: Notice that (1) q (r v) = 0 [Why?] (2) q r = r 2 v r (r v)r r + [r (r v)] r = r 2 v r (r v)r = 0 [Why?] (3) q v = r 2 v v (r v)r v + [r (r v)] v = r 2 v 2 (r v) 2 + (v r) (r v) = r 2 v 2 (r v) 2 r v 2 = 0 [Why?] By (1), (2), and (3), we obtain q = 0. [Why?] Thus dw dt = 0, and so w is a constant vector, independent of time t. ABD section: Kepler s Laws of planetary motion 97

98 An algebraic equation of the positional vector r We see that r w = r r r r (v b) k = r [Why do we have the 2nd and 3rd equalities?] Thus r = b2 k + r w. (r v) b k = r b2 k To help analysing this equation further, let us calculate the length of w: w 2 = w w = r r r = 1 + b2 v 2 k 2 [Why do we have the 3rd equality?] (v b) (v b) r (v b) + 2 k 2 2 ( kr 2 b2 kr = 1 + 2b2 v 2 k 2 2 k ) = 1 + 2αb2 r k 2 ABD section: Kepler s Laws of planetary motion 98

99 A trivial case: when b = 0. What about if b 0? Since b = 0, we see that w = 1 and r w = r. [What does the latter means?] Thus r w, i.e. m is moving on a straight line through M (at 0) parallel to w. In fact, m stays on the ray starting from (and including 0) and in the direction of w. Of course, m could move away from M, or towards M, or initially away from M and then towards M depending on its initial velocity. Physically, this is rather easy to explain: because r v = b = 0 means that v r... In that case, it s still true that r always lies in a fixed plane containing 0 [Why?] ABD section: Kepler s Laws of planetary motion 99

100 When b 0: So r lies in the plane through 0, perpendicular to b the plane of the ecliptic. Set ε = w and q = b2, which are constants, with ε 0 and q > 0. k Suppose that ε > 0. [What happen if ε = 0?] Set i = w w. Then i is a constant unit vector with w = εi. The equation r = b2 k + r w becomes ( q ) ( q ) r = q + εr i = ε ε i i + r i = ε ε i + r i Let l be the line through the point q ε i and perpendicular to i (as well as to b). Let d be the distance from r to l. Then l is a fixed line (and d depends on t). The equation now means that r is always on the same side of l as 0, and r = εd ABD section: Kepler s Laws of planetary motion 100

101 Deriving Kepler s First Law: We have shown that there are a fixed line on the ecliptic plane, not going through M (at 0), and, a constant ε > 0 such that m (at r) moves in such a way that the distance from m to M = ε times the distance from m to l. This is the description of a conic section with M at a focus and l a directrix; moreover, in the case of a hyperbola, m is on the branch that is at the same side of l as M. Fig.: Ellipse, ε < 1 Fig.: Parabola, ε = 1 Fig.: Hyperbola, ε > 1 Where should the case ε = 0 fall? ABD section: Kepler s Laws of planetary motion 101

102 Deriving Kepler s First Law (cont.): This proves Kepler s First Law. Note that ε < 1 α < 0, whose physical meaning is that the kinetic energy of m < its gravitational potential exerted by M which obviously must be the case for the planets in the solar system. ABD section: Kepler s Laws of planetary motion 102

103 Deriving Kepler s Second Law Consider the case where b 0. Set j := b b, a constant unit vector. If A(t) is the area that (the segment joining M and) the particle mass m has swept up to time t, then A(t + h) A(t) is the net area swept between time t and time t + h (the red region), which is the area of the triangle with vertices O, r(t), and r(t + h). That is As h 0, the approximation gets better. Hence, [A(t + h) A(t)]j 1 r(t) (r(t + h) r(t)) 2 ( ) A A(t + h) A(t) 1 r(t + h) r(t) (t)j = lim j = lim r(t) h 0 h h 0 2 h = 1 2 r(t) r (t) = 1 2 r(t) v(t) = 1 2 b. Thus A (t) = b 2, which is a constant. Hence, the area that m swept during a time interval (t 1, t 2 ) is A(t 2 ) A(t 1 ) = t2 t 1 A (t) dt = b 2 (t 2 t 1 ). ABD section: Kepler s Laws of planetary motion 103

104 Kepler s Third Law? Kepler s Third Law states: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. It only makes sense in the case where b 0 and ε < 1. Also, it holds true only under the assumption that M is significantly larger than m. Because of (the formula in the proof of) the Second Law, all we have to do is to compute the area bounded by the ellipse. For example, by determining the semi-major and semi-minor axes of the ellipse, etc. Although it is not difficult, as not much of what we learnt would be involved in that calculation, it will be left for those who are interested to figure out by themselves. ABD section: Kepler s Laws of planetary motion 104