Vector-Valued Functions - PDF Free Download

Vector-Valued Functions 1 Parametric curves 8 ' 1 6 1 4 8 1 6 4 1 ' 4 6 8 Figure 1: Which curve is a graph of a function? 1 4 6 8 1

8 1 6 4 1 ' 4 6 8 Figure : A graph of a function: = f() 8 ' 1 6 4 1 1 4 6 8 Figure 3: A graph of a function: = f()

5 t=π/ 4 3 β a t=0 1 α 1 3 4 5 6 Figure 4: Is it a graph of a function? 3

5 t=π/ 4 3 β a t=0 1 α 1 3 4 5 6 Figure 5: This curve is not a graph of a function. This curve is a circle of radius a centred at (α, β) ( α) + ( β) = a It is a parametric curve which can be represented b the eqs = α + a cos t, = β + a sin t, 0 t π. 4

5 t=π/ 4 3 β a t=0 1 α 1 3 4 5 6 Figure 6: This curve is not a graph of a function. This curve is a circle of radius a centred at (α, β) ( α) + ( β) = a It is a parametric curve which can be represented b the eqs = α + a cos t, = β + a sin t, 0 t π. In general in -space a parametric curve is = f(t), = g(t), t 0 t t 1. The graph of a function = f() is a parametric curve = t, = f(t). The graph of a function = f() is a parametric curve = f(t), = t. 5

t=0 z t=10 t=100 t=1 orientation Figure 7: A parametric curve in 3-space = f(t), = g(t), z = h(t), t 0 t t 1. These parametric eqs generate a curve in 3-space called the parametric curve represented b the eqs, and as t increases the point moves in a specific direction which defines the curve orientation. A parametric curve can be considered as a path of a point particle with the parameter t identified with time. A parametric curve in n-space is represented b n eqs 1 = f 1 (t), = f (t),, n = f n (t), t 0 t t 1. 6

Eample 1. ( α) a + ( β) b = 1. 7

Eample 1. ( α) a + ( β) b = 1. 4 3.5.0 1.5 1.0 1 0.5 1 3 4 0.5 1.0 1.5.0.5 Figure 8: These are ellipses. This ellipse can be represented b the eqs = α + a cos t, = β + b sin t, 0 t π. 8

Eample 1. Ellipse Eample. ( α) a + ( β) b = 1. ( α) ( β) = 1. a b ( α) ( β) = 1. a b 9

Eample 1. Ellipse Eample. ( α) a + ( β) b = 1. ( α) ( β) = 1. a b ( α) ( β) = 1. a b 3 3 1 1-4 6-4 6 Figure 9: These are hperbolas. The right (black) branch of the first hperbola can be represented b the eqs = α + a cosh t, = β + b sinh t, < t <. because cosh t = et + e t > 0, sinh t = et e t, cosh t sinh t = 1. 10

Eample 1. Ellipse Eample. Hperbolas ( α) a + ( β) b = 1. Eample 3. ( α) a ( α) a ( β) b = 1. ( β) b = 1. = a. = a. 11

Eample 1. Ellipse ( α) + a Eample. Hperbolas Eample 3. ( β) b = 1. ( α) ( β) = 1. a b ( α) ( β) = 1. a b = a. = a. 8 6 1 4 0.5 1.0 1.5-1 - -1 1 - Figure 10: These are parabolas. The can be represented b the eqs = t, = at, = 1 a t, = t, < t <. Ellipses, hperbolas and parabolas are conic sections. 1

Eample 4. = a cos t, = a sin t, z = vt, < t <. = a sin t, = a cos t, z = vt, < t <. 13

Eample 4. = a cos t, = a sin t, z = vt, < t <. = a sin t, = a cos t, z = vt, < t <. Figure 11: These are circular helies. 14

Eample 5. = 0 + v t, = 0 + v t, z = z 0 + v z t, < t <. 15

Eample 5. Line through r 0 = ( 0, 0, z 0 ) in the direction of the vector v = (v, v, v z ) = 0 + v t, = 0 + v t, z = z 0 + v z t, < t <. Eample 5b. Find parametric equations for the line through (, 0, 0) and (0, 4, 3). Figure 13: This is the line through (, 0, 0) and (0, 4, 3). Let t = 0 correspond to (, 0, 0), and let t = 1 corresponds to (0, 4, 3). Then, (0) = 0 =, (1) = 0 + v = 0 = v =, Thus (0) = 0 = 0, (1) = 0 + v = 4 = v = 4, z(0) = z 0 = 0, z(1) = z 0 + v z = 3 = v z = 3. = t, = 4t, z = 3t, < t <. 17

Eample 5. Line through r 0 = ( 0, 0, z 0 ) in the direction of the vector v = (v, v, v z ) = 0 + v t, = 0 + v t, z = z 0 + v z t, < t <. Eample 6. = 0 1 v t, = 0 1 v t, z = z 0 1 v zt, < t <. 18

Eample 5. Line through r 0 = ( 0, 0, z 0 ) in the direction of the vector v = (v, v, v z ) = 0 + v t, = 0 + v t, z = z 0 + v z t, < t <. Eample 6. The same line as in Eample 5 = 0 1 v t, = 0 1 v t, z = z 0 1 v zt, < t <. Eample 7. The same line as in Eamples 5 and 6 = 0 1 v (t t 0 ), = 0 1 v (t t 0 ), z = z 0 1 v z(t t 0 ). Eample 8. This is a segment of the same line as in Eamples 5, 6 and 7 = 0 + v t 3, = 0 + v t 3, z = z 0 + v z t 3, 1 t 1. 1

man parametric eqs representing the same curve. Make a change t = γ(τ) where γ(τ) is a one-to-one function on the interval τ 0 τ τ 1 of interest. t t t 1 8 1.0 6 0.5 4 t 0-1.0-0.5 0.5 1.0-0.5 τ τ 0 τ 1-0.5 0.5 1.0 1.5.0.5 τ -1.0 Figure 14: γ(τ 0 ) = t 0, γ(τ 1 ) = t 1 ; t = τ 3 ; The parameterisations are considered as equivalent if γ (τ) 0 for τ 0 τ τ 1. If γ (τ) > 0 the curves have the same orientation. 4

man parametric eqs representing the same curve. Make a change t = γ(τ) where γ(τ) is a one-to-one function on the interval τ 0 τ τ 1 of interest. t t t 1 8 1.0 6 0.5 4 t 0-1.0-0.5 0.5 1.0-0.5 τ τ 0 τ 1-0.5 0.5 1.0 1.5.0.5 τ -1.0 Figure 15: γ(τ 0 ) = t 0, γ(τ 1 ) = t 1 ; t = τ 3 ; The parameterisations are considered as equivalent if γ (τ) 0 for τ 0 τ τ 1. If γ (τ) > 0 the curves have the same orientation. Eample. = cos t, = sin t, t = τ = = cos τ, = sin τ, t = τ π = = sin τ, = cos τ. 5

Vector-Valued Functions z = f(t), = g(t), z = h(t) r(t) = ( (t), (t), z(t) ) k r(t) r(t) = f(t) i + g(t) j + h(t) k Thus, r is a function of t: r = r(t). i j This is a vector-valued function of a real variable t. (t), (t), z(t) are components of r(t). The graph of r(t) is the parametric curve C described b the components of r(t). r(t) is called the radius vector for the curve C. The domain of r(t) is the intersection of domains of (t), (t), z(t). It is called the natural domain of r(t). Eample. r(t) = t 1 i + ln(3 t) j + 1 t k. D( r) = [1, ) (, 3). 6

3 Vector form of a line segment If r 0 is a vector with its initial point at O then the line through r 0 t(r1 - r 0 ) r O r 1 the terminal point of r 0 and parallel to v is r = r(t) = r 0 + t v. So, r(0) = r 0, r(1) = r 1 = r 0 + v r(t) = r 0 + t( r 1 r 0 ) = (1 t) r 0 + t r 1 This is a two-point form of a line. If 0 t 1, then r = (1 t) r 0 + t r 1 represents the line segment from r 0 to r 1. 7

4 Calculus of Vector-valued Functions ( ) ( ) ( ) lim r(t) = lim (t) i + lim (t) j + lim z(t) k, t a t a t a t a where all the three limits must eist. A vector-valued function is continuous at t = a (or on an interval I) if all its components are. Then lim r(t) = r(a) a I. t a Derivative of r with respect to t is the following vector r (t) = (t) i + (t) j + z (t) k. r(t) is differentiable if all its components are. Notations d dt r(t), d r dt, r (t), r d, r(t), r, dt [ r(t)]. Derivative Rules 1. d dt[ a r1 (t) + b r (t) ] = a r 1 (t) + b r (t), a, b are const. d dt[ f(t) r(t) ] = f (t) r(t) + f(t) r (t), 8

Derivative of r with respect to t is the following vector r (t) = (t) i + (t) j + z (t) k. r(t) is differentiable if all its components are. Derivative Rules 1. d dt[ a r1 (t) + b r (t) ] = a r 1 (t) + b r (t), a, b are const. d dt[ f(t) r(t) ] = f (t) r(t) + f(t) r (t), 8 r(t+ϵ) r(t+ϵ)-r(t) 8 r'(t) 6 6 4 r(t) ɛ 0 4 r(t) C C 0. 0.4 0.6 0.8 1.0 0. 0.4 0.6 0.8 1.0 If r (t) 0 and it is positioned with its initial point at the terminal point of r then r is tangent to C and points in the direction of increasing parameter. In mechanics it is velocit of a particle moving along C. Eample. r(t) = cos π t i + 3 1 + 3e t j + r (t) =? ln t 0 e 3u u du k. 9

5 Tangent lines to graphs of vector-valued functions 8 r ' (t 0 ) Let P be a point on the graph C of a VV function r(t), and let 6 4 P r(t 0 ) C r (t 0 ) 0 where r(t 0 ) is the radius vector from O to P. Then, r (t 0 ) is a tangent vector to C at P, and the line through P that is 0. 0.4 0.6 0.8 1.0 parallel to r (t 0 ) is the tangent line to C at P. The tangent line is given b the vector eq R(t) = r 0 + t v 0, v 0 = r (t 0 ), r 0 = r(t 0 ). The unit tangent vector is T = v v = r r. Eample. r(t) = a sin t i + a cos t j + v t k, r (t) =?, r (t) =?, T (t) =?, R(t) with t0 = π 3?, 30

6 Derivatives of dot and cross products 8 r r 1 = ( 1, 1, z 1 ), r = (,, z ). 6 4 r 1 r 1 r = 1 + 1 + z 1 z = r 1 r cos α = r r 1 α 0. 0.4 0.6 0.8 1.0 r 1 r = +( 1 z z 1 ) i r 1 r ( 1 z z 1 ) j α r r 1 +( 1 1 ) k i j k = 1 1 z 1 = r r 1 z r 1 r = r 1 r sin α, r 1 r r 1 and r. 31

d dt ( r 1 r ) = d r 1 dt r + r 1 d r dt, d dt ( r 1 r ) = d r 1 dt r + r 1 d r dt. Theorem. If r(t) is a constant then r(t) r (t) = 0. Proof: d dt ( r r) = d dt r = 0, d d r ( r r) = dt dt r + r d r dt Thus, r(t) r (t), and T (t) T (t). = r d r dt, Eample. A curve on the surface of a sphere that is centred at the origin has r(t) =const, thus r(t) r (t). 3

7 Integrals of Vector-valued Functions b a r(t)dt = ( b ( (t)dt) b ( i + (t)dt) b j + z(t)dt) k a a a Eample 1. 1 Rules of integration b a 0 ( 3t + 1 i + e t j + 3 sin(πt) k )dt =? b b (c 1 r 1 (t) + c r (t))dt = c 1 r 1 (t)dt + c r (t)dt An antiderivative for a v-v function r(t) is a v-v function R(t) such that R (t) = r(t) = r(t)dt = R(t) + C a a Eample. ( 1 t + 1 i + t cos(t 1) j )dt =? 33

Integration properties d r(t)dt = r(t) dt and r (t)dt = r(t) + C Fundamental Theorem of Calculus b a r(t)dt = R(t) b a = R(b) R(a) Eample 3. 1 Eample 4. Find r(t) given that 0 ( 1 t + 1 i + t cos(t 1) j ) t =? r (t) = ( 1 t+1, t cos(t 1) ), r(1) = (1, 3) 34

8 Arc Length Parametrisation We sa that r(t) is smoothl parameterised or that r(t) is a smooth function of t if r (t) is continuous and r (t) 0 for an allowed value of t. Geometricall it implies that the tangent vector r (t) varies continuousl along the curve. For this reason a smoothl parameterised function is said to have a continuousl turning tangent vector. Eample 1. Eample. r(t) = a cos t i + a sin t j + vt k, r (t) =? r(t) = t i + t 3 j, r (t) =? 35

Change of Parameter 3.0 3.0 3.0.5.5.5.0 r(t).0 s r(s).0 r(ϕ) ϕ 1.5 1.5 1.5 1.0 1.0 1.0 0.5 0.5 0.5 0.5 1.0 1.5.0.5 3.0 0.5 1.0 1.5.0.5 3.0 0.5 1.0 1.5.0.5 3.0 Figure 16: Parameters: time; distance travelled; angle φ A change of parameter in a v-v function r(t) is a substitution t = g(τ) that produces a new v-v function r(g(τ)) having the same graph as r(t). Eample. Find a change of parameter t = g(τ) for the circle r(t) = cos t i + sin t j, such that the circle is traced 0 t π (a) counterclockwise; (b) clockwise as τ increases over the interval [0, 1]. Chain rule: d r dτ = d r dt dt dτ, t = g(τ). A change of parameter t = g(τ) in which r(g(τ)) is smooth if r(t) is smooth is called a smooth change of parameter. It requires dt/dτ 0 τ and dt/dτ continuous. If dt/dτ > 0 it is positive change of parameter, if dt/dτ < 0 it is negative. 36

Arc Length as a Parameter t=a z The arc length L of a parametric curve = (t), = (t), z = z(t), a t b: t=b L = b a ( d dt ) + ( d dt ) + ( dz dt) dt In a vector form r(t) = (t) i + (t) j + z(t) k d r dt = d dt i + d dt j + dz dt k = Eample. = L = b a d r (d dt = dt d r b dt dt = r (t) dt. a ) + ( d dt = a cos t = a sin t z = vt 0 t π L =? ) ( dz ) + dt 37

The length of arc measured along the curve from some fied reference point can serve as a parameter. 1. Select an arbitrar point on the curve C to serve as a reference point O. s=- - dir s=-1 s=0 O s=1 s= + dir. Choose one direction along C to be positive, and the other to be negative. 3. If P is a point on C, let s be the signed arc length along C from O to P, where s is positive/negative if P is in +/- parts of C. 4. = (s), = (s), z = z(s) is called an arc length parametrisation of the curve. It depends on the reference point and direction. Eample. = a cos t, = a sin t, 0 t π 38

Finding arc length parameterisation 8 6 4 O r(t 0 ) s P r(t) C s = t t 0 d r du du Since ds dt = d r dt > 0 it is a positive change of parameter from t to s. 0. 0.4 0.6 0.8 1.0 Eample 1. r = cos t i + sin t j + t k Eample. A line through r 0 and parallel to v Properties of arc length parameterisation 1. t, ds dt = d r dt > 0 (it is speed).. s, the tangent vector has length 1: d r ds = ds ds = 1 3. If d r dt = 1 t, then t0, s = t t 0 is the arc length parameter that as the reference point at t = t 0. 39