10 VECTOR FUNCTIONS
VECTOR FUNCTIONS Later in this chapter, we are going to use vector functions to describe the motion of planets and other objects through space. Here, we prepare the way by developing the calculus of vector functions.
VECTOR FUNCTIONS 10.8 Derivatives and Integrals of Vector Functions In this section, we will learn how to: Develop the calculus of vector functions.
DERIVATIVES The derivative r of a vector function is defined in much the same way as for real-valued functions.
DERIVATIVE Equation 1 dr r( t h) r( t) r '( t) lim dt h0 h if this limit exists.
DERIVATIVE The geometric significance of this definition is shown as follows.
SECANT VECTOR If the points P and Q have position vectors r(t) and r(t + h), then the vector r(t + h) r(t). PQ represents This can therefore be regarded as a secant vector.
DERIVATIVES If h > 0, the scalar multiple (1/h)(r(t + h) r(t)) has the same direction as r(t + h) r(t). As h 0, it appears that this vector approaches a vector that lies on the tangent line.
TANGENT VECTOR For this reason, the vector r (t) is called the tangent vector to the curve defined by r at the point P, provided: r (t) exists r (t) 0
TANGENT LINE The tangent line to C at P is defined to be the line through P parallel to the tangent vector r (t).
UNIT TANGENT VECTOR We will also have occasion to consider the unit tangent vector: Tt () '( t) r r '( t)
DERIVATIVES The following theorem gives us a convenient method for computing the derivative of a vector function r: Just differentiate each component of r.
DERIVATIVES Theorem 2 If r(t) = f(t), g(t), h(t) = f(t) i + g(t) j + h(t) k, where f, g, and h are differentiable functions, then: r (t) = f (t), g (t), h (t) = f (t) i + g (t) j + h (t) k
DERIVATIVES Proof r '( t) 1 lim [ r( t t) r( t)] t 0 t 1 lim f ( t t), g( t t), h( t t), f ( t), g( t), h( t) t 0 t f ( t t) f ( t) g( t t) g( t) h( t t) h( t) lim,, t 0 t t t f ( t t) f ( t) g( t t) g( t) h( t t) h( t) lim, lim, lim t t t t0 t0 t0 f '( t), g '( t), h'( t)
DERIVATIVES Example 1 a. Find the derivative of r(t) = (1 + t 3 ) i + te t j + sin 2t k b. Find the unit tangent vector at the point where t = 0.
DERIVATIVES Example 1 a According to Theorem 2, we differentiate each component of r: r (t) = 3t 2 i + (1 t)e t j + 2 cos 2t k
DERIVATIVES Example 1 b As r(0) = i and r (0) = j + 2k, the unit tangent vector at the point (1, 0, 0) is: T(0) r '(0) j 2k r '(0) 1 4 1 2 j 5 5 k
DERIVATIVES Example 2 For the curve r( t) t i (2 t) j, find r (t) and sketch the position vector r(1) and the tangent vector r (1).
DERIVATIVES Example 2 We have: 1 r '( t) i j 2 t and 1 r '(1) i j 2
DERIVATIVES Example 2 The curve is a plane curve. Elimination of the parameter from the equations x t, y = 2 t gives: y = 2 x 2, x 0
DERIVATIVES Example 2 The position vector r(1) = i + j starts at the origin. The tangent vector r (1) starts at the corresponding point (1, 1).
DERIVATIVES Example 3 Find parametric equations for the tangent line to the helix with parametric equations x = 2 cos t y = sin t z = t at the point (0, 1, π/2).
DERIVATIVES Example 3 The vector equation of the helix is: r(t) = 2 cos t, sin t, t Thus, r (t) = 2 sin t, cos t, 1
DERIVATIVES Example 3 The parameter value corresponding to the point (0, 1, π/2) is t = π/2. So, the tangent vector there is: r (π/2) = 2, 0, 1
DERIVATIVES Example 3 The tangent line is the line through (0, 1, π/2) parallel to the vector 2, 0, 1. So, by Equations 2 in Section 10.5, its parametric equations are: x 2t y 1 z t 2
DERIVATIVES The helix and the tangent line in the Example 3 are shown.
SECOND DERIVATIVE Just as for real-valued functions, the second derivative of a vector function r is the derivative of r, that is, r = (r ). For instance, the second derivative of the function in Example 3 is: r (t) = 2 cos t, sin t, 0
DIFFERENTIATION RULES The next theorem shows that the differentiation formulas for real-valued functions have their counterparts for vector-valued functions.
DIFFERENTIATION RULES Theorem 3 Suppose: u and v are differentiable vector functions c is a scalar f is a real-valued function
DIFFERENTIATION RULES Theorem 3 Then, d 1. [ u( t) v( t)] u' ( t) v' ( t) dt d 2. [ c u( t )] c u' ( t ) dt d 3. [ f ( t ) u( t )] f '( t ) u( t ) f ( t ) u' ( t ) dt
DIFFERENTIATION RULES Theorem 3 d 4. [ u( t) v( t)] u' ( t) v( t) u( t) v' ( t) dt d 5. [ u( t) v( t)] u' ( t) v( t) u( t) v' ( t) dt d 6. [ u( f ( t))] f '( t) u' f ( t) (Chain Rule) dt
DIFFERENTIATION RULES This theorem can be proved either: Directly from Definition 1 By using Theorem 2 and the corresponding differentiation formulas for real-valued functions
DIFFERENTIATION RULES The proof of Formula 4 follows. The remaining are left as exercises.
FORMULA 4 Let Proof u(t) = f 1 (t), f 2 (t), f 3 (t) v(t) = g 1 (t), g 2 (t), g 3 (t) Then, u( t) v( t) f ( t) g ( t) f ( t) g ( t) f ( t) g ( t) 1 1 2 2 3 3 3 i1 f ( t) g ( t) i i
FORMULA 4 Proof So, the ordinary Product Rule gives: 3 d d [ u( t) v( t)] fi( t) gi( t) dt dt 3 i1 3 i1 i1 d dt [ f ( t ) g ( t )] i [ f '( t) g ( t) f ( t) g '( t)] 3 3 f '( t) g ( t) f ( t) g '( t) i i i i i1 i1 u' ( t) v( t) u( t) v' ( t) i i i i i
DIFFERENTIATION RULES Example 4 Show that, if r(t) = c (a constant), then r (t) is orthogonal to r(t) for all t.
DIFFERENTIATION RULES Example 4 Since r(t) r(t) = r(t) 2 = c 2 and c 2 is a constant, Formula 4 of Theorem 3 gives: d 0 [ r( t) r( t)] dt r '( t) r( t) r( t) r '( t) 2 r'( t) r( t)
DIFFERENTIATION RULES Thus, r (t) r(t) = 0 This says that r (t) is orthogonal to r(t).
DIFFERENTIATION RULES Geometrically, this result says: If a curve lies on a sphere with center the origin, then the tangent vector r (t) is always perpendicular to the position vector r(t).
INTEGRALS The definite integral of a continuous vector function r(t) can be defined in much the same way as for real-valued functions except that the integral is a vector.
INTEGRALS However, then, we can express the integral of r in terms of the integrals of its component functions f, g, and h as follows.
INTEGRALS b a r() t dt n i 1 * i lim r( t ) n t n n * i * i i i1 i1 lim f ( t ) t g( t ) t j n n h( t * ) t k i i1
INTEGRALS Thus, b a r(t)dt b a f (t)dt i b a g(t)dt j b a h(t)dt k This means that we can evaluate an integral of a vector function by integrating each component function.
INTEGRALS We can extend the Fundamental Theorem of Calculus to continuous vector functions: a b r (t) dt R ( t) R ( b) R ( a) a b Here, R is an antiderivative of r, that is, R (t) = r(t). We use the notation r(t) dt for indefinite integrals (antiderivatives).
INTEGRALS Example 5 If r(t) = 2 cos t i + sin t j + 2t k, then r(t)dt 2cost dt where: C is a vector constant of integration 2 0 i sint dt 2sint i cost j t 2 k C r( t) dt [2sin t i cos t j t k] 2 2i j k 4 j 2t dt 2 2 0 k