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Vector differentiation and the vector

Vector differentiation and the vector - Chapter 2 Vector...

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Chapter 2 Vector differentiation and the vector differential operator Syllabus topics covered: 1. Vector fields 2. Grad, div and curl operators in Cartesian coordinates. Grad, div, and curl of products etc. Here we cover differentiation of vectors. This differs from the gradient introduced in the previous chapter, which is a vector quantity that arises from differentiating a scalar. 2.1 Vector functions of one or more variables (See Thomas 13.1) In many physical contexts one is interested in vectors that vary with position or time. For example, the position of a point can be described by a vector r . Thus, if we consider a moving particle, its position can be described as a function of time t by the vector r ( t ) , and its rate of change with respect to t will be the velocity (which has magnitude and direction, i.e. is a vector: its magnitude is the speed). The position vector is then a function of one variable. Another context is where we have a vector defined at each point, say F ( r ) = F ( x , y , z ) and a curve with a parameter u , say, so its points are ( x ( u ) , y ( u ) , z ( u )) . Then we can define a vector function of u , F ( u ) = F (( x ( u ) , y ( u ) , z ( u )) . We can deal with this and the moving particle case as follows. A vector function of a scalar u , F ( u ) , can be defined by specifying its components as functions of u : F ( u ) = ( f 1 ( u ) , f 2 ( u ) , f 3 ( u )) . The derivative d F / d u of F with respect to u is then: d F d u = parenleftbigg d f 1 d u , d f 2 d u , d f 3 d u parenrightbigg . This simply goes back to the fundamental definition of a derivative: d F d u = lim δ u 0 F ( u + δ u ) F ( u ) δ u . 16
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Clearly one can compute higher derivatives, such as d 2 F / d u 2 , by differentiating the components of F the required number of times. Example 2.1. If r ( t ) is the position vector of a particle, as a function of time t , then d r / d t is the velocity v of the particle. Also d v / d t d 2 r / d t 2 is the particle’s acceleration. Example 2.2. The continuous parameter t can take all real values. Write down the derivatives d r / d t and d 2 r / d t 2 for the vector r = ( sin t ) i + t j . Also, sketch the curve whose parametric equation is r = r ( t ) . The first and second derivatives are d r d t = ( cos t ) i + j , d 2 r d t 2 = ( sin t ) i . The sketch is shown in Fig. 2.1. -1 1 π 2 π π 2 π t = 0 t = π / 2 t = π Figure 2.1: Sketch of the curve defined parametrically by r = ( sin t ) i + t j It is easy to prove, by writing out the components and collecting terms, that if F and G are vector functions of u , then d ( F . G ) d u = F . d G d u + d F d u . G . Proof: d F . G d u = d d u ( f 1 g 1 + f 2 g 2 + f 3 g 3 ) = f 1 d g 1 d u + f 2 d g 2 d u + f 3 d g 3 d u + d f 1 d u g 1 + d f 2 d u g 2 + d f 3 d u g 3 = F . d G d u + d F d u . G . Q . E . D . Exercise 2.1. Sketch the curves whose parametric equations are 17
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(a) r = ( 3sin π t ) i +( 2cos π t ) j (b) r = ( cos π t ) j (c) r = t i + t 2 k ( t ), and write down the derivatives d r / d t and d 2 r / d t 2 where they are defined.
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