Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 1 The Calculus of Vec- Summary: tors 1. Calculus of Vectors: Limits and Derivatives 2. Parametric representation of Curves r(t) = [x(t), y(t), z(t)] Suggested Reading: Griffiths: Chapter 1, Section 1.2.2, page 13. Kreyszig, Chapter 8, Sections 8.6-8.8, pages 464-471 Wangsness, Chapter 1, Section 1.8, page 11
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 2 The Calculus of Vectors: Differentiation Before we can evaluate the derivatives of a vector function we need to define the notion of the limit for vector quantities. The limit as t t 0 lim[ V (t)] = l t t 0 exists only if the corresponding limits for each component exist lim t t0 [V x (t)] = l x (t 0 ) lim[ V (t)] = l lim t t0 [V y (t)] = l y (t 0 ) t t 0 lim t t0 [V z (t)] = l z (t 0 ) (1) The vector function V (t) is said to be continuous at t = t 0 if it is defined in some neighborhood of t 0 and if lim[ V (t)] = V (t 0 ) t t 0 The derivative with respect to t of V (t) is defined as [ d V dt [ V (t)] = V (t + t) ] V (t) (t) lim t 0 t which should look like the familiar definition of the derivative of any scalar function of t.
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 3 Instead of the variable t we could substitute any other variable, say p. For the moment we will restrict ourselves to scalar variables (or parameters) such as t and p. From the definition of the limit of a vector function (see above), the derivative of a vector function V (t), i.e., V (t) is defined if and only if the derivatives of each of the vectors components is defined. Then, V (t) = [V 1(t), V 2(t), V 3(t)] All of the usual rules of differentiation apply to V (t) (as to its components). In particular, (cv ) = cv where c is a scalar constant independent of t, ( u + v) = u + v ( u v) = ( u v) + ( u v ) ( u v) = ( u v) + ( u v ), etc. We could also write d ( u v) = (d u dt dt etc. v) + ( u d v dt ),
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 4 Example 1: A particle moves in such a way that its position vector as a function of time, t, is given in Cartesian coordinates by: r(t) = [x(t), y(t), z(t)] = [r 0 cos ωt, r 0 sin ωt, 0], where r 0 and ω are constants independent of t. (a) Calculate v(t) and show that v r, this is the linear velocity of the particle. (b) Find a(t), the linear acceleration (c) Find ω(t), the angular velocity Solution: (a) v(t) = d r dt = [ r 0 ω sin ωt, r 0 ω cos ωt, 0] (2) = r 0 ω sin ωtî + r 0 ω cos ωtĵ To show that v(t) is perpendicular to r(t), simply take the scalar product: r(t) v(t) = r 2 0ω sin ωt cos ωt+r 2 ω cos ωt sin ωt = 0 Therefore, v(t) is perpendicular to r(t).
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 5 (b) a(t) = d v(t) dt = d2 r(t) dt 2 = [ r 0 ω 2 cos ωt, r 0 ω 2 sin ωt, 0](3) a(t) = ω 2 r I.e., a(t) = ω 2 r(t), a is proportional to r, but the direction of a is opposite to that of r (c) v = ω r r v = r ( ω r) = ( r r) ω ( r ω) r We showed above (in part (a)) that v r, therefore, since both r and v lie in the x,y-plane, the vector r v must be perpendicular to this plane, i.e., it must be in the z-direction. The three vectors r, v and r v are mutually perpendicular so that r ω = 0 and, finally, ω = 1 r 2( r v) is the angular velocity having a magnitude ω and with direction r and v. In fact, r, v and ω form a right handed triple (like î, ĵ and ˆk).
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 6 Angular Momentum: You may recall from mechanics that the angular momentum about the origin of a particle at position r with linear momentum p is l = r p = r (m v) = m r v = mr 2 ω = I ω where I = mr 2 is the Moment of Inertia. The Parametric Representation of a Curve in Space In the previous example we used a parametric representation of the position of the particle in space. The parameter we used was the time, t. It will often be convenient to define some scalar parameter (such as t) and then write a path in space in terms of this parameter as r(t) = [x(t), y(t), z(t)] The parameter need not be time, however. It may simply be some convenient way to define a particular path. The example above was basically r(t) = [cos t, sin t, 0] which is the parametric representation of a circle (usually for this type of planar motion, one would
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 7 write x = x(y), in this case it would be x 2 = 1 y 2, or x 2 + y 2 = 1. In terms of the parameter t this would be x 2 = cos 2 t = 1 sin 2 t or cos 2 t+sin 2 t = 1. Similarly, the equation r(t) = [a cos t, b sin t, 0] is the equation of an ellipse. What is: r(t) = [a cos t, a sin t, ct] Notice: if c = 0 then this is the equation of a circle with radius a. By examining how the vector r(t) traces out the curve one can easily see that the tangent to any curve C at a point C on that curve is simply given by the derivative of the parametric expression for the curve with respect to the parameter, evaluated at the point P: since d r(t) dt û = d r/dt d r/dt + t) r(t) = lim { r(t } t 0 t
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 8
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 9 Partial Derivatives of a Function of Several Variables You should already be familiar with the concept of partial derivatives of a function of several variables. For a function f(x, y, z) of three coordinates, x, y and z, to take the partial derivative of f with respect to x at a point P : (x 0, y 0, z 0 ), you hold the other two coordinates fixed (i.e., you consider the intersection of the function with the planes at y = y 0 and z = z 0, and then take the derivative with respect to x). f x = lim {f(x + x, y 0, z 0 ) f(x, y 0, z 0 ) } x 0 x Return to the example of the equation of a plane discussed last time. Here we had an expression of the form F ( r) = F (x, y, z) = ˆn r d. If this plane cuts the three axes at x = 1, y = 2 and z = 3, then we find that n x = d, 2n y = d and 3n z = d (these relations are obtained by substituting in the coordinates at the points where the plane cuts the axes). Using ˆn ˆn = 1 since it is a unit vector, we find that d 2 (1 + 1/4 + 1/9) = 1 or d = ±6/7.
Physics 2460 Electricity and Magnetism I, Fall 2007, Lecture 3 10 Since F ( r) = n x x + n y y + n z z 6/7 we have that F x = n x = 6/7 F y = n y = 3/7 F z = n z = 2/7 Thus, the components of the unit normal to the plane defined by our function F ( r), ˆn = [n x, n y, n z ] = [6.7, 3/7, 2/7] are simply the partial derivatives of F ( r), i.e., the normal to the plane is given by ˆn = [ F (x, y, z), x F (x, y, z), y F (x, y, z) ] z This interesting result suggests that the set of partial derivatives of a function F (x, y, z) is a vector.