Calculus: Several Variables Lecture 27

Transcription

1 alculus: Several Variables Lecture 27 Instructor: Maksim Maydanskiy Lecture 27 Plan 1. Work integrals over a curve continued. (15.4) Work integral and circulation. Example by inspection. omputation via parametrization. Examples. 2. Independence of path and conservative vector fields. Independence of work form path taken. Potential energy and scalar potential of a vector field. Force from potential. Gradient vector fields. Multivariable Main Theorem of alculus, version 1 AKA Fundamental Theorem for Line Integrals. hecking for conservativeness. Finding scalar potential. Sufficient condition for conservativenes. Integral of a (tangential component of) vector field over a curve or Work integral. Physically The work integral will give the answer to the following question: if a force given at each pint by F (x, y, z), acts on a particle as it moves over a path, what is the work it does on the particle? Note: I warned you in the first lecture there will be some interpretation from physics. We are now coming to the more physics-related part of the course. I will try to define the concepts from physics that I use. If I forget one and you haven t encountered it or don t remember it, please ask. Reminder about/definition of work done by a force: If a constant force F acts on a body that travels in a straight line over displacement δ r, the work done by the force on the body is F r. In particular, if the body is not moving the work done is zero. If the body is moving perpendicular to the direction of the force, the work done is zero. It is only the tangential component of the force that does work. Physical reason to define work this way is that it measures energy transferred to the body via the force. For the purposes of this math class, we can take this simply as definition, without going further into the question of why it is a good definition for physicists. See, however, the discussion at questions/1984/why-does-holding-something-up-cost-energy-while-no-work-is-being-done. 1

2 alculus: Several Variables Lecture 27 Mathematically: The integral of tangential component will give a natural way to integrate a vector field over a curve by turning it into a scalar along the curve. To see that this question is answered by an integral, observe that if we decompose into two pieces 1 and 2 then the work done by F over is the sum of work over 1 and work over 2. So this is an additive quantity. Our strategy for answering this question is thus the same as for all additive quantities: decompose into small pieces and see what answer you get for each small piece, then sum up (integrate) the results. In that spirit, let ) i be a small piece of S. What volume of water crosses through it? The two approximations we are going to make reduce us precisely to the case of the constant force acting over straight line displacement which we described above: 1. Since ) i is small, it is well approximated by tangent line T ) at any of its points i i. That is, we will make an approximation by pretending that ) i is actually flat, lying in the tangent line at i. 2. Since ) i is small for all the points on ) i, we have F (x, y, z) F ( i ). Under these approximations, the work done by F () over ) i is approximated by F ( i ) δ r i = F ( i ) T ( i ) s i. Here T ( i ) is the unit tangent to at i and ds i is the length of i. Summing up over i we get W i F ( i ) T ( i )δs i, which is the Reiman sum for th scalar line integral ) F T ds. Important note: It is important here that we take a unit tangent pointing along the displacement, not opposite to it. Otherwise we geus the work we are trying to compute. A choice of such a unit tangent is called orientation of. So work integrals are defined not over curves, but over oriented curves. Thus we expect that the total work should be ) F T ds, and we make it into a definition. Definition: (compare (Adams & Essex, p. 88)): The line integral of tangential component of vector field F along an oriented curve (or, simply, line integral of F over ; or, informally, work integral of F over ) is defined to be F T ds ) Notation: The book uses notation d r = T ds. Note that mathematically the flux integral is really a special type of a scalar integral over we first make a scalar function out of F on by doting it with T, and then integrate the result. Example: Find the work integral or F (x, y) = ( y, x) over the circle in the x y plane of radius a, centered at (, ), and oriented counterclockwise. 2

3 alculus: Several Variables Lecture 27 Solution: Geometrically, the vector field F is obtained from radial vector field r(x, y) = (x, y) by 9 degree counterclockwise rotation. So it is tangent to the circle, positively proportional to T. We get F T = F hatt cos θ = F = a. So the integral is F T ds = ads = al() = a2πa = 2πa2. Algebraically, for any point P = (x, y) on the circle of radius a, T (x, y) = 1( y, x), a so F (P ) T (P ) 1 a (y2 + x 2 ) = a, so F T ds = ) ads = a(length) = ) 2πa2. Example: Find the work integral or F (x, y) = ( y x, ) over the circle in the x 2 +y 2 x 2 +y 2 x y plane of radius a, centered at (, ), and oriented counterclockwise. Solution: Geometrically, the vector field F is tangent to the circle, positively proportional to T, and of length 1. We get F T = F hatt cos θ = F = 1. So the integral a a is F T ds = ads = 1L() = 1 2πa = 2π, independently of the radius. a a Notation: If a curve is closed (in human language, if it s a loop), then the work integral int F T ds is also written F T ds and is called circulation of F around. Similarly a flux integral through a closed surface S is sometimes denoted by S F NdS. In this case one assumes that the unit normal N is chosen to point out of S 1 Work integral via parametrization. We can use a parametrization to compute work integrals. If we have a parametrization, we can use it for both finding uint tangent and for computing resulting scalar integral over the curve. Note that in the first procedure we take T = v(t) and divide it by its length, v(t), while in the second procedure we have to multiply f( r(t)) by the Jacobian stretching factor which is precisely v(t). These two operations cancel out. We get: F T tmax ds = F ( r(t)) v(t) ) Note: A parametrization gives orientation as well, by choosing the tangent direction at each point of which is positively proportional to velocity at that point. Example: Find work of F = (y 2, 2xy) over path y = x 2 in the plane from (, ) to (1, 1). Solution: Let r(t) = (t, t 2 ), for t [, 1]. Then ) F T ds = tmax F ( r(t)) v(t) = 1 ((t 2 ) 2, 2t(t 2 ) (1, 2t) = 1 t 4 + 4t 4 = 1 Example: Find work of F = (y 2, 2xy) over path y = x in the plane from (, ) to (1, 1). 1 A non-trivial theorem ensures that this is always well defined. 3

4 alculus: Several Variables Lecture 27 Solution: Let r(t) = (t, t), for t [, 1]. Then F T ds = tmax F ( r(t)) v(t) = 1 (t 2, 2t 2 ) (1, 1) = ) 1 3t 2 = 1 Example: Find work of F = (y 2, 2xy) over the unit circle (oriented counterclockwise). Solution: We parametrize x(t) = cos(t), y(t) = sin(t). We get ) F T ds = tmax F ( r(t)) v(t) = 2π (sin 2 t, 2 sin t cos t) ( sin t, cos t) = 2π sin 3 t+2 sin t cos 2 d We can evaluate this integral by hand, or we can use a symmetry trick: with respect to the map t t + π the function we are integrating is odd, so the integral is zero. One way to show this rigorously is to split the integral into two parts exchanged by t t + π and check that the contributions cancel out: 2π π sin 3 t + 2 sin t cos 2 = π sin 3 t + 2 sin t cos 2 t + sin 3 t + 2 sin t cos 2 t + 2π π π where u = t π, so that sin u = sin t, cos u = cos t and du =. sin 3 t + 2 sin t cos 2 t = sin 3 u 2 sin u cos 2 udu = Independence of path and conservative vector fields. Independence of work form path taken. scalar potential of a vector field. Potential energy and We see that in the above example two different paths from (, ) to (1, 1) gave the same answer for work integral of force F = (y 2, 2xy). One can check a few other paths as well and see that for this force work performed (, ) to (1, 1) is always 1. Similarly, work performed by F from (, ) to (2, 2) is always 8. One can check other points and other paths the result will always depend on endpoints, but not on the path taken between these endpoints. As a special case, integral over any loop will be zero. On the other hand we saw that for F (x, y) = ( y, x) the integral over unit circle was not zero. So this is not something which happens for all F. So, which F have this property? One way to state the answer comes from our physical interpretation: some forces do work which is independent of path. Such forces come from a potential or potential energy and are called conservative, because if a particle moves under the influence of such a conservative force, its total energy (kinetic plus potential) is conserved throughout the motion. Here is a prototypical example: 4

5 alculus: Several Variables Lecture 27 Example: Find work done by force field F (x, y, z) = gmk moving a particle of mass m from the origin to location (x f, y f, z f ) along any path Solution: Notice F d r ( = (,, mg) dx(t) r(t) = (x(t), y(t), z(t)). Hence: tmax F d r =, dy(t) [ d ( mg z(t)) ] ), dz(t) = mg dz(t) = d ( mg z(t)) for any = ( mg z(t max ) ( mg z( )) = mgz f = mgz f by (usual, single variable) Main Theorem of alculus. Observe that the resulting function f(x, y, z) = mgz is minus what a physicist would call potential for this gravitational force, which is U(x, y, z) = mgz. The reason is that physical potential measures the amount of work a particle can do on an external system (it is called potential because it is potential to do work ); while we found the work done by the force on the particle. Hence the minus sign. Observe also that if we chose a different starting point (x, y, z ) (rather than the origin), we would get that the work done is mg(z z ). So the potential is only defined up to a constant. Similar things are true in general. We summarize this in the following definition: Definition: Suppose D is an open connected domain and let F be a smooth vector field on D. Suppose that the work integral of F between any two points p and q does not depend on path. Then we can define function f(p), called scalar potential of F by picking any starting point p and declaring f(p) = F r for any path from p to p. Observe: If f is potential of F then for any path from p to q F r = f(p) f(q) The reason is as follows: f(q) is integral from the point p to q over some and so any path. Take such a path, attach to it path. The composite path is a path from p to p, so the integral over it is f(p). Thus the difference, which is integral over, is f(p) f(q). Example For F (x, y, z) = mgk we know f(x, y, z) = mgz is a potential, and F r = mgz(p) (mgz(q)). Example We suspect that F (x, y) = (y 2, 2xy) is conservative. We can compute integral along straight line segment from (, ) to (x f, y f ) given by r(t) = (tx f, ty f ) to be 1 F (tx f, ty f ) (x f, y f ) = 1 (t2 yf 2x f + 2t 2 x f y f y f ) = x f yf 2. It looks like f(x, y) = xy2 should be the potential for F. 5

6 alculus: Several Variables Lecture 27 Force from potential. Gradient vector fields. Multivariable Main Theorem of alculus, version 1 AKA Fundamental Theorem for Line Integrals. So, we can construct the potential from a conservative vector field. What about the reverse? In physical terms, if we know potential energy, can we reconstruct the force? In mathematical terms, can we roecover F from scalar potential U(x, y, z)? Theorem: (Fundamental Theorem for Line Integrals) f(x, y, z) is a scalar potential for F (x, y, z) if and only if F (x, y, z) = f(x, y, z). In particular, work integrals of F are independent of path precisely when F is a gradient of some U. The reason: Suppose f is potential for F. Take to be a small straight segment in direction û = T through point p. Then on one hand, F T ds F ûl(). On the other hand, since U is a potential for F, F T ds = f(end) f(start) f ul(). We conclude F û = f u for all u, so F = f. In the other direction, suppose that F is gradient of U. We compute for a parametrized path: F ( r(t)) d r d r = f = D f = d d r (f( r(t))) The last equality is the definition of directional derivative as slice derivative derivative of restriction of a function to a path (in the definition we had for D v f the slice was straight, but in fact as long any path with same velocity works). Integrating, we get: F d r = tmax [ d ( f( r(t))) ] = f( r(t max )) f( r( )) = f(q) f(p) Example: F (x, y, z) = mgk = ( mgz) = f(x, yz). Example: F (x, y) = (y 2, 2xy) = (xy 2 ) = f(x, y, z), and so indeed, F is conservative, and f(x) = xy 2 is its scalar potential. So, in summary: conservative vector fields are gradient vector fields, and f r = f(p) f(q) 6