Multivariable Calculus Lecture #11 Notes

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1 Multivariable alculus Lecture #11 Notes In this lecture, we ll discuss changing coordinates more generally in multiple integrals. We ll also discuss the idea of integration along a curve and the application of this idea in conjunction with vector fields to define the wor done by a variable force along a specified curve (line integrals). This, in turn, will lead to a discussion of conservative vector fields vs. non-conservative vector fields, potential functions, and the Fundamental Theorem of Line Integrals. hange of Variables in Multiple Integrals In the previous two lectures we used direct geometric observations to derive methods for setting up double integrals in polar coordinates and for setting up triple integrals in cylindrical and spherical coordinates. There are some circumstances in which we may want to invent our own coordinates in order to set up and calculate integrals. The approach is analogous to the u-substitution method in single-variable calculus. There are three aspects to such a change: (a) changing the integrand, (b) changing the measure, and (c) changing the integral limits. Transforming a double integral Suppose you need to calculate an integral f ( x, y) da = f ( x, y) dxdy over a region D in R for which D standard ( xy, ) artesian coordinates may not be the ideal choice. Further suppose that you have decided on x= xuv (, ) using new coordinates ( uv, ) and that there is a coordinate transformation expressed by y= yuv (, ). We may wish to express this as Fuv (, ) = ( xuv (, ), yuv (, )) where ( uv, ) represents points in a region D in the uv-plane that is transformed in a one-to-one manner to the region D in the xy-plane. Transforming the integrand into the new ( uv, ) coordinates is straightforward substitution, but we need to better understand how the area element will transform. The ey to understanding this is to focus on the parameterized curves obtained by varying one parameter at a time. If we start at a point ( uv, ) D and vary u only (and thin of this variable u lie time ), then a curve in F D will be traced out and the vector will be lie a velocity vector that is tangent to this curve. A small u F change u will then yield an approximate displacement u u. Similarly, if we vary v only (and again F thin of this variable v lie time ), then a curve in D will be traced out and the vector will be tangent to v F this curve. A small change v will then yield an approximate displacement v v. These two displacement vectors will determine a small approximate parallelogram in D that is in one-to-one correspondence with the rectangle in D with side lengths u and v. If we relate the area u v of the rectangle in D and the area of the corresponding approximate parallelogram in D, we can use a linear algebra fact that this is given by the (absolute value of) the determinant of the matrix whose columns are the vectors that F F determine the parallelogram. That is A det u v det u v u v u v = y y. We introduce the u v D 1 Revised November 6, 017

2 notation det u v = ( uv, ) y y for the Jacobian determinant. The matrix itself J F u v Jacobian matrix associated with this coordinate transformation. u v = u v is the The above calculation enables us to relate a small area A = u v in D to the corresponding area in D by A u v. For the purposes of integration, this enables us, in the limit, to relate the area elements by ( uv, ) dxdy = dudv. Basically, the absolute value of the Jacobian determinant acts as a scaling factor ( uv, ) ( uv, ) for small areas that enables us to relate the measures in the respective regions. Using these facts, we conclude that D f ( x, y) dxdy = f ( x( u, v), y( u, v)) dudv D ( uv, ). For the purpose of calculation it is sometimes easier to deal with the inverse coordinate transformation. Understanding the Jacobian determinant to be a local area scaling factor, it follows that the reciprocal. ( uv, ) = ( uv, ) Example 1: The change from artesian coordinates to polar coordinates is accomplished via the transformation x= rcos θ given by y = rsin θ. Its Jacobian matrix is r θ cos θ r sin θ y y = sin rcos and the Jacobian determinant is θ θ r θ = r(cos θ+ sin θ ) = r, so dxdy = drdθ= r drdθ as expected. (, r θ) (, r θ) Example : Find the area of the region D bounded by the ellipse with semi-axes a and b and given by equation x y + = 1. a b Solution: Area ( D ) da dxdy x y = = D. If we express the equation of the ellipse as 1 D + =, this a b x a = u x = au suggests a coordinate change by rescaling the axes, i.e. y or = v b y = bv. The bounding curve of the transformed region D is then the unit circle with equation u + v = 1. a 0 The Jacobian determinant is = det = ab ( uv, ) 0 b, so the area is Area( D) = dxdy = dudv = ab dudv = ab dudv = ab Area( D ) = πab D D ( uv, ) D D the circular unit dis D is just π (no integration necessary). since the area of Revised November 6, 017 1, i.e.

3 Example 3: alculate the integral x y da = x y dxdy D where D is the parallelogram region bounded by D the lines y = x, y = x+, y = x+, and y = x+ 6. Solution: We can realize these boundary lines as level sets of coordinate functions by rewriting them as u = x+ y x+ y = 0, x+ y =, x+ y =, and x+ y = 6 and defining the coordinate change by v= x+ y. ( uv, ) 1 The Jacobian determinant is then = det = + 1 = Therefore = 1 ( uv, ) 3. To transform the u v x = integrand, we solve for 3 u v. The boundary curves (going counterclocwise around the boundary) + y = 3 are: (1) x+ y = u = ; () x+ y = 0 v= 0 ; (3) x+ y = 6 u = 6 ; and x+ y = v=. The resulting region D is therefore just a rectangle and the integral becomes v= u= 6 6 u v u+ v v= u= x y dxdy = 0 3 dudv 81 ( u 3uv v ) dudv D v= u= = v= 0. u= Inner Integral = u= u= ( u 3uv + v ) du = 4u u v uv 4 (196 16) 3 (36 4) v (6 ) v 30 48v 8v u= + = + = + u= v Outer Integral = ( ) = v v v dv v v v = v= 0 v= = = 81 ( ) = 81 Transforming a triple integral Suppose you need to calculate an integral f ( x, y, z) dv = f ( x, y, z) dxdydz over a region B in R 3 for B which standard ( xyz,, ) artesian coordinates may not be the ideal choice. Further suppose that you have decided on using new coordinates ( uvw,, ) and that there is a coordinate transformation expressed by x= xuvw (,, ) y= yuvw (,, ). We may wish to express this as Fuvw (,, ) = ( xuvw (,, ), yuvw (,, ), zuvw (,, )) where ( uvw,, ) z = zuvw (,, ) represents points in a region B in uvw-space that is transformed in a one-to-one manner to the region B in xyzspace. Transforming the integrand into the new ( uvw,, ) coordinates is straightforward substitution, but we need to better understand how the volume element will transform. Again, the ey to understanding this is to focus on the parameterized curves obtained by varying one parameter at a time. If we start at a point ( uvw,, ) B and vary u only (and thin of this variable u lie time ), then a F curve in B will be traced out and the vector will be lie a velocity vector that is tangent to this curve. A u F small change u will then yield an approximate displacement u u. If we vary v only (and again thin F of this variable v lie time ), then a curve in B will be traced out and the vector will be tangent to this v B 3 Revised November 6, 017

4 F curve. A small change v will then yield an approximate displacement v v. If we vary w only (and F again thin of this variable w lie time ), then a curve in B will be traced out and the vector will be w F tangent to this curve. A small change w will then yield an approximate displacement w w. These three displacement vectors will determine a small approximate parallelepiped in B that is in one-to-one correspondence with the rectangular solid region in B with side lengths u, v, and w. If we relate the volume u v w of the rectangle in B and the volume of the corresponding approximate parallelepiped in B, we can use a linear algebra fact that this is given by the (absolute value of) the determinant of the matrix whose columns are the vectors that determine the parallelepiped or, equivalently, the triple scalar product. That is u v w F F F y y y V det u v w det u v w u v w =. We introduce the notation u v w z z z u v w u v w u v w = det for the Jacobian determinant. The matrix itself J F = is the ( uvw,, ) u v w u v w z z z z z z u v w u v w Jacobian matrix associated with this coordinate transformation. The above calculation enables us to relate a small area V = u v w in B to the corresponding area in B by V u v w. For the purposes of integration, this enables us, in the limit, to relate the volume ( uvw,, ) elements by dxdydz = dudvdw. The absolute value of the Jacobian determinant acts as a ( uvw,, ) ( uvw,, ) scaling factor for small volumes that enables us to relate the measures in the respective regions. Using these facts, we conclude that f ( x, y, z) dxdydz = f ( x( u, v, w), y( u, v, w), z( u, v, w)) dudvdw B B ( uvw,, ). For the purpose of calculation it is sometimes easier to deal with the inverse coordinate transformation. Understanding the Jacobian determinant to be a local volume scaling factor, it follows that 1 ( uvw,, ) =, i.e. the reciprocal. ( uvw,, ) Example 4: onsider the spherical coordinate change given by the equations determinant is: 4 x =ρcos θsin φ y =ρsin θsin φ. The Jacobian z =ρcos φ Revised November 6, 017

5 ρ φ θ cos θsin φ ρcos θcos φ ρsin θsin φ = det = det sin θsin φ ρsin θcos φ ρcos θsin φ ( ρ, φ, θ) ρ φ θ cos sin 0 z z z φ ρ φ ρ φ θ = cos θsin φ( ρ cos θsin φ ) +ρcos θcos φ( ρcos θsin φcos φ ) +ρsin θsin φρ [ sin θ(sin φ+ cos φ)] So =ρ sin φ[cos θ(sin φ+ cos φ ) + sin θ ] =ρ sin φ[cos θ+ sin θ ] = ρ sin φ = = ρ φ θ= ρ φ ρ φ θ ρφθ (,, ) dv dxdydz d d d sin d d d as expected. Example 5: Find the volume of the region B bounded by the ellipsoid with semi-axes a, b, and c and given by x y z equation + + = 1. a b c Solution: Volume ( B ) = dv = dxdydz. If we express the equation of the ellipse as B B x y z + + = 1, this suggests a coordinate change by rescaling the axes, i.e. a b c The bounding surface of the transformed region B is then the unit sphere with equation a 0 0 The Jacobian determinant is = det 0 b 0 = abc, so the volume is ( uvw,, ) 0 0 c Volume( B) = dxdydz = B dudvdw = abc dudvdw = abc dudvdw B ( uvw,, ) B B 4 3 = abc Volume( B ) = πabc since the volume of the unit sphere B is just π (1) = π (no integration necessary) 3 3 x a = u x = au y = v b or y = bv z c = w z = cw u + v + w = 1. Integration along a curve We now turn our attention to situations where we may want to measure some quantity that s defined for points along a curve. As with all of our previous integral constructions there is a range of situations where might need to do this. A motivating example mass of a wire Suppose a wire is located in either R or R 3 with endpoints A and B and following a curve between these endpoints. Further suppose that there is a mass density function σ ( xy, ) or σ ( xyz,, ) defined for all points along the wire and measured in appropriate units such as grams/centimeter (a linear density). If we partition the curve into small pieces where the length of the -th piece is s, then we can estimate the mass of this piece as m s( x ) s where x is a sample point chosen within this -th piece. Summing these, we get that the total mass is given (approximately) by Mass( ) = m s( ) s x. As we refine the partition and pass to the limit as the mesh = max[ diam( )] tends to zero and if this limit exists independent of any. 5 Revised November 6, 017

6 choices, then we define lim ( ) s ( ) ds 0 s x = s x. We can, for the sae of simplicity when woring with these integrals, write dm = s ds which allows us to write Mass ( ) = dm = s ds. There is, of course, nothing special about using mass as the quantity to be measured. If, for example, σ ( xyz,, ) measured the electric charge density at points along a charged wire, then we would write harge ( ) = dq = s( x, y, z ) ds σ xyz measured the population density at points along a road, then. If (,, ) Population = dp = s x y z ds. In this manner, we can measure the total amount of we would write ( ) (,, ) any quantity defined along such a curve from its associated density function. Many of the same applications that we described in terms of double integrals or triple integrals can also formulated for curves. Here is a list of some of these applications with minimal derivation: 1) If σ ( xy, ) or σ ( xyz,, ) is a mass density function for an object that occupies a curve, then the total mass of. this object is given by: Mass ( ) = dm = s ds There is, of course, nothing special about using mass as the quantity to be measured. If, for example, σ ( xyz,, ) measured the electric charge density at points along a charged wire, then we would write. If ( xyz,, ) harge ( ) = dq = s( x, y, z ) ds then we would write Population ( ) = dp = s( x, y, z ) ds σ measured the population density at points along a road,. In this manner, we can measure the total amount of any quantity defined along such a curve from its associated density function. ) The arclength of a curve can be calculated as: Length( ) = ds. We ll elaborate on this shortly when we discuss methods of calculation for these integrals. 3) In the case of a function f defined along a curve, we can define the (unweighted) average value of this function over this curve implicitly by the relation: (Length of ) f = f ds. Therefore the average value is given by: f = f ds. Once again, the (unweighted) average is calculated by integrating the Length ( ) function over its domain and then dividing by the geometric content of the domain. 4) The centroid or geometric center of a curve in R 3 is the point ( xyz,, ) where x ds x = and y ds y = and z = z ds. Length ( ) Length ( ) Length ( ) We can also do this for curves in the xy-plane using just the first two expressions. The centroid need not actually lie on the curve. 5) In the case of a function of three variables f defined along a curve, if we have an associated density function σ defined on this curve that permits us to weigh some parts more than others, we can define the weighted average of the function implicitly by the condition that (Mass of ) f = f dm. wtd 6 Revised November 6, 017

7 This gives the definition f wtd f dm f s ds = =. Mass( ) s ds 6) The center of mass of a region in R 3 with associated mass density function σ ( xyz,, ) is the point ( xcm, ycm, z cm) where x cm x dm x s ds = = Mass ( ) s ds, y cm y dm y s ds = =, Mass( ) s ds z cm z dm z s ds = = Mass( ) s ds It is worth noting that in the case of constant density the centroid and center of mass will coincide.. Example 6: Find the center of mass of the half-circular region of radius 1 with x + y = 1 with y 0 if the density is (a) constant; and (b) σ ( xy, ) = y. Solution: In either case, the symmetry of the curve as well as the density function enables us to conclude that y sds s y ds y ds x cm = 0. In case (a), the density is constant, so ycm = = = = y. The sds s ds Length( ) denominator is just π, and we can relate arclength to the central angle θ by ds = Rdθ= dθ. So the numerator becomes y ds π (sin ) d π [ cos ] y ds = θ θ= θ 0 0 =. So y cm = = π Length( ) y sds ( ) y y ds In case (b), ycm = =. Again using y = sin θ and ds = dθ, we have: sds ( y) ds π π Denominator = y ds d [ ] 0 0 Numerator = ( y y ) ds π (sin sin ) d 4 0 y ( y) ds 8 π Therefore 8 π ycm = = =.567 ( y) ds π 4π 4 ( ) = ( sin θ) θ= θ+ cosθ = π = θ θ θ= = = 7 π 8 π greatest toward lower y-values, we expect the center of mass to be lower in (b).. omparing this with part (a), since the density is General Method: If the given curve is be parameterized by r () t where a t b and if v() t = r () t is the ds velocity vector, we can use the fact that the speed is v () t = to write ds = v() t dt = v dt. This will enable dt us to pull bac a given integral to produce an ordinary definite integral in the parameter t. That is, t= b f ( x) ds = f ( r()) t v () t dt. t= a Example 7: Find the length and mass of a wire configured along the portion of the parabola y = x from the point ( 1,1) to the point (, 4) if the density is given by σ ( xy, ) = y. x= t Solution: We can parameterize the curve by y = t with 1 t. The parameterization function is r () t = tt,. The velocity is v ( t) = 1, Using the above method, the length will be t, and the speed is v () t = 1+ 4t. ( ) = = ( ) = Length ds v t dt t dt Revised November 6, 017

8 The mass is given by 4 Mass ( ) = s ds = y ds = t 1+ 4 t dt Vector Fields in R and R 3 One of the most important concepts in such disparate fields as physics, ecological modeling, and economics is the concept of vector field. The term itself is relatively self-explanatory. To every point in whatever space we are situated, we assign a vector. This might be done via formulas or perhaps defined by the situation, e.g. the velocity vector at a point in space associated with the wind at any given moment. If we tae a artesian point of view, we might define a vector field F in R by assigning to each point ( xy, ) a corresponding vector F ( xy, ) = Pxy (, ), Qxy (, ) where we ll refer to the functions Pxy (, ) and Qxy (, ) as the component functions of this vector field. We will often require such conditions as continuity or differentiability of the component functions, but this depends on the application. For a vector field F in R 3 we would assign to each point ( xyz,, ) a corresponding vector F ( xyz,, ) = Pxyz (,, ), Qxyz (,, ), Rxyz (,, ). Example 8: We define the radial vector field r ( xy, ) = xy, in R. This simply assigns to every point its own position vector but relocated to the given point. This vector field forms a good starting point for constructing other interesting vector fields. In R 3 we would have r ( xyz,, ) = xyz,,. r, Example 9: We define the unit radial vector field rˆ = = xy = x, y in R. This r x + y x + y x + y assigns to every point (except the origin) a unit vector pointing radially outward. In R 3 we would have r,, rˆ = = xyz = x, y, z. As you may surmise, this is a r x + y + z x + y + z x + y + z x + y + z vector field that s easy to understand conceptually but doesn t mesh well with artesian coordinates. Example 10: We can define a vector field ˆθ in R that assigns to each point (except the origin) a unit vector in the direction of increasing polar angle θ. First, note that if v = ab,, then we can rotate this vector counterclocwise (retaining its magnitude) to get the vector ba,. Using this idea, we can define ˆ yx, y x θ= =, x + y x + y x + y. Example 11: We now from physics that the gravitational attraction associated with a mass M located at the origin on another mass m will be directed radially inward toward the origin with magnitude given by the GMm inverse square law where R is the distance between the masses. Using the above constructions with R appropriate scaling and reversal of sign, we can express the gravitational force by: GMm GMm r GMm GMm x, y, z F= ˆ r = = r = 3 3 R r r r ( x + y + y ) Example 1: Given any differentiable function Vxy, (, ) the gradient of this function actually defines a vector V V field that assigns to every point ( xy, ) R the gradient vector at that point, i.e. Vxy (, ) =,. You x y 8 Revised November 6, 017

9 should recall that the gradient vector at any given point will be perpendicular to the level set passing through that point and will be directed toward increasing values. For a differentiable function Vxyz, (,, ) the gradient 3 of this function actually defines a vector field that assigns to every point ( xyz,, ) R the gradient vector at V V V that point, i.e. Vxyz (,, ) =,,. At any point this vector will be perpendicular to the level x y z surface passing through that point. Wor done by a variable force along a curve (line integrals) In physics, wor done by a constant force F over a distance s is just the product F s. If the force is a vector, then only the tangential part of this vector F T will contribute to the wor, i.e. FT s. Now suppose that F is a variable force field acting along a path that goes from a starting point A to an ending point B. If we partition the path into small pieces where the length of the -th piece is s, then we can estimate the wor done along this small segment as W FT s where F T is evaluated at a sample point x within this -th piece. Summing these, we get that the total wor is given (approximately) by W = W FT s the partition and pass to the limit as the mesh = max[ diam( )] tends to zero we get F s F ds lim T T 0 =. As we refine. This is the fundamental definition of wor, but it is often expressed in other forms. Suppose we parameterize the curve in such a way that the velocity vector never vanishes, i.e. eep moving. If the parameterization function is r () t = xt (), yt (), zt () where a t b, the velocity vector will then be dr dx dy v () t = =, in R dr dx dy dz or v () t = =,, in R 3, the speed will be v () t and we can rewrite dt dt dt dt dt dt dt ds the fact that v () t = as ds = v() t dt = v dt. The unit tangent vector can be calculated as T = v dt v, and dr F T = FT. We can also formally express the fact that () t dt = v as dr = v dt = dx, dy in R or dr = v dt = dx, dy, dz in R 3. Using these relations, we can write: b v b Wor = F T ds = FT ds = dt = dt = d F v a Fv a F r v If F ( xy, ) = Pxy (, ), Qxy (, ), then this can be expressed as Wor = F d r = P ( x, y ) dx + Q ( x, y ) dy, though we will often express this simply as P dx + Q dy. If F ( xyz,, ) = Pxyz (,, ), Qxyz (,, ), Rxyz (,, ), then this can be expressed as Wor = F d r = P ( x, y, z ) dx + Q ( x, y, z ) dy + R ( x, y, z ) dz, though we will often express this simply as P dx + Q dy + R dz. Example 13: alculate the wor done by the force (5, 3). F = xy, x along the straight line path from (1, ) to 9 Revised November 6, 017

10 Solution: Using the above formalism, we can express the wor in this case as t t t dt t dt t t xy dx x dy. The path can be x= 1+ 4t parameterized as y = 5t where 0 t 1. We calculate dx = 4dt dy = 5dt, and substitution into the integral gives [(1 + 4 )( 5 )(4) (1 + 4 ) ( 5)] = ( ) = = 39. Example 14: alculate the wor done by the same force F along the path consisting of the = xy, x horizontal segment 1 from (1, ) to (5, ) followed by the vertical segment from (5, ) to (5, 3). Solution: xy dx x dy = ( xy dx x dy) + ( xy dx x dy). = 1 1 We can parameterize the first segment using x as both parameter and coordinate, we get 1 5 x = 5 x= 1 1 ( xy dx x dy ) = xdx = x = 5 1 = 4 For the second segment, we can use y as both parameter and coordinate, [ ] ( xy dx x 3 dy) y= 5dy 5y 3 = = = 15 y= Therefore = 1 = = 149 xy dx x dy x= x y =, dx = dx dy = 0, and x = 5 y = y, dx = 0 dy = dy, and we get Note that for this vector field, the wor integral gives different results for different paths connecting the same two endpoints. onservative Vector Fields and the Fundamental Theorem of Line Integrals It s essential to eep in mind that when calculating a path integral of the form F dr (nown as a line integral even though it has nothing to do with lines), the value of this integral from one point to another will generally depend on which path is taen. The idea is derived from physics and basically means that the wor (energy) done by a vector field (representing a force) as you travel from one point to another may well depend on the path followed. Definition: A vector field F is called conservative if the wor integral depends only on the endpoints of the path. F dr is independent of path, i.e. it We calculated a wor integral above following two different paths and found two different values for the wor. This raises the question of when might the wor integral be independent of path? We ll see that, in fact, conservative vector fields and gradient vector fields are the same thing. It s easy to see why a gradient vector field is conservative based on the following: Fundamental Theorem of Line Integrals: Let be a smooth curve given by the vector function r ( t), a t b and let F = V where Vxy (, ) (or Vxyz) (,, ) is a differentiable function of two (or three) variables whose gradient F = V is continuous on the curve. Then: d F dr = V dr = V( r( b)) V( r( a)) = V(end) V(start) 10 Revised November 6, 017

11 The function V is generally called a potential function, and the Fundamental Theorem of Line Integrals essentially says that the wor done by a conservative vector field in following a given path is the potential difference. The fact that a gradient vector field is conservative should be clear from the statement of this theorem. In the case of a gradient vector field, the wor depends only on the values of the potential function at the endpoints not on any particular path followed from the starting point to the endpoint. Proof of the Fundamental Theorem of Line Integrals: This is just a blending of the Fundamental Theorem of alculus and the hain Rule. In the case of F = V where V= Vxyz (,, ) in R 3, we have: d V V V V V V b V dx V dy V dz V dr =,, dx, dy, dz dx dy dz dt = x y z + + = x y z a + + dt dt z dt b d = ( V ( r( t )) dt = V ( r( b )) V ( r( a )) a dt There are two other important aspects to this topic, namely: (a) How do you now when a given vector field is a gradient (conservative) vector field? (b) If you now that a vector field is conservative (gradient), how do you find a potential function? We can partially answer the first question by invoing lairaut s Theorem (equality of mixed partials). Specifically, if F V V = V=, = Pxy (, ), Qxy (, ) where Vxy (, ) is sufficiently differentiable, then: P V V V V Q = = = = = yx xy P Q So it would have to be the case that =. If not, then the vector field could not be a gradient vector field. V V V In the case where F = V=,, = PQR,,, there are three such relations that would have to hold: z = = = = = = yx xy P V V V V R P R = = = = = = z z zx xz z z Q V V V V R Q R = = = = = = z z zy yz z z P V V V V Q P Q These are nown as the exactness conditions or the test for exactness. If any of these three conditions fails to be the case, then the vector field could not be a gradient vector field. These calculations provide necessary conditions for a vector field to be conservative, but they do not provide sufficient conditions. For that we ll need either Green s Theorem (in R ) or Stoes Theorem (in R 3 ). However, if we can find an everywhere differentiable potential function, then this will be sufficient. This brings us to the second question: How do we find a potential function after we have established that the conditions above have been met? This is really just a matter of finding antiderivatives and doing a little detective wor, though often it comes down simply to guess and chec. 11 Revised November 6, 017

12 3 3 For example, suppose we are given the vector field F= xy,3x y + 8y = xy i+ (3x y + 8 y) j. Before doing anything else, we chec to see if the required condition is met: P 3 Q P Q = ( xy ) = 6xy and = (3x y + 8 y) = 6xy, so = and we re good to go! V 3 V We re looing for a function Vxy (, ) such that = P( x, y) = xy and = Qxy (, ) = 3 xy + 8 y. 3 The first condition implies that Vxy (, ) = xy+ gy ( ) where g( y ) is an arbitrary function of y alone. V Differentiation then gives that = 3 xy + g ( y) = Qxy (, ) = 3xy + 8y, so we must have g ( y) = 8y. Therefore g( y ) must be of the form g( y) = 4y + where is an arbitrary constant. The potential function 3 must then necessarily be of the form Vxy (, ) = xy+ 4y +. We actually only need one potential function, so 3 we just tae the arbitrary constant to be = 0 and we use Vxy (, ) = xy+ 4y. It s important to note that you could have looed at both of the components of F and guessed a potential function, but if you do this you must tae the partial derivatives to chec that it gives the correct gradient. 3 Example 15: Find the wor done by the vector field F= xy i+ (3x y + 8 y) j along some wild and crazy path from the starting point (1,1) to the endpoint (,3). [Had we given a specific path, the method would still be the same.] Solution: First, you have to chec whether or not the vector field is conservative (gradient). If it isn t, then you have no choice but to parameterize the given path. However, in this case, we ve already shown that this vector 3 field is the gradient of the potential function Vxy (, ) = xy+ 4y. Therefore, by the Fundamental Theorem of Line Integrals, we have: d F dr = V dr = V(end) V(start) = V(,3) V(1,1) = [ ] [1 + 4] = = 139 Examples involving vector fields and paths in R 3 wor pretty much the same way except that you have to chec three conditions in order to verify whether a given vector field could possibly be a gradient vector field, and then you have to use a bit more deduction or careful guessing and checing to find the potential function. Example 16: alculate the wor done by the vector field F= yz i+ ( xz + 6 y) j+ (xyz + 5) along any path from the point (0, 0, 0) to the point (3,1, 1). Solution: We first chec that the exactness conditions hold: P Q = z = x ; P R = yz = z x, and Q R = xz = z V V V So F = V V =,, = yz, xz + 6 y,xyz + 5. = yz V ( x, y, z) = xyz + f ( y, z) z V f f So = xz + = xz + 6y = 6 y f ( y, z) = 3 y + g( z) V = xyz + 3 y + g( z) V So = xyz + g ( z) = xyz + 5 g ( z) = 5 g( z) = 5 z V ( x, y, z) = xyz + 3y + 5z will do. z d By the Fundamental Theorem of Line Integrals, F dr = V dr = V(3,1, 1) V(0,0,0) = 1 0 = 1. In the next lecture we ll develop a list of statements equivalent to a vector field being conservative. We ll also state and prove Green s Theorem to help connect these facts. Notes by Robert Winters and Renée hipman 1 Revised November 6, 017