Vector Calculus PDF

Chapter 1 Vector Calculus 1.1 Vector function of a scalar variable Vector function of a scalar variable t is a function F~ = F~ (t) which uniquely associates a vector F~ for each scalar t. If F~ (t) = F1 (t)î + F2 (t)ĵ + F3 (t)k̂ then dF~ dF1 dF2...

Chapter 1 Vector Calculus 1.1 Vector function of a scalar variable Vector function of a scalar variable t is a function F~ = F~ (t) which uniquely associates a vector F~ for each scalar t. If F~ (t) = F1 (t)î + F2 (t)ĵ + F3 (t)k̂ then dF~ dF1 dF2 dF3 = î + ĵ + k̂ dt dt dt dt 1.2 Application: velocity and acceleration Let the position vector of a point P (x, y, z) in space be ~r = xî + y ĵ + z k̂. If x, y and z are all functions of a single parameter t then ~r(t) = x(t)î + y(t)ĵ + z(t)k̂ (1.2.1) As t varies, the point P traces a curve in space and the vector equation of the curve is given by the equation (1.2.1). d~r dx dy dz = î + ĵ + k̂ dt dt dt dt is a vector along the tangent to the curve at P. d~r If t is the time variable then ~v = gives the velocity of the particle at any time t dt 2 d~v d ~r and ~a = = 2 gives the acceleration of the particle at any time t. dt dt Example 1.2.1. A particle moves along the curve x = t2 , y = −t3 and z = t4 where t is the time variable. Find its velocity and acceleration at t = 1. 1 2 Solution. The vector equation of the particle moving along the curve x = t2 , y = −t3 and z = t4 is given by ~r(t) = t2 î − t3 ĵ + t4 k̂ (1.2.2) Velocity, ~v , of the particle is given by d~r ~v = dt dt2 dt3 dt4 = î − ĵ + k̂ dt dt dt = 2tî − 3t2 ĵ + 4t3 k̂ Acceleration, ~a, of the particle is given by d~v ~a = dt d(2t) d(3t2 ) d(4t3 ) = î − ĵ + k̂ dt dt dt = 2î − 6tĵ + 12t2 k̂ At t = 1 we have ~v = 2î − 3ĵ + 4k̂ and ~a = 2î − 6ĵ + 12k̂ Example 1.2.2. A particle moves along the curve x = e−t , y = 2 cos 3t and z = 2 sin 3t where t is the time variable. Find its velocity and acceleration at any time. Further find the magnitude of the velocity and acceleration at t = 0. Solution. The vector equation of the particle moving along the curve x = e−t , y = 2 cos 3t and z = 2 sin 3t is given by ~r(t) = e−t î + 2 cos 3tĵ + 2 sin 3tk̂ (1.2.3) Velocity, ~v , of the particle is given by d~r ~v = dt d d d = (e−t )î + (2 cos 3t)ĵ + (2 sin 3t)k̂ dt dt dt −t = −e î − 6 sin 3tĵ + 6 cos 3tk̂ Acceleration, ~a, of the particle is given by d~v ~a = dt d d d = (−e−t )î + (−6 sin 3t)ĵ + (6 cos 3t)k̂ dt dt dt −t = e î − 18 cos 3tĵ − 18 sin 3tk̂  3 At t = 0 we have ~v = −î + 6k̂ and ~a = î − 18ĵ p | ~v | = (−1)2 + 62 √ = 37 p | ~a | = (1)2 + (−18)2 √ = 325 1.3 Gradient, divergence, curl and Laplacian Definition 1.3.1. The gradient of a scalar function φ(x, y, z), denoted by grad φ, is defined as ∂φ ∂φ ∂φ grad φ = î + ĵ + k̂ ∂x ∂y ∂z Note. The gradient of a scalar function is a vector quantity. Definition 1.3.2. The divergence of a vector function F~ = F1 î + F2 ĵ + F3 k̂, denoted by div F~ , is defined as ∂F1 ∂F2 ∂F3 div F~ = + + ∂x ∂y ∂z Note. The divergence of a vector function is a scalar quantity. Definition 1.3.3. The curl of a vector function F~ = F1 î + F2 ĵ + F3 k̂, denoted by curl F~ , is defined as       ∂F 3 ∂F 2 ∂F 1 ∂F 3 ∂F 2 ∂F 1 curl F~ = − î + − ĵ + − k̂ ∂y ∂z ∂z ∂x ∂x ∂y Note. The curl of a vector function is a vector quantity. Definition 1.3.4. The Laplacian of a scalar function φ(x, y, z) is defined as ∂2φ ∂2φ ∂2φ + + 2 ∂x2 ∂y 2 ∂z For our convenience, we shall now introduce the following two operators: vector differential operator: The vector differential operator, denoted by ∇ (read as del or nabla), is defined as ∂ ∂ ∂ ∇= î + ĵ + k̂ ∂x ∂y ∂z Laplacian operator: The Laplacian operator, denoted by ∇2 , is defined as ∂2 ∂2 ∂2 ∇2 = + + ∂x2 ∂y 2 ∂z 2 4 Note. ∇ and ∇2 are operators and they are neither scalar nor vector quantities. If φ(x, y, z) is a scalar function and F~ = F1 î + F2 ĵ + F3 k̂ is a vector function then using these operators we write   ∂φ ∂φ ∂φ ∂ ∂ ∂ î + ĵ + k̂ = î + ĵ + k̂ φ = ∇φ ∂x ∂y ∂z ∂x ∂y ∂z    ∂F1 ∂F2 ∂F3 ∂ ∂ ∂  + + = î + ĵ + k̂ · F1 î + F2 ĵ + F3 k̂ = ∇ · F~ ∂x ∂y ∂z ∂x ∂y ∂z       i j k ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 ∂ ∂ ∂ − î + − ĵ + − k̂ = ∂x ∂y ∂z ∂y ∂z ∂z ∂x ∂x ∂y F1 F2 F3   ∂ ∂ ∂ = î + ĵ + k̂ × (F1 î + F2 ĵ + F3 k̂) ∂x ∂y ∂z = ∇ × F~ ∂2φ ∂2φ ∂2φ ∂2 ∂2 ∂2   + + 2 = + + φ = ∇2 φ ∂x2 ∂y 2 ∂z ∂x2 ∂y 2 ∂z 2 With the help of the operators ∇ and ∇2 we can thus conveniently express and easily remember grad φ, div F~ , curl F~ and Laplacian of φ as follows: grad φ = ∇φ div F~ = ∇ · F~ curl F~ = ∇ × F~ Laplacian of φ = ∇2 φ Example 1.3.5. If φ = xyz then find grad φ at the point (1,-1,2)  5 Solution. Given φ = xyz. We know that grad φ = ∇φ   ∂ ∂ ∂ = î + ĵ + k̂ φ ∂x ∂y ∂z ∂φ ∂φ ∂φ = î + ĵ + k̂ ∂x ∂y ∂z ∂ ∂ ∂ = (xyz)î + (xyz)ĵ + (xyz)k̂ (because φ = xyz) ∂x ∂y ∂z = yz î + xz ĵ + xy k̂ At the point (1,-1,2) we have grad φ = (−1)2î + (1)2ĵ + 1(−1)k̂ = −2î + 2ĵ − k̂ Example 1.3.6. If φ = x2 + y 2 + z 2 + 2xyz then find |∇φ| at the point (2,-1,1) Solution. Given φ = x2 + y 2 + z 2 + 2xyz. We know that grad φ = ∇φ   ∂ ∂ ∂ = î + ĵ + k̂ φ ∂x ∂y ∂z ∂φ ∂φ ∂φ = î + ĵ + k̂ ∂x ∂y ∂z ∂ 2 ∂ 2 ∂ = (x + y 2 + z 2 + 2xyz)î + (x + y 2 + z 2 + 2xyz)ĵ + (x2 + y 2 + z 2 + 2xyz)k̂ ∂x ∂y ∂z = (2x + 2yz)î + (2y + 2xz)ĵ + (2z + 2xy)k̂ At the point (2,-1,1) we have grad φ = [2(2) + 2(−1)1]î + [2(−1) + 2(2)1]ĵ + [2(1) + 2(2)(−1)]k̂ = 2î + 2ĵ − 2k̂ and p |∇φ| = 22 + 22 + (−2)2 √ =2 3 Example 1.3.7. If φ = 2x3 y 2 z 4 and F~ = ∇φ then find div F~ and curlF~ at the point (1,-1,1) 6 Solution. Given φ = 2x3 y 2 z 4 and F~ = ∇φ. Therefore,   ∂ ∂ ∂ F~ = î + ĵ + k̂ φ ∂x ∂y ∂z ∂φ ∂φ ∂φ = î + ĵ + k̂ ∂x ∂y ∂z (1.3.1) ∂ ∂ ∂ = (2x3 y 2 z 4 )î + (2x3 y 2 z 4 )ĵ + (2x3 y 2 z 4 )k̂ ∂x ∂y ∂z 2 2 4 3 4 3 2 3 = 6x y z î + 4x yz ĵ + 8x y z k̂ We know that div F~ = ∇ · F~   ∂ ∂ ∂ = î + ĵ + k̂ · (6x2 y 2 z 4 î + 4x3 yz 4 ĵ + 8x3 y 2 z 3 k̂) ∂x ∂y ∂z (1.3.2) ∂ ∂ ∂ = (6x2 y 2 z 4 ) + (4x3 yz 4 ) + (8x3 y 2 z 3 ) ∂x ∂y ∂z 2 4 3 4 3 2 2 = 12xy z + 4x z + 24x y z and curlF~ = ∇ × F~ i j k ∂ ∂ ∂ = ∂x ∂y ∂z F1 F2 F3 i j k ∂ ∂ ∂ = ∂x ∂y ∂z 2 2 4 3 4 3 2 3 6x y z 4x yz 8x y z     ∂ 3 2 3 ∂ 3 4 ∂ 2 2 4 ∂ 3 2 3 = (8x y z ) − (4x yz ) î + (6x y z ) − (8x y z ) ĵ ∂y ∂z ∂z ∂x   ∂ 3 4 ∂ 2 2 4 + (4x yz ) − (6x y z ) k̂ ∂x ∂y = 16x3 yz 3 − 16x3 yz 3 î + 24x2 y 2 z 3 − 24x2 y 2 z 3 ĵ + 12x2 yz 4 − 12x2 yz 4 k̂


= ~0 (1.3.3) At the point (1,-1,1) we have div F~ = 12 + 16 + 24 = 40 and curlF~ = ~0  7 Definition 1.3.8. The directional derivative of φ(x, y, z) at the point P0 (x0 , y0 , z0 ) in the direction of ~a is the number (∇φ)P0 · â Example 1.3.9. Find the directional derivative of φ = xy 2 + yz 3 at the point (2, −1, 1) in the direction î + 2ĵ + 2k̂. Solution. The directional derivative of φ = xy 2 + yz 3 at the point P0 ≡ (2, −1, 1) in the direction of ~a = î + 2ĵ + 2k̂ is given by (∇φ)P0 · â. We know that   ∂ ∂ ∂ ∇φ = î + ĵ + k̂ φ ∂x ∂y ∂z ∂φ ∂φ ∂φ = î + ĵ + k̂ ∂x ∂y ∂z ∂ ∂ ∂ = (xy 2 + yz 3 )î + (xy 2 + yz 3 )ĵ + (xy 2 + yz 3 )k̂ ∂x ∂y ∂z 2 3 2 = y î + (2xy + z )ĵ + 3yz k̂ Therefore, (∇φ)P0 = î − 3ĵ − 3k̂ ~a î + 2ĵ + 2k̂ î + 2ĵ + 2k̂ Note that â = =√ = |~a| 12 + 22 + 22 3 î + 2ĵ + 2k̂ 11 Thus, the required directional derivative is (∇φ)P0 ·â = (î−3ĵ −3k̂)·( )=−. 3 3 Note. The directional derivative of φ at any point is maximum along ∇φ at that point. 1.4 Vector identities Theorem 1.4.1. If φ(x, y, z) is a scalar function, F~ = F1 î + F2 ĵ + F3 k̂ and G ~ = G1 î + G2 ĵ + G3 k̂ are vector functions then the following hold. 1. ∇ · (∇φ) = ∇2 φ 2. ∇ × (∇φ) = ~0 3. ∇ · (∇ × F~ ) = 0 4. ∇ × (∇ × F~ ) = ∇(∇ · F~ ) − ∇2 F~ 5. ∇ · (φF~ ) = ∇φ · F~ + φ(∇ · F~ ) 8 6. ∇ × (φF~ ) = ∇φ × F~ + φ(∇ × F~ ) 7. ∇ · (F~ × G) ~ =G ~ · (∇ × F~ ) − F~ · (∇ × G) ~ Proof. 1). Consider the LHS,      ∂ ∂ ∂ ∂ ∂ ∂ ∇ · (∇φ) = î + ĵ + k̂ · î + ĵ + k̂ φ ∂x ∂y ∂z ∂x ∂y ∂z     ∂ ∂ ∂ ∂φ ∂φ ∂φ = î + ĵ + k̂ · î + ĵ + k̂ ∂x ∂y ∂z ∂x ∂y ∂z ∂2φ ∂2φ ∂2φ = + + ∂x2 ∂y 2 ∂z  2 2 ∂2  ∂ ∂ = + + φ ∂x2 ∂y 2 ∂z = ∇2 φ 5). Consider the LHS,   h ∂ ∂ ∂ i ∇ · (φF~ ) = î + ĵ + k̂ · φ(F1 î + F2 ĵ + F3 k̂) ∂x ∂y ∂z   h ∂ ∂ ∂ i = î + ĵ + k̂ · φF1 î + φF2 ĵ + φF3 k̂ (1.4.1) ∂x ∂y ∂z ∂ ∂ ∂ = (φF1 ) + (φF2 ) + (φF3 ) ∂x ∂y ∂z Now consider first term of RHS,   ~ ∂ ∂ ∂ ∇φ · F = î + ĵ + k̂ φ · (F1 î + F2 ĵ + F3 k̂) ∂x ∂y ∂z   ∂φ ∂φ ∂φ = î + ĵ + k̂ · (F1 î + F2 ĵ + F3 k̂) (1.4.2) ∂x ∂y ∂z ∂φ ∂φ ∂φ = F1 + F2 + F3 ∂x ∂y ∂z Consider second term of RHS,   ∂ ∂ ∂ φ(∇ · F~ ) = φ î + ĵ + k̂ · (F1 î + F2 ĵ + F3 k̂) ∂x ∂y ∂z   ∂F1 ∂F2 ∂F3 =φ + + (1.4.3) ∂x ∂y ∂z ∂F1 ∂F2 ∂F3 =φ +φ +φ ∂x ∂y ∂z  9 From equations (1.4.2) and (1.4.3), we get       ~ ~ ∂φ ∂F1 ∂φ ∂F2 ∂φ ∂F3 ∇φ · F + φ(∇ · F ) = F1 + φ + F2 + φ + F3 + φ ∂x ∂x ∂y ∂y ∂z ∂z (1.4.4) ∂ ∂ ∂ = (φF1 ) + (φF2 ) + (φF3 ) ∂x ∂y ∂z From equations (1.4.1) and (1.4.4), we get ∇ · (φF~ ) = ∇φ · F~ + φ(∇ · F~ ). 1.5 Solenoidal and irrotational vector fields Definition 1.5.1. A vector field F~ is said to be solenoidal if div F~ = 0 at all points. Example 1.5.2. 1. Every constant vector field is solenoidal. 2. F~ = yz î + xz ĵ + xy k̂ is a solenoidal vector field. 3. F~ = 2xyz î + x2 z 3 ĵ − yz 2 k̂ is a solenoidal vector field. Definition 1.5.3. A vector field F~ is said to be irrotational if curl F~ = ~0 at all points. Example 1.5.4. 1. Every constant vector field is an irrotational vector field. 2. F~ = x2 î + y ĵ + z 3 k̂ is an irrotational vector field. 3. F~ = 6x2 y 2 z 4 î + 4x3 yz 4 ĵ + 8x3 y 2 z 3 k̂ is an irrotational vector field. ~ and B Example 1.5.5. If A ~ are irrotational then A ~×B ~ is solenoidal. ~ and B Solution. Let A ~ be irrotational vector fields. This implies ~ =∇×B ∇×A ~ = ~0 (1.5.1) We know that ~ × B) ∇ · (A ~ =B~ · (∇ × A)~ −A ~ · (∇ × B) ~ ~ · ~0 − A =B ~ · ~0 (by equation (1.5.1)) =0 Example 1.5.6. Determine the constants a,b and c if F~ = (axy + z 3 )î + (bx2 + z)ĵ + (bxz 2 + cy)k̂ is an irrotational vector field. 10 Solution. Let F~ = (axy + z 3 )î + (bx2 + z)ĵ + (bxz 2 + cy)k̂ be an irrotational vector field. This implies ~0 = ∇ × F~ i j k ∂ ∂ ∂ = ∂x ∂y ∂z axy + z 3 bx + z bxz + cy 2 2     ∂ 2 ∂ 2 ∂ 3 ∂ 2 = (bxz + cy) − (bx + z) î + (axy + z ) − (bxz + cy) ĵ ∂y ∂z ∂z ∂x   ∂ 2 ∂ 3 + (bx + z) − (axy + z ) k̂ ∂x ∂y = (c − 1)î + (3z 2 − bz 2 )ĵ + (2bx − ax)k̂ This implies c − 1 = 0, 3z 2 − bz 2 = 0 and 2bx − ax = 0. Thus c = 1, b = 3 and a = 6. Example 1.5.7. Find the value of a if the vector field F~ = (ax2 y + yz)î + (xy 2 − xz 2 )ĵ + (2xy − 2x2 y 2 )k̂ has zero divergence. find the curl of the vector field when it has zero divergence. Solution. Let F~ = (ax2 y + yz)î + (xy 2 − xz 2 )ĵ + (2xy − 2x2 y 2 )k̂ has zero divergence. Then 0 = div F~ = ∇ · F~   ∂ ∂ ∂ = î + ĵ + k̂ · [(ax2 y + yz)î + (xy 2 − xz 2 )ĵ + (2xy − 2x2 y 2 )k̂] ∂x ∂y ∂z ∂ ∂ ∂ = (ax2 y + yz) + (xy 2 − xz 2 ) + (2xy − 2x2 y 2 ) ∂x ∂y ∂z = 2axy + 2xy This implies 2xy(a + 1) = 0 which intern implies a = −1. Now by taking a = −1 we get curl F~ = ∇ × F~ i j k ∂ ∂ ∂ = ∂x ∂y ∂z −x2 y + yz xy 2 − xz 2 2xy − 2x2 y 2     ∂ 2 2 ∂ 2 2 ∂ 2 ∂ 2 2 = (2xy − 2x y ) − (xy − xz ) î + (−x y + yz) − (2xy − 2x y ) ĵ ∂y ∂z ∂z ∂x   ∂ 2 2 ∂ 2 + (xy − xz ) − (−x y + yz) k̂ ∂x ∂y = [(2x − 4x2 y) − (−2xz)]î + [y − (2y − 4xy 2 )]ĵ + [(y 2 − z 2 ) − (z − x2 )]k̂ = (2x − 4x2 y + 2xz)î + (−y − 4xy 2 )ĵ + (y 2 − z 2 − z + x2 )k̂ = 2x(1 − 2xy + z)î − y(1 + 4xy)ĵ + (x2 + y 2 − z 2 − z)k̂  11 Figure 1.1: Cylindrical coordinates 1.6 Curvilinear coordinates 1.6.1 Cylindrical coordinates Let P (x, y, z) be any point in space. The projection of P onto the xy-plane is Q(x, y). If (r, θ) is a polar coordinates of the point Q then (r, θ, z) is a cylindrical coordinates of the point P. The following equations relate cartesian coordinates and cylindrical coordinates. x = r cos θ y = r sin θ z=z p y r= x2 + y 2 tan θ = z=z x Definition 1.6.1. A curvilinear coordinate system is said to be orthogonal if the coordi- nate surfaces intersect at right angles. Theorem 1.6.2. Cylindrical coordinate system is an orthogonal curvilinear coordinate system. Proof. Let P (x, y, z) be any point in space and let (r, θ, z) be its cylindrical coordinates. ~ := − Let R → OP then we have ~ = xî + y ĵ + z k̂ R = (r cos θ)î + (r sin θ)ĵ + z k̂ If T~r , T~θ and T~z are the vectors at the point P in the direction of tangent to r, θ and 12 z-curves respectively then ~ ∂R T~r = ∂r ∂ h i = (r cos θ)î + (r sin θ)ĵ + z k̂ ∂r      ∂ ∂ ∂ = (r cos θ) î + (r sin θ) ĵ + (z) k̂ ∂r ∂r ∂r = cos θ î + sin θ ĵ ~ ∂R T~θ = ∂θ ∂ h i = (r cos θ)î + (r sin θ)ĵ + z k̂ ∂θ      ∂ ∂ ∂ = (r cos θ) î + (r sin θ) ĵ + (z) k̂ ∂θ ∂θ ∂θ = (−r sin θ)î + (r cos θ)ĵ and ~ ∂R T~z = ∂z ∂ h i = (r cos θ)î + (r sin θ)ĵ + z k̂ ∂z      ∂ ∂ ∂ = (r cos θ) î + (r sin θ) ĵ + (z) k̂ ∂z ∂z ∂z = k̂ Note that h i h i T~r · T~θ = cos θ î + sin θ ĵ · (−r sin θ)î + (r cos θ)ĵ = cos θ(−r sin θ) + sin θ(r cos θ) = −r cos θ sin θ + r cos θ sin θ =0 h i h i T~r · T~z = cos θ î + sin θ ĵ · z k̂ h i h i = cos θ î + sin θ ĵ + 0k̂ · 0î + 0ĵ + z k̂ =0  13 and h i h i ~ ~ Tθ · Tz = (−r sin θ)î + (r cos θ)ĵ · z k̂ h i h i = (−r sin θ)î + (r cos θ)ĵ + 0k̂ · 0î + 0ĵ + z k̂ =0 Thus the vectors T~r , T~θ and T~z are mutually perpendicular which implies the coordinate surfaces intersect at right angles and hence the cylindrical coordinate system is orthogonal. 1.7 Vector integral calculus 1.7.1 Line integral Definition 1.7.1. If F~ = F1 î+F2 ĵ +F3 k̂ is a vector function and C is a piecewise smooth curve (in the domain of F~ ) given by~r(t) = x(t)î + y(t)ĵ + z(t)k̂ (a ≤ t ≤ b) then the line Rb d integral of F~ over C is F~ · d~r = F~ (~r(t)) · (~r(t)) dt. R C a dt Example 1.7.2. If F~ = (3x2 + 6y)î − 14yz ĵ + 20xz 2 k̂ and C is the curve given by ~r(t) = tî + t2 ĵ + t3 k̂ (0 ≤ t ≤ 1) then find F~ · d~r R C Solution. Given F~ = (3x2 + 6y)î − 14yz ĵ + 20xz 2 k̂ (1.7.1) and ~r(t) = tî + t2 ĵ + t3 k̂ (0 ≤ t ≤ 1) (1.7.2) We know that Z Zb   ~ ~ d F · d~r = F (~r(t)) · (~r(t)) dt (1.7.3) dt C a Curve Parametric equations 2 2 2 x +y =a x = a cos t, y = a sin t (0 ≤ t ≤ 2π) x2 y 2 + 2 =1 x = a cos t, y = b sin t (0 ≤ t ≤ 2π) a2 b y = f (x) x = t, y = f (t) x = f (y) x = f (t), y = t Straight line segment joining (x0 , y0 , z0 ) and (x1 , y1 , z1 ) x = (1 − t)x0 + tx1 , y = (1 − t)y0 + ty1 and z = (1 − t)z0 + tz1 (0 ≤ t ≤ 1) Table 1.1: Curves and their parametric equations 14 Now consider F~ (~r(t)) = (3t2 + 6t2 )î − 14t2 (t3 )ĵ + 20t(t3 )2 k̂ (by putting x = t, y = t2 and z = t3 in equation (1.7.1)) (1.7.4) 2 5 7 = 9t î − 14t ĵ + 20t k̂ Differentiating equation (1.7.2) w.r.t t, we get d (~r(t)) = î + 2tĵ + 3t2 k̂ (1.7.5) dt From equations (1.7.4) and (1.7.5), we get d F~ (~r(t)) · (~r(t)) = (9t2 î − 14t5 ĵ + 20t7 k̂) · (î + 2tĵ + 3t2 k̂) dt = 9t2 − 14t5 (2t) + 20t7 (3t2 ) (1.7.6) = 9t2 − 28t6 + 60t9 From equations (1.7.3) and (1.7.6), we get Z Zb   d F~ · d~r = ~ F (~r(t)) · (~r(t)) dt dt C a Z1  2 9t − 28t6 + 60t9 dt  = 0 1 t3 t7 t10 = 9 − 28 + 60 3 7 10 0 9 28 60 = − + 3 7 10 =3−4+6 =5 Note. If F~ is a vector function and C1 , C2 ,... , Cn are smooth curves(in the domain of ~ F point of Ci Ris the initial point of Ci+1 (1 ≤ i ≤ n − 1) then R ) such that R the terminal ~ ~ ~ F · d~r = F · d~r + F · d~r + · · · + F~ · d~r R C C1 C2 Cn Example 1.7.3. If F~ = (3x2 + 6y)î − 14yz ĵ + 20xz 2 k̂ and C is the curve made of straight line segments from origin to (1, 0, 0); from (1, 0, 0) to (1, 1, 0) and (1, 1, 0) to (1, 1, 1), then find F~ · d~r R C  15 Figure 1.2: A smooth curve Figure 1.3: A piecewise smooth curve Figure 1.4: A simple closed curve Figure 1.5: A closed(but not simple) curve Figure 1.6: Curves in Example 1.7.3 Figure 1.7: Circle in Example 1.7.4 16 Solution. Given F~ = (3x2 + 6y)î − 14yz ĵ + 20xz 2 k̂ (1.7.7) and C = C1 + C2 + C3 where C1 is the straight line segment from origin to (1, 0, 0), C2 is the straight line segment from (1, 0, 0) to (1, 1, 0) and C3 is the straight line segment from (1, 1, 0) to (1, 1, 1). Note that Z Z Z Z F · d~r = F · d~r + F · d~r + F~ · d~r ~ ~ ~ (1.7.8) C C1 C2 C3 The parametric equations of the curve C1 are x = t, y = 0 and z = 0 (0 ≤ t ≤ 1) so that ~r(t) = tî (0 ≤ t ≤ 1) (1.7.9) We know that Z Z1   ~ ~ d F · d~r = F (~r(t)) · (~r(t)) dt (1.7.10) dt C1 0 Clearly F~ (~r(t)) = 3t2 î (by putting x = t, y = 0 and z = 0 in equation (1.7.7)) (1.7.11) Differentiating equation (1.7.9) w.r.t t, we get d (~r(t)) = î (1.7.12) dt From equations (1.7.11) and (1.7.12), we get d F~ (~r(t)) · (~r(t)) = 3t2 (1.7.13) dt From equations (1.7.10) and (1.7.13), we get Z Z1   d F~ · d~r = F~ (~r(t)) · (~r(t)) dt dt C1 0 Z1 = 3t2 dt (1.7.14) 0 1 t3 =3 3 0 =1  17 The parametric equations of the curve C2 are x = 1, y = t and z = 0 (0 ≤ t ≤ 1) so that ~r(t) = î + tĵ (0 ≤ t ≤ 1) (1.7.15) We know that Z Z1   ~ ~ d F · d~r = F (~r(t)) · (~r(t)) dt (1.7.16) dt C2 0 Clearly F~ (~r(t)) = [3 + 6t]î (by putting x = 1, y = t and z = 0 in equation (1.7.7)) (1.7.17) Differentiating equation (1.7.15) w.r.t t, we get d (~r(t)) = ĵ (1.7.18) dt From equations (1.7.17) and (1.7.18), we get d F~ (~r(t)) · (~r(t)) = 0 (1.7.19) dt From equations (1.7.16) and (1.7.19), we get Z Z1   ~ ~ d F · d~r = F (~r(t)) · (~r(t)) dt dt C2 0 Z1 (1.7.20) = {0}dt 0 =0 The parametric equations of the curve C3 are x = 1, y = 1 and z = t (0 ≤ t ≤ 1) so that ~r(t) = î + ĵ + tk̂ (0 ≤ t ≤ 1) (1.7.21) We know that Z Z1   ~ ~ d F · d~r = F (~r(t)) · (~r(t)) dt (1.7.22) dt C3 0 Clearly F~ (~r(t)) = 9î − 14tĵ + 20t2 k̂(by putting x = 1, y = 1 and z = t in equation (1.7.7)) (1.7.23) 18 Differentiating equation (1.7.21) w.r.t t, we get d (~r(t)) = k̂ (1.7.24) dt From equations (1.7.23) and (1.7.24), we get d F~ (~r(t)) · (~r(t)) = 20t2 (1.7.25) dt From equations (1.7.22) and (1.7.25), we get Z Z1   d F~ · d~r = F~ (~r(t)) · (~r(t)) dt dt C2 0 Z1 = 20t2 dt 0 " 1 # (1.7.26) t3 = 20 3 0   1 = 20 3 20 = 3 From equations (1.7.8), (1.7.14), (1.7.20) and (1.7.26) we get Z 20 F~ · d~r = 1 + 0 + 3 C 23 = 3 Note. If F~ represents the force acting on a particle moving along R a piecewise smooth curve C with the initial point A and the terminal point B then F~ · d~r gives the work C done in moving a particle from the point A to the point B along the curve C in the force field F~. Example 1.7.4. Find the work done in moving a particle once around the circle C in the xy-plane if the circle has center at the origin and radius 3, given that the force field is F~ = (2x − y + z)î + (x + y − z 2 )ĵ + (3x − 2y)k̂. Solution. Given F~ = (2x − y + z)î + (x + y − z 2 )ĵ + (3x − 2y)k̂ (1.7.27)  19 and C is the circle in the xy-plane with center at the origin and radius 3. Note that the parametric equations of C are x = 3 cos t, y = 3 sin t and z = 0 (0 ≤ t ≤ 2π) so that ~r(t) = 3 cos t î + 3 sin t ĵ (0 ≤ t ≤ 2π) (1.7.28) TheR work done in moving a particle once around the circle C in the force field F~ is given by F~ · d~r. C We know that Z Z2π   d F~ · d~r = ~ F (~r(t)) · (~r(t)) dt (1.7.29) dt C 0 Clearly F~ (~r(t)) = [2(3 cos t) − 3 sin t]î + (3 cos t + 3 sin t)ĵ + [3(3 cos t) − 2(3 sin t)]k̂ (by putting x = 3 cos t, y = 3 sin t and z = 0 in equation (1.7.27)) (1.7.30) = (6 cos t − 3 sin t)î + (3 cos t + 3 sin t)ĵ + (9 cos t − 6 sin t)k̂ Differentiating equation (1.7.28) w.r.t t, we get d (~r(t)) = −3 sin tî + 3 cos tĵ (1.7.31) dt From equations (1.7.30) and (1.7.31), we get d F~ (~r(t)) · (~r(t)) = (6 cos t − 3 sin t)(−3 sin t) + (3 cos t + 3 sin t)(3 cos t) dt = −18 cos t sin t + 9 sin2 t + 9 cos2 t + 9 cos t sin t (1.7.32) = 9 − 9 cos t sin t 20 From equations (1.7.30) and (1.7.32), we get Z Z2π   ~ ~ d F · d~r = F (~r(t)) · (~r(t)) dt dt C 0 Z2π = (9 − 9 cos t sin t)dt 0 Z2π Z2π = 9 dt − 9 cos t sin t dt 0 0 Z2π Z2π 9 =9 dt − 2 cos t sin t dt 2 0 0 2π 9 2 = 9 t|2π 0 − sin t 2 0 9 = 9(2π) − (0) 2 = 18π Therefore the required work done is 18π units. 1.7.2 Path independence Definition 1.7.5. The line integral F~ · d~r is said to be path independent in a domain R C D if F~ · d~r = F~ · d~r for any two paths(or curves) C1 and C2 in D, with the same R R C1 C2 initial point A and terminal point B. Theorem 1.7.6. If F~ = ∇φ for some function φ in a domain D then the line integral ~ R F · d~r is path independent in D. C Proof. Let F~ = ∇φ for some function φ in a domain D, C be a path in D with initial point A and terminal point B and let ~r(t) = x(t)î + y(t)ĵ + z(t)k̂ (a ≤ t ≤ b) be the parametric representation of C.(This in particular implies A ≡ (x(a), y(a), z(a)) and B ≡ (x(b), y(b), z(b)).) Note that F~ = ∇φ   ∂ ∂ ∂ = î + ĵ + k̂ φ ∂x ∂y ∂z ∂φ ∂φ ∂φ = î + ĵ + k̂ ∂x ∂y ∂z  21 Figure 1.8: Two paths Figure 1.9: A simple closed curve and d dx dy dz (~r(t)) = î + ĵ + k̂ dt dt dt dt Now, by the definition, we have Z Zb   d F~ · d~r = F~ (~r(t)) · (~r(t)) dt dt C a Zb     ∂φ ∂φ ∂φ dx dy dz = î + ĵ + k̂ · î + ĵ + k̂ dt ∂x ∂y ∂z dt dt dt a Zb   ∂φ dx ∂φ dy ∂φ dz = + + dt ∂x dt ∂y dt ∂z dt a Zb dφ = dt (by chain rule) dt a = φ(x(t), y(t), z(t))|ba = φ(x(b), y(b), z(b)) − φ(x(a), y(a), z(a)) = φ(B) − φ(A) F~ · d~r depends only on the end points and not on the path joining R Thus the line integral C them and hence it is path independent. Theorem 1.7.7. The line integral F~ · d~r is path independent in a domain D if and only R C ~ H if F · d~r = 0 for any simple closed curve C in D. C 22 F~ · d~r be path independent in a domain D and let C : R Proof. Let the line integral C AF BEA be a simple closed curve in D as shown in the Figure 1.9. Clearly, C1 : AF B and C2 : AEB are two paths joining the points A and B. Therefore by hypothesis, we get Z Z F · d~r = F~ · d~r ~ C1 C2 i.e., Z Z F~ · d~r = F~ · d~r AF B AEB This implies, Z Z 0= F~ · d~r − F~ · d~r AF B AEB Z Z = F~ · d~r + F~ · d~r AF B BEA I = F~ · d~r AF BEA I = F~ · d~r C F~ · d~r = 0 for any simple closed curve C in D. Let C1 : AF B H Conversely, suppose that C and let C2 : AEB be two paths joining the points A and B as shown in the Figure 1.8. Clearly, C : AF BEA is a simple closed curve. Therefore, by hypothesis, we get I F~ · d~r = 0 C i.e., I F~ · d~r = 0 AF BEA  23 This implies I 0= F~ · d~r AF BEA Z Z = F~ · d~r + F~ · d~r AF B BEA Z Z = F~ · d~r − F~ · d~r AF B Z ZAEB = F~ · d~r − F~ · d~r C1 C2 F~ · d~r = F~ · d~r and hence the line integral F~ · d~r is independent of paths in R R R Thus C1 C2 C D. 1.7.3 Conservative vector field Definition 1.7.8. A vector field F~ is said to be conservative on a domain D if F~ ·d~r = 0 H C for any simple closed curve C in D. Definition 1.7.9. If F~ = ∇φ for some scalar function φ on D then φ is called a scalar potential function for F~. Theorem 1.7.10. If F~ = F1 î+F2 ĵ +F3 k̂ is a vector field on a domain D such that F1 , F2 and F3 have continuous first order partial derivatives then the following are equivalent: 1. F~ is conservative on D; 2. curl F~ = ~0 on D; 3. F~ = ∇φ for some differentiable scalar function φ on D. Example 1.7.11. Prove that the vector field F~ = (2xy+z 3 )î+x2 ĵ+3xz 2 k̂ is conservative. Find the scalar potential for F~ and hence find the work done in moving a particle from A(1, −2, 1) to B(3, 1, 4) in the force field. Solution. Given F~ = (2xy + z 3 )î + x2 ĵ + 3xz 2 k̂ (1.7.33) 24 We first prove F~ is conservative. By Theorem 1.7.10, it is enough to prove that curl F~ = ~0. Consider curl F~ = ∇ × F~ i j k ∂ ∂ ∂ = ∂x ∂y ∂z 2xy + z 3 x 3xz 2 2       ∂ 2 ∂ 2 ∂ 3 ∂ 2 ∂ ∂ 3 = (3xz ) − (x ) î + (2xy + z ) − (3xz ) ĵ + (x2) − (2xy + z ) k̂ ∂y ∂z ∂z ∂x ∂x ∂y = 0 î + 0 ĵ + 0 k̂ = ~0 Thus, by Theorem 1.7.10, F~ = ∇φ for some differentiable scalar function φ i.e., F~ = ∇φ   ∂ ∂ ∂ = î + ĵ + k̂ φ (1.7.34) ∂x ∂y ∂z ∂φ ∂φ ∂φ = î + ĵ + k̂ ∂x ∂y ∂z From equations (1.7.33) and (1.7.34) we get ∂φ = 2xy + z 3 (1.7.35) ∂x ∂φ = x2 (1.7.36) ∂y ∂φ = 3xz 2 (1.7.37) ∂z On integrating equation (1.7.35) w.r.t. x, we get x2 φ=2 y + z 3 x + f (y, z) 2 i.e., φ = x2 y + z 3 x + f (y, z) (1.7.38) Differentiating equation (1.7.38) partially w.r.t. y, we get ∂φ ∂f = x2 + (1.7.39) ∂y ∂y From equations (1.7.36) and (1.7.39), we get ∂f =0 (1.7.40) ∂y  25 On integrating equation (1.7.40) w.r.t.y we get f (y, z) = g(z) (1.7.41) From equations (1.7.38) and (1.7.41), we get φ = x2 y + z 3 x + g(z) (1.7.42) Differentiating equation (1.7.42) partially w.r.t. z, we get ∂φ = 3xz 2 + g 0 (z) (1.7.43) ∂z From equations (1.7.37) and (1.7.43), we get g 0 (z) = 0 (1.7.44) On integrating equation (1.7.44) w.r.t.z we get g(z) = C (1.7.45) From equations (1.7.42) and (1.7.45), we get φ = x2 y + xz 3 + C (1.7.46) Which is the required scalar potential function. Now the work done in moving a particle from A(1, −2, 1) to B(3, 1, 4) in the force field F~ is given by Z F~ · d~r = φ(B) − φ(A) C = φ((3, 1, 4)) − φ((1, −2, 1)) = [(32 )1 + 3(43 ) + C] − [12 (−2) + 1(13 ) + C] = 202