mvc - answers

ZPEM2309 Engineering Mathematics 2A MVC Tutorial Problem Set 2 (Sections 3.23.7) ANSWERS page 1 of 12

1. SLS 3.133.16 Answer: See SLS (below):CHAPTER 3 Vector Differentiai'ion

3.13. Let A = (2x2y - x4)i + (crY - y sinx)j + (x2 cos y)k. Find: (a) ~~, (b) :. Solution

(a) aA = ~(2x2y _ x4)i + ~(e'Y _ y sinx)j + ~(x2 cosy)k ax ax ax ax

= (4xy - 4x3)i + (yeTY - ycosx)j + 2xcosyk

(b) aA = ~(2x2y _ x4)i + ~V)' - ysinx)j +~(~ cosy)k ay ay ay ay

= 22i + (xe'Y - sinx)j - ~ sinyk

3.14. Let A be the vector in Problem 3.13. Find:

Solution

(a) a2 A a 4 3. a 'Y ). a ax2 = a/ xy - 4x )1 + ax (ye - ycosx j + ax (2x cos y)k

= (4y -12~)i + (y2e'Y + ysinx)j + 2cosyk

(b) a2A = ~(2~)i +~(xe'Y _ sinx)j - ~(~ siny)k ay2 ay ay ay

= 0 + x2e'Yj - ~ cosyk = x2e'Yj - x2 cosyk

a2A a2A 3.15. Let A be the vector in Problem 3.13. Find: (a) axay' (b) ayax'

Solution

-=- - =-(2x )I+-(xe -smx)J -- x-smy)k (a) a2 A a (aA) a 2. a X). a ( 2 . axay ax ay ax ax ax

= 4xi + (xyeT)' + e\)' - cos x)j - 2x sin yk

(b) - = - - = -(4xy - 4x3)1 + -(ye\) - ycosx)J + -(2xcosy)k a2 A a (aA) a . a. . a ayax ay ax ay ay ay

= 4xi + (xyCl' + e'Y - cosx)j - 2xsinyk

Note that a2 Ajayax = a2 Ajaxay, that is, the order of differentiation is immaterial. This is true in general if A has continuous partial derivatives of the second order at least.

3.16. Suppose cfJ(x, y, z) = xy2Z and A = xzi - xy2j + yz2k. Find + (cfJA) at the point (2, -1, 1). ax az

Solution

A = (,,-y2z)(xzi - xlj + yik) = ~lZ2i - ~lti + xy3Z3k ~(A) = ~(x2lii - ~lti + xy3Z3k) = 2x2lzi - x2lj + 3xy3Z2k az az

~(A) = ~(2X2lzi - ~lj + 3xlz2k) = 4Ay2zi - 2xy4j + 3lz2k axaz ax

a::az (A) = :x (4xy2Zi - 2xy4j + 3lz2k) = 4lzi - 2lj

If x = 2, y = -1, and z = 1, this becomes 4( -1)2(1)i - 2( _1)4j = 4i - 2j.


2. Calculate ux ,vy , u,

uy ,

vx , u and 2u for the following vector fields u = (u,v,w), and also match them to

one of the figures (e.g. (i) is shown in figure b; note that the vector lengths may have been scaled):

(i) u = (1,0,0)

(ii) u = (1,2,0)(iii) u = (x,0,0)

(iv) u = (2 y,2,0)(v) u = (3,2x,0)

(vi) u = (y,x,0)(vii) u = (x,y,0)

(viii) u = (y2,0,0)

(ix) u = (x,y,0)

x

y

4 2 0 2 4

4

2

0

2

4

x

y

4 2 0 2 4

4

2

0

2

4

x

y

4 2 0 2 4

4

2

0

2

4

x

y

4 2 0 2 4

4

2

0

2

4

x

y

4 2 0 2 4

4

2

0

2

4

x

y

4 2 0 2 4

4

2

0

2

4

x

y

4 2 0 2 4

4

2

0

2

4

x

y

4 2 0 2 4

4

2

0

2

4

x

y

4 2 0 2 4

4

2

0

2

4

a b c

d e f

g h i

Answer:

Q Fig field du/dx dv/dy div(u) du/dy dv/dx curl(u) lap(u)(i) b (1,0,0) 0 0 0 0 0 (0,0,0) (0,0,0)(ii) d (81,2,0) 0 0 0 0 0 (0,0,0) (0,0,0)(iii) a (x,0,0) 1 0 1 0 0 (0,0,0) (0,0,0)(iv) f (28y,2,0) 0 0 0 81 0 (0,0,1) (0,0,0)(v) i (3,2x,0) 0 0 0 0 2 (0,0,2) (0,0,0)(vi) e (y,8x,0) 0 0 0 1 81 (0,0,82) (0,0,0)(vii) h (x,y,0) 1 1 2 0 0 (0,0,0) (0,0,0)(viii) c (y^2,0,0) 0 0 0 2y 0 (0,0,82y) (2,0,0)(ix) g (x,8y,0) 1 81 0 0 0 (0,0,0) (0,0,0)

3. If A = (x2z2,2y2z2,xy2z), find

i. Aii. A (NB: worked solution in SLS is wrong)

iii. ( A)

iv. ( A)

v. ( A) (NB: working in SLS is wrong)

vi. 2A


Answer:

i. A = 2xz2 4yz2 + xy2 (see SLS problem 4.15)

ii. A = (2xyz+ 4y2z,2x2z y2z,0) (see SLS 4.23 for working but note the corrections below!)

iii. ( A) = 2yz 2yz = 0 (as it must be, by identity (x))

iv. ( A) = (2xz2 4yz2 + xy2) = (2z2 + y2,2xy 4z2,4xz 8yz) (using result from (i))

v. ( A) = (y2 2x2,2xy+ 4y2,2xz 8yz) (using result from (ii); see SLS 4.24 with corrections below!)

vi. 2A = ( A) ( A) = (2z

2 + y2,2xy 4z2,4xz 8yz) (y2 2x2,2xy+ 4y2,2xz 8yz) = (2z2 +y2 y2 + 2x2,2xy 4z2 2xy 4y2,4xz 8yz 2xz+ 8yz) = (2z2 + 2x2,4z2 4y2,4xz 2xz) (using resultsfrom (iv) and (v))

4. Show that 2 f =

2 fx2 +

2 fy2 +

2 fz2 in Cartesian coordinates. Answer:

2 f = ( f ) = ( fx ,

fy ,

fz ) =

2 fx2 +

2 fy2 +

2 fz2 .

5.Prove the following vector identities by writing them out in Cartesian components.

(i) (+ ) =+ (i.e. grad is a linear operator)

(ii) (A + B) = A + B (i.e. div is a linear operator)

(iii) (A + B) = A + B (i.e. curl is a linear operator)

(iv) (A) = () A + ( A) (the product rule for div)

(v) (A) = () A + ( A) (the product rule for curl)

(ix) () = 0 (the curl of a gradient field is zero, i.e. is irrotational)

(x) ( A) = 0 (the divergence of a curl field is zero, i.e. A is solenoidal)


Answer: Worked solutions are provided in Spiegel, Lipschutz & Spellman:

(i) See SLS problem 4.2a:

(ii) See SLS problem 4.18a:

(iii) See SLS problem 4.25a:

(iv) See SLS problem 4.18b:


(v) See SLS problem 4.25b:

(ix) See SLS problem 4.27a:

(x) See SLS problem 4.27b:

Do the following problems from Spiegel, Lipschutz & Spellman 2nd Edition (SLS) without looking at the workedsolutions, and then cross-check with the worked solutions where available. (Those with a * are more advanced.)

6. SLS 4.74 (where r =x2 + y2 + z2 is the distance from the origin) Answer:


7. SLS 4.76 (where r = (x,y,z) is the position vector; hint: check the table of vector identities and see section 3.3 inthe notes) Answer: Using identity (vi), v= ( r) = ( r) = 0 since is constant and r is curl-free.This is a special case of the results proved in SLS 4.88 and SLS 4.104. Note from the example in section 3.3 in the notesthat v is a solid-body rotation (with angular velocity vector ) so it makes intuitive sense that v has zero divergence.(Weve just proven that v is solenoidal. It can therefore be written as the curl of a vector potential. So heres another question:

show that G = r2/2 is a vector potential for v = r, where r =

x2 + y2 + z2. This is a neat result: this vector

potential is antiparallel to the rotation vector .)

8. SLS 4.93 (hint: check the table of vector identities) Answer:Need to show that () = 0 (irrotational).

Method 1:By the chain rule, =

12

2 (verify by components that this is true).

So by identity (ix), () =12 (

2) = 0.

Method 2:Let A =.Thus we need to show that (A) = 0.By identity (v), (A) = () A + ( A).But by identity (ix), A = () = 0.


Therefore () = () (). But this is the zero vector because it is the cross-product of two parallelvectors.So () = 0, i.e. is irrotational.

Method 3:The long way (by components) not recommended!

9. SLS 4.98 (where v= v; hint: check the table of vector identities) Answer: This follows directly from identity (viii)by setting A = B = v and recalling that v v = v

2.

10. SLS 4.22 Answer: See SLS:

CHAPTER 4 Gradient, Divergence, Curl

4.19. Prove Vo = O. Solution Let cP = r-3 and A = r in the result of Problem 4.18(b). Then Vo(r-3r) = (Vr-3)or + (r-3)Vr

= -3r-5rr + 3r-3 = 0, using Problem 4.4.

4.20. Prove Vo (UVV - VVU) = UV2V - VV2 U. Solution From Problem 4.18(b), with cp = U and A = VV,

V(UVV) == (VU)o(VV) + U(VoVV) = (VU)(VV) + UV2V Interchanging Uand V yields

Then, subtracting, Vo(VVU) = (VV)(VU) + VV2U.

Vo(UVV) - Vo(VVU) = Vo(UVV - VVU)

= (VU) 0 (VV) + UV2V - [(VV)(VU) + VV2 U] = UV2V- VV2 U

4.21. A fluid moves so that its velocity at any point is vex, y, z). Show that the loss of fluid per unit volume per unit time in a small parallelepiped having center at P(x, y, z) and edges parallel to the coordinate axes and having magnitude b.x, b..y, b.z respectively, is given approximately by div v = Vo v.

z

x Fig. 4-3

Solution Referring to Fig. 4-3,

x component of velocity v at P = VI 1 aVI x component of v at center of face AFED = VI - --.c,.xapprox. 2 ax 1 aVI x component of v at center of face GHCB = VI + --.c,.x approx. 2 ax

y


Then (1) volume of fluid crossing AFED per unit time = (VI - av I .c,.x).c,. y .c,.z, , (2) volume of fluid crossing GHCB per unit time = (VI + .c,.x ).c,.y.c,.z.

Loss in volume per unit time in x direction = (2) - (1) = : .c,.x.c,.y.c,.z.

Similarly, loss in volume per unit time in y direction = aV2 fu.c,.y.c,.z ay loss in volume per unit time in z direction = a;: .c,.x.c,.y.c,.z.

Then, total loss in volume per unit volume per unit time

(iNI iN2 iN3) -+-+- .c,.x.c,.y.c,., z ax ay az . = =dlvv=Vv .c,.x.c,.y.c,.z

This .is true exactly ?nly in the limit as the. parallelepiped shrinks to P, i.e. as .c,.x, .c,.y, and.c,.z approach zero. If no of anywhere, then V 0 v = O. This is called the continuity equation for an incompressible flUid. Smce flUid IS neither created nor destroyed at any point, it is said to have no sources or sinks. A vector such

, as v whose divergence is zero is sometimes called solenoidal.

4.22. Determine the constant a so that the following vector is solenoidal. V = (-4x - 6y + 3z)i + ( - 2x + y - 5z)j + (5x + 6y + az)k

Solution Avector V is solenoidal if its divergence is zero.

a a a vo V = ax(-4x - 6y+ 3z) + ay (-2x+y - 5z) + a/5x+ 6y + az) = -4+ 1 + a = -3 + a. Then VoV=-3+a=Owhena=3.

The Curl

4.23. Suppose A = x2z2j - 2y2z2j + xy2zk. Find V x A (or curl A) at the point P = (1, -1, 1). Solution

i j k a a a VxA= ax ay az xZZ2 -2y 2Z2 xy2Z

= [; (xy2Z) - ! (-2l z2)} - [r - ! (X2Z2)} [r (-2lz2) + ; (x2l) Jk = (2xyz + 4yz2)i - (lz - Wz)j

Thus V x A(P) = 2i + j.

l

11. * SLS 4.32 Answer: See SLS:


aH aE . 2 "a2u 4.31. Suppose V E = 0, V H = 0, V x E = - at' V x H = at Show that E and H satIsfy V u = at2'

Solution

( aH) a a (aE) 1E v x (V x E) = V x Tt = - at (V x H) = - at Tt = - ift2

2 2 1E By Problem 4.29, V x (V x E) = - V2E + V(V E) = -V E. Then V E = ift2. (aE) a a (aH) a

2H Similarly, V x (V x H) = V x Tt = at (V x E) = at -Tt = - at2 '

" ' 2 2 _ 1H But V x (V x H) = - V2H + V(V H) = - V H. Then V H - iji2' The given equations are related to MaxweU's equations of electromagnetic theory. The equation

a2u a2u 1u a2u -+-+-=-ax2 ay2 az2 at2 is called the wave equation.

Miscellaneous Problems 4.32. A vector V is called irrotational if curl V = O. (a) Find constants a, b, and c so that

V = (-4x - 3y + az)i + (bx + 3y + 5z)j + (4x + cy + 3z)k is irrotational. (b) Show that V can be expressed as the gradient of a scalar function.

Solution (a) curl V = V x V

k a

VxV= i a ax

j a ay az

-4x-3y+az bx+3y+5z 4x+cy+3z a a a a

ax az j ay b + 3y+5z

az i-4x+cy+ 3z -4x - 3y + az 4x + cy + 3z

a a + ay k

-4x - 3y + az bx + 3y + 5z = (c - 5)i - (4 - a)j + (b + 3)k.

This equals the zero vector when a = 4, b = -3, and c = 5. So V = (-4x - 3y + 4z)i + (-3x + 3y + 5z)j + (4x + 5y + 3z)k.

a4>. a4>. a4>k Th (b) Assume V = V 4> = ax I + ay J + az' en a4> = -4x - 3y + 4z ax ' a4> = -3x+3y+5z ay a4> = 4x + 5y + 3z az

(1)

(2)

(3)


Integrating (1) partially with respect to x keeping y and z constant, we obtain 4> = -2.? - 3xy + 4xz + f(Y, z)

where f( y, z) is an arbitrary function of y and z. Similarly, we obtain from (2) and (3) 3 4> = -3xy +"2l + 5yz + g(x, z)

and 3 4> = 4xz + 5yz + "2Z2 + h(x, y).

Comparison of (4), (5), and (6) shows that there will be a common value of 4> if we choose

so that 3 3 4> = - 2.? + "2l + "2 Z2 - 3xy + 4xz + 5yz

Note that we can add any constant to 0/. In general, if V x V = 0, then we can find 4> so that V = V 4>.

(4)

(5)

(6)

A vector field V, which can be obtained from a scalar field 4>, so that V = V 4> is called a conservative vector field and 4> is called the scalar potential. Note conversely that, ifY = V 4>, then V x V = 0 (see Problem 4.27a).

4.33. Show that if cp(x, y, z) is any solution of Laplace's equation, then Vcp is a vector that is both solenoidal and irrotational. '

Solution By hypothesis, 4> satisfies Laplace's equation V2 4> = 0, that is, V (V4 = O. Then V4> is solenoidal (see Problems 4.21 and 4.22).

From Problem 4.27a, V x (V4 = 0, so that V4> is also irrotational.

4.34. Give a possible definition of grad B.

Solution Assume B = B1i + B2j + B3k. Formally, we can define grad B as

The quantities ii, ij, and so on, are called unit dyads. (Note that ij, for example, is not the same as ji.) A quantity of the form

is called a dyadic and the coefficients all, a12, ... are its components. An array of these nine components in the form

al2 al3] a22 a23 a32 a33

l I

!

ZPEM2309 Engineering Mathematics 2A MVC Tutorial Problem Set 2 (Sections 3.23.7) ANSWERS page 8 of 12CHAPTER 4 Gradient, Divergence, Curl


Solution









k a

VxV= i a ax

j a ay az




a a + ay k

-4x - 3y + az bx + 3y + 5z = (c - 5)i - (4 - a)j + (b + 3)k.



(1)

(2)

(3)




and 3 4> = 4xz + 5yz + "2Z2 + h(x, y).


so that 3 3 4> = - 2.? + "2l + "2 Z2 - 3xy + 4xz + 5yz


(4)

(5)

(6)









al2 al3] a22 a23 a32 a33

l I

!

12. SLS 4.33 Answer: See SLS:



Solution









k a

VxV= i a ax

j a ay az




a a + ay k

-4x - 3y + az bx + 3y + 5z = (c - 5)i - (4 - a)j + (b + 3)k.



(1)

(2)

(3)




and 3 4> = 4xz + 5yz + "2Z2 + h(x, y).


so that 3 3 4> = - 2.? + "2l + "2 Z2 - 3xy + 4xz + 5yz


(4)

(5)

(6)









al2 al3] a22 a23 a32 a33

l I

!

13. SLS 4.84 Answer: It can be seen by inspection that A = 0, i.e. A is solenoidal.

14. SLS 4.85 Answer: A = 4x+ 8y2z+ 3x3 3x 8y2z 2x3 = x+ x3 6= 0, i.e. A is not solenoidal.

From identity (iv), (A) = () A + ( A). Here = xyz2, so = (yz

2,xz2,2xyz)

so (A) = (yz2,xz2,2xyz) (2x2 + 8xy2z,3x3y 3xy,4y2z2 2x3z) + xyz2(x+ x3)

= (2x2yz2 + 8xy3z3) + (3x4yz2 3x2yz2) (8xy3z3 + 4x4yz2) + (x2yz2 + x4yz2)= 0, i.e. B = xyz

2A is solenoidal.This same result could have been found directly, without using identity (iv).

15. SLS 4.88 (hint: check the table of vector identities) Answer:Need to show that (U V) = 0 (solenoidal).Define A =U and B =V.Thus we need to show that (A B) = 0.By identity (vi), (A B) = B ( A) A ( B).But by identity (ix), A = (U) = 0 and B = (V) = 0.Therefore (U V) = 0, i.e. U V is solenoidal (nondivergent).Note: this question is essentially the same as SLS 4.104, and SLS 4.76 is a special case of this result.

16. SLS 4.92 Answer: A = ( axy z3, (a 2)x2, (1 a)xz2 ) = ( 0, 3z2 (1 a)z2, 2(a 2)x ax ) =

( 0, (a 4)z2, (a 4)x ) = 0 when a = 4.17. SLS 4.104 (hint: check the table of vector identities) Answer:Need to show that (A B) = 0 (solenoidal).By identity (vi), (A B) = B ( A) A ( B).But A = 0 and B = 0 (since we are told they are irrotational).Therefore (A B) = 0, i.e. A B is solenoidal (nondivergent) if A and B if are irrotational.


Note: this question is essentially the same as SLS 4.88, and SLS 4.76 is a special case of this result.

18. SLS 4.105 (where r = (x,y,z) is the position vector and r =x2 + y2 + z2) Answer:

16.2

Q18


19. * SLS 4.102 Answer:

XXX 3.73.7


20. * SLS 4.103 (where r= (x,y,z) is the position vector and r=x2 + y2 + z2 is its magnitude, i.e. the distance from

the origin, and express as a function of r, so the last part of the question asks you to find such that (a) = 0 atsome positive radius a) Answer:

16.2

Q20

21. * SLS 4.106 (In part (a) r = (x,y,z) is the position vector. For part (b), look at section 3.3 and think about what aconstant curl represents; this will lead you to a result for V which differs from that given in SLS but you should beable to check that it is correct and that the difference from the SLS answer is an irrotational field, i.e. they are bothvalid vector potentials.) Answer:

mvc - answers

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