Homework Assignment 1 Solution Set

PHYCS 4420

23 January, 2004

Problem 1

a (Griffiths 1.11(b))f(x, y, z) = x2y3z4

~∇f =∂f

∂xx +

∂f

∂yy +

∂f

∂zz

= 2xy3z4x + 3x2y2z4y + 4x2y3z3z

b (Griffiths 1.15(c))

~v(x, y, z) = y2x + (2xy + z2)y + 2yzz

~∇ · ~v =∂vx

∂x+

∂vy

∂y+

∂vz

∂z= 2x + 2y

c (Griffiths 1.18(c))

~v(x, y, z) = y2x + (2xy + z2)y + 2yzz

~∇× ~v = (∂vz

∂y− ∂vy

∂z)x + (

∂vx

∂z− ∂vz

∂x)y + (

∂vy

∂x− ∂vx

∂y)z

= (2z − 2z)x + (0− 0)y + (2y − 2y)z= 0.

Note: You may have recognized that ~v(x, y, z) = ~∇f , where f = xy2 + yz2.Thus, by Griffiths’ equation 1.44, ~∇× ~v = 0.

Problem 2 (Griffiths 1.12)

h(x, y) = 10(2xy − 3x2 − 4y2 − 18x + 28y + 12)

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a The stationary points of h are found where ~∇h = 0. Thus, to find thepeak we set

~∇h = 0

(−60x + 20y − 180)x + (−80y + 20x + 280)y = 0,

which, of course, can only be true if

−3x + y − 9 = 0

andx− 4y + 14 = 0.

Solving this system of simultaneous equations gives only one solution at thepoint (−2, 3); i.e., 3 miles north and 2 miles west of South Hadley. This positionmust correspond to the top of the hill. We know it is a maximum (rather thananother type of stationary point) because of the form of h(x, y) (it’s quadratic,and the coefficients of x2 and y2 in h(x, y) are both negative).

b

h(−2, 3) = 10(2(−2)(3)− 3(−2)2 − 4(3)2 − 18(−2) + 28(3) + 12= 720ft

c

~∇h(1, 1) = (−60(1) + 20(1)− 180)x + (−80(1) + 20(1) + 280)y= −220x + 220y

The slope of the hill is the magnitude of the gradient, which is 220√

2ft./mile.It points uphill in the direction of the gradient, which is to the north-west.

Problem 3 (Griffiths 1.21)

a We should obey the parentheses and compute the product they containfirst. However, ~∇ is an operator, so we should leave it open to act on the rightas it is written. Thus,

( ~A · ~∇) ~B = (Ax∂

∂x+ Ay

∂

∂y+ Az

∂

∂z) ~B

= (Ax∂Bx

∂x+Ay

∂Bx

∂y+Az

∂Bx

∂z)x+(Ax

∂By

∂x+Ay

∂By

∂y+Az

∂By

∂z)y+(Ax

∂Bz

∂x+Ay

∂Bz

∂y+Az

∂Bz

∂z)z.

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b

r =xx + yy + zz√x2 + y2 + z2

Since the functional form of rı is identical for ı ∈ {x, y, z}, we need only computeone component of (r · ~∇)r (the others will be the same with the variables simplyinterchanged).

((r · ~∇)r)x = rx∂rx

∂x+ ry

∂rx

∂y+ rz

∂rx

∂z

=1√

x2 + y2 + z2(x(

y2 + z2

(x2 + y2 + z2)32) + y(

−xy

(x2 + y2 + z2)32) + z(

−xz

(x2 + y2 + z2)32))

=1

(x2 + y2 + z2)2(x(y2 + z2)− xy2 − xz2)

= 0

Thus, (r · ~∇)r = 0.

Problem 4 (Griffiths 1.24(a) - optional)

~A = xx + 2yy + 3zz

~B = 3yx− 2xy

We’d like to prove product rule (iv) from the text, namely

~∇ · ( ~A× ~B) = ~B · (~∇× ~A)− ~A · (~∇× ~B).

First, let’s expand the left hand side:

~A× ~B = (0− (−6xz))x + (9yz − 0)y + (−2x2 − 6y2)z.

So, taking the divergence gives

~∇ · ( ~A× ~B) = 6z + 9z = 15z.

Now for the right hand side. First we’ll compute the curls:

~∇× ~A = (0− 0)x + (0− 0)y + (0− 0)z = 0,

~∇× ~B = (0− 0)x + (0− 0)y + (−2− 3)z = −5z.

So the right hand side is

~B · (~∇× ~A)− ~A · (~∇× ~B) = 0− 3z(−5) = 15z.

Problem 5 (Griffiths 1.25)

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b

T (x, y, z) = sin x sin y sin z

∇2T = − sinx sin y sin z − sinx sin y sin z − sinx sin y sin z

= −3T (x, y, z)

d

~v(x, y, z) = x2x + 3xz2y − 2xzz

∇2~v = (∇2vx)x + (∇2vy)y + (∇2vz)z= 2x + 6xy

Problem 6 (Griffiths 1.31)

T (x, y, z) = x2 + 4xy + 2yz3

~∇T = (2x + 4y)x + (4x + 2z3)y + (6yz2)z

We’d like to prove that ∫ b

a

(~∇T ) · d~l = T (b)− T (a)

regardless of the path we choose. First, let’s find the right hand side so we knowwhat to look for on the left.

T (b)− T (a) = T (1, 1, 1)− T (0, 0, 0)= (1)2 + 4(1)(1) + 2(1)(1)3 = 7

a

(0, 0, 0) → (1, 0, 0) → (1, 1, 0) → (1, 1, 1)

Along this path we have three separate line integrals.∫ (1,1,1)

(0,0,0)

(~∇T ) · d~l =∫ 1

0

~∇T · dxx∣∣∣y=0=z

+∫ 1

0

~∇T · dyy∣∣∣x=1,z=0

+∫ 1

0

~∇T · dzz∣∣∣x=1=y

=∫ 1

0

2xdx +∫ 1

0

4dy +∫ 1

0

6z2dz

= 1 + 4 + 2 = 7

4

c

(0, 0, 0) → (1, 1, 1); z = y2, y = x

Now we have three integrals for a different reason (sort of). There’s just onesmooth path, but our path integral gets split into components. In each integralwe substitute expressions for each variable in terms of the variable of integration.∫ (1,1,1)

(0,0,0)

(~∇T ) · d~l =∫ 1

0

(2x + 4y)dx +∫ 1

0

(4x + 2z3)dy +∫ 1

0

6yz2dz

=∫ 1

0

6xdx +∫ 1

0

(4y + 2y6)dy +∫ 1

0

6z52 dz

= 3 + (2 +27) +

127

= 7

Problem 7 (Griffiths 1.33)

~v(x, y, z) = xyx + 2yzy + 3zxz

Stokes’ Theorem says that∫S

(~∇× ~v) · d ~A =∫

P

~v · d~l.

The right hand side of this equation is easy. Following the arrows in Figure1.34, and starting from the origin, we have three line integrals:∫

P

~v·d~l =∫ 2

0

(xyx+2yzy+3zxz)·dyy+∫

P ′(xyx+2yzy+3zxz)·(dyy+dzz)+

∫ 0

2

(xyx+2yzy+3zxz)·(−dzz).

The second line integral must be split into two integrals: one in the y-directionand one in the z-direction. Also, in each integral we should replace the non-integrated variables with appropriate values. In the first path we hold x = z = 0,so the integral vanishes. In the third path we have x = y = 0, so it also vanishes.However, in the second path we have x = 0 and z = 2− y. Thus,∫

P

~v · d~l =∫ 2

0

2yzdy +∫ 0

2

2y(2− y)dy +∫ 2

0

3zxdz +∫ 0

2

3zx(−dz)

=∫ 2

0

(2y2 − 4y)dy

=163− 8 = −8

3.

5

For the left hand side of Stokes’ Theorem we note that the positive x directionis the direction of d ~A. This is given by the right hand rule according to thepath we took above. Thus, we need to find∫

S

(~∇× ~v) · d ~A =∫∫

S

(~∇× ~v) · (dydzx),

where~∇× ~v = (0− 2y)x + (0− 3z)y + (0− x)z.

There are a few ways to proceed with the double integral from here. Let’sintegrate over z first, since vx doesn’t depend on z. The z boundary, however,depends on y. Thus,∫

S

(~∇× ~v) · d ~A =∫ 2−y

0

∫ 2

0

dzdy(−2yx− 3zy − xz) · x

=∫ 2

0

dy(2− y)(−2y)

= −83.

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