Solutions to Homework Problems on Page 77
1. :
(a) v = hx; yi. div v = 1 + 1 = 2. To get a representative graph of this
vector eld execute the following commands in Mathematica:
<<Graphics`PlotField`
v = {x, y};
PlotVectorField[v, {x, -2, 2}, {y, -2, 2}]
The output is shown in Figure 1.
(b) v = hx; ,yi. div v = 1 , 1 = 0. See Figure 2.
(c) v = h px2x+y2 ; , px2y+y2 . Then div v = @x@ ( px2x+y2 ) + @y@ (, px2y+y2 )
2
2
= p(yx2,+xy2 )3 .
To determine this divergence in Mathematica execute
<<Calculus`VectorAnalysis`
SetCoordinates[Cartesian[x, y, z]];
v = {x/Sqrt[x^2+y^2], -y/Sqrt[x^2+y^2], 0};
Simplify[Div[v]]
See Figure 3.
y
@ cos x
(d) v = hsin y; cos xi. Then, div v = @ sin
@x + @y = 0. See Figure 4.
p
(e) v = hln x2 + y2 ; yi. Then, div v = 1 + x2 +x y2 . See Figure 5.
2. Let f be any function of x, y, and z . By denition
f = r (rf ):
@f @f
But rf = h @f
@x ; @y ; @z i. Hence,
@ @f
@ @f
@ @f
( @x ) + @y
( @y ) + @z
( @z ) = @@xf + @@yf + @@zf :
r (rf ) = @x
2
2
2
2
2
2
3. (a) Let f = 3x2 , 4y2. Then f = ,2. (Try Laplacian[3 x^ 2 - 4
y^ 2] in Mathematica.
(b) Let f = xy . Then f = x2y2 .
(c) Let f = sin x2 +1 y2 . Then f = (x2 +4y2 )3 (x2 + y2 ) cos x2 +1 y2 , sin x2 +1 y2 .
(d) Let f = ln(x2 + y2 ). Then, f = 0.
(e) Let f = px21+y2 . Then, f = (x2 +4y2 )2 .
1
(f) Let f = arctan yx . Then, f = 0. (Try Simplify[Laplacian[ArcTan[y/x]]]
in Mathematica.
5. (a) div (v) = divv + v r.
v1 ) @ (v2 )
+ @y +
Proof: L.H.S. = div (bf v) = div (hv1 ; v2; v3i) = @(@x
@ (v3 )
@z . We now use the product rule of dierentiation to get
@
@v1
@
@v1
@
@v3
v1 + )
+
(
v2 + )
+
(
v3 + );
L.H.S. =( @x
@x
@x
@y
@x
@z
which can be regrouped as
@v1
@x
+
@v2
@x
+
@v3
@z
+
The above expression is equivalent to the R.H.S.
2
@
@
@
v1 +
v2 +
v3 :
@x
@y
@z
Figure 1: Vector eld v = hx; yi.
3
Figure 2: Vector eld v = hx; ,yi.
4
Figure 3: Vector eld v = h px2x+y2 ; , px2y+y2 i.
5
Figure 4: Vector eld v = hsin y; cos xi.
6
p
Figure 5: Vector eld v = hln
7
x2 + y 2 ; y i.