# Gradient, divergence and curl (Cartesian coordinates)

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```					                     ENGI 4430 Advanced Calculus for Engineering
Faculty of Engineering and Applied Science
Problem Set 4 Solutions
[Gradient, divergence and curl (Cartesian coordinates)]

1.   Find the divergence and curl of the vector field F = x y 2 ˆ + x 2 y ˆ + z k .
i         j     ˆ

⎡ x y2    ⎤
⎡ ∂ ∂ ∂ ⎤⎢ 2           ⎥
div F = ⎢          ⎥⎢x y       ⎥ = 1y + x 1 + 1 ⇒
2     2

⎣∂x ∂ y ∂z ⎦ ⎢ z       ⎥
⎣         ⎦
div F = x 2 + y 2 + 1

ˆ    ∂
i          x y2
∂x
⎡ 0−0 ⎤
∂
curl F =    ˆ
j           2
x y    = ⎢ 0−0 ⎥ ⇒
⎢             ⎥
∂y
⎢ 2 xy − 2 xy ⎥
⎣             ⎦
ˆ    ∂
k           z
∂z
curl F = 0
F is therefore an irrotational field.

xˆ yˆ z ˆ
2.   Find the divergence and curl of the vector field F =       i + j+ k.
y    z  x

T               T
⎡∂ ∂ ∂ ⎤ ⎡x y z⎤                         1 1 1
div F = ⎢          ⎥ i⎢      ⎥               =    + +      ⇒
⎣∂x ∂ y ∂z ⎦ ⎣ y z x ⎦                   y z x

1 1 1
div F =    + +
x y z
ENGI 4430                                      Problem Set 4 Solutions                           Page 2 of 5

2 (continued)
∂
ˆ
i             x y −1
∂x                     ⎡ 0 − ( − y z −2 ) ⎤
∂                      ⎢                  ⎥
curl F =     ˆ
j             y z −1     = ⎢ 0 − ( − z x −2 ) ⎥ ⇒
∂y                     ⎢                  ⎥
⎢ 0 − ( − x y −2 ) ⎥
∂                      ⎣                  ⎦
ˆ
k             z x −1
∂z
T
⎡ y z x ⎤
curl F = ⎢ 2 2 2 ⎥
⎣z x y ⎦

[Note: the cyclic symmetry between x, y, z allows any two parts of the calculations for
divergence and curl to be deduced from the other part.]

3.       A temperature distribution for a region within 75 metres of the origin is given by
10 000 − x 2 − y 2
T ( x, y , z ) =
z + 100
(a) Find the gradient of the temperature function T.

T
⎡ ∂T        ∂T       ∂T ⎤
T
⎡ −2 x      −2 y  x 2 + y 2 − 10000   ⎤
∇T = ⎢                               =   ⎢                                     ⎥
⎣ ∂x        ∂y       ∂z ⎥           ⎢ z + 100 z + 100     ( z + 100 )
2
⎦           ⎣                                     ⎥
⎦

(b) Find the [instantaneous] rate at which the temperature is changing at the point
(50, 50, 0) in the same direction as the vector ˆ − ˆ .
i j

P is the point (50, 50, 0)
Direction a = ˆ − ˆ
i j       ⇒ a = a = 1+1 =                          2
⎡ 1⎤
a    2⎢ ⎥
⇒ u =
ˆ     =    −1
a   2 ⎢ ⎥
⎢ 0⎥
⎣ ⎦
T
⎡ −2(50) −2(50) 502 + 502 − 10000 ⎤
∇T       = ⎢                                 ⎥
⎣ 0 + 100 0 + 100   (0 + 100) 2
P
⎦
= − 1[ 2 2 1]
T
= ⎡ −1 −1 − 1 ⎤
T
⎣         2⎦                2
ENGI 4430                                  Problem Set 4 Solutions                                Page 3 of 5

3 (b) (continued)
⎡2⎤     ⎡ 1⎤
1⎢ ⎥ 2⎢ ⎥             2
⇒ Du T           = ∇T Pi u = − ⎢ 2 ⎥i   ⎢ −1 ⎥ = − 4 ( 2 − 2 + 0 ) ⇒
ˆ         P
ˆ
2       2
⎢1⎥
⎣ ⎦     ⎢ 0⎥
⎣ ⎦
Du T
ˆ        P
= 0

[The vector ˆ − ˆ is at right angles to the gradient vector (and therefore
i j
tangential to a level surface) at the point (50, 50, 0).]

[That is, does the gradient vector point directly towards or directly away from the
origin at every point?]

1
At P (50, 50, 0), r = [ 50 50 0 ]                    but ∇ T       = −     [ 2 2 1 ] ≠ kr
T                                      T
P
2
for any scalar k. Therefore,

4.       Find the equations of the tangent plane and the normal line to the sphere
x2 + y2 + z2 = 9 at the point (−2, 1, 2).

⎡ 2x ⎤      ⎡x⎤
f ( x, y , z ) = x 2 + y 2 + z 2 = 9      ⇒ ∇f = ⎢ 2 y ⎥ = 2 ⎢ y ⎥
⎢     ⎥     ⎢ ⎥
⎢ 2z ⎥
⎣     ⎦     ⎢z⎦
⎣ ⎥
Use n = [ x y z ]
T

P ( −2, 1, 2 )      ⇒ n = [ −2 1 2]                           ⎡⇒ n =        4 + 1 + 4 = 3⎤
T
⎣                          ⎦
Tangent plane at P:
r in = ain    ⇒ [ x y z ] i[ − 2 1 2 ] = [ − 2 1 2 ] i[ − 2 1 2 ]                             ⇒
T            T             T             T

− 2x + y + 2z = 9
ENGI 4430                                   Problem Set 4 Solutions                                    Page 4 of 5

4 (continued)

Normal line at P:
⎡ x ⎤ ⎡ −2 ⎤       ⎡ −2 ⎤
r = a + t v with v = n                ⎢ y ⎥ = ⎢ 1⎥ + t ⎢ 1⎥
⇒ ⎢ ⎥ ⎢            ⎥    ⎢     ⎥
⎢ z ⎥ ⎢ 2⎥
⎣ ⎦ ⎣         ⎦    ⎢ 2⎥
⎣     ⎦
[from which is it easy to deduce that this normal line passes through the origin.]
One form of the equations of the normal line is

x+2   y −1 z − 2
=     =
−2    1     2

5.    Find the angle between the elliptic paraboloid z = 3x 2 + 2 y 2 and the parabolic
cylinder 7 y 2 = 2 x + z at the point (1, 1, 5), to the nearest 0.01 degree.

The elliptic paraboloid has its axis of
symmetry along the z-axis.

The vertex line of the parabolic
cylinder is 2 x + z = 0 , y = 0 .

[One should check that (1, 1, 5) does
indeed lie on both surfaces.]

Define
f ( x, y , z ) = 3 x 2 + 2 y 2 − z = 0
⇒ ∇f = [ 6 x 4 y − 1 ]
T

⇒ ∇f                  = [ 6 4 −1 ]         = n1
T

(1,1,5)
and
g ( x, y , z ) = − 2 x + 7 y 2 − z = 0
⇒ ∇g = [ − 2 14 y − 1 ]                       ⇒ ∇g             = [ − 2 14 − 1 ]         = n2
T                                           T

(1,1,5)
θ = angle between surfaces = angle between ∇f                                 and ∇g
(1,1,5)            (1,1,5)
[ 6 4 − 1 ] i[ − 2 14 − 1 ] = −12 + 56 + 1 =
T                 T
n1 in 2                                                                      +45
⇒ cos θ =                    =
n1 n2         ( 36 + 16 + 1) × ( 4 + 196 + 1) 53 × 201                      10 653
⇒ cos θ ≈ .4360                   ⇒       θ = 64.15° ( 2 d.p.) ( ≈ 64°9 ')
ENGI 4430                                Problem Set 4 Solutions                                     Page 5 of 5

Calculate the directional derivative of φ ( r ) = x ln y − e x / z at the point (8, 1, −2)
3
6.
in the direction of the vector a = 12 ˆ + 2 ˆ − k .
i     j ˆ

T
x z −3          ⎡                  −3         x                  x z −3   ⎤
φ ( r ) = x ln y − e            ⇒ ∇φ = ⎢ ln y − z −3e x z                          −4
+ 3xz e             ⎥
⎣                             y                           ⎦
T
⎡     e−1                    ⎤
T
3                       ⎡ 1    3 ⎤
⇒ ∇φ 8,1, −2 = ⎢ 0 −               8 + e −1 ⎥               = ⎢    8
(      ) ⎣      −8               2     ⎦                 ⎣ 8e   2e ⎥
⎦
149
a = [ 12 2 − 1 ]            ⇒ a = 144 + 4 + 1 = 149                ⇒ a =               [ 12 2 − 1 ]
T                                                                            T
ˆ
149
T
⎡ 1    3 ⎤    149
) ⎣ 8e 8 2e ⎥ i 149 [ 12 2 − 1 ]
⇒ Da φ 8,1, −2 = ⎢
T
(                ⎦
149 ⎛ 3        3 ⎞
=      ⎜ + 16 − ⎟ ⇒
149 ⎝ 2e        2e ⎠
16 149
Da φ               =
(8,1, −2)         149

7.      Find the family of streamlines associated with the velocity field
v ( x, y ) = y ˆ + x ˆ
i     j
and find the streamline through the point (0, –1).

First let us verify that the equation of continuity is satisfied:
∂u ∂ v        ∂        ∂
div v =       +      =     ( y) + ( x) = 0 + 0 = 0
∂x ∂ y       ∂x       ∂y
Stream function ψ ( x, y ) :
∂ψ                   ∂ψ                             x2 y 2
= + v = x and         = −u = − y      ⇒ ψ =         −    =C
∂x                  ∂y                              2    2
Therefore the family of streamlines is the family of rectangular hyperbolae
x2 − y 2 = A
( x, y ) = ( 0, − 1)        ⇒ 02 − ( −1) = A                ⇒ x2 − y2 = −1
2
or
y 2 − x2 = 1

Back to the index of solutions

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