Handbook of Differential Equation

SOLUTION
QUESTION ONE
Finding the nominal of the function f (x, y, z) at a point p (-1,2,0).
f ( x , y , z )=x y3+ 3 xy −cos 3 z
f (−1,2,0 )=−1∗03 +3∗−1∗2−cos 3∗0
¿ 0−6−cos 0
¿−7
QUESTION TWO
Evaluating the directional directive of a scalar function f (x, y, z) at a point p (1, -2,1) in the
direction of u (2/3 ,1/3,2/3).
f ( x , y , z)=x y3 z2 +2 z y2−x2
∇ f =i df
dx + j df
dy +k df
dz
¿ i ( y3 z2−2 x )+ j ( 3 x y2 z2 + 4 zy ) +k ( 2 x y3 z+ 2 y2 )
¿ i ((−2)3 12 −2∗1 )+ j ( 3∗1∗(−2 )2∗12 + 4∗1∗−2 )+ k (2∗1∗(−2 )3∗1+2 (−2 )2 )
∇ f =−10 i+ 4 j−8 k
a= 2
3 i + 1
3 j+ 2
3 k
Directional derivative=∇ f a
|a|
¿ ( −10 i+ 4 j−8 k ) ( 2
3 i+ 1
3 j+ 2
3 k )
√ ( 2
3 )
2
+( 1
3 )
2
+ ( 2
3 )
2
¿
−20
3 + 4
3 −16
3
1
¿−32
3
QUESTION THREE
The divergence of the vector v at point p (1, 3, -1) will be obtained as:
⃗
v=i ( x3 y2−z ) + j ( yz ) + x2 z

¿ ( ⃗ v )= d ⃗ v1
dx + d ⃗ v2
dy + d ⃗ v3
dz
¿ 3 x2+ z + x2
¿ 3 ¿ 12−1+ (−1 )2=3
QUESTION FOUR
Given the expression for the force as F (x, y, z) the curl of the force and curl of curl of f (x, y,
z) will be obtained as:
⃗
F= ( 2 x y2− yx ) i+ ( 2 y x2 +2 y z2−xz ) j+ ( 2 z y2−xy ) k
curl ( ⃗ F )=
| i j k
d
dx
d
dy
d
dz
F1 F2 F3
|
¿
| i j k
d
dx
d
dy
d
dz
2 x y2− yx 2 y x2+ 2 y z2−xz 2 z y2−xy
|
¿ ( 4 zy−x ) i− (− y ) j+ ( 4 yx−z−4 xy+ x ) k
¿ ( 4 zy−x ) i+ ( y ) j+ ( x−z ) k
curl ( curl ( ⃗ F ) ) =
| i j k
d
dx
d
dy
d
dz
4 zy−x y x−z
|
¿ 0− ( 1−4 y ) j+ ( 4 z ) k
¿ ( 1−4 y ) j+ ( 4 z ) k
QUESTION FIVE
Divergence and curl of the given spherical-symmetric vector field.
⃗
F ( x , y , z )= ⃗ r
r3
⃗
F ( x , y , z ) =ur
1
r2 sin θ cos φ+uθ
1
r2 sinθ sin φ+ uφ
1
r2 cos θ
¿ ( ⃗ F ( x , y , z ) )= 1
r2
d
dr ( r2 ur ) + 1
r∗sin θ
d
dθ ( uθ sinθ ) + 1
r sin θ
d
dφ ( uφ )
¿ ( ⃗ F ( x , y , z ) )= 1
r2
d
dr ( sinθ cos φ ) + 1
r3∗sinθ
d
dθ ( sin2 θ sin φ ) + 1
r sin θ
d
dφ ( 1
r2 cos θ)

¿ ( ⃗ F ( x , y , z ) )=0+ 1
r3∗sin θ 2cos θ sin θ+ 0
¿ ( ⃗ F ( x , y , z ) )= 1
r3 2cos θ
QUESTION SIX
An incompressible fluid velocity is given by:
⃗
v=i ( cos xyzt )+ j( y2 t )
i. For an incompressible fluid:
Divergence:
¿ ( ⃗ v ) = d ⃗ v1
dx + d ⃗ v2
dy + d ⃗ v3
dz =0
Gradient:
dv
dt =0
ii. Difference between gradient and divergence
Gradient of a scalar function or filed is a vector representing both the magnitude and
direction of the maximum space rate (derivative with respect to spatial coordinates) of
increase of the function field (Chapekar, 2019). On the other hand, divergence of a
vector field at a given point is the net outward flux per unit volume as the volume tends to
zero at that point (Chapekar, 2019).
iii. Is vector physically acceptable?
No, this because fluids are always compressible.
QUESTION SEVEN
i. Condition for divergence and curl to be zero.
¿ ( ⃗ v ) = d ⃗ v1
dx + d ⃗ v2
dy + d ⃗ v3
dz =a+ d
¿ ( ⃗ v )=0 when a+d =0 i. e a=−d

curl ( curl ( ⃗ F ) )=
| i j k
d
dx
d
dy
d
dz
ax+ by cx +dy 0 |=0i+0 j+ ( c−b ) k
curl ( curl ( ⃗ F ) )=0 when c−b=0 i . e c=b
ii. Flow’s velocity potential
i (u )+ j ( v ) +k ( w )=i d ∅
dx + j d ∅
dy + k d ∅
dz
∅= 1
2 (a x2+2 bxy −a y2)
u= d ∅
dx =ax +by
v= d ∅
dy =bx−ay but a=−d , c=b
v= d ∅
dy =cx + dy
w= d ∅
dz =0
Therefore, the velocity function is given by:
∅= 1
2 (a x2+2 bxy −a y2)
iii. Matlab plots
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