Download Vector Integral and Differential Calculus (ACM 20150) – Assignment 4

Vector Integral and Differential Calculus
(ACM 20150) – Assignment 4
Issue Date: Monday 17th October 2016
Due Date: Wednesday 26th October 2016
1. (a) If u = x3 x̂−xy ŷ +yz 2 ẑ and φ = zx−z 3 y, compute u·∇φ and u×∇φ.
Evaluate your results at the point (1, −1, 1).
(b) Let u = (yz + 4xy)x̂ + (xz + 2x2 − 3z 2 )ŷ + (xy − 6yz)ẑ. Compute
∇ × u. Find ψ(x, y, z) such that u = ∇ψ.
(c) Find the unit outward drawn normal to the surface (x − 2)2 + (y − 1)2 +
z 2 = 25 at the point (−2, 4, 0).
(d) Find the equation for the tangent plane and the equation of the normal
line to the surface xz 2 + x2 y = z − 1 at the point (1, −1, 1).
(a) We have,
∇φ = z x̂ − z 3 ŷ + (x − 3z 2 y)ẑ.
hence
u · ∇φ = x3 x̂ − xy ŷ + yz 2 ẑ · z x̂ − z 3 ŷ + (x − 3z 2 y)ẑ ,
= zx3 + xyz 3 + yz 2 (x − 3z 2 y).
Evaluating at (1, −1, 1), this is
(1)(1) + (1)(−1)(1) + (−1)(1) [1 − 3(−1)] = −4.
Moreover,
x3 x̂ − xy ŷ + yz 2 ẑ × z x̂ − z 3 ŷ + (x − 3z 2 y)ẑ ,
x̂
ŷ
ẑ
3
2
,
yz
= x −xy
z −z 3 (x − 3z 2 y) = x̂ −xy x − 3z 2 y + yz 5 − ŷ x3 x − 3z 2 y − yz 3
+ ẑ −x3 z 3 + xyz .
u × ∇φ =
Evaluating at (1, −1, 1), this is
x̂ [(1)(1 + 3) − 1] − ŷ [(1 + 3) + 1] + ẑ (−1 − 1) = 3x̂ − 5ŷ − 2ẑ.
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Vector Integral and Differential Calculus
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(b) We have
x̂
ŷ
ẑ
∂x
∂y
∂z
∇×u=
yz + 4xy xz + 2x2 − 3z 2 xy − 6yz
.
Hence,
∇ × u = x̂ ∂y (xy − 6yz) − ∂z xz + 2x2 − 3z 2
− ŷ [∂x (xy − 6yz) − ∂z (yz + 4xy)]
ẑ ∂x xz + 2x2 − 3z 2 − ∂y (yz + 4xy) ,
= x̂ [x − 6z − (x − 6z)] − ŷ [y − y] + ẑ [z + 4x − (z + 4x)] ,
= 0.
We now work out the scalar field ψ such that u = ∇ψ:
ψx = yz + 4xy
ψy = xz + 2x2 − 3z 2
ψz = xy − 6yz
=⇒
=⇒
=⇒
ψ = xyz + 2x2 y + f (y, z),
ψ = xyz − 3yz 2 + 2x2 y + g(x, z),
ψ = xyz − 3yz 2 + h(x, y).
Simply by inspection, h(x, y) = 2x2 y, hence
ψ = xyz − 3yz 2 + 2x2 y + Const.
Check:
ψx = yz + 4xy,
ψy = xz − 3z 2 + 2x2 ,
ψz = xy − 6yz.
(c) Let
φ(x, y, z) = (x − 2)2 + (y − 1)2 + z 2 − 25.
Hence, φ = 0 describes a sphere of radius 5 centred at (2, 1, 0). A normal
to the sphere is
∇φ = 2(x − 2)x̂ + 2(y − 1)ŷ + 2z ẑ,
hence
n̂ =
∇φ
2(x − 2)x̂ + 2(y − 1)ŷ + 2z
=p
.
|∇φ|
4(x − 2)2 + 4(y − 1)2 + 4z 2
Note that φ(−2, 4, 0) = (−4)2 + 32 + 0 − 25 = 0, hence (−2, 4, 0) lies on
the sphere. At this point,
∇φ = 2(−4)x̂ + 2(3)ŷ = −8x̂ + 6ŷ,
and
n̂ =
∇φ
=
|∇φ|
q
1
100
(−8x̂ + 6ŷ) =
1
10
(−8x̂ + 6ŷ) = 51 (−4x̂ + 3ŷ) .
We know from class notes that this normal vector points outwards.
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Vector Integral and Differential Calculus
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(d) Define
φ = xz 2 + x2 y − z + 1.
The equation φ = 0 defines a surface with outward-pointing normal n̂ =
∇φ/|∇φ|. Hence,
x̂(z 2 + 2xy) + ŷx2 + ẑ(2zx − 1)
.
n̂ = p
(z 2 + 2xy)2 + x4 + (2zx − 1)2
Note, φ(1, −1, 1) = 0, hence (1, −1, 1) lives in the surface. At this point,
n̂ =
x̂(−1) + ŷ + ẑ
−x̂ + ŷ + ẑ
√
√
=
.
···
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Tangent plane: Point on plane: r0 = (1, −1, 1); general point r = (x, y, z);
general outward-pointing normal: (−1, 1, 1). Equation:
(r − r0 ) · n = 0,
or
(x−1)nx +(y+1)ny +(z−1)nz = 0 =⇒ (x−1)(−1)+(y+1)(1)+(z−1)(1) = 0.
Tidy up:
x − y − z = 1, or z = x − y − 1.
Normal line:
r(t) = r0 + tn̂,
= (1, −1, 1) + (−1, 1, 1)t,
or
−x + 1 = y + 1 = z − 1.
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Vector Integral and Differential Calculus
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2. (a) Find ∇rn , where n is an integer.
(b) Compute ∇f (r).
(c) Compute ∇2 f (r).
Here r = (x, y, z), r = |r|, and f (·) is a smooth functions of a single variable.
Answer clue for (c): if f (r) = ln r, then ∇2 f (r) = 1/r2 .
First, consider
p
x2 + y 2 + z 2 ,
∂
x̂ (x2 + y 2 + z 2 )1/2 + · · · ,
∂x
2x
1
x̂ 2 2
+ ··· ,
(x + y 2 + z 2 )1/2
x̂x
+ ··· ,
|r|
r
.
|r|
∇|r| = ∇
=
=
=
=
Now, using the ordinary rules of calculus,
(a) ∇|r|n = n|r|n−1 ∇|r| = n|r|n−1 (r/|r|) = nr|r|n−2 .
(b) Extend this result to
r
∇f (r) = f 0 (r) .
r
(c) We have the following string of results:
f 0 (r)
r ,
∇ · (∇f (r)) = ∇ ·
r
f 0 (r)
f 0 (r)
(∇ · r) + r · ∇
,
r
r
0 0 3f 0 (r)
f (r) r
,
+r·
r
r
r
00
3f 0 (r)
rf (r) − f 0 (r)
+r
,
r
r2
3f 0 (r)
1
+ f 00 (r) − f 0 (r),
r
r
2f 0 (r)
+ f 00 (r).
r
2
∇ f (r) =
=
=
=
=
=
Check: f (r) = ln r, f 0 (r) = 1/r, f 00 (r) = −1/r2 , hence
∇2 ln r =
2
1
1
− 2 = 2.
2
r
r
r
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Vector Integral and Differential Calculus
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y
(1,1)
x
Figure 1:
H
3. (a) Evaluate A · dx around the closed curve in Figure 1 if A = (x − y)x̂ +
(x + y)ŷ.
R
(b) Evaluate A · dx along the curve x2 + y 2 = 1, z = 1, in the positive
direction from (0, 1, 1) to (1, 0, 1), if
A = (yz + 2x)x̂ + xz ŷ + (xy + 2z)ẑ.
Hint: Use the following parametrization of the curve:
(cos ϕ, sin ϕ, 1).
x(ϕ) =
(a) Break up closed curve into segments I and II. On segment I, we have xI (t) =
(x, y) = (x, x2 ) = (t, t2 ), hence
dxI =
dxI
dt = (1, 2t)dt.
dt
We have
A · dxI = [(x − y), (x + y)] · (1, 2t)dt,
= (x − y)dt + (x + y)2t dt,
= (t − t2 )dt + 2t(t + t2 )dt.
The parameter t goes from t = 0 to t = 1. Hence,
Z
Z 1
A · dx =
(t − t2 ) + 2t(t + t2 ) dt.
I
0
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Vector Integral and Differential Calculus
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This is a straightforward integral that evaluates to 4/3. Next, we do segment
II, whereon we hhave xII (t) = (x, y) = (x, x1/2 ) = (t, t1/2 ), hence
dxII =
dxII
dt = 1, 12 t−1/2 dt.
dt
We have
A · dxII = [(x − y), (x + y)] 1, 21 t−1/2 dt,
= (x − y) + 12 (x + y)t−1/2 ,
= (t − t1/2 )dt + 21 (t + t1/2 )t−1/2 dt,
= tdt − t1/2 dt + 12 t1/2 dt + 12 dt,
= tdt − 12 t1/2 dt + 21 dt.
The parameter t goes from t = 1 to t = 0. However, this is the same as
going from t = 0 to t = 1 provided we put an overall minus sign in front of
the integral at the end:
Z
Z 1
A · dx =
t − 12 t1/2 + 12 dt.
II
0
Hence,
Z
A · dx = −
II
1
2
t2 −
1 2 3/2
t
23
1
+ 12 t 0 = − 1 − 13 + 12 = − 32 .
Put the two results together:
Z
I
Z
A · dx =
A · dx = A · dx +
I
II
4
3
−
2
3
= 23 .
(b) Solution The parametrization is x(ϕ) = (cos ϕ, sin ϕ, 1). Hence,
dx =
dx
dϕ = (− sin ϕ, cos ϕ, 0)dϕ.
dϕ
The starting point of the curve is (0, 1, 1), hence (x, y) = (0, 1). But (x, y) =
(cos ϕ, sin ϕ), hence ϕ = π/2 at the starting point. Similarly, the end point
of the curve is (1, 0, 1), hence (x, y) = (1, 0). But (x, y) = (cos ϕ, sin ϕ),
hence ϕ = 2π.
Caution: The final parameter value is ϕ = 2π (not ϕ = 0). The reason is
that the sense of the integral is in the positive direction. Therefore, when
projected into the xy plane, the curve looks like a portion of a circle with an
anticlockwise sense, i.e. going from ϕ = π/2 to ϕ = 2π.
Thus,
A · dx =
=
=
=
=
[(yz + 2x), xz, (xy + 2z)] · (− sin ϕ, cos ϕ, 0) dϕ,
(yz + 2x)(− sin ϕ)dϕ + xz cos ϕdϕ,
(sin ϕ + 2 cos ϕ)(− sin ϕ)dϕ + cos2 ϕ dϕ,
since z = 1,
2
2
(cos ϕ − sin ϕ)dϕ − 2 cos ϕ sin ϕ dϕ,
cos 2ϕ dϕ − sin 2ϕ dϕ.
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Vector Integral and Differential Calculus
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Hence,
Z
Z
2π
Z
cos(2ϕ)dϕ −
A · dx =
1
2
sin(2ϕ)dϕ,
π/2
π/2
=
2π
2π
2π
1
sin(2ϕ) + 2 cos(2ϕ) ,
π/2
π/2
= 0 + 21 [cos(4π) − cos(π)] ,
= 1.
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Vector Integral and Differential Calculus
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4. Let u(x, t) be a three-dimensional vector field that also depends on time, t.
Assume that u satisfies Euler’s equation,
∂u
+ u · ∇u = −∇p,
∂t
∇ · u = 0,
where p(x, t) is a scalar field.
(a) Define ω = ∇ × u and show that Euler’s equation can be re-written as
∂u
+ ω × u = −∇ p + 21 u2 .
∂t
You may assume the following vector identity:
(a · ∇) a = (∇ × a) × a + ∇
1 2
a
2
,
for any smooth vector field a.
(b) By operating on both sides of the equation in (a) with the curl, show
that Euler’s equation can further be re-written as
∂ω
+ (u · ∇) ω = (ω · ∇) u.
∂t
This is the so-called vortex-stretching equation.
You may assume the following vector identity:
∇ × (a × b) = (b · ∇) a − (a · ∇) b + a (∇ · b) − b (∇ · a)
for any two smooth vector fields a and b.
For (a), write (u · ∇)u as ω × u + ∇(u2 /2) as instructed, to obtain
∂u
+ ω × u = −∇ p + 12 u2 .
∂t
For (b), take the curl of this equation, using curl grad = 0 to obtain
∂ω
+ ∇ × (ω × u) = 0.
∂t
Using the vector identity given, namely
∇ × (a × b) = (b · ∇) a − (a · ∇) b + a (∇ · b) − b (∇ · a) ,
we have
∇ × (ω × u) = (u · ∇) ω − (ω · ∇) u + ω (∇ · u) − u (∇ · ω) .
Now, ∇ · u = 0 by assumption. Also, ∇ · ω = div curl u = 0, hence
∇ × (ω × u) = (u · ∇) ω − (ω · ∇) u,
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Vector Integral and Differential Calculus
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and Euler’s equation is thus
∂ω
+ (u · ∇) ω = (ω · ∇) u.
∂t
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