ANALYSIS IN MANY VARIABLES II (MATH 2031)
Solutions to Problem Sheet
Lecturer: P. Mansfield
Michaelmas 2006
1. For the function f (x, y) = cos(x + y) exp(x − y) calculate ∂f /∂x, ∂f /∂y, ∂ 2 f /∂x2 ,
∂ 2 f /∂y 2 , ∂ 2 f /∂x∂y, ∂ 2 f /∂y∂x. Use your results to show that ∂ 2 f /∂x2 = −∂ 2 f /∂y 2 and
∂ 2 f /∂x∂y = ∂ 2 f /∂y∂x.
Solution ∂f /∂x = − sin(x + y) exp(x − y) + cos(x + y) exp(x − y), ∂f /∂y = − sin(x +
y) exp(x − y) − cos(x + y) exp(x − y), ∂ 2 f /∂x2 = −2 sin(x + y) exp(x − y), ∂ 2 f /∂y 2 =
2 sin(x + y) exp(x − y), ∂ 2 f /∂x∂y = −2 cos(x + y) exp(x − y), ∂ 2 f /∂y∂x = −2 cos(x +
y) exp(x − y). By inspection ∂ 2 f /∂x2 = −∂ 2 f /∂y 2 and ∂ 2 f /∂x∂y = ∂ 2 f /∂y∂x.
2. Let F (t) be the value of the function f (x, y, z) = cos(xy) z restricted to the helix x =
cos(t), y = sin(t) , z = t which is parametrised by t and −∞ < t < ∞. Calculate dF/dt
as a function of t (i) directly by substituting the equations of the helix into f (x, y, z) to
calculate F (t) as a function of t and then differentiating, and (ii) using the chain-rule.
Note how similar these approaches are.
Solution (i) Substitution gives F = cos(cos(t) sin(t)) t. Differentiating: dF/dt =
sin(cos(t) sin(t)) sin2 (t) t − sin(cos(t) sin(t)) cos2 (t) t + cos(cos(t) sin(t)). (ii) Using the
chain-rule: dF/dt = (∂f /∂x)(dx/dt)+(∂f /∂y)(dy/dt)+(∂f /∂z)(dz/dt) = sin(xy) yz sin(t)
− sin(xy) xz cos(t) + cos(xy) = sin(cos(t) sin(t)) sin2 (t) t − sin(cos(t) sin(t)) cos2 (t) t +
cos(cos(t) sin(t)).
√
√
3. If u = x2 + y 2 + x and v = x2 + y 2 − x find ∂u/∂x, ∂u/∂y, ∂v/∂x and ∂v/∂y. Solve
for x and y in terms of u and v and then find ∂x/∂u, ∂x/∂v, ∂y/∂u and ∂y/∂v. Is ∂x/∂u
equal to 1/(∂u/∂x) and if not, why not?
√
√
√ 2
2 ) + 1, ∂u/∂y = y/ x2 + y 2 , ∂v/∂x = (x/ x2 + y 2 ) − 1
x
+
y
Solution
∂u/∂x
=
(x/
√
√
=√−v/ x2 + y 2, ∂v/∂y = y/ x2 + y 2. To solve for y, x as functions of u, v note u + v =
2 x2 + y 2 so u = (u+v)/2+x implying that√x = (u−v)/2 and that (u+v)2/4 = x2 +y 2 =
(u − v)2 /4q+ y 2 so y 2 = uv giving
y = ± uv. Hence ∂x/∂u = 1/2, ∂x/∂v = −1/2,
q
∂y/∂u = v/u/2, ∂y/∂v = u/v/2. So in general ∂x/∂u is not equal to 1/(∂u/∂x),
which is because different variables are held fixed in each differential coefficient, i.e. ∂x/∂u
is the derivative of x wrt u keeping v fixed, whilst ∂u/∂x is the derivative of u wrt x keeping
y fixed.
4. The variables u, v are related to x, y by u = c x + s y and v = −s x + c y where the
constants s and c satisfy s2 + c2 = 1. What is this transformation geometrically? If
f (x, y) = F (u, v) show that ∂ 2 f /∂x2 + ∂ 2 f /∂y 2 = ∂ 2 F/∂u2 + ∂ 2 F/∂v 2 .
Solution The transformation between x, y and u, v, represents a rotation of rectangular coordinates through an angle θ where c = cos θ and s = sin θ. By the chain-rule
∂/∂x = c ∂/∂u − s ∂/∂v and ∂/∂y = s ∂/∂u + c ∂/∂v so
∂2f
∂
∂
∂2f
+
= c
−s
2
2
∂x
∂y
∂u
∂v
!2
∂
∂
+c
F+ s
∂u
∂v
2
∂2
∂2
∂2
2 ∂
−
cs
−
cs
+
s
= c
∂u2
∂u∂v
∂v∂u
∂v 2
2
!
!2
F
F
2
∂2
∂2
∂2
∂2F
∂2F
2 ∂
+ s
+
cs
+
cs
+
c
F
=
+
.
∂u2
∂u∂v
∂v∂u
∂v 2
∂u2
∂v 2
√
√
5. The variables ξ, τ are related to x, t by ξ = (x − v t)/ 1 − v 2 and τ = (t − v x)/ 1 − v 2
where the constant v has absolute value less than one. If f (x, t) = F (ξ, τ ) show that
∂ 2 f /∂x2 − ∂ 2 f /∂t2 = ∂ 2 F/∂ξ 2 − ∂ 2 F/∂τ 2 . The deep meaning of this will be explored in
the Mathematical Physics option.
!
2
Solution
Using the chain-rule
∂
∂
1
v
∂
√
=√
−
∂x
1 − v 2 ∂ξ
1 − v 2 ∂τ
so that
∂
∂
∂
v
1
√
= −√
+
∂t
1 − v 2 ∂ξ
1 − v 2 ∂τ
∂2f
∂2f
−
∂x2
∂t2


!2
!2

∂
∂
∂
1  ∂
−
v
+
F
−
−v
F
=

1 − v 2  ∂ξ
∂τ
∂ξ
∂τ
2
∂2F
∂2F
∂2F
2 ∂ F
−
v
−
v
+
v
∂ξ 2
∂ξ∂τ
∂τ ∂ξ
∂τ 2
)
2
∂2F
∂2F
∂2F
1
2∂ F
v
−
v
−
v
+
−
1 − v2
∂ξ 2
∂ξ∂τ
∂τ ∂ξ
∂τ 2
)
1
=
1 − v2
(
(
2
1
∂2F
2 ∂ F
2
=
(1
−
v
)
+
(v
−
1)
1 − v2
∂ξ 2
∂τ 2
(
=
)
∂2F
∂2F
−
.
∂ξ 2
∂τ 2
6. The variables u, v are related to x, t by u = x + t and v = x − t. If f (x, y) = F (u, v)
show that ∂ 2 f /∂x2 −∂ 2 f /∂t2 = k∂ 2 f /∂u∂v, provided that the second derivatives of F are
continuous, where k is some constant whose value you should find. Use this to show that
(i) f (x, t) = (x + t)2 + exp((x − t)3 ) satisfies ∂ 2 f /∂x2 − ∂ 2 f /∂t2 = 0, (ii) more generally
any f (x, t) of the form of a sum of a function of the single variable x + t and a function of
the single variable x−t, i.e. f (x, t) = H(x+t)+G(x−t), satisfies ∂ 2 f /∂x2 −∂ 2 f /∂t2 = 0.
Solution
Using the chain-rule
∂
∂x
∂2f
∂2f
−
=
∂x2
∂t2
=
=
∂
∂u
+
∂
∂v
∂
∂
+
∂u ∂v
and
!2
∂
∂x
=
∂
∂u
−
∂
∂v
∂
∂
−
F−
∂u ∂v
so
!2
F
∂2
∂2
∂2
∂2
∂2
∂2
∂2F
∂2
∂2
+
+
+
F
−
−
+
−
F
=
4
.
∂u2 ∂v∂u ∂u∂v ∂v 2
∂u2 ∂v∂u ∂u∂v ∂v 2
∂u∂v
!
!
So k = 4. (i) For f (x, t) = (x + t)2 + exp((x − t)3 ) the transformed function is F = u2 +
exp(v 3 ) so ∂ 2 f /∂x2 −∂ 2 f /∂t2 = 4{∂ 2 /∂u∂v}(u2+exp(v 3 )) = 8∂u/∂v+4{∂ / ∂u}(3v 2 exp(v 3 )) =
0. (ii) More generally ∂ 2 H(u)/∂u∂v = 0 because ∂H(u)/∂v = 0 and ∂ 2 G(v)/∂u∂v =
∂ 2 G(v)/∂v∂u = 0 because ∂G(v)/∂u = 0 so ∂ 2 f /∂x2 − ∂ 2 f /∂t2 = 4{∂ 2 /∂u∂v}(H(u) +
G(v))= 4∂ 2 H(u)/∂u∂v + 4∂ 2 G(v)/∂u∂v = 0
7. If a = sin2t e1 +et e2 −(t3 −5t) e3 , find (a) da/dt, (b)||da/dt||, (c) d2 a/dt2 , (d)||d2a/dt2 ||
at t = 0.
t
2
Solution (a) da/dt = 2 cos(2t)e√
5)e3 . At t = 0 this is 2e1 + e2 + 5e3 .
1 + e e2 − (3t − √
(b) which has length ||da/dt|| = 22 + 12 + 52 = 30, (c) d2 a/dt2 = −4 sin(2t)e1 +
et e2 − 6te3 , at t = 0 this is e2 , (d) so ||d2 a/dt2 || = 1 at t = 0.
8. Find a unit vector tangent to the space curve x = t3 , y = t, z = t2 at t = 2.
Solution Setting x(t) = t3 e1 + te2 + t2 e3 then dx(t)/dt = 3t2 e1 √
+ e2 + 2te3 is parallel
to the tangent. At t = 2 this is 12e1 +
e
+
4e
which
has
length
161 so a unit vector
3
√ 2
in this direction is (12e1 + e2 + 4e3 )/ 161
9.
10.
11. Compute the gradient, ∇f , for the following functions:
√
(a)
f (x, y, z) = x2 + y 2 + z 2 ,
(b)
√
√ 2
2 + z 2 , ∂f /∂y = y/ x2 + y 2 + z 2 , and ∂f /∂z =
x
+
y
Solution
∂f
/∂x
=
x/
√
√
z/ x2 + y 2 + z 2 , so ∇f = (xe1 + ye2 + ze3 )/ x2 + y 2 + z 2 = x /||x ||.
f (x, y, z) = xy + yz + xz,
Solution ∂f /∂x = y + z, ∂f /∂y = x + z, ∂f /∂z = y + x, so ∇f = (y + z)e1 +
(x + z)e2 + (x + y)e3 .
(c)
f (x, y, z) = 1/(x2 + y 2 + z 2 ).
Solution ∂f /∂x = −2x/(x2 +y 2 +z 2 )2 , ∂f /∂y = −2y/(x2 +y 2 +z 2 )2 , and ∂f /∂z =
−2z/(x2 + y 2 + z 2 )2 so ∇f = −2(xe1 + ye2 + ze3 )/(x2 + y 2 + z 2 )2 = −2x /||x ||4.
12. Compute the divergence, ∇ · A, of the following vector fields:
(a)
A(x, y, z) = yze1 + xze2 + xye3 ,
Solution
∇·A=
(b)
∂(yz) ∂(xz) ∂(xy)
+
+
= 0 + 0 + 0 = 0,
∂x
∂y
∂z
A(x, y, z) = (x2 + y 2 + z 2 ) (3e1 + 4e2 + 5e3 ),
Solution
∇·A =
∂(3(x2 + y 2 + z 2 )) ∂(4(x2 + y 2 + z 2 )) ∂(5(x2 + y 2 + z 2 ))
+
+
= 6x+8y+10z ,
∂x
∂y
∂z
Alternatively write A = ||x ||2B with B = 3e1 + 4e2 + 5e3 a constant vector and
use ∇ · (||x ||2B) = (∇(||x ||2)) · B + ||x ||2∇ · B with
∇(||x ||2)) = e1
∂(x2 + y 2 + z 2 )
∂(x2 + y 2 + z 2 )
∂(x2 + y 2 + z 2 )
+ e2
+ e3
∂x
∂y
∂z
= e1 2x + e2 2y + e3 2z = 2x
(c)
and ∇ · B = 0.
A(x, y, z) = (x + y)e1 + (y + z)e2 + (z + x)e3 .
Solution
∇·A=
∂(x + y) ∂(y + z) ∂(z + x)
+
+
= 1 + 1 + 1 = 3,
∂x
∂y
∂z
13. Compute the curl, ∇ × A, of each of the vector fields, A, in the previous question.
(a)
A(x, y, z) = yze1 + xze2 + xye3 ,
Solution
∇ × A = e1
!
!
∂(yz) ∂(xy)
∂(xz) ∂(yz)
∂(xy) ∂(xz)
+ e2
+ e3
−
−
−
∂y
∂z
∂z
∂x
∂x
∂y
= e1 (x − x) + e2 (y − y) + e3 (z − z) = 0 ,
!
A(x, y, z) = (x2 + y 2 + z 2 ) (3e1 + 4e2 + 5e3 ),
(b)
Solution
∇ × A = e1
∂(5(x2 + y 2 + z 2 )) ∂(4(x2 + y 2 + z 2 ))
−
∂y
∂z
+e2
∂(3(x2 + y 2 + z 2 )) ∂(5(x2 + y 2 + z 2 ))
−
∂z
∂x
!
+e3
∂(4(x2 + y 2 + z 2 )) ∂(3(x2 + y 2 + z 2 ))
−
∂x
∂y
!
!
= e1 (10y − 8z) + e2 (6z − 10x) + e3 (8x − 6y) ,
(c)
Alternatively write A = ||x ||2B with B = 3e1 + 4e2 + 5e3 a constant vector and use
∇ × (||x ||2B) = (∇(||x ||2)) × B + ||x ||2∇ × B with ∇(||x ||2)) = 2x and ∇ × B = 0.
A(x, y, z) = (x + y)e1 + (y + z)e2 + (z + x)e3 .
Solution
e1
∇×A=
!
!
∂(x + y) ∂(z + x)
∂(y + z) ∂(x + y)
∂(z + x) ∂(y + z)
+ e2
+ e3
−
−
−
∂y
∂z
∂z
∂x
∂x
∂y
= e1 (0 − 1) + e2 (0 − 1) + e3 (0 − 1) = −(e1 + e2 + e3 ) ,
14. If f (r) is a differentiable function of r = ||x ||, show that
grad f (r) = f ′ (r) x /r,
(a)
Solution
grad f (r) = e1
= e1
Now r =
√
∂f (r)
∂f (r)
∂f (r)
+ e2
+ e3
∂x
∂y
∂z
df ∂r
df ∂r
df ∂r
+ e2
+ e3
dr ∂x
dr ∂y
dr ∂z
x2 + y 2 + z 2 so
1 2x
x
∂r
=
=
∂x
2 r
r
and similarly
∂r
y
= ,
∂y
r
so
∂r
z
=
∂z
r
x
df
y
z
e1 + e2 + e3
grad f (r) =
dr
r
r
r
= f ′ (r)x /r
!
(b)
curl [f (r)x ] = 0.
Solution
Using curl (f V) = (∇f ) × V + f curl V, the result to part (a) and
∂z ∂y
−
∂y ∂z
curl x = e1
!
+ e2
!
∂x ∂z
∂y ∂x
+ e3
−
−
∂z
∂x
∂x ∂y
!
= 0,
gives curl (f x ) = (∇f ) × x + f curl x = f ′ (r) xr × x + 0 = 0 since x × x = 0.
15. Let x be the position vector in 3-dimensions, with r = ||x ||, and let a be a constant
vector. Show that
(a)
div x = 3,
Solution
∇·x =
(b)
∂x ∂y ∂z
+
+
= 1+ 1 + 1 = 3,
∂x ∂y ∂z
curl x = 0,
Solution
curl x = e1
(c)
∂z ∂y
−
∂y ∂z
!
+ e2
!
∂x ∂z
∂y ∂x
+ e3
−
−
∂z
∂x
∂x ∂y
!
= 0,
grad r = x /r,
Solution
1 2x
x
∂r
=
= ,
∂x
2 r
r
∂r
y
= ,
∂y
r
∂r
z
=
∂z
r
so
grad r = e1 x/r + e2 y/r + e3 z/r = x /r
(d)
div (r n x ) = (n + 3) r n ,
Solution Use div (f V) = (∇f ) · V + f div V with f (x ) = r n so that using 8 (a)
∇f = nr n−1 x /r = nr n−2 x and V = x so
div (r n x ) = nr n−2 x · x + r n div x = nr n + 3r n = (n + 3)r n ,
(e)
grad (a · x ) = a,
Solution
a · x = a1 x + a2 y + a3 z so
∂(a · x )
= a1 ,
∂x
∂(a · x )
= a2 ,
∂y
hence grad (a · x ) = a1 e1 + a2 e2 + a3 e3 = a
∂(a · x )
= a3
∂z
(f)
div (a × x ) = 0,
Solution
a × x = e1 (a2 z − a3 y) + e2 (a3 x − a1 z) + e3 (a1 y − a2 x) so
div (a × x ) =
(g)
∂(a2 z − a3 y) ∂(a3 x − a1 z) ∂(a1 y − a2 x)
+
+
= 0,
∂x
∂y
∂z
curl (a × x ) = 2 a,
Solution
curl (a × x ) = e1
+e2
∂(a1 y − a2 x) ∂(a3 x − a1 z)
−
∂y
∂z
!
!
∂(a2 z − a3 y) ∂(a1 y − a2 x)
∂(a3 x − a1 z) ∂(a2 z − a3 y)
+ e3
−
−
∂z
∂x
∂x
∂y
!
= e1 (a1 + a1 ) + e2 (a2 + a2 ) + e3 (a3 + a3 ) = 2a ,
curl (r 2 a) = 2 (x × a),
(h)
(i)
(j)
(k)
Solution Use curl (f V) = (grad f ) × V + f curl V with f = r 2 = ||x ||2 and
∇(||x ||2) = 2x so that curl (r 2 a) = 2x × a + r 2 curl a = 2x × a,
∇2 (1/r) = 0 ,
if r 6= 0,
Solution
∇2 (1/r) ≡ ∇ · ∇(1/r). From 8 (a) with f = 1/r use ∇(1/r) =
3
−x /r . From 9 (d) ∇ · (x /r 3 ) = (−3 + 3)/r 3 = 0 so
∇2 (1/r) = 0, but this is
only valid for r 6= 0 since the calculation involves division by r.
∇2 (log r) = 1/r 2 ,
if r 6= 0,
Solution
∇2 (log r) ≡ ∇ · ∇(log r). From 8 (a) with f = log r ∇(log r) =
2
x /r . From 9 (d) ∇ · (x /r 2 ) = (−2 + 3)/r 3 = 1/r 2 so
∇2 (log r) = 1/r 2 , but as
in part (i) this is only valid for r 6= 0 since the calculation involves division by r.
div [(a · x )x ] = 4 a · x ,
Solution Using div f V = (∇f ) · V + f div V, with f = a · x so ∇f = a from part
(e) and V = x so div x = 3 from part (a) hence div [(a·x )x ] = a·x +a·x 3 = 4a·x ,
(l)
div [x × (x × a)] = 2 a · x ,
Solution Use x × (x × a) = x (x · a) − a(x · x ) and then from part (k) use
div (x (x · a)) = 4x · a and
∇ · (||x ||2a) = (∇(||x ||2)) · a + ||x ||2∇ · a = 2x · a
so
div [x × (x × a)] = div (x (x · a)) − ∇ · (||x ||2a) = 4x · a − 2x · a
curl (a × x /r 3 ) = 3 (a · x )x /r 5 − a/r 3 ,
(m)
Solution
curl a × x /r 3 = (curl (a × x )) /r 3 + (∇r −3 ) × (a × x )
Using part (g) curl (a × x ) = 2a together with x × (a × x ) = a(x · x ) − x (a · x )
and ∇r −3 = −3x /r 5 (from 8(a)) gives
curl a × x /r 3 =
(n)
3
2a
−
(ar 2 − x (a · x )) = 3 (a · x )x /r 5 − a/r 3
r3
r5
Exam question June 2002 (Section A): calculate the curl of (a · x ) x .
Solution
curl ((a · x ) x ) = (∇(a · x )) × x + a · x curl x = a × x
using part parts (e) and (b).
16. If x is the position vector, x = xe1 + ye2 + ze3 , a is a constant vector, F = (a · x ) x
and G = r 2 a, show that
(a)
div F = 2 div G = 4 a · x ,
Solution
Using the product rule
div F = ∇ · ((a · x ) x ) = (∇(a · x )) · x + (a · x ) ∇ · x
As in 9(e): a · x = a1 x + a2 y + a3 z so
∂(a · x )
= a1 ,
∂x
∂(a · x )
= a2 ,
∂y
∂(a · x )
= a3
∂z
hence grad (a · x ) = a1 e1 + a2 e2 + a3 e3 = a and as in 9(a)
∇·x =
∂x ∂y ∂z
+
+
= 1+ 1 + 1 = 3,
∂x ∂y ∂z
so
div F = ∇ · ((a · x ) x ) = a · x + 3a · x = 4a · x .
Using the product rule
div G = (∇(r 2 )) · a + r 2 ∇ · a
and ∇(||x ||2) = 2x , ∇ · a = 0 so that div G = 2 a · x ,
(b)
curl G = −2 curl F = 2 x × a,
Solution
Use curl (f V) = (grad f ) × V + f curl V with f = r 2 = ||x ||2 and ∇(||x ||2) = 2x so
that
curl G = curl (r 2 a) = 2x × a + r 2 curl a = 2x × a .
Now use this product rule with f = a · x and V = x so that
∇ × ((a · x ) x ) = (∇(a · x )) × x + (a · x )∇ × x
As in 9(b)
curl x = e1
∂z ∂y
−
∂y ∂z
!
!
∂x ∂z
∂y ∂x
+ e3
−
−
∂z
∂x
∂x ∂y
+ e2
!
= 0,
so
∇ × ((a · x ) x ) = a × x = −x × a
(c)
div curl F = div curl G = 0,
Solution div curl F = div (a × x ) = 0 as in 9(f):
a × x = e1 (a2 z − a3 y) + e2 (a3 x − a1 z) + e3 (a1 y − a2 x) so
div (a × x ) =
(d)
∂(a2 z − a3 y) ∂(a3 x − a1 z) ∂(a1 y − a2 x)
+
+
= 0,
∂x
∂y
∂z
Since curl G = −2 curl F this implies that div curl G = 0,
curl curl G = −2 curl curl F = −4 a.
Solution
curl curl G == 2curl (x × a) = −4a
using a × x = −x × a and 9(g)
curl (a × x ) = e1
+e2
∂(a1 y − a2 x) ∂(a3 x − a1 z)
−
∂y
∂z
!
!
∂(a2 z − a3 y) ∂(a1 y − a2 x)
∂(a3 x − a1 z) ∂(a2 z − a3 y)
+ e3
−
−
∂z
∂x
∂x
∂y
= e1 (a1 + a1 ) + e2 (a2 + a2 ) + e3 (a3 + a3 ) = 2a ,
Since curl G = −2 curl F this implies that
curl curl G = −2 curl curl F = −4 a .
!
17. What is the divergence of the vector function A(x ) = r x + ∇ r where x is the position
vector in 3 dimensions and r = ||x ||? What is the corresponding result in n dimensions?
Solution ∇ · A = ∇ · (r x + ∇ r) = (∇r) · x + r∇ · x + ∇ · ∇r. Use ∇r = x /r so that
∇ · ∇r = (∇ · x )/r + x · ∇(1/r) = 3/r + x · (−r −2 ∇r) = 3/r − x · x /r 3 = 2/r. Hence
∇ · A = x · x /r + 3r + 2/r = 4r + 2/r. In n dimensions ∇ · x = n which changes the
result to (n + 1)r + (n − 1)/r.
18. Exam question June 2001 (Section A):
(a) Give a representation of the vector function A(x, y) = ye1 as a collection of arrows in the region of the (x, y)-plane bounded by (x1 , y1) = (−2, 2), (x2 , y2 ) =
(2, 2), (x3 , y3) = (2, −2), (x4 , y4) = (−2, −2).
Solution
(b) Calculate the curl of the vector field A(x, y) = (−ye1 + xe2 )/(x2 + y 2) defined everywhere in the (x, y)-plane except at the origin.
Solution
∇×A = e1
!
!
∂
∂
∂0
∂
∂ −y
x
x
y
∂0
+e3
+e2
−
−
+
2
2
2
2
2
2
2
∂y ∂y x + y
∂z x + y
∂x
∂x x + y
∂y x + y 2
= e3
2x2
1
2y 2
1
−
+
−
x2 + y 2 (x2 + y 2)2 x2 + y 2 (x2 + y 2 )2
!
= 0,
(c) Give the unit vector normal to the surface of equation ax + by = cz, where a, b, c,
are three real constants.
Solution ax + by = cz is the equation of a plane, the general form of which would
be x · V = k where V is orthogonal
−c), so a unit
√ to the plane.
√ Here V = (a, b,√
2
2
2
2
2
2
vector normal to the surface is (a/ a + b + c , b/ a + b + c , −c/ a2 + b2 + c2 ).
Alternatively: in general ∇f is normal to the level-surface f = k, here f = ax+by−cz
so ∇f√ = (a, b, −c) and
normal to the surface is ∇f /||∇f ||
√ again a unit vector
√
2
2
2
2
2
2
2
= (a/ a + b + c , b/ a + b + c , −c/ a + b2 + c2 ).
!
(d) Let x be the position vector in 3-dimensions and a be a constant vector. Show that
div [ x × (x × a)] = 2a · x .
Solution Use x ×(x ×a) = x (x ·a)−a(x ·x ) so div [ x ×(x ×a)] = div (x (x ·a)−a(x ·
x )) = (∇·x )(x ·a)+x ·∇(x ·a)−a·∇(x ·x ). This is 3(x ·a)+(x ·a)−2(x ·a) = 2(x ·a)
19. Exam question June 2002 (Section A): Give the unit vector normal to the surface of
equation x2 /a2 + y 2 /b2 + z 2 /c2 = 4 where a, b, c are three real constants.
Solution From Core A: ∇f (x ) is orthogonal to the level surface f = const. at the
point x , so take f = x2 /a2 + y 2/b2 + z 2 /c2 , then ∇f (x ) = e1 2x/a2 + e2 2y/b2 + e3 2z/c2
is normal to the surface at x . A unit vector normal
to the surface is therefore n ≡
q
∇f (x )/||∇f (x )|| = (e1 x/a2 + e2 y/b2 + e3 z/c2 )/ x2 /a4 + y 2/b4 + z 2 /c4
√
√
What is the unit vector normal to a sphere of radius 2 at the point (x, y, z) = ( 2, 0, 2)?
Solution When a = b = c = 1 the ellipsoid in the first√part √
of the question becomes
√ a
2,
0,
sphere
of
radius
2,
so
substituting
this
and
(x,
y,
z)
=
(
2)
into
n
gives
(e
√
√
√ 1 2+
e3 2)/2, which is a unit vector along the radial direction at (x, y, z) = ( 2, 0, 2), as it
should be.
20. Exam question June 2002 (Section A): What is the derivative of the scalar function
φ(x, y, z) = xcosz − y in the direction V = e1 + e2 + e3 ? What is the gradient at the
point (x, y, z) = (0, 1, π/2)? In which direction is φ increasing the most when moving
away from this point?
21. For which values of (x, y) are the following continuous:
(a)
x/(x2 + y 2 + 1),
Solution Use the theorem that if f and g are continuous at a point then so are
f +g, f g and f /g if g does not vanish there. Taking f = x and g = y all the functions
can be built up by repeated applications of this result, so all the functions will be
continuous whenever their denominators are non-zero so this function is continuous
everywhere since x2 + y 2 + 1 > 0,
(b)
x/(x2 + y 2 ),
Solution
(c)
(x + y)/(x − y),
Solution
(d)
everywhere except the origin,
everywhere except the line y = x,
x3 /(y − x2 )?
Solution
everywhere except the parabola y = x2 .
22. Which of the following sets are open:
(a)
{(x, y, z) : x > 0},
Solution An open set has to contain a neighbourhood of each of its points so this
is open because for any point we can construct a sphere centred on that point and
lying entirely within the set, even points arbitrarily close to x = 0,
(b)
{(x, y, z) : y ≥ 0},
Solution this is not open because it contains points for which y = 0, and any
sphere with such a point as its centre contains points outside the set,
(c)
{(x, y, z) : 1 > (x2 + y 2)/z},
Solution 1 > (x2 + y 2)/z implies z > x2 + y 2 or z < 0 this is open because for
any point we can construct a sphere centred on that point and lying entirely within
the set, even points arbitrarily close to the paraboloid z = x2 + y 2 or to z = 0
(d)
{(x, y, z) : 1 ≥ (x2 + y 2 )/z}?
Solution this is not open because it contains points for which z = x2 + y 2 , and
any sphere with such a point as its centre contains points outside the set.
23. Determine the points of R2 at which the function f (x, y) = |x2 − y 2 | is
(a) differentiable; (b) continuously differentiable.
Solution f (x, y) = |x2 − y 2 | is x2 − y 2 for x2 − y 2 = (x − y)(x + y) > 0 and y 2 − x2
for x2 − y 2 = (x − y)(x + y) < 0. For the first case (x − y)(x + y) > 0 for x − y > 0 and
x + y > 0 or x − y < 0 and x + y < 0, call this part of R2 region 1. For the second case
(x − y)(x + y) < 0 for x − y > 0 and x + y < 0 or x − y < 0 and x + y > 0, call this part
of R2 region 2. Within both regions the function is a polynomial in x and y and so has
continuous partial derivatives, hence the function is continuously differentiable there. In
neighbourhoods of points on the line y = ±x the function is no longer a polynomial and
we have to be more careful so use the definition of the partial derivative: for y = x = a
∂f
|(a + h)2 − a2 |
|2ah + h2 |
= lim
= lim
h→0
∂x h→0
h
h
If a 6= 0 we can neglect the h2 term in comparison to the ah piece so we get
lim
h→0
which does not exist because
the limit is
|h|
h
|2ah|
|2a| |h|
= lim
h→0
h
h
is ±1 depending on the sign of h. However if a = 0 then
|h2 |
= 0.
h→0 h
lim
also does not exist on this line except at the origin, and
Similar arguments show that ∂f
∂y
also that neither partial derivative exists on the line y = −x, except at the origin. Since
the p.d.s do not exist on the lines y = ±x away from 0 the function cannot be differentiable
there. It might be differentiable at 0 but cannot be continuously differentiable there,
because the p.d.s are not continuous at 0 (since they do not exist on the lines). Now
investigate the differentiablity of f at the origin. Consider
f (h) − f (0) − h · ∇f
|h2 − h22 |
R
=
= 1
||h||
||h||
||h||
Now
h2
h2
|h21 − h22 |
≤ 1 + 2 ≤ ||h|| + ||h||
||h||
||h|| ||h||
we have that |R/||h||| < ǫ whenever ||h|| < δ by taking δ = ǫ/2 so limh→0 R/||h|| = 0
and f is differentiable at the origin.
24. Define f : R2 → R by f (0) = 0 whilst for x 6= 0:
f (x ) =
x3
.
x2 + y 2
Calculate the partial derivatives of f with respect to x and y at x = 0 using their
definitions as limits.
Solution
and
∂f
f (he1 ) − f (0)
h−0
= lim
= lim
=1
h→0
h→0
∂x
h
h
f (he2 ) − f (0)
0−0
∂f
= lim
= lim
=0
h→0
h→0
∂y
h
h
Define R at the origin by R = f (h) − f (0) − h · ∇f as usual and show that the limit as
h tends to 0 of R/||h|| is not zero, so that f is not differentiable at the origin.
R
f (h) − f (0) − h · ∇f
h3
h1
−h1 h22
=
= 2 1 2 3/2 − 0 − 2
=
||h||
||h||
(h1 + h2 )
(h1 + h22 )1/2
(h21 + h22 )3/2
now consider how this behaves as the origin is approached along the line h2 = mh1 then
R
−m2 h31
−m2
=
=
||h||
(1 + m2 )3/2 h31
(1 + m2 )3/2
which remains constant as the origin is approached. Since this is not zero f cannot be
differentiable at the origin.
On the line through the origin, x = bt, (with b a constant vector), f becomes a function
of the single variable t, f (bt). Write b = e1 b1 + e2 b2 and use this to write f (bt) explicitly
as a function of t. Show that this function is differentiable at the origin, i.e. df /dt exists
at t = 0 despite f (x ) not being differentiable at 0.
Solution
f (bt) =
b31
(b1 t)3
=
t
(b1 t)2 + (b2 t)2
b21 + b22
which is t multiplied by a constant (this also holds at 0 since f is defined to vanish there),
and so is differentiable at the origin.
25. Exam question June 2001 (Section B): You are given the following family of scalar functions labelled by a real parameter λ : Φλ (x, y, z) = (y − λ)cosx + zxy.
(a) What are their derivatives in the direction V = e1 + 2(e2 + e3 )?
(b) Which member of the family has its gradient at the point ( π2 , 1, 1) equal to π2 (e1 +
e2 + e3 )?
(c) Calling this particular member of the family Φλ0 , in which direction is Φλ0 decreasing
most rapidly when starting at the point ( π2 , 1, 1)?
26. Exam question June 2003 (Section A): Find the directional derivative of the function
φ(x, y, z) = xy 2 z 3 at the point P = (1, 1, 1) in the direction from P towards Q = (3, 1, −1).
Starting from P, in which direction is the directional derivative maximum and what is
the value of this maximum?
Solution The directional derivative of φ at P in the direction from P towards Q =
(3, 1, −1) is n·∇φ(P) where n is a unit vector in this direction, i.e. n = (Q−P)/||Q−P||.
2 3
3
2 2
Now ∇φ
√ = e1 y z + e√2 2xyz + e3 3xy z , so ∇φ(P) = e1 + e2 2 + e3 3, and n = (e1 2 −
2 so the required directional derivative is (e1 + e2 2 + e3 3) · (e1 −
e3 2)/√ 8 = (e1 − e3 )/ √
e3 )/ 2 which equals − 2. The directional derivative is a maximum in the
√ direction of
e
√1 + e2 2 + e3 3, i.e. parallel to e1 + e2 2 + e3 3, and its value then is ||∇φ|| = 1 + 4 + 9 =
14.
27. In which direction does the function f (x, y) = x2 − y 2 increase fastest at the point (a)
(1, 0), (b) (−1, 0), (c) (2, 1) ? Illustrate with a sketch.
28. Let f (x, y) = (x2 − y 2)/(x2 + y 2 ).
(a) In which direction is the directional derivative of f at (1, 1) equal to zero?
(b) What about at an arbitrary point (x0 , y0 ) in the first quadrant?
(c) Describe the level curves of f and discuss them in the light of the result in (b).
29. A marble is released from the point (1, 1, c − a − b) on the elliptic paraboloid defined by
z = c − ax2 − by 2 , where a, b, c are positive real numbers and the z-coordinate is vertical.
In which direction does the marble begin to roll?
30. Find the vector equations of tangent and normal lines in R2 to the following curves at
the given points
(a)
x2 + 2y 2 = 3 at (1, 1),
(a)
xy = 1 at (2, 1/2),
(a)
x2 − y 2 = 3 at (2, 1).
31. Find the equations of the tangent planes R3 to the following surfaces at the given points
x2 + xy 2 + y 3 + z + 1 = 0 at (2, −3, 4),
(a)
(a)
x2 y 2 + xz − 2y 3 = 10 at (2, 1, 4),
(a)
sin(xy) + sin(yz) + sin(zx) = 1 at (1, π/2, 0).
32. Assuming that F (0, 0) = 0, find conditions on the function F which allow the equation
F (F (x, y), y) = 0 to be solved for y as a function of x near the origin. (Hint consider the
function H(x, y) = F (F (x, y), y).)
Solution Putting H(x, y) = F (F (x, y), y) allows us to write F (F (x, y), y) = 0 as
H(x, y) = 0. This level curve of H passes through the origin because F (0, 0) = 0. The
6= 0
implicit function theorem implies that if H is differentiable at a point at which ∂H
∂y
then we can find a differentiable g(x) so that the curve can be re-written as y = g(x) in
a neighbourhood of that point. By the chain-rule
∂H
∂F (x, y) ∂F (X, y)
∂F (X, y)
=
|X=F (x,y) +
|X=F (x,y)
∂y
∂y
∂X
∂y
so we need
∂F (X, y)
∂F (x, y) ∂F (X, y)
|X=F (x,y) +
|X=F (x,y) 6= 0 .
∂y
∂X
∂y
33. If y = 1 + xy 5 show that y may be written in the form y = f (x) in a neighbourhood of
(0, 1) and find the gradient of the graph of f at the point (0, 1).
Solution y = 1+xy 5 is a level curve of H = y −1−xy 5 with H = 0 and passes through
(0, 1). H is polynomial in x and y and so has continuous derivatives and is therefore
differentiable. The implicit function theorem then implies that y may be written in the
6= 0 there. Now ∂H
= 1 − 5xy 4 = 1 at
form y = f (x) in a neighbourhood of (0, 1) if ∂H
∂y
∂y
(0, 1) so the condition is satisfied. The implicit function theorem also gives the gradient
of the function as
∂H ∂H
g ′ (0) = −
/
,
∂x ∂y
now
∂H
∂x
= −y 5 = −1 at (0, 1) so g ′(0) = 1.
34. Show that the equation xy 3 − y 2 + 1 = 0 can be written in the form y = f (x) in a
neighbourhood of the point (0, 1), and in the form y = g(x) in a neighbourhood of the
point (0, −1). Is it true that f (x) and g(x) are of the same form?
Solution xy 3 − y 2 + 1 = 0 is a level curve of H = xy 3 − y 2 + 1 with H = 0 and passes
through (0, ±1). H is polynomial in x and y and so has continuous derivatives and is
= 3xy 2 − 2y, so at (0, ±1), ∂H
6= 0. The implicit function
therefore differentiable. ∂H
∂y
∂y
theorem then implies that y may be written in the form y = f (x) in a neighbourhood of
the point (0, 1), and in the form y = g(x) in a neighbourhood of the point (0, −1). Note
that at x = 0 we must have f (0) = 1 but g(0) = −1 so f (x) and g(x) cannot be of the
same form. In fact solving xy 3 − y 2 + 1 = 0 for y gives three solutions, since it is a cubic
equation and one passes through (0, 1) another through (0, −1).
35. Determine whether or not the equation x2 + y + sin(xy) = 0 can be written in the form
y = f (x) or in the form x = g(y) in some small open disc about the origin for some
suitable continuously differentiable functions f, g.
Solution x2 + y + sin(xy) = 0 is a level curve of H = x2 + y + sin(xy) with H = 0 and
passes through the origin. H has continuous partial derivatives and so is differentiable.
∂H
= 1 + x cos(xy) is 1 at the origin and ∂H
= 2x + y cos(xy) vanishes at the origin, so by
∂y
∂x
the implicit function theorem y may be written in the form y = f (x) in a neighbourhood
of the origin with differentiable f , but x cannot be written in the form x = g(y).
36. The vector a has components (ar ) = (1, 1, 1) and the vector b has components (br ) =
(2, 3, 4). In the following expressions state which indices are free and which are dummy,
and give the numerical values of the expressions for each value that the free variable takes
(e.g. for ar − br the free variable is r and it takes the values 1, 2, 3 so a1 − b1 = −1,
a2 − b2 = −2, a3 − b3 = −3,)
(a) ar + br ,
(b) ar br ,
(c) ar bs ar ,
(d) ar bs ar bs − ar br as bs .
Solution (a) r is a free index so we need to specify 3 pieces of information. ar + br
is the r-th component of the vector a + b so we can give the information in the form
(ar + br ) = (3, 4, 5) or as a1 + b1 = 3, a2 + b2 = 4, a3 + b3 = 5.
(b) r is a dummy-index and ar br = a1 b1 + a2 b2 + a3 b3 = 9, which is the scalar product
of a and b.
(c) r is a dummy variable, and s is free so to specify ar bs ar we need to specify 3 pieces of
information, these are the components of the vector b a · a. Now ar ar = a21 + a22 + a23 = 3
so (ar bs ar ) = (3bs ) = (6, 9, 12)
(d) r and s are both dummy variables. ar bs ar bs −ar br as bs = (ar ar )(bs bs )−(ar br )(as bs )
now ar ar = 3 and bs bs = 4 + 9 + 16 = 29 so ar bs ar bs − ar br as bs = 3 × 29 − 92 = 6.
37. If δrs is the three-dimensional Kronecker delta, evaluate
(a) δrs δsr δpq δpq ,
Solution
(b) δrs δsk δkl δlr ,
(c) δrs δqr δpq δsp .
(a) δrs δsr δpq δpq = δrr δpq δpq = 3 δpp = 3 × 3 = 9.
(b)δrs δsk δkl δlr = δrk δkl δlr = δrl δlr = δrr = 3.
(c) δrs δqr δpq δsp = δqs δpq δsp = δps δsp = δss = 3.
38. If δrs is the three-dimensional Kronecker delta, simplify
(a) (δrp δsq − δrq δsp ) ap bq ,
Solution
(δrp δsq − δrq δsp ) ap bq = δrp δsq ap bq − δrq δsp ap bq = ar bs − as br
(b) (δrp δsq − δrq δsp ) δpq ,
Solution
(δrp δsq − δrq δsp ) δpq = δrp δsq δpq − δrq δsp δpq = δrs − δrs = 0
39. If δrs is the Kronecker delta in n-dimensions, calculate
(a) δrr ,
Solution
δrr = 1 + 1 + .. + 1 (n terms), so δrr = n
(b) δrs δrs ,
Solution
δrs δrs = δrr = n
(c) δrs δst δtr
Solution
i.e.
∂F
∂y
6= 0 and
∂F
∂x
+ 1 6= 0 δrs δst δtr = δrs δsr = δrr = n.
40. Starting from ǫijk ǫist = δjs δkt − δjt δks simplify as much as possible:
(a) ǫijk ǫijp ,
Solution
Substituting s = j and t = p into the givn formula:
ǫijk ǫijp = δjj δkp − δjp δkj = 3 δkp − δkp = 2 δkp
(b) ǫijk ǫijk ,
Solution
Substituting p = k into the previous result
ǫijk ǫijk = 2 δkk = 6
(c) ǫijk ǫipq ǫprs .
Solution
ǫijk ǫipq ǫprs = (δjp δkq − δjq δkp ) ǫprs
= δjp δkq ǫprs − δjq δkp ǫprs = δkq ǫjrs − δjq ǫkrs
or
ǫijk ǫipq ǫprs = ǫijk (ǫpqi ǫprs )
= ǫijk (δqr δis − δqs δir ) = ǫijk δqr δis − ǫijk δqs δir
= ǫsjk δqr − ǫrjk δqs
41. Calculate ǫijj .
Solution ǫijk vanishes whenever two indices take the same value, so ǫijj also vanishes,
being a sum of terms in each of which the last two indices take the same value.
42. Show, using index notations, that
(a)
a × (b × c) + b × (c × a) + c × (a × b) = 0,
Solution
The i-th component of
a × (b × c) is
ǫijk aj (ǫkrs br cs ) = (ǫijk ǫrsk ) aj br cs =
(δir δjs − δis δjr ) aj br cs = δir δjs aj br cs − δis δjr aj br cs = as bi cs − ar br ci
so the i-th component of
a × (b × c) + b × (c × a) + c × (a × b) is
(as bi cs − ar br ci ) + (bs ci as − br cr ai ) + (cs ai bs − cr ar bi ) =
ai (cs bs − br cr ) + bi (as cs − cr ar ) + ci (bs as − ar br ) = 0
since the names of dummy indices can be changed.
(b)
(a × b) × (c × d) = [d · (a × b)] c − [c · (a × b)] d
= [a · (c × d)] b − [b · (c × d)] a,
Solution
The i-th component of (a × b) × (c × d) is
ǫijk (ǫjrs ar bs ) (ǫkpq cp dq ) = ǫijk ǫjrs ǫkpq ar bs cp dq
= (ǫjki ǫjrs ) ǫkpq ar bs cp dq = δkr δis ǫkpq ar bs cp dq − δks δir ǫkpq ar bs cp dq
= ǫrpq ar bi cp dq − ǫspq ai bs cp dq
which is the same as the i-th component of
[a · (c × d)] b − [b · (c × d)] a
because this is
[ ar (ǫrpq cp dq )] bi − [ bs (ǫspq cp dq )] ai
We can also expand the i-th component of (a × b) × (c × d) as
ǫijk (ǫjrs ar bs ) (ǫkpq cp dq ) = ǫijk ǫjrs ǫkpq ar bs cp dq
(ǫijk ǫpqk ) ǫjrs ar bs cp dq = δip δjq ǫjrs ar bs cp dq − δiq δjp ǫjrs ar bs cp dq
= ǫqrs ar bs ci dq ǫprs ar bs cp di
which is the same as the i-th component of
[d · (a × b)] c − [c · (a × b)] d
because this is
[ dq (ǫqrs ar bs )] ci − [ cp (ǫprs ar bs )] di .
(c)
(d)
(e)
(a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c),
a × [a × (a × b)] = a2 (b × a),
(a × b) · (c × d) + (b × c) · (a × d) + (c × a) · (b × d) = 0.
43. Exam question June 2002 (Section A): Evaluate ǫijk ǫijl xk xl .
44. Exam question June 2001 (Section A): Evaluate ǫijk ∂i ∂j (xl xl )1/2 where ǫijk is a totally
antisymmetric 3-rank tensor in three dimensions.
45. Exam question June 2003 (Section A): Calculate ∂i (ǫijk ǫjkl xl ).
Solution Use ∂i xj = δij in ∂i (ǫijk ǫjkl xl ) = ǫijk ǫjkl ∂i xl = ǫijk ǫjkl δil = ǫijk ǫjki=
ǫijk ǫijk = δii δjj − δij δji = 3 × 3 − δii = 9 − 3= 6.
46. The functions f , g are scalars, while A and B are vector functions. Verify the following
identities using index notation:
(a)
grad (f g) = f grad g + g grad f ,
(b)
grad (A · B) = A × curl B + B × curl A + (B · ∇)A + (A · ∇)B,
(c)
(d)
(e)
(f)
(g)
(h)
div (f A) = f div A + (gradf ) · A,
curl (f A) = f curl A + (grad f ) × A,
div (A × B) = B · curl A − A · curl B,
curl (A × B) = (div B)A − (div A)B + (B · ∇)A − (A · ∇)B,
div curl A = 0,
curl curl A = grad div A − ∇2 A.
47. Let V : R2 → R2 be defined by (V) = (x cos y, x sin y). At which points is it a local
diffeomorphism?