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```							Answers for Tutorial 2
[M283 - Field theory and PDEs]

Thursday 18th October

1. The gradients of the following functions:
(a) T (x, y, z) = x2 + xy + 3z 3
(b) T (x, y, z) = x2 y 3 + z 4
(c) T (x, y, z) = ez sin x + ln(y)
are respectively
(a) ∇T = (2x + y, x, 9z 2 )
(b) ∇T = (2xy 3 , 3x2 y 2 , 4z 3 )
(c) ∇T = (ez cos x, 1/y, ez sin x).

2. The unit inward pointing normal to the sphere deﬁned by the equation T (x, y, z) =
x2 + y 2 + z 2 = 1 at a general point (x, y, z) is (note the minus sign)

∇T        (2x, 2y, 2z)          (x, y, z)
ˆ
n=−           = −√ 2       2 + 4z 2
= −√ 2          .
| ∇T |     4x + 4y               x + y2 + z2

At point a = (1, 0, 0) it can be recast as:

ˆ
n = −(1, 0, 0) .

Hence, the cartesian equation for the tangent plane which touches the sphere at that
point is given by

ˆ
(r − a) · n = 0
(r − (1, 0, 0)) · (1, 0, 0) = 0
(x − 1, y, z) · (1, 0, 0) = 0 .

Or
x=1.

3. The unit outward pointing normal to the ellipsoid T (x, y, z) = x2 + 2y 2 + 3z 2 = 6 at a
general point (x, y, z) is (note that the sign is now positive)

∇T       (2x, 4y, 6z)         (x, 2y, 3z)
ˆ
n=          =√ 2        2 + 36z 2
=√ 2              .
| ∇T |   4x + 16y              x + 4y 2 + 9z 2

At point a = (1, 1, 1) it can be recast as:

(1, 2, 3)
ˆ
n=        √       .
14

1
Hence, the cartesian equation for the tangent plane which touches the ellipsoid at that
point is given by

ˆ
(r − a) · n = 0
(1, 2, 3)
(r − (1, 1, 1)) · √        =0
14
(x − 1) + 2(y − 1) + 3(z − 1) = 0

Or
x + 2y + 3z = 6 .

4. The directional derivative of the function
1
φ(x, y, z) = x2 + y 2 +
z
at a general point (x, y, z) in the direction given by the vector b = (−1, −1, 1) writes
as
b                         (−1, −1, 1)
Db Φ = ∇Φ ·          = (2x, 2y, −1/z 2 ) ·    √        .
|b|                               3
At the point a = (1, 0, 1), it reduces to

b                 (−1, −1, 1)  −2 − 1   √
Db Φ(a) = ∇Φ(a) ·       = (2, 0, −1) ·    √        = √     =− 3.
|b|                      3         3

5. Given that r = (x, y, z), r = |r|, and that r = r/|r|, we see that
(a) ∇(r 2 ) = (∂/∂x, ∂/∂y, ∂/∂z)(x2 + y 2 + z 2 ) = (2x, 2y, 2z).
This can be further simpliﬁed as

(x, y, z)               r
∇(r 2 ) = 2 x2 + y 2 + z 2 √                    = 2| r |     = 2r r
x2 + y 2 + z 2          |r|

(b) ∇ r12 = (∂/∂x, ∂/∂y, ∂/∂z)(x2 + y 2 + z 2 )−1 = − (x(2x,2y,2z))2 .
2 +y 2 +z 2

This can be further simpliﬁed as
√ 2
1          x + y2 + z2     (x, y, z)                     r       2
∇ 2 = −2 √ 2          2 + z2 √ 2
= −2| r |         5 = − 3r
r           x +y          x +y    2 + z24               |r|      r

6. The divergence and curl of the function

v = (2xy, −yz, 3zx) .

are respectively

∇ · v = (∂/∂x, ∂/∂y, ∂/∂z) · (2xy, −yz, 3zx) = 2y − z − 3x ,

2
(which is a scalar ﬁeld) and

∇ × v = (∂/∂x, ∂/∂y, ∂/∂z) × (2xy, −yz, 3zx)
∂           ∂            ∂         ∂
= ˆ
i      (3zx) − (−yz) − ˆ   j    (3zx) − (2xy)
∂y         ∂z            ∂x        ∂z
ˆ ∂ (−yz) − ∂ (2xy)
+ k
∂x           ∂y
= yˆ − 3zˆ − 2xk
i      j    ˆ                                                (1)

which is a vector ﬁeld.

7. The vector ﬁeld
j ˆ
F1 = (14x + 3y)ˆ + 3xˆ + k ,
i
has a curl given by

∇ × F1 = (∂/∂x, ∂/∂y, ∂/∂z) × (14x + 3y, 3x, 1)
∂        ∂            ∂        ∂
= ˆ
i      (1) − (3x) − ˆ  j    (1) − (14x + 3y)
∂y      ∂z           ∂x       ∂z
ˆ ∂ (3x) − ∂ (14x + 3y)
+ k
∂x         ∂y
ˆ
= 0ˆ − 0ˆ − (3 − 3)k = 0 .
i    j                                                      (2)

It can thus be expressed as the gradient of a scalar ﬁeld since for any scalar ﬁeld f
∇ × ∇f = 0 This scalar ﬁeld is f (x, y, z) = 7x2 + 3yx + z. Indeed, we check that

∇f = (∂/∂x, ∂/∂y, ∂/∂z)(7x2 + 3xy + z) = (14x + 3y, 3x, 1) = F1 .

3

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