2E2 Tutorial Sheet 21 Third Term, Solutions1
and so on. For (c)
20 April 2004
p
1. (1) Find a unit p
normal to z = x2 + y 2 at (6, 8, 10). Note that this surface is f = 0
where f = z − x2 + y 2
p
Solution: The normal to f =pc is given by ∇f , here f = z − x2 + y 2 and the
surface is f = 0. Writing ρ = x2 + y 2 we have
x
y
∇f = − i − j + k
ρ
ρ
(1)
(2)
√
This has length |∇f | = 2 so the unit normal is
n̂ =
3
∇f
4
1
=− √ i− √ j+ √ k
|∇f |
5 2
5 2
2
(3)
and if you add up the contibution from v1 , v2 and v3 you get zero.
3. (2) What is the curl of
(a) v1 (x)i + v2 (y)j + v3 (z)k
(b) xyz(xi + yj + zk)
Solution: Well, for (a), the answer is zero because, in calculating the curl, the v
component only gets differenciated by y and z and so on. For (b) we have
i
j
k ∂
∂
∂
2
2
2
∇ × (x yz, xy z, xyz ) = ∂x
∂y
∂z
x2 yz xy 2 z xyz 2 = (xz 2 − xy 2 , yx2 − yz 2 , zy 2 − zx2 ).
The actual shape of the surface is a cone with its point at the origin.
−3yz
∂
z
= 2
∂y (x2 + y 2 + z 2 )3/2
(x + y 2 + z 2 )5/2
(a) v1 (y, z)i + v2 (x, z)j + v3 (x, y)k
(8
In other words the numerator is the multiple of the original numerator by the variabl
we are differenciating by. If you work it out
(b) xyz(xi + yj + zk)
∂
y
−3yz
= 2
∂z (x2 + y 2 + z 2 )3/2
(x + y 2 + z 2 )5/2
(c) (x2 + y 2 + z 2 )−3/2 (xi + yj + zk)
Solution: Well, for (a) the answer is zero becuase in calculating div v1 is differenciated
with respect to x but only depends on y and z, v2 is differenciated with respect to
y but only depends on x and z and v3 is differenciated with respect to z but only
depends on y and z. For (b) we have
∇ · (x2 yz, xy 2 z, xyz 2 ) = 2xyz + 2xyz + 2xyz = 6xyz
1
(7
For (c) note that
2. (2) What is the div of
where we have used
(6
(c) (x2 + y 2 + z 2 )−3/2 (xi + yj + zk)
so, at (6, 8, 10), ρ = 10 and the vector, ∇f , normal to the surface is
3
4
∇f = − i − j + k
5
5
∂
x
2x2
1
3
=
−
∂x (x2 + y 2 + z 2 )3/2
(x2 + y 2 + z 2 )3/2 2 (x2 + y 2 + z 2 )5/2
∂x2 yz
= 2xyz
∂x
4. (1) Show ∇ × (∇f ) = 0.
Solution: As for ∇ × (∇f ) = 0, well
∇f =
so
Conor Houghton, houghton@maths.tcd.ie and http://www.maths.tcd.ie/~houghton/ 2E2.html
1
and this is the same thing and so the curl is zero.
(4)
(5)
(9
∇ × (∇f ) =
∂2f
∂2f
−
∂zy ∂yz
∂f
∂f
∂f
i+
j+
k
∂x
∂y
∂z
i+
∂2f
∂f
−
∂xz ∂zx
(10
j+
∂2
∂f
−
∂yx ∂xy
and again that is zero by changing the order of differenciation.
2
k
(11
For completeness, here is the other one, ∇ · (∇ × v) = 0. So,
∂v1 ∂v3
∂v2 ∂v1
∂v3 ∂v2
i+
j+
k
−
−
−
∇×v =
∂y
∂z
∂z
∂x
∂x
∂y
(12)
Now,
∇ · (∇ × v) =
∂ 2 v3 ∂ 2 v2 ∂ 2 v1 ∂ 2 v3 ∂ 2 v2 ∂ 2 v1
−
+
−
+
−
∂yx
∂xz
∂zy
∂xy
∂xz
∂yz
(13)
and that is zero if you take into account the fact that the order of differenciation
doesn’t matter.
5. (2) Show curl(f v) = (grad f ) × v + f curl v.
Solution: Another one to do by brute force. The vector f v = (f v1 , f v2 , f v3 ) and the
curl is
i
j
k ∂
∂
∂
(14)
∇ × (f v1 , f v2 , f v3 ) = ∂x ∂y
∂z f v1 f v2 f v3 and so there are two parts corresponding to differentiating the vi , this gives a f curl v,
and differenciating the f
∂f
∂f
∂f
∂f
∂f
∂f
∇ × (f v1 , f v2 , f v3 ) = f ∇ × v +
v3 −
v2 ,
v1 −
v3 ,
v2 −
v1 (15)
∂y
∂z
∂z
∂x
∂x
∂y
and, by writing it out, it is easy to see that the second part is (grad f ) × v.
3