Download We stress that f(x, y, z) is a scalar-valued function and ∇f is a vector

MATHEMATICS FOR ENGINEERS & SCIENTISTS
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We stress that f (x, y, z) is a scalar -valued function and ∇f is a vector -valued function.
All of the above works in any number of dimensions. For instance, consider the following 2-dimensional
example:
Example 1.11. The height of a slope above sea level is f (x, y) = 2y 2 + x2 (in some unit of distance)
at coordinates (x, y). Find the rate of change of height (i.e. the rate of ascent) when starting at (1, 2)
and moving at unit speed towards the south-east.
We need to calculate the directional derivative of f at (1, 2) in the direction i − j. Now
∇f = (2x, 4y)
so evaluating this at the point gives
∇f (1, 2) = 2i + 8j.
Hence the required directional derivative is
6
(i − j)
= −√
(2i + 8j) • √
1+1
2
√
and so the rate of change of height is −6/ 2 ' −4.24.
Natural question: Given a scalar function f (x, y, z) and a point p, in which direction from p does
f have the largest directional derivative, and what is this largest value? For instance, in the previous
example, in which direction is the steepest ascent, and how steep is it?
Recall that the directional derivative of f at p in the direction v is
v
= |∇f (p)| cos θ
∇f (p) •
|v|
where θ is the angle between ∇f (p) and v. This clearly has maximum value when cos(θ) = 1 so we
need θ = 0 and hence v is parallel to ∇f (p). Thus, the steepest increase is we move off in the direction
∇f (p and the maximum value is |∇f (p|.
Example 1.12. If f (x, y) = − cos(xy), in what direction should one travel from
maximise the rate of change of f ?
π
2,1
in order to
We calculate
∇f (x, y) = (y sin(xy), x sin(xy))
and
π
, 1 = i + j.
2
2
So we have to travel in the direction i + π2 j and the rate of change is
r
π π2
.
i + j = 1 +
2
4
∇f
π
1.5. Tangent plane to a surface.
First, we give a brief reminder about planes, lines and surfaces in 3-dimensional space.
Planes. The equation of the plane through the point p = (a, b, c) and normal to n = n1 i + n2 j + n3 k
is given by
(x − p) • n = 0.
Putting in the coordinates x = (x, y, z), turns this into the familiar form
n1 x + n2 y + n3 z = C
where C = n1 a + n2 b + n3 c. For example, the equation of the plane through (2, −1, 5) which is normal
to 4i + 3j − k is
4x + 3y − z = 4 × 2 + 3(−1) − 5 = 0.
Lines. The equation of the line through p = (a, b, c) in the direction v = v1 i + v2 j + v3 k is, in
parametric form, given by
x = p + λv
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MATHEMATICS FOR ENGINEERS & SCIENTISTS
where λ is an arbitrary parameter. Putting in the coordinates x = (x, y, z) makes this
(x, y, z) = (a, b, c) + λ (v1 , v2 , v3 )
= (a + λv1 , b + λv2 , c + λv3 ).
Rearranging by solving for λ gives the alternative form:
λ=
x−a
y−b
z−c
=
=
.
v1
v2
v3
For example, the equation of the line through (2, −1, 5) in the direction 4i + 3j − k is
x = (2, −1, 5) + λ (4, 3, −1)
or equivalently,
y+1
z−5
x−2
=
=
.
4
3
−1
Surfaces. We will consider surfaces with equations of the form f (x, y, z) = C where C is a constant.
These are the sometimes called level surfaces of the function f . You are probably familiar with a
number of surfaces written in this form:
Example 1.13.
The surface of the sphere of radius r
and centred at the origin has equation
x2 + y 2 + z 2 = r2 .
Example 1.14.
The cylinder of radius r centred along
the z-axis has equation
x2 + y 2 = r2 .
Example 1.15.
The paraboloid obtained by rotating a
parabola about the z-axis has equation
x2 + y 2 − z = 0.
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A surface has a tangent plane and normal line at each point. We will next see how to write down the
equations of this plane and line.
Let S be a surface with equation f (x, y, z) = C and let r(t) = (x(t), y(t), z(t)) be a curve on S. Then
f (r(t)) = C for all t, so differentiating with respect to t using the chain rule gives
df
0=
dt
∂f dx ∂f dy ∂f dz
0=
+
+
∂x dt
∂y dt
∂z dt
dr
0 = ∇f •
dt
But the tangent plane to S at p is made up of all vectors tangent to curves on S through p, so it
follows that
Proposition 1.16. Let S be a surface with equation f (x, y, z) = C and let p be a point of S. Then
∇f (p) is the normal vector to S at p.
Equivalently, the tangent plane to S at p is the plane through p which is perpendicular to ∇f (p).
Example 1.17. A sheet of metal is shaped to make a surface with equation f (x, y, z) = x2 +y 2 +yz = 0.
A length of metal rod is to be welded to the sheet at two points with one end at (2, −1, 5) meeting the
sheet at right angles. Find the other point of the sheet at which the metal rod should be attached and
the tangent plane to the surface at (2, −1, 5).
First check that (2, −1, 5) is on the surface: f (2, −1, 5) = 4 + 1 − 5 = 0.
Now ∇f = 2xi + (2y + z)j + yk and so
∇f (2, −1, 5) = 4i + 3j − k.
Also, the normal line is the line through (2, −1, 5) in the direction 4i + 3j − k, and so has equation
y+1
z−5
x−2
=
=
4
3
−1
or, in parametric form,
(x, y, z) = (2, −1, 5) + λ (4, 3, −1) = (2 + 4λ, −1 + 3λ, 5 − λ).
We need to find the values of λ for which these points lie on S, i.e. for which
2
2
(2 + 4λ) + (−1 + 3λ) + (−1 + 3λ) (5 − λ) = 0
that is, 22λ2 + 26λ = 0. This has two solutions: λ = 0 which corresponds to the point (2, −1, 5) and
13
which corresponds to the other intersection point, namely
λ = − 11
52
39
13
30 50 68
2 − , −1 − , 5 +
= − ,− ,
.
11
11
11
11 11 11
Furthermore, the tangent plane through (2, −1, 5) is perpendicular to ∇f (2, −1, 5) = 4i + 3j − k and
so has equation
4x + 3y − z = 4 × 2 + 3(−1) − 5 = 0.
Actually, the surface x2 + y 2 + yz = 0 is a cone
and looks something like the picture here.
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MATHEMATICS FOR ENGINEERS & SCIENTISTS
1.6. Div and Curl.
Recall that for a general function f (x) = f (x, y, z), we have the gradient
∂f ∂f ∂f
∂f
∂f
∂f
∇f =
,
,
=i
+j
+k .
∂x ∂y ∂z
∂x
∂y
∂z
We can think of
∇=
∂ ∂ ∂
,
,
∂x ∂y ∂z
=i
∂
∂
∂
+j
+k
∂x
∂y
∂z
as a vector in it’s own right - a kind of hybrid of vector and differentiation which operates on the
scalar-valued function f to give the vector-valued function ∇f . We will now see two other important
ways of combining ∇ with functions. As well as scalar-valued functions of vectors (often called scalar
fields) such as f (x) = f (x, y, z) above, we can also consider vector-valued functions of vectors:
A (x) = (A1 (x, y, z), A2 (x, y, z), A3 (x, y, z)) .
Such functions (often called vector fields) are extremely important in many areas of Engineering and
Science. For instance, consider a gravitational field; at every point x = (x, y, z) in space, there is a
gravitational force. This is a vector with each component being a function of x, y and z.
Now, we can combine two vectors a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) using the dot product:
a • b = a1 b1 + a2 b2 + a3 b3 .
This leads us to make the following definition.
Definition 1.18. Given a vector-valued function A (x) = (A1 (x, y, z), A2 (x, y, z), A3 (x, y, z)), we define it’s divergence to be
∂A1
∂A2
∂A3
∂ ∂ ∂
,
,
• (A1 , A2 , A3 ) =
+
+
.
div A = ∇ • A =
∂x ∂y ∂z
∂x
∂y
∂z
Example 1.19.
(i) Let A (x) = x2 + yz, xyz 2 , y 2 − z 2 . Then div A is
∂ ∂ ∂
∇•A=
,
,
• x2 + yz, xyz 2 , y 2 − z 2
∂x ∂y ∂z
∂
∂
∂
=
x2 + yz +
xyz 2 +
y2 − z2
∂x
∂y
∂z
2
= 2x + xz − 2z.
(ii) Let A (x) = yz sin x, x + z 2 cos2 y, xyz − tan z . Then div A is
∂ ∂ ∂
∇•A=
,
,
• yz sin x, x + z 2 cos2 y, xyz − tan z
∂x ∂y ∂z
∂
∂
∂
=
(yz sin x) +
x + z 2 cos2 y +
(xyz − tan z)
∂x
∂y
∂z
= yz cos x − 2z 2 cos y sin y + xy − sec2 z.
Another way to combine two vectors a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) is via the cross-product:
i
j k a × b = a1 a2 a3 = (a2 b3 − a3 b2 ) i + (a3 b1 − a1 b3 ) j + (a1 b2 − a2 b1 ) k.
b1 b2 b3 Definition 1.20. Given a vector-valued function A (x) = (A1 (x, y, z), A2 (x, y, z), A3 (x, y, z)), we define it’s curl to be
i
j
k ∂
∂A3
∂A2
∂A1
∂A3
∂A2
∂A1
∂
∂ curl A = ∇ × A = ∂x ∂y
=
−
i
+
−
j
+
−
k.
∂z ∂y
∂z
∂z
∂x
∂x
∂y
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Example 1.21.
(i) Let A (x) = x2 + yz, xyz 2 , y 2 − z 2 . Then curl A is
i
j
k ∂
∂
∂ ∇×A=
∂x
∂y
∂z x2 + yz xyz 2 y 2 − z 2 !
!
!
∂ y2 − z2
∂ x2 + yz
∂ xyz 2
∂ xyz 2
∂ y2 − z2
∂ x2 + yz
=
i+
j+
k
−
−
−
∂y
∂z
∂z
∂x
∂x
∂y
= (2y − 2xyz) i + yj + yz 2 − z k
= 2y − 2xyz, y, yz 2 − z .
(ii) Let A (x) = yz sin x, x + z 2 cos2 y, xyz − tan z . Then curl A is
i
j
k
∂
∂
∂
∇×A=
∂x
∂y
∂z
yz sin x x + z 2 cos2 y xyz − tan z !
∂ (xyz − tan z) ∂ x + z 2 cos2 y
∂ (yz sin x) ∂ (xyz − tan z)
=
−
i+
−
j
∂y
∂z
∂z
∂x
!
∂ x + z 2 cos2 y
∂ (yz sin x)
−
k
+
∂x
∂y
= xz − 2z cos2 y i + (y sin x − yz) j + (1 − z sin x) k
= xz − 2z cos2 y, y sin x − yz, 1 − z sin x .
Summarising how our three vector calculus operators work:
- grad takes scalar-valued functions to vector-valued functions,
- div takes vector-valued functions to scalar-valued functions,
- curl takes vector-valued functions to vector-valued functions.
We don’t have the time to go into the many applications of these new objects but will end this section
here by indicating why they’re so useful. In very vague terms:
- as we have seen, the gradient grad f measures how a scalar field f (x) increases,
- the divergence div A measures how a vector field A (x) expands or contracts,
- the curl curl A measures how a vector field A (x) rotates.
Understanding these is vital, for instance, in the study of Electromagnetism, Fluid Dynamics, Kinematics, General Relativity,... See e.g. the Wikipedia pages on div, grad and curl for more details.
1.7. Critical points - local maxima and minima.
df
= 0 are called critical points or
Recall that for a function f (x) of one variable, the points at which
dx
stationary points. The tangent to the graph at these points is horizontal.
There is a simple test for local maxima and minima:
df
d2 f
• If
= 0 and
< 0 then it’s a local maximum.
dx
dx2
2
df
d f
• If
= 0 and
> 0 then it’s a local minimum. For example, f (x) = x2 at x = 0.
dx
dx2
d2 f
df
= 0 then the test is inconclusive.
= 0 and
• If
dx
dx2
For example,
• f (x) = −x2 has a local maximum at x = 0 since f 0 (0) = 0 and f 00 (0) = −2 < 0.
• f (x) = x2 has a local minimum at x = 0 since f 0 (0) = 0 and f 00 (0) = 2 > 0.
• f (x) = −x4 , x4 , x3 at x = 0 have a local maximum, local minimum, and point of inflexion
respectively.