Download Lecture 4

Part 4
Directional Derivatives
Printed version of the lecture Differential Geometry on 13. September 2013
Tommy R. Jensen, Department of Mathematics, KNU
4.1
Overview
Contents
1
Directional Derivatives
1
2
Operation of a Vector Field
4
3
Conclusion
5
4.2
1
Directional Derivatives
Derivative of a function
Definition
Let f : R3 → R be a differentiable function.
Let p ∈ R3 be a point.
Let v p be a tangent vector to R3 in p, that is, v p ∈ Tp (R3 ).
We write
d
f (p + tv)|t=0 .
dt
The number v p [ f ] is called the derivative of f with respect to v p .
vp[ f ] =
We also say that v p [ f ] is a directional derivative.
1
The directional derivative v p [ f ] informs us about the change in the value of f as we move away from p
in the direction of v p .
4.3
An example of computing a directional derivative
Let f = xy2 z: R3 → R. Let p = (1, 1, 0) and v = (1, 0, −3).
Then p + tv = (1, 1, 0) + t(1, 0, −3) = (1 + t, 1, −3t).
Evaluate f in this new point:
f (p + tv) = f (1 + t, 1, −3t) = (1 + t)(−3t) = −3t − 3t 2 .
d
Differentiate with respect to t : dt
f (p + tv) = −3 − 6t.
Setting t = 0, we get the answer v p [ f ] = −3.
We deduce that f is decreasing when moving away from p in the direction of v.
To express the answer differently we could write
(1, 0, −3)(1,1,0) [xy2 z] = −3.
But we don’t want to ask for trouble. . .
4.4
How to differentiate composite functions
The chain rule
Let f : R3 → R be a differentiable function.
Let a : R → R,t 7→ a(t) b : R → R,t 7→ b(t) c : R → R,t 7→ c(t) be differentiable functions of one
variable t.
Then
f (a, b, c) : R → R,t 7→ f (a(t), b(t), c(t))
is a differentiable function of one variable t, and
d
∂f
da ∂ f
db ∂ f
dc
f (a, b, c) =
(a, b, c) +
(a, b, c) +
(a, b, c) .
dt
∂ x1
dt ∂ x2
dt ∂ x3
dt
4.5
How to compute directional derivatives more efficiently
Lemma 3.2
Let v p = (v1 , v2 , v3 ) p be a tangent vector to R3 in p.
Then
v p [ f ] = ∑ vi
∂f
(p).
∂ xi
Proof of Lemma 3.2
Let p = (p1 , p2 , p3 ). Then p + tv = (p1 + tv1 , p2 + tv2 , p3 + tv3 ).
Now use the chain rule:
2
vp[ f ] =
df
(p + tv)|t=0
dt
∂f
d
=
∑ ∂ xi (p + tv) dt (pi + tvi )|t=0
=
∑ ∂ xi (p)vi .
∂f
4.6
The same example as before
Let f = xy2 z: R3 → R. Let p = (1, 1, 0) and v = (1, 0, −3).
Now we first find the partial derivatives of f :
∂f
= y2 z,
∂x
∂f
= 2xyz,
∂y
∂f
= xy2 .
∂z
∂f
(p) = 0,
∂y
∂f
(p) = 1.
∂z
At the point p = (1, 1, 0) we calculate:
∂f
(p) = 0,
∂x
By Lemma 3.2:
v p [ f ] = 0 + 0 + (−3) · 1 = −3.
4.7
The main properties of directional derivatives
Theorem 3.3
Let f : R3 → R and g : R3 → R be functions, let v p and w p be tangent vectors to R3 in p, and let a, b ∈ R
be numbers. Then
(1) (av p + bw p )[ f ] = av p [ f ] + bw p [ f ],
(2) v p [a f + bg] = av p [ f ] + bv p [g],
(3) v p [ f g] = v p [ f ] · g(p) + f (p) · v p [g].
(1) and (2) mean that v p [ f ] is linear in both the tangent vector v p and the function f .
(3) means that the derivative satisfies the usual product rule for differentiation (also called Leibniz rule).
4.8
Proof of Theorem 3.3
Proof of v p [a f + bg] = av p [ f ] + bv p [g]
Let v = (v1 , v2 , v3 ).
From Lemma 3.2 we have
v p [a f + bg]
=
=
=
=
=
=
∂ (a f + bg)
(p)
∂ xi
∂f
∂g
∑ vi (a ∂ xi (p) + b ∂ xi (p))
∂f
∂g
∑(vi a ∂ xi (p) + vi b ∂ xi (p))
∂f
∂g
∑ vi a ∂ xi (p) + ∑ vi b ∂ xi (p)
∂f
∂g
a ∑ vi
(p) + b ∑ vi
(p)
∂ xi
∂ xi
av p [ f ] + bv p [g].
∑ vi
3
4.9
A picture
It is confusing, but nicer to look at than a proof.
4.10
2
Operation of a Vector Field
The operation of a vector field on a function
Definition
Let V be a vector field, and let f : R3 → R be a function.
Let V [ f ] be the function defined by
V [ f ] : R3 → R, p 7→ V (p)[ f ].
We explain V [ f ] as the function which to each point p in R3 gives us the derivative of f in p with respect
to the tangent vector V (p).
Example
For the natural frame field U1 ,U2 ,U3 we get by Lemma 3.2
U1 [ f ](p) = U1 (p)[ f ] = (1, 0, 0) p [ f ] =
1·
∂f
∂f
∂f
∂f
(p) + 0 ·
(p) + 0 ·
(p) =
(p).
∂x
∂y
∂z
∂x
4.11
4
Basic properties of operations of vector fields
Corollary 3.4
Let V and W be vector fields on R3 , let f , g, h : R3 → R be functions, and let a, b ∈ R be numbers. Then
(1) ( fV + gW )[h] = fV [h] + gW [h],
(2) V [a f + bg] = aV [ f ] + bV [g],
(3) V [ f g] = V [ f ] · g + f ·V [g].
Proof of V [a f + bg] = aV [ f ] + bV [g]
Let p be any point in R3 .
Def.
V [a f + bg](p)
=
V (p)[a f + bg]
Theorem 3.3
=
aV (p)[ f ] + bV (p)[g]
Def.
=
aV [ f ](p) + bV [g](p)
=
(aV [ f ] + bV [g])(p)
4.12
Example
Let V be the vector field V = x2U1 + yU3 , and let f be the function f = x2 y − y2 z.
V[f]
=
(x2U1 + yU3 )[x2 y − y2 z]
=
x2U1 [x2 y] − x2U1 [y2 z] + yU3 [x2 y] − yU3 [y2 z]
=
x2 (2xy) − x2 · 0 + y · 0 − y(y2 )
=
2x3 y − y3 .
4.13
3
Conclusion
Here it ends
END OF THE LECTURE!
4.14
5
Next time:
Curves in R3
4.15
4.16
6