GRADIENTS,
DIRECTIONAL DERIVATIVES
So you are back in Yosemite and feel like doing a bit of
serious hiking. You've pulled up the contour map to the right on your iPhone and located
yourself at the black dot on the 9200' contour. One natural question to
ask is for the slope of the terrain from where you in the particular
direction you set off to walk.
If you are feeling particularly energetic, you
might even try to take the steepest path to the top of the
mountain - we might call that the Path of Steepest Ascent'. This will
mean that at each point on your path you always take the direction of
greatest slope. But if you are on the 9200' contour, that direction of
greatest slope must be perpendicular to the contour, because if you
walk uphill in a direction that is not perpendicular, then you are
'cutting across' the mountain and not climbing uphill as fast as you
can.
|
 |
Now, mathematically, the terrain is the graph of $z
= f(x, y)$ and if $(x(t), y(t))$
is a curve in the $xy$-plane,
then the graph
of
r$(t) = \langle
x(t), y(t), f(x(t), y(t)) \rangle$
is a space curve lying on the graph of $f$. It's the
mathematical way of describing a path in Yosemite, say. Different
curves provide different information. But first we make a crucial
Definition:
For
a function $f(x,
y)$ the
GRADIENT
of $f$ is the
vector function
$\nabla f(x, y)
= f_x(x, y)\,$i$
\, +\, f_y(x, y)\,$j
whose
respective components are $f_x$
and $f_y$.
|
For
example, when $f(x,
y) = x^2 - y^2$, then $(\nabla f)(x, y)
= 2x\,$i$
- 2y\,$j.
Composition suggests the Chain Rule:
$ \displaystyle{\frac{d
}{d
t} f(x(t), y(t)) } = \displaystyle{\frac{\partial
f}{\partial x} \frac{d x}{d t}\, +\, \frac{\partial f}{\partial
y}\frac{d y}{d t}} = x'(t) f_x + y'(t)
f_y $,
which in
turn suggests the dot product. For if we set c$(t) = \langle
x(t), y(t) \rangle,$ then c$'(t)
=\langle x'(t), y'(t) \rangle$ and
$\displaystyle{\frac{d
}{d t}f(}$c$(t)) = \displaystyle{ \frac{d }{d t} f(x(t),
y(t)) = x'(t)f_x + y'(t) f_y =
(\nabla f)(}$c$(t))
\cdot $c$'(t)$.
An example
from a familiar setting will help.
|
CURVE 1:
set
$f(x, y) =
2+ x^2 - y^2$, c$(t)
= \langle \cos t, \, \sin t\rangle$.
The graph
of c is a
circle in the $xy$-plane,
while the graph of
$f($c$(t)) =
f(\cos t, \sin t) = 2 + \cos 2t$
is the
intersection of a
hyperbolic paraboloid and a circular cylinder as shown to the right. Now
$\displaystyle{\frac{d
}{dt} f(}$c$(t)) =
-2\sin 2t,}$
while
c$'(t) =
\langle -\sin t, \cos t\rangle$,
$\nabla f($c$(t)) =
2\cos t$ i $ -2 \sin t$ j.
But then
$(\nabla f )($c$(t))
\cdot $c$'(t)
= -4 \cos t \sin t =
-2 \sin 2t$,
confirming that
$\displaystyle{\frac{d
}{d t} f(}$c$(t))
= (\nabla f )($c$(t))
\cdot $c$'(t)$.
|
 |
Now the
tangent vector to the space curve can be calculated:
r$'(t) =
\langle x'(t), y'(t), \displaystyle{\frac{d }{d t}
f(x(t),
y(t)) \rangle } = \langle x'(t), y'(t), (\nabla f)(}$c$(t)) \cdot $c$'(t)) \rangle$.
It is
shown in orange in the previous graphic.
Second Definition:
- For a UNIT
vector v$= h\,$i $+k\,$j
the DIRECTIONAL
DERIVATIVE of $z = f(x, y)$
at $(a, b)$
in the direction v
is
- $Df$v$(a,
b) = \displaystyle{\lim_{t
\to 0} \frac{f(a+th, b+tk) - f(a, b)}{t}.}$
Of
course, $Df$i$(a, b) = f_x(a, b)$
and $Df$j$(a, b) =
f_y(a, b)$.
|
Since
$Df$v$(a,
b) = \displaystyle{\lim_{t
\to 0}\frac{f(a+th, b+tk) - f(a, b)}{t} = \frac{d
}{d t} f(a+th, b+tk) \Big|_{t = 0}},$
this
suggests what the next choice of curve should be.
CURVE 2.
Let
c$(t) = (a+th)$i + $(b+tk)$j
be the line through $(a,
b)$ in the direction of a vector v$ = h\,$i +$k\,$j. When the
surface to the right is the graph of $f$, the graph
of
$f($c$(t)) =
f(a+th, b+tk)$
is the curve shown in
orange on the surface. On the other hand,
c$(0) = (a, b)$,
c$'(0)
= h$ i $+ k$j$ = $v,
so
$Df$v$(a,
b) = \nabla f(a, b)\cdot $v.
This
tells us how to compute all directional derivatives
$Df$v$(a, b)$.
|
 |
Finally, we can see what the gradient vector has
to do with contours. What's a contour? Well, its a curve c$(t) =
\langle x(t), y(t) \rangle$ in the $xy$-plane such that $f($c$(t)) = k$
where $k$ is some constant. The graph to the left below shows one
contour, while the one to the right shows two contours realized as
curves in the
$xy$-plane.
|
CURVE 3:
 |
 |
But if we
differentiate the equation $f($c$(t)) = k$, then
$\displaystyle\frac{d
}{d t} f(}$c$(t))
= (\nabla f)($c$(t))\cdot
$c$'(t)
= 0.$
Now c$'(t)$ is the
tangent vector to the contour, so the fact that $(\nabla f)($c$(t)\cdot
$c$'(t) = 0$ means that the $\nabla f$ is perpendicular at c$(t)$ to
the contour. That's what the right hand graph shows. But why are some
vectors longer than others? Look on the graph where those short vectors
occur, it's close to the $y$-axis where the surface isn't very steep,
whereas the vectors get longer and longer the more we move along the
contour away from the $y$-axis. But the surface gets steeper and
steeper as one moves away from the $y$-axis. In other words the slope
increases!