MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
The gradient
Tangent
planes to level
surfaces
MATH 209—Calculus, III
Volker Runde
University of Alberta
Edmonton, Fall 2011
Directional derivatives, I
MATH 209—
Calculus,
III
Volker Runde
Recall. . .
Directional
derivatives
Let z = f (x, y ) be a function of two variables. Then:
The gradient
Tangent
planes to level
surfaces
∂f
∂x :
slope of the tangent line to the graph of f in the
direction of the x-axis;
∂f
∂y :
slope of the tangent line to the graph of f in the
direction of the y -axis.
Question
What about the slope of a tangent line to the graph of f in the
direction of an arbitrary line in the xy -plane?
Directional derivatives, II
MATH 209—
Calculus,
III
Volker Runde
Definition
The directional derivative of f at (x0 , y0 ) in the direction of the
unit vector u = ha, bi is the limit
Directional
derivatives
The gradient
Tangent
planes to level
surfaces
Du f (x0 , y0 ) = lim
h→0
f (x0 + ha, y0 + hb) − f (x0 , y0 )
h
if it exists.
Example
What is Du f (1, 2), for f (x, y ) = x 3 − 3xy + 4y 2 and u given by
the angle θ = π6 , i.e.,
*√
+
D
π
πE
3 1
u = cos , sin
=
,
?
6
6
2 2
Directional derivatives, III
MATH 209—
Calculus,
III
Volker Runde
An application of the chain rule
Directional
derivatives
Set g (t) := f (x0 + ta, y0 + tb), so that
The gradient
Tangent
planes to level
surfaces
f (x0 + ha, y0 + hb) − f (x0 , y0 )
h
g (h) − g (0)
= lim
h→0
h
dg
=
(0).
dt
Du f (x0 , y0 ) = lim
h→0
Directional derivatives, IV
MATH 209—
Calculus,
III
Volker Runde
An application of the chain rule (continued)
Directional
derivatives
With g (t) = f (x, y ), x = x0 + ta, and y = y0 + tb, the chain
rule yields
The gradient
Tangent
planes to level
surfaces
dg
(0)
dt
∂f
dx
∂f
dy
=
(x0 , y0 ) (0) +
(x0 , y0 ) (0)
∂x
dt
∂y
dt
∂f
∂f
=
(x0 , y0 )a +
(x0 , y0 )b.
∂x
∂y
Du f (x0 , y0 ) =
Directional derivatives, V
MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
The gradient
Tangent
planes to level
surfaces
Theorem
If f is differentiable, then Du f exists for any unit vector
u = ha, bi and is computed as
Du f (x0 , y0 ) =
∂f
∂f
(x0 , y0 )a +
(x0 , y0 )b.
∂x
∂y
Directional derivatives, V
MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
Example (resumed)
For f (x, y ) = x 3 − 3xy + 4y 2 and u =
D√
3 1
2 ,2
E
, we have
The gradient
Tangent
planes to level
surfaces
∂f
= 3x 2 − 3y
∂x
and
∂f
− 3x + 8y ,
∂y
so that
√
√
3
1
13 − 3 3
Du f (1, 2) = (3 − 6)
+ (−3 + 16) =
.
2
2
2
Definition of the gradient
MATH 209—
Calculus,
III
Volker Runde
Definition
Directional
derivatives
The gradient of f is the vector function defined by
The gradient
Tangent
planes to level
surfaces
∇f (x, y ) = hfx (x, y ), fy (x, y )i = fx (x, y )i + fy (x, y )j.
Important
If f is differentiable, and u is a unit vector, then
Du f (x, y ) = ∇f (x, y ) · u.
Examples, I
MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
Example
Find the directional derivative of f (x, y ) = ln x + e (x−1)y at
(1, 0) in the direction of v = −i + 2j.
As
The gradient
Tangent
planes to level
surfaces
fx =
1
+ ye (x−1)y
x
and
fy = (x − 1)e (x−1)y ,
we have ∇f √
(1, 0) = h1, 0i.
Since |v| = 5, the unit vector u in the direction of v is
u = √15 v = − √15 i + √25 j.
All in all:
1 2
1
Du f (1, 0) = h1, 0i · − √ , √
= −√ .
5 5
5
Geometric meaning of the gradient, I
MATH 209—
Calculus,
III
Volker Runde
A geometric observation
Directional
derivatives
Let f be a differentiable function of two variables, let u be a
unit vector, and let θ be the angle between u and ∇f .
We have:
The gradient
Tangent
planes to level
surfaces
Du f = ∇f · u
= |∇f ||u| cos θ
= |∇f | cos θ.
This becomes maximal if cos θ = 1, i.e., θ = 0, which means
that u has the same direction as ∇f .
Geometric meaning of the gradient, II
MATH 209—
Calculus,
III
Volker Runde
Theorem
Directional
derivatives
Let f be a differentiable function of two variables, and let u be
a unit vector. Then Du f (x, y ) is maximal when u has the same
direction as ∇f (x, y ). This maximal value of Du f (x, y ) is
|∇f (x, y )|.
The gradient
Tangent
planes to level
surfaces
Important
Everything we said about directional derivatives and gradients
of functions of two variables remains valid for functions of three
(or more) variables.
Examples, II
MATH 209—
Calculus,
III
Example
Volker Runde
Directional
derivatives
The gradient
Tangent
planes to level
surfaces
The temparature in a room at (x, y , z) is given by
T =
1+
x2
80
,
+ 2y 2 + 3z 2
with T in centigrade and x, y , z in meters.
In which direction increases the temparature fastest at
(1, 1, −2)? What is the maximum rate of change?
The gradient is
∇T =
∂T
∂T
∂T
i+
j+
k,
∂x
∂y
∂z
Examples, III
MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
The gradient
Tangent
planes to level
surfaces
Example (continued)
. . . i.e.,
160x
320y
i−
j
(1 + x 2 + 2y 2 + 3z 2 )2
(1 + x 2 + 2y 2 + 3z 2 )2
480z
−
k
2
(1 + x + 2y 2 + 3z 2 )2
160
=
(−xi − 2y j − 3zk).
2
(1 + x + 2y 2 + 3z 2 )2
∇T = −
It follows that
5
∇T (1, 1, −2) = (−i − 2j + 6k).
8
Examples, IV
MATH 209—
Calculus,
III
Volker Runde
Example (continued)
Directional
derivatives
This vector has the same direction as h−1, −2, 6i.
The unit vector in this direction is
The gradient
Tangent
planes to level
surfaces
1
2
6
u = − √ i − √ j + √ k,
41
41
41
which gives the direction of the maximum increase.
The maximum increase is
|∇T (1, 1, −2)| =
5√
41 ≈ 4 ◦ C/m.
8
More geometric meaning of the gradient, I
MATH 209—
Calculus,
III
Another geometric observation
Volker Runde
Directional
derivatives
The gradient
Tangent
planes to level
surfaces
Let S be a level surface of a function F (x, y , z), i.e., given by
the equation F (x, y , z) = C with a constant C .
Let P(x0 , y0 , z0 ) be a point on S, and let
r(t) = hx(t), y (t), z(t)i be a curve in S passing through P.
Then F (x(t), y (t), z(t)) = C , so that
dC
dt
∂F dx
∂F dy
∂F dz
=
+
+
∂x dt
∂y dt
∂z dt
0
= ∇F · r (t).
0=
More geometric meaning of the gradient, II
MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
The gradient
Tangent
planes to level
surfaces
Another geometric observation (continued)
Let t0 be such that r(t0 ) = hx(t0 ), y (t0 ), z(t0 )i.
Then
0 = ∇F (x0 , y0 , z0 ) · r0 (t0 ).
This means that ∇F (x0 , y0 , z0 ) is perpendicular to the tangent
vector r0 (t0 ) to any curve r(t) passing through P.
Definition of tangent plane and normal line
MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
Definition
The tangent plane to the level surface S given by
F (x, y , z) = C at P(x0 , y0 , z0 ) is the plane passing through P
with normal vector ∇F (x0 , y0 , z0 ), i.e., with the equation
The gradient
Tangent
planes to level
surfaces
∂F
(x0 , y0 , z0 )(x − x0 )
∂x
∂F
∂F
(x0 , y0 , z0 )(y − y0 ) +
(x0 , y0 , z0 )(z − z0 ) = 0.
+
∂y
∂z
The normal line to S through P is the line passing through P
perpendicularly to the tangent plane, i.e., with the symmetric
equations
x − x0
y − y0
z − z0
=
=
.
Fx (x0 , y0 , z0 )
Fy (x0 , y0 , z0 )
Fz (x0 , y0 , z0 )
Examples, IV
MATH 209—
Calculus,
III
Volker Runde
Example
Let E be the ellipsoid given by
x 2 (y − 1)2
+
+ (z + 2)2 = 4.
2
4
Directional
derivatives
The gradient
Tangent
planes to level
surfaces
Find the tangent plane and the normal line to E at
P(−2, 3, −1).
Set
x 2 (y − 1)2
F (x, y , z) =
+
+ (z + 2)2 .
2
4
Then E is the level surface F (x, y , z) = 4.
We have
∂F
= x,
∂x
∂F
1
= (y − 1),
∂y
2
and
∂F
= 2(z + 2).
∂z
Examples, V
MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
Example (continued)
The gradient
We obtain:
Tangent
planes to level
surfaces
∂F
(−2, 3, −1) = −2,
∂x
∂F
(−2, 3, −1) = 1,
∂y
and
∂F
(−2, 3, −1) = 2.
∂z
Examples, VI
MATH 209—
Calculus,
III
Volker Runde
Directional
derivatives
Example (continued)
Hence:
the tangent plane has the equation
The gradient
Tangent
planes to level
surfaces
−2(x + 2) + (y − 3) + 2(z + 1) = 0,
i.e.,
−2x + y + 2z = 5;
the normal line has the symmetric equations
−
x +2
z +1
=y −3=
.
2
2