14.6 Directional Derivativess & Gradient Vectors

Chapter 14 – Partial Derivatives
14.6 Directional Derivatives and the Gradient Vector
Objectives:
 Determine the directional
derivative of a vector
 Determine the gradient of a
vector
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14.6 Directional Derivatives and the Gradient Vector
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Definition – Directional Derivative
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Visualization

Directional Derivatives
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Theorem
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Note

If the unit vector u makes an angle  with the
positive x-axis, then we can write
u = <cos , sin  > and the formula in theorem 3
becomes
Du f  x, y   f x  x, y  cos  f y  x, y  sin 
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Example 1- pg. 920 # 4

Find the directional derivative of f at the
given point in the direction indicated by
the angle θ.
f ( x, y )  x y  y , (2,1),  
2
3
4
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Example 2

Find the directional derivative of f at the
given point in the direction indicated by
the angle θ.
f ( x, y )  x sin( xy ), (2, 0),  
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
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
The gradient of a multi-variable function has a
component for each direction.

Be careful not to confuse the coordinates and the
gradient. The coordinates are the current location,
measured on the x-y-z axis.

The gradient is a direction to move from our current
location, such as move up, down, left or right.

The gradient at any location points in the direction
of greatest increase of a function.
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
We can now rewrite the directional
derivative as
which expresses the directional derivative
in the direction of u as the scalar
projection of the gradient vector onto u.
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
What is the difference between the two?
◦ The gradient is a vector quantity that points in
the direction of greatest increase and has a
magnitude equal to the rate of change in that
direction.
◦ The directional derivative is a scalar quantity
giving the rate of change in a specified
direction. It is equal to the dot product of the
gradient with a unit vector in the desired
direction.
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
What is the difference between the two?
◦ The gradient uses all of the partials.
◦ A directional derivative uses only one partial.
◦ The gradient is a vector orthogonal to a level
curve (surface, etc.).
◦ A directional derivative measures how function
values change with respect to movement in a
particular direction in the domain...and it is not a
vector
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Example 3 – pg. 943 # 8

2

y
1
f ( x, y ) 
, P(1,2), u  2i  5 j
x
3
b) Evaluate the gradient at the point P.
c) Find the rate of change of f at P in the
direction of vector u.
a)
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Definition – Directional Derivative

If we use vector notation, then the
definition is more compact.
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
and can be rewritten as
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Example 4
2 3 6
f ( x, y, z )  x  yz , P(1,3,1), u  , ,
7 7 7
b) Evaluate the gradient at the point P.
c) Find the rate of change of f at P in the
direction of vector u.
a)
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Maximizing the Directional Derivative

Visualization
◦ Maximizing the Directional Derivative
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Example 5

Find the maximum rate of change of f at
the given point and the direction in which
it occurs.
p
q
f ( p, q)  qe  pe , (0,0)
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Example 6

Find the maximum rate of change of f at
the given point and the direction in which
it occurs.
f ( x, y, z)  tan(x  2 y  3z), (5,1,1)
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
The equation ∇F(x0,y0,z0)∙r’(t0) says that
the gradient vector at P, ∇F(x0,y0,z0),is
perpendicular to the tangent vector r’(t0)
to any curve C on S that passes through P.
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Definition

If ∇F(x0,y0,z0)≠0, it is thus natural to
define the tangent plane to the level surface F(x, y, z) = k
at P(x0, y0, z0) as the plane that passes through P and has
normal vector ∇F(x0,y0,z0). This equation of a tangent
plane can be written as
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Normal Line


The normal line to S at P is the line:
◦ Passing through P
◦ Perpendicular to the tangent plane
The direction of the normal line is given by the gradient
vector
∇F(x0,y0,z0)
and its symmetric equations are
x  x0
y  y0
z  z0


Fx ( x0 , y0 , z0 ) Fy ( x0 , y0 , z0 ) Fz ( x0 , y0 , z0 )
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Example 7

Find the equations of (a) the tangent plane
and (b) the normal line to the given
surface at the specified point.
x  z  4 arctan(yz), (1   ,1,1)
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Group work

1. Pg. 944 # 28

Find the directions in which the
directional derivative of the function
below at the point (0,2) has the value of 1.
 xy
f ( x, y)  ye
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Group work
2. Pg. 944 # 32
 The temperature at a point (x, y, z) is given by

T ( x, y, z)  200e
 x 2 3 y 2 9 z 2
Where T is measured in oC and x, y, z in meters.
a) Find the rate of change of temperature at the point
P(2,-1,2) in the direction towards the point (3, -3, 3).
b) In which direction does the temperature increase the
fastest at P?
c) Find the maximum rate of increase at P.

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Group work

3. Pg. 944 # 34

Suppose you are climbing a hill whose shape is given by the
equation z  1000 0.005x2  0.01y 2, where x, y, and z are
measured in meters, and you are standing at the point with
coordinates (60, 40, 996). The positive x-axis points east and
the positive y-axis points another.
a)
b)
c)
If you walk due south, will you start to ascend or descend? At
what rate?
If you walk northwest, will you start to ascend or descend? At
what rate?
In which direction is the slope the largest? What is the rate of
ascent in that direction? At what angle above the horizontal does
the path in that direction begin?
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Group work


4. Pg. 945 # 50
2
2
If g ( x, y )  x  y  4 x find the gradient vector
g (1,2) and use it to find the tangent line to the level
curve g(x, y)=1 at the point (1,2). Sketch the level
curve, the tangent line, and the gradient vector.
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More Examples
The video examples below are from section
on your own time for extra instruction.
Each video is about 2 minutes in length.
◦ Example 4
◦ Example 5
◦ Example 8
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14.6 Directional Derivatives and the