01 Directional Derivatives and Gradient

8/18/2019 01 Directional Derivatives and Gradient

1/130

Directional Derivatives and the Gradient Vector

Institute of Mathematics

University of the PhilippinesDiliman

Directional Derivatives & Gradient


2/130

Recall: Derivatives

Let y = f (x).



3/130

Recall: Derivatives

Let y = f (x).

dy

dx = f (x) = lim

h→0f (x + h) − f (x)

h



4/130

Recall: Derivatives

Let y = f (x).

dy

dx = f (x) = lim

h→0f (x + h) − f (x)

h

f (x) represents the rate of change of y = f (x) with respectto x



5/130

Recall: Partial Derivatives

Let z = f (x, y).

∂z

∂x = f x(x, y) = lim

h→0f (x + h, y) − f (x, y)

h

∂z

∂y = f y(x, y) = lim

h→0f (x, y + h) − f (x, y)

h



6/130

Recall: Partial Derivatives

Let z = f (x, y).

∂z

∂x = f x(x, y) = lim

h→0f (x + h, y) − f (x, y)

h

∂z

∂y = f y(x, y) = lim

h→0f (x, y + h) − f (x, y)

h

f x represents the rate of changeof f with respect to x (indirection parallel to the x-axis)

when y is fixedf y represents the rate of changeof f with respect to y (indirection parallel to the y-axis)when x is fixed



7/130

Directional Derivative

Goal:



8/130


Goal: Study the rate of change of f in an arbitrary direction.



9/130



Suppose we want to find the rate of change of f at (x0, y0) inthe direction of a unit vector u = a, b.



10/130



Suppose we want to find the rate of change of f at (x0, y0) inthe direction of a unit vector u = a, b. That is, the slope of the tangent line to C at P .



11/130




If Q(x,y,z) is anotherpoint on C and P , Q arethe projections of P andQ on the xy-plane, resp.,then the vector P Q isparallel to u



12/130




If Q(x,y,z) is anotherpoint on C and P , Q arethe projections of P andQ on the xy-plane, resp.,then the vector P Q isparallel to u so for somescalar h,

P Q = hu = ha, hb .



13/130


Therefore,

x − x0 = ha ⇒ x = x0 + hay − y0 = hb ⇒ y = y0 + hb



14/130


Therefore,


The rate of change of f (with respect to distance) in thedirection of u at the point (x0, y0) is

limh→0∆f

h



15/130


Therefore,



limh→0∆f

h = limh→0f (x, y)

− f (x0, y0)

h



16/130


Therefore,



limh→0∆f

h = limh→0f (x, y)

− f (x0, y0)

h

= limh→0

f (x0 + ha, y0 + hb) − f (x0, y0)h



17/130


Definition

Let f be a function of x and y. The directional derivative of f at a point (x, y) along a unit vector u = a, b, denoted byDuf (x, y), is given by

Duf (x, y) := limh→0

f (x + ha, y + hb) − f (x, y)h

,

if this limit exists.



18/130


Definition



f (x + ha, y + hb) − f (x, y)h

,


Remarks:1

If u = ı̂ = 1, 0, thenDı̂f (x, y) = lim

h→0f (x + h, y) − f (x, y)

h



19/130


20/130


Definition



f (x + ha, y + hb) − f (x, y)h

,


Remarks:1

If u = ı̂ = 1, 0, thenDı̂f (x, y) = lim

h→0f (x + h, y) − f (x, y)

h = f x(x, y).

2 If u = ˆ = 0, 1, thenDˆ f (x, y) = lim

h→0

f (x, y + h) − f (x, y)

hDirectional Derivatives & Gradient


21/130


Definition



f (x + ha, y + hb) − f (x, y)h

,


Remarks:1

If u = ˆı = 1

,0, then

Dı̂f (x, y) = limh→0

f (x + h, y) − f (x, y)h

= f x(x, y).

2 If u = ˆ = 0, 1, thenDˆ f (x, y) = lim

h→0

f (x, y + h) − f (x, y)

h

= f y(x, y).



22/130

Example 1

Example

Use the definition to find the directional derivative of f (x, y) = x2 + 3y2 along u = √ 3

2 ,−1

2 at the point (1,−1).


l


23/130

Example 1

Example


2 ,−1


Duf (1,−1)


E l 1


24/130

Example 1

Example


2 ,−1


Duf (1,−1) = limh→0

f (1 +√

32 h,−1 − h

2) − f (1,−1)

h


E l 1


25/130

Example 1

Example

Use the definition to find the directional derivative of f (x, y) = x2 + 3y2 along u = √ 32 ,−1


Duf (1,−1) = limh→0

f (1 +√

32 h,−1 − h

2) − f (1,−1)

h

= limh→0

1

h

(1 + √ 32 h

2

+ 3

−1 − h

2

2− 4


E l 1


26/130

Example 1

Example



Duf (1,−1) = limh→0

f (1 +√

32 h,−1 − h

2) − f (1,−1)

h

= limh→0

1

h

(1 + √ 32 h

2

+ 3

−1 − h

2

2− 4

= lim

h→0

1

h1 +

√ 3h +

3

4

h2 + 31 + h + h2

4−

4


E l 1


27/130

Example 1

Example



Duf (1,−1) = limh→0

f (1 +√

32 h,−1 − h

2) − f (1,−1)

h

= limh→0

1

h

(1 + √ 32 h

2

+ 3

−1 − h

2

2− 4

= lim

h→0

1

h1 +

√ 3h +

3

4

h2 + 31 + h + h2

4−

4= lim

h→01

h

√ 3h +

3

2h2 + 3h


E l 1


28/130

Example 1

Example



Duf (1,−1) = limh→0

f (1 +√

32 h,−1 − h

2) − f (1,−1)

h

= limh→0

1

h

(1 + √ 32 h

2

+ 3

−1 − h

2

2− 4

= lim

h→0

1

h1 +

√ 3h +

3

4

h2 + 31 + h + h2

4−

4= lim

h→01

h

√ 3h +

3

2h2 + 3h

= limh→

0√

3 + 3

2h + 3


Example 1


29/130

Example 1

Example



Duf (1,−1) = limh→0

f (1 +√

32 h,−1 − h

2) − f (1,−1)

h

= limh→0

1

h

(1 + √ 32 h

2

+ 3

−1 − h

2

2− 4

= lim

h→0

1

h1 +

√ 3h +

3

4

h2 + 31 + h + h2

4−

4= lim

h→01

h

√ 3h +

3

2h2 + 3h

= limh

→0

√ 3 +

3

2h + 3 =

√ 3 + 3




30/130


Theorem

If f is a differentiable function of x and y, then f has a directional derivative at point (x0, y0) in the domain of f , in the direction of any unit vector u = a, b and

Duf (x0, y0) = f x(x0, y0)a + f y(x0, y0)b.




31/130


Theorem



Proof. Define g(h) = f (x0 + ha, y0 + hb).




32/130


Theorem



Proof. Define g(h) = f (x0 + ha, y0 + hb). Then

g(0) = lim

h→0

g(h) − g(0)

h




33/130


Theorem




g(0) = lim

h→0

g(h) − g(0)

h = lim

h→0

f (x0 + ha, y0 + hb) − f (x0, y0)

h




34/130


Theorem




g(0) = lim

h→0

g(h) − g(0)

h = lim

h→0

f (x0 + ha, y0 + hb) − f (x0, y0)

h

= Duf (x0, y0).




35/130


Theorem




g(0) = lim

h→0

g(h) − g(0)

h = lim

h→0

f (x0 + ha, y0 + hb) − f (x0, y0)

h

= Duf (x0, y0).

Let g(h) = f (x, y) where x = x0 + ha, y = y0 + hb. By Chain Rule,

g(h)




36/130


Theorem




g(0) = lim

h→0

g(h) − g(0)

h = lim

h→0

f (x0 + ha, y0 + hb) − f (x0, y0)

h

= Duf (x0, y0).


g(h) =

∂f

∂x

dx

dh +

∂f

∂y

dy

dh




37/130


Theorem




g(0) = lim

h→0

g(h) − g(0)

h = lim

h→0

f (x0 + ha, y0 + hb) − f (x0, y0)

h

= Duf (x0, y0).


g(h) =

∂f

∂x

dx

dh +

∂f

∂y

dy

dh = f x(x, y)a + f y(x, y)b




38/130

Theorem




g(0) = lim

h→0

g(h) − g(0)

h = lim

h→0

f (x0 + ha, y0 + hb) − f (x0, y0)

h

= Duf (x0, y0).


g(h) =

∂f

∂x

dx

dh +

∂f

∂y

dy


h = 0, x = x0, y = y0 ⇒




39/130

Theorem




g(0) = lim

h→0

g(h) − g(0)

h = lim

h→0

f (x0 + ha, y0 + hb) − f (x0, y0)

h

= Duf (x0, y0).


g(h) =

∂f

∂x

dx

dh +

∂f

∂y

dy


h = 0, x = x0, y = y0 ⇒ g(0) = f x(x0, y0)a + f y(x0, y0)b




40/130

Theorem




g(0) = lim

h→0

g(h) − g(0)

h = lim

h→0

f (x0 + ha, y0 + hb) − f (x0, y0)

h

= Duf (x0, y0).


g(h) =

∂f

∂x

dx

dh +

∂f

∂y

dy


h = 0, x = x0, y = y0 ⇒ g(0) = f x(x0, y0)a + f y(x0, y0)b = Duf (x0, y0)


Example 2


41/130

p

Example

Given f (x, y) = x2

− 2xy − y2

. Find the directional derivative of f at P (1, 2) in the direction from P to Q(4,−1).


Example 2


42/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y)


Example 2


43/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y


Example 2


44/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2)


Example 2


45/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2


Example 2


46/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y)


Example 2


47/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y


Example 2


48/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y

⇒ f y(1, 2)


Example 2


49/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y

⇒ f y(1, 2) =

−6


Example 2


50/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y

⇒ f y(1, 2) =

−6


Example 2


51/130

Example

Given f (x, y) = x2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y

⇒ f y(1, 2) =

−6

Since PQ = 3,−3 is not a unit vector, take u =


Example 2


52/130

Example

Given f (x, y) = x

2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y

⇒ f y(1, 2) =

−6

Since PQ = 3,−3 is not a unit vector, take u = √

22 ,−

√ 2

2


Example 2


53/130

Example

Given f (x, y) = x

2

− 2xy − y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y

⇒ f y(1, 2) =

−6


22 ,−

√ 2

2

Duf (1, 2) = f x(1, 2)

√ 2

2

+ f y(1, 2)

−

√ 2

2


Example 2


54/130

Example

Given f

(x, y

) = x2

− 2xy

−y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y

⇒ f y(1, 2) =

−6


22 ,−

√ 2

2

Duf (1, 2) = f x(1, 2)

√ 2

2

+ f y(1, 2)

−

√ 2

2

= −2√

2

2

− 6

−

√ 2

2


Example 2


55/130

Example

Given f

(x, y

) = x2

− 2xy

−y2


f x(x, y) = 2x − 2y ⇒ f x(1, 2) = −2f y(x, y) =

−2x

− 2y

⇒ f y(1, 2) =

−6


22 ,−

√ 2

2

Duf (1, 2) = f x(1, 2)

√ 2

2

+ f y(1, 2)

−

√ 2

2

= −2√

2

2

− 6

−

√ 2

2

= 2√

2


Example 3


56/130

Example

Determine the directional derivative of f (x, y) = xy2

− cos(xy)at any point, in the direction of the vector 3 ı̂ − 4ˆ .


Example 3


57/130

Example


− cos(xy)at any point, in the direction of the vector 3 ı̂ − 4ˆ .The unit vector in the same direction as the given vector is

u = 3,

−4 3,−4


Example 3


58/130

Example



u = 3,

−4 3,−4 =

3,−

4 32 + (−4)2


Example 3


59/130

Example



u = 3,

−4 3,−4 =

3,−

4 32 + (−4)2 =

3

5 ,−4

5


Example 3


60/130

Example



u = 3,

−4 3,−4 =

3,−

4 32 + (−4)2 =

3

5 ,−4

5

Hence,

Duf (x, y) = f x(x, y)3

5

+ f y(x, y)−45


Example 3


61/130

Example



u = 3,

−4 3,−4 =

3,−

4 32 + (−4)2 =

3

5 ,−4

5

Hence,

Duf (x, y) = f x(x, y)3

5

+ f y(x, y)−45

=y2 + y sin(xy)


Example 3


62/130

Example



u = 3,

−4 3,−4 =

3,−

4 32 + (−4)2 =

3

5 ,−4

5

Hence,

Duf (x, y) = f x(x, y)3

5

+ f y(x, y)−45

=y2 + y sin(xy)

35


Example 3


63/130

Example



u = 3,

−4 3,−4 =

3,−

4 32 + (−4)2 =

3

5 ,−4

5

Hence,

Duf (x, y) = f x(x, y)3

5

+ f y(x, y)−45

=y2 + y sin(xy)

35

+ (2xy + x sin(xy))



64/130



65/130

(x0, y0)

u

a

b

θ

If the unit vector u = a, b, makes an angle θ with the x-axis,then




66/130

(x0, y0)

u

a

b

θ


a = u cos θ




67/130

(x0, y0)

u

a

b

θ


a = u cos θ = cos θ




68/130

(x0, y0)

u

a

b

θ


a = u cos θ = cos θb = u sin θ




69/130

(x0, y0)

u

a

b

θ


a = u cos θ = cos θb = u sin θ = sin θ




70/130

(x0, y0)

u

a

b

θ


a = u cos θ = cos θb = u sin θ = sin θ

and the formula in the previous theorem becomes

Duf (x, y) = f x(x, y)cos θ + f y(x, y)sin θ.


Example 4


71/130

Example

Find the directional derivative of f (x, y) = ex2

y − 2x + y at thepoint (−1, 0) in the direction of the unit vector given by θ = π

3.


Example 4


72/130

Example



3.

From the previous formula,

Duf (x, y) = f x(x, y)cos θ + f y(x, y)sin θ


Example 4


73/130

Example



3.



=

2xyex2y − 2


Example 4


74/130

Example

Find the directional derivative of f (x, y) = ex2y

− 2x + y at thepoint (−1, 0) in the direction of the unit vector given by θ = π

3.



=

2xyex2y − 2

cos

π

3


Example 4


75/130

Example



3.



=

2xyex2y − 2

cos

π

3 +x2ex

2y + 1


Example 4

E l


76/130

Example



3.



=

2xyex2y − 2

cos

π

3 +x2ex

2y + 1

sin π

3


Example 4

E l


77/130

Example



3.



=

2xyex2y − 2

cos

π

3 +x2ex

2y + 1

sin π

3

Hence,

Duf (−1, 0)


Example 4

E l


78/130

Example



3.



=

2xyex2y − 2

cos

π

3 +x2ex

2y + 1

sin π

3

Hence,

Duf (−1, 0) = −2


Example 4

E l


79/130

Example



3.



=

2xyex2y − 2

cos

π

3 +x2ex

2y + 1

sin π

3

Hence,

Duf (−1, 0) = −2

1

2


Example 4

Example


80/130

Example



3.



=

2xyex2y − 2

cos

π

3 +x2ex

2y + 1

sin π

3

Hence,

Duf (−1, 0) = −2

1

2

+ 2


Example 4

Example


81/130

Example



3.



=

2xyex2y − 2

cos

π

3 +x2ex

2y + 1

sin π

3

Hence,

Duf (−1, 0) = −2

1

2

+ 2

√ 3

2


Example 4

Example


82/130

Example



3.



=

2xyex2y − 2

cos

π

3 +x2ex

2y + 1

sin π

3

Hence,

Duf (−1, 0) = −2

1

2

+ 2

√ 3

2

= −1 +

√ 3.


The Gradient Vector


83/130

From the previous theorem,

Duf (x, y) = f x(x, y)a + f y(x, y)b


The Gradient Vector


84/130



= f x(x, y), f y(x, y) · a, b


The Gradient Vector


85/130



= f x(x, y), f y(x, y) · a, b= f x(x, y), f y(x, y) · u


The Gradient Vector


86/130





The Gradient Vector


87/130




The vector f x(x, y), f y(x, y) appears not only in the formulafor directional derivative but in many applications as well.


The Gradient Vector


88/130




The vector f x(x, y), f y(x, y) appears not only in the formulafor directional derivative but in many applications as well.

This vector is called the gradient of f , denoted grad f or ∇

f .


The Gradient Vector

Definition


89/130

Definition

If f is a function of x and y, then the gradient of f is thevector function ∇f defined as

∇f (x, y) = f x(x, y), f y(x, y)


The Gradient Vector

Definition


90/130

Definition

If f is a function of x and y, then the gradient of f is thevector function ∇f defined as

∇f (x, y) = f x(x, y), f y(x, y)

Thus, the directional derivative of f in the direction of a unitvector u = a, b can be written as

Duf (x, y) = ∇f (x, y) · u.

Example 5

Use gradient to find the directional derivative in Example 1.


Functions of Three Variables

Definition

Th di ti l d i ti f f( ) t ( ) i th


91/130

The directional derivative of f (x,y,z) at (x0, y0, z0) in the

direction of the unit vector u = a,b,c islimh→0

f (x0 + ha, y0 + hb, z0 + hc) − f (x0, y0, z0)h

provided this limit exists.



Definition

Th di ti n l d i ti f f( ) t ( ) i th


92/130





Definition

The gradient vector of f (x,y,z) is

∇f (x,y,z) =

f x(x,y,z), f y(x,y,z), f z(x,y,z)



Definition

The directional derivative of f(x y z) at (x y z ) in the


93/130





Definition

The gradient vector of f (x,y,z) is

∇f (x,y,z) =

f x(x,y,z), f y(x,y,z), f z(x,y,z)

Duf (x,y,z) = f x(x,y,z)a + f y(x,y,z)b + f z(x,y,z)c



94/130

The Gradient Vector

Theorem


95/130

Suppose f is a differentiable function of x and y. The

maximum value of Duf is ∇f and it occurs when u is in the same direction as ∇f . The minimum value of Duf is −∇f and it occurs when u is in the opposite direction as ∇f


The Gradient Vector

Theorem

S f i diff i bl f i f d Th


96/130



Proof. Let θ be the angle between ∇f and u.


The Gradient Vector

Theorem

S f i diff ti bl f ti f d Th


97/130




Duf (x, y) = ∇f · u

Di ti l D i ti s & G di t

The Gradient Vector

Theorem



98/130




Duf (x, y) = ∇f · u= ∇f u cos θ

Di ti l D i ti & G di t

The Gradient Vector

Theorem



99/130



Proof.

Let θ be the angle between ∇f and u.Duf (x, y) = ∇f · u

= ∇f u cos θ=

∇f

cos θ


The Gradient Vector

Theorem

Suppose f is a differentiable function of x and y The


100/130



Proof.


= ∇f u cos θ=

∇f

cos θ

Result follows since the maximum value of cos θ is 1 and isattained when θ = 0, i.e., ∇f and u are in the samedirection


The Gradient Vector

Theorem

Suppose f is a differentiable function of x and y The


101/130



Proof.


= ∇f u cos θ=

∇f

cos θ

Result follows since the maximum value of cos θ is 1 and isattained when θ = 0, i.e., ∇f and u are in the samedirection(minimum value occurs at cos θ = −1, i.e., θ = π).


Example 6

Example


102/130

Find the minimum directional derivative of f (x,y,z) = xy2 − x2z + y3 − 1 at the point (3, 0, 2).


Example 6

Example


103/130


∇f (x,y,z)


Example 6

Example


104/130


∇f (x,y,z) = f x(x,y,z), f y(x,y,z), f z(x,y,z)=

y2 − 2xz,


Example 6

Example


105/130



y2 − 2xz, 2xy + 3y2,


Example 6

Example


106/130



y2 − 2xz, 2xy + 3y2, − x2



107/130


108/130

Example 6

Example


109/130



y2 − 2xz, 2xy + 3y2, − x2

⇒ ∇f (3, 0,−2) = 12, 0,−9

Hence, the minimum directional derivative of f at (3, 0, 2) is


Example 6

Example


110/130



y2 − 2xz, 2xy + 3y2, − x2

⇒ ∇f (3, 0,−2) = 12, 0,−9

Hence, the minimum directional derivative of f at (3, 0, 2) is

−∇f (3, 0,−2)



111/130

Example 7

Example

Suppose that the shape of a hill above sea level is given by the


112/130

equation h(x, y) = 1000 − 0.02x2 − 0.01y2, where x, y, andh(x, y) are measured in meters. The positive x−axis points eastand the positive y−axis points north. Suppose you are standingat a point with coordinates (50, 80, 886).

1

If you walk due west, will you start to ascend or descend?At what rate?

2 If you walk due southeast, will you start to ascend ordescend? At what rate?

3 In which direction is the elevation changing fastest (inwhat direction should you proceed initially in order toreach the top of the hill fastest)? What is the rate of ascentin that direction?



113/130

Example 7

∇h(x, y)


114/130


Example 7

∇h(x, y) = −0.04x,


115/130


Example 7

∇h(x, y) = −0.04x, − 0.02y


116/130


Example 7

∇h(x, y) = −0.04x, − 0.02y


117/130

⇒ ∇h(50, 80) = −0.04(50),−0.02(80)


Example 7

∇h(x, y) = −0.04x, − 0.02y


118/130

⇒ ∇h(50, 80) = −0.04(50),−0.02(80) = −2,−1.6



119/130

Example 7

∇h(x, y) = −0.04x, − 0.02y


120/130

⇒ ∇h(50, 80) = −0.04(50),−0.02(80) = −2,−1.61 Due west: u = −1, 0


Example 7

∇h(x, y) = −0.04x, − 0.02y


121/130

⇒ ∇h(50, 80) = −0.04(50),−0.02(80) = −2,−1.61 Due west: u = −1, 0 ⇒ Duh(50, 80)


Example 7

∇h(x, y) = −0.04x, − 0.02y


122/130

⇒ ∇h(50, 80) = −0.04(50),−0.02(80) = −2,−1.61 Due west: u = −1, 0 ⇒ Duh(50, 80) = 2


Example 7

∇h(x, y) = −0.04x, − 0.02y( ) ( ) ( )


123/130

⇒ ∇h(50, 80) = −0.04(50),−0.02(80) = −2,−1.61 Due west: u = −1, 0 ⇒ Duh(50, 80) = 2(ascend)


Example 7

∇h(x, y) = −0.04x, − 0.02yh( ) ( ) ( )


124/130

⇒ ∇h(50, 80) = −0.04(50),−0.02(80) = −2,−1.61 Due west: u = −1, 0 ⇒ Duh(50, 80) = 2(ascend)2 Due SE:


Example 7

∇h(x, y) = −0.04x, − 0.02y∇h(50 80) 0 04(50) 0 02(80) 2 1 6


125/130

⇒ ∇h(50, 80) = −0.04(50),−0.02(80) = −2,−1.61 Due west: u = −1, 0 ⇒ Duh(50, 80) = 2(ascend)2 Due SE: u = 1√

21,−1


Example 7

∇h(x, y) = −0.04x, − 0.02y∇h(50 80) 0 04(50) 0 02(80) 2 1 6


126/130


21,−1 ⇒ Duh(50, 80)


Example 7

∇h(x, y) = −0.04x, − 0.02y∇h(50 80) 0 04(50) 0 02(80) 2 1 6


127/130


21,−1 ⇒ Duh(50, 80) = −

√ 2

5


Example 7

∇h(x, y) = −0.04x, − 0.02y⇒ ∇h(50 80) 0 04(50) 0 02(80) 2 1 6


128/130


21,−1 ⇒ Duh(50, 80) = −

√ 2

5 (descend)


Example 7

∇h(x, y) = −0.04x, − 0.02y⇒ ∇h(50 80) 0 04(50) 0 02(80) 2 1 6


129/130


21,−1 ⇒ Duh(50, 80) = −

√ 2

5 (descend)

3 The maximum rate of change occurs in the direction of ∇h(50, 80) and the rate of ascent is given by∇h(50, 80) =

(−2)2 + (−1.6)2 = 2.5612.


Example 7

∇h(x, y) = −0.04x, − 0.02y⇒ ∇h(50 80) 0 04(50) 0 02(80) 2 1 6


130/130


21,−1 ⇒ Duh(50, 80) = −

√ 2

5 (descend)

3 The maximum rate of change occurs in the direction of ∇h(50, 80) and the rate of ascent is given by∇h(50, 80) =

(−2)2 + (−1.6)2 = 2.5612.

Remark

We say that at (x0, y0), f will increase most rapidly in thedirection of ∇f (x0, y0), or that ∇f (x0, y0) points in thedirection of the steepest ascent.


Documents

01 Directional Derivatives and Gradient