01 Directional Derivatives and Gradient.pdf

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Directional Derivatives and the Gradient Vector

Lucky Galvez

Institute of Mathematics University of the Philippines

Diliman

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Recall: Partial Derivatives

Letz=f(x, y).

∂z

∂x = fx(x, y) = limh→0

f(x+h, y)−f(x, y)

h ∂z

∂y = fy(x, y) = limh→0

f(x, y+h)−f(x, y)

h

R fx represents the rate of change of f with respect tox when y is fixed

R fy represents the rate of change of f with respect toy when x is fixed

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Recall: Partial Derivatives

Letz=f(x, y).

∂z

∂x = fx(x, y) = limh→0

f(x+h, y)−f(x, y)

h ∂z

∂y = fy(x, y) = limh→0

f(x, y+h)−f(x, y)

h

R fx represents the rate of change of f with respect tox when y is fixed

R fy represents the rate of change of f with respect toy when x is fixed

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Directional Derivative

Suppose we want to find the rate of change off at (x0, y0) in

the direction of a unit vectoru=ha, bi.

That is, the slope of the tangent line toC atP.

If Q(x, y, z) is another point on C and P0, Q0

are the projections of P

and Q on the xy-plane, resp., then the vector

~

P0_Q0 _{is parallel to} _u _so for some scalarh,

~

P0_Q0₌_hu₌_h_{ha, hb}_i_.

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Directional Derivative

the direction of a unit vectoru=ha, bi. That is, the slope of the tangent line toC atP.

~

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Directional Derivative

~

P0_Q0 _{is parallel to} _u

so for some scalarh,

~

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Directional Derivative

~

P0_Q0 ₌_hu₌_h_{ha, hb}_i_.

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Directional Derivative

Therefore,

x−x0 =ha ⇒ x=x0+ha

y−y0=hb ⇒ y=y0+hb

The rate of change off (with respect to distance) in the direction ofu is

lim h→0

∆f

h = hlim→0

f(x, y)−f(x0, y0)

h

= lim h→0

f(x0+ha, y0+hb)−f(x0, y0)

h

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Therefore,

lim h→0

∆f h

= lim h→0

f(x, y)−f(x0, y0)

h

= lim h→0

h

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Therefore,

lim h→0

∆f

h = hlim→0

f(x, y)−f(x0, y0)

h

= lim h→0

h

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Directional Derivative

Therefore,

lim h→0

∆f

h = hlim→0

f(x, y)−f(x0, y0)

h

= lim h→0

h

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Definition

Thedirectional derivative off(x, y) at (x0, y0) in the

direction of a unit vectoru=ha, bi is

Duf(x0, y0) = lim h→0

h

provided this limit exists.

Remarks:

1 If u= ˆı=h1,0i, then

Dˆıf = lim h→0

f(x0+h, y0)−f(x0, y0)

h =fx(x0, y0).

2 _If _u_{= ˆ}_₌h₀_,₁i_{, then}

Dˆf = lim h→0

f(x0, y0+h)−f(x0, y0)

h =fy(x0, y0).

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Directional Derivative

Definition

h

Remarks:

Dˆıf = lim h→0

f(x0+h, y0)−f(x0, y0)

h

=fx(x0, y0).

2 _If _u_{= ˆ}_₌h₀_,₁i_{, then}

Dˆf = lim h→0

f(x0, y0+h)−f(x0, y0)

h =fy(x0, y0).

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Definition

h

Remarks:

Dˆıf = lim h→0

f(x0+h, y0)−f(x0, y0)

h =fx(x0, y0).

2 _If _u_{= ˆ}_₌h₀_,₁i_{, then}

Dˆf = lim h→0

f(x0, y0+h)−f(x0, y0)

h =fy(x0, y0).

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Directional Derivative

Definition

h

Remarks:

Dˆıf = lim h→0

f(x0+h, y0)−f(x0, y0)

h =fx(x0, y0).

2 If u= ˆ=h0,1i, then

Dˆf = lim h→0

f(x0, y0+h)−f(x0, y0)

h

=fy(x0, y0).

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Definition

h

Remarks:

Dˆıf = lim h→0

f(x0+h, y0)−f(x0, y0)

h =fx(x0, y0).

2 If u= ˆ=h0,1i, then

Dˆf = lim h→0

f(x0, y0+h)−f(x0, y0)

h =fy(x0, y0).

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Directional Derivative

Theorem

If f is a differentiable function of x andy, then f has a directional derivative at any point(x0, y0) in the domain of f,

in the direction of any unit vectoru=ha, bi and

Duf(x0, y0) =fx(x0, y0)a+fy(x0, y0)b.

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

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

Letg(h) =f(x, y) wherex=x0+ha,y=y0+hb. By Chain Rule,

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

dh =fx(x, y)a+fy(x, y)b

Note thath= 0⇒x=x0, y=y0 sog0(0) =fx(x0, y0)a+fy(x0, y0)b. Hence,

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Directional Derivative

Theorem

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

Then

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

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Directional Derivative

Theorem

g0(0) = lim h→0

g(h)−g(0)

h

= lim h→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

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Directional Derivative

Theorem

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

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Directional Derivative

Theorem

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

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Directional Derivative

Theorem

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy dh

=fx(x, y)a+fy(x, y)b

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Directional Derivative

Theorem

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

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Directional Derivative

Theorem

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

Note thath= 0⇒x=x0, y=y0

sog0(0) =fx(x0, y0)a+fy(x0, y0)b. Hence,

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Directional Derivative

Theorem

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

Note thath= 0⇒x=x0, y=y0 sog0(0) =fx(x0, y0)a+fy(x0, y0)b.

Hence,

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Directional Derivative

Theorem

g0(0) = lim h→0

g(h)−g(0)

h = limh→0

h

= Duf(x0, y0).

g0(h) =∂f

∂x dx dh+

∂f ∂y

dy

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Directional Derivative

Example

EvaluateDuf(1,2) if f(x, y) =x2−2xy−y2 ifu=

D√

2 2 ,−

√

2 2

E .

Solution. First, we compute the partial derivatives

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

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Example

D√

2 2 ,−

√

2 2

E .

fx(x, y)

= 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

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Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y

⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

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Directional Derivative

Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2)

=−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

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Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

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Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y)

=−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

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Directional Derivative

Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y

⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

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Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2)

=−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

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Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

(36)

Directional Derivative

Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

(37)

Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2 ! = 2 √ 2

(38)

Example

D√

2 2 ,−

√

2 2

E .

fx(x, y) = 2x−2y ⇒ fx(1,2) =−2

fy(x, y) =−2x−2y ⇒ fy(1,2) =−6

Hence,

Duf(1,2) = fx(1,2)

√ 2 2

!

+fy(1,2) −

√ 2 2

!

= −2 √

2 2

!

−6 −

√ 2 2

!

= 2√2

(39)

Directional Derivative

Example

Determine the directional derivative off(x, y) =xy2−cos(xy) at any point, in the direction of the vector 3ˆı−4ˆ.

Solution. The unit vector in the same direction as the given

vector is

u= h3,−4i k h3,−4i k =

h3,−4i

p

32_{+ (}₋₄₎2 =

3 5,−

4 5

Hence,

Duf(x, y) = fx(x, y)

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

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Example

vector is

u= h3,−4i k h3,−4i k

= p h3,−4i 32_{+ (}₋₄₎2 =

3 5,−

4 5

Hence,

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

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Directional Derivative

Example

vector is

u= h3,−4i k h3,−4i k =

h3,−4i

p

32_{+ (}₋₄₎2

=

3 5,−

4 5

Hence,

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

(42)

Example

vector is

u= h3,−4i k h3,−4i k =

h3,−4i

p

32_{+ (}₋₄₎2 =

3 5,−

4 5

Hence,

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

(43)

Directional Derivative

Example

vector is

u= h3,−4i k h3,−4i k =

h3,−4i

p

32_{+ (}₋₄₎2 =

3 5,−

4 5

Hence,

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

(44)

Example

vector is

u= h3,−4i k h3,−4i k =

h3,−4i

p

32_{+ (}₋₄₎2 =

3 5,−

4 5

Hence,

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

(45)

Directional Derivative

Example

vector is

u= h3,−4i k h3,−4i k =

h3,−4i

p

32_{+ (}₋₄₎2 =

3 5,−

4 5

Hence,

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

(46)

Example

vector is

u= h3,−4i k h3,−4i k =

h3,−4i

p

32_{+ (}₋₄₎2 =

3 5,−

4 5

Hence,

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

(47)

Directional Derivative

Example

vector is

u= h3,−4i k h3,−4i k =

h3,−4i

p

32_{+ (}₋₄₎2 =

3 5,−

4 5

Hence,

3 5

+fy(x, y)

−4 5

= y2+ysin(xy)

3 5

+ (2xy+xsin(xy))

−4 5

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(x0, y0)

u

a b

θ

If the unit vectoru=ha, bi, makes an angleθ with thex-axis, then

cosθ= a

kuk =a and sinθ=

b

kuk =b and the formula in the previous theorem becomes

Duf(x, y) =fx(x, y) cosθ+fy(x, y) sinθ.

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Directional Derivative

(x0, y0)

u

a b

θ

cosθ

= a

kuk =a and sinθ=

b

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(x0, y0)

u

a b

θ

cosθ= a kuk

=a and sinθ= b kuk =b and the formula in the previous theorem becomes

(51)

Directional Derivative

(x0, y0)

u

a b

θ

cosθ= a kuk =a

and sinθ= b kuk =b and the formula in the previous theorem becomes

(52)

(x0, y0)

u

a b

θ

cosθ= a

kuk =a and sinθ

= b

(53)

Directional Derivative

(x0, y0)

u

a b

θ

cosθ= a

kuk =a and sinθ=

b

kuk

=b

and the formula in the previous theorem becomes

(54)

(x0, y0)

u

a b

θ

cosθ= a

kuk =a and sinθ=

b

kuk =b

and the formula in the previous theorem becomes

(55)

Directional Derivative

(x0, y0)

u

a b

θ

cosθ= a

kuk =a and sinθ=

b

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Example

Find the directional derivative off(x, y) =ex2y−2x+y at the point (−1,0) in the direction of the unit vector given byθ= π₃.

Solution: From the previous formula,

Duf(x, y) = fx(x, y) cosθ+fy(x, y) sinθ

=

2xyex2y−2

cosπ 3 +

x2ex2y+ 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(57)

Directional Derivative

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y+ 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(58)

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y+ 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(59)

Directional Derivative

Example

=

2xyex2y−2

cosπ 3

+

x2ex2y+ 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(60)

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y + 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(61)

Directional Derivative

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y + 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(62)

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y + 1

sinπ 3

Hence,

Duf(−1,0)

= −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(63)

Directional Derivative

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y + 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(64)

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y + 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2 √

3 2

!

= −1 + √

3

(65)

Directional Derivative

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y + 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(66)

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y + 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

(67)

Directional Derivative

Example

=

2xyex2y−2

cosπ 3 +

x2ex2y + 1

sinπ 3

Hence,

Duf(−1,0) = −2

1 2

+ 2

√ 3 2

!

= −1 + √

3

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The Gradient Vector

From the previous theorem,

Duf(x, y) = fx(x, y)a+fy(x, y)b

= hfx(x, y), fy(x, y)i · ha, bi = hfx(x, y), fy(x, y)i ·u

The vectorhfx(x, y), fy(x, y)i appears not only in the formula for directional derivative but in many applications as well.

This vector is called thegradientof f, denoted grad f or∇f.

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The Gradient Vector

Duf(x, y) = fx(x, y)a+fy(x, y)b = hfx(x, y), fy(x, y)i · ha, bi

= hfx(x, y), fy(x, y)i ·u

(70)

The Gradient Vector

Duf(x, y) = fx(x, y)a+fy(x, y)b = hfx(x, y), fy(x, y)i · ha, bi = hfx(x, y), fy(x, y)i ·u

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The Gradient Vector

Duf(x, y) = fx(x, y)a+fy(x, y)b = hfx(x, y), fy(x, y)i · ha, bi = hfx(x, y), fy(x, y)i·u

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The Gradient Vector

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The Gradient Vector

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The Gradient Vector

Definition

Iff is a function ofx andy, then the gradient of f is the vector function∇f defined as

∇f(x, y) =hfx(x, y), fy(x, y)i

Thus, the directional derivetive off in the direction of a unit vectoru=ha, bi can be written as

Duf(x, y) =∇f(x, y)·u

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The Gradient Vector

Definition

Iff is a function ofx andy, then the gradient of f is the vector function∇f defined as

∇f(x, y) =hfx(x, y), fy(x, y)i

Thus, the directional derivetive off in the direction of a unit vectoru=ha, bi can be written as

Duf(x, y) =∇f(x, y)·u

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Example

Find the directional derivative off(x, y) =x2lny at the point (2,1) in the direction of 2ˆı+ ˆ.

Solution: We normalize the given vector

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

Next, compute the gradient

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y)

= hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i

=

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,

x2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,

4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u

=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Example

u=√h2,1i

22_{+ 1}2 =

₂ √

5, 1

√

5

∇f(x, y) = hfx(x, y), fy(x, y)i =

2xlny,x

2

y

⇒ ∇f(2,1) = h0,4i

Hence,

Duf(2,1) =∇f(2,1)·u=h0,4i · ₂

√

5, 1

√

5

=√4

5

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Functions of Three Variables

Definition

Thedirectional derivative off(x, y, z) at (x0, y0, z0) in the

direction of the unit vectoru=ha, b, ci is

lim h→0

f(x0+ha, y0+hb, z0+hc)−f(x0, y0, z0)

h

Definition

The gradient vector off(x, y, z) is

∇f(x, y, z) =hfx(x, y, z), fy(x, y, z), fz(x, y, z)i

Duf(x, y, z) = fx(x, y, z)a+fy(x, y, z)b+fz(x, y, z)c = ∇f(x, y, z)·u

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Functions of Three Variables

Definition

lim h→0

f(x0+ha, y0+hb, z0+hc)−f(x0, y0, z0)

h

Definition

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Functions of Three Variables

Definition

lim h→0

f(x0+ha, y0+hb, z0+hc)−f(x0, y0, z0)

h

Definition

Duf(x, y, z) = fx(x, y, z)a+fy(x, y, z)b+fz(x, y, z)c

= ∇f(x, y, z)·u

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Functions of Three Variables

Definition

lim h→0

f(x0+ha, y0+hb, z0+hc)−f(x0, y0, z0)

h

Definition

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The Gradient Vector

Theorem

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

maximum value of Duf isk∇fk and it occurs when u is in the

same direction as ∇f.

Proof. Let θbe the angle between ∇f and u.

Duf(x, y) = ∇f ·u

= k∇fkkukcosθ

= k∇fkcosθ

and the result follows since the maximum value of cosθis 1 and is attained whenθ= 0, i.e.,∇f andu are in the same direction.

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The Gradient Vector

Theorem

Proof. Let θbe the angle between∇f and u.

= k∇fkkukcosθ

= k∇fkcosθ

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The Gradient Vector

Theorem

= k∇fkkukcosθ

= k∇fkcosθ

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The Gradient Vector

Theorem

= k∇fkkukcosθ

= k∇fkcosθ

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The Gradient Vector

Theorem

= k∇fkkukcosθ

= k∇fkcosθ

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The Gradient Vector

Theorem

= k∇fkkukcosθ

= k∇fkcosθ

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Example

Suppose that the height of a hill above sea level is modelled by

h(x, y) = 1000−0.02x2−0.01y2. Standing at the point (50,80), in which direction is the elevation changing fastest? What is the maximum rate of change of the elevation?

Solution: The maximum rate of change occurs in the

direction of∇h(50,80).

∇h(x, y) = h−0.04x, −0.02yi

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

and the maximum rate of change ofh (Duh) is

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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Example

∇h(x, y) = h−0.04x, −0.02yi

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

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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Example

∇h(x, y)

= h−0.04x, −0.02yi

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

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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Example

∇h(x, y) = h−0.04x,

−0.02yi

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

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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Example

∇h(x, y) = h−0.04x, −0.02yi

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

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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Example

∇h(x, y) = h−0.04x, −0.02yi ⇒ ∇h(50,80) = h−0.04(50),−0.02(80)i

=h−2,−1.6i

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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Example

∇h(x, y) = h−0.04x, −0.02yi

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

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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Example

∇h(x, y) = h−0.04x, −0.02yi

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

k∇h(50,80)k

=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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Example

∇h(x, y) = h−0.04x, −0.02yi

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

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2

= 2.5612.

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Example

∇h(x, y) = h−0.04x, −0.02yi

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

k∇h(50,80)k=p(−2)2_{+ (}₋₁_.₆₎2 _{= 2}_.₅₆₁₂_.

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The Gradient Vector

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Exercises

1 _{Find the directional derivative of the given function at the given}

point in the direction of the vectorv.

a. f(r, θ) =e−r_sin_θ_{, 0}_,π

3

, v= 3ˆı−2ˆ

b. g(x, y, z) =_y₊x_z, (4,1,1),v=h1,2,3i

2 _If_f₍_{x, y}_{) =}_xey_{, find the rate of change of}_f _at _P₍₂_,_{0) in the}

direction fromP toQ 1₂,2. In what direction is f changing fastest? What is the maximum rate of change of f?

3 _{Find all points for which the direction of fastest change of} f(x, y) =x2₊_y2₋₂_x₋₄_y _{is ˆ}_ı_{+ ˆ}__.

4 _If_f _and_g _{are diffentiable functions of}_x_and_y_{, show that} ∇(f g) =f∇g+g∇f and∇

f g

=g∇f−f∇g

g2 .

5 Letf be a function ofxandy and consider the pointsA(1,3), B(3,3), C(1,7) and D(6,15). Given that DAB~ f(1,3) = 3 and DAC~ f(1,3) = 6, find the directional derivative off at Ain the

direction of the vectorAD~ .

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Exercises

a. f(r, θ) =e−r_sin_θ_{, 0}_,π

3

, v= 3ˆı−2ˆ b. g(x, y, z) =_y₊x_z, (4,1,1),v=h1,2,3i

f g

=g∇f−f∇g

g2 .

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Exercises

a. f(r, θ) =e−r_sin_θ_{, 0}_,π

3

f g

=g∇f−f∇g

g2 .

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Exercises

a. f(r, θ) =e−r_sin_θ_{, 0}_,π

3

f g

=g∇f−f∇g

g2 .

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Exercises

a. f(r, θ) =e−r_sin_θ_{, 0}_,π

3

_f

g

=g∇f−f∇g

g2 .

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Exercises

a. f(r, θ) =e−r_sin_θ_{, 0}_,π

3

_f

g

=g∇f−f∇g

g2 .

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References

1 _{Stewart, J.,} _{Calculus, Early Transcendentals}_{, 6 ed., Thomson}

Brooks/Cole, 2008

2 _{Dawkins, P.,}_{Calculus 3}_{, online notes available at}

http://tutorial.math.lamar.edu/