01 Directional Derivatives and Gradient - Handout

8/9/2019 01 Directional Derivatives and Gradient - Handout

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Directional Derivatives and the Gradient Vector

Math 55 - Elementary Analysis III

Institute of Mathematics

University of the PhilippinesDiliman

Math 55 Directional Derivatives & Gradient 1/ 21


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Recall: Partial Derivatives

Let z=f(x, y).

z

x = fx(x, y) = lim

h0f(x+h, y)f(x, y)

h

z

y = fy(x, y) = lim

h0f(x, y+h)f(x, y)

h

fx represents the rate of changeoffwith respect to x when y is

fixed

fy represents the rate of changeoffwith respect to y when x isfixed



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Recall: Partial Derivatives

Find z

x.

1 z=xy2 sin3(3x).2 z=xy +yx + log5(4x

2 y).3 z= tan1(3x

y2) + y

x

.

1z

x=y2 3 sin2(3x) cos(3x)3

2

z

x =yxy1 +yx ln y+

1

(4x2 y) l n 58x

3z

x=

1

1 + (3xy2)23 +1

2

yx

1/2 yx2.



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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 vector u=a, b. That is, the slope ofthe tangent line to C at P.

If Q(x,y,z) is anotherpoint on C and P, Q

are the projections ofPand Q on the xy-plane,resp., then the vectorPQ is parallel to u so

for some scalar h,

PQ =hu=ha, hb .



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Therefore,

xx0 =ha x= x0+hayy0 =hb y=y0+hb

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

limh0

f

h

= limh0

f(x, y)f(x0, y0)h

= limh0

f(x0+ha, y0+hb)f(x0, y0)h



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Definition

The directional derivative off(x, y) at (x0, y0) in thedirection of a unit vector u=a, b is

Duf(x0, y0) = limh0

f(x0+ha, y0+hb)f(x0, y0)h

provided this limit exists.

Remarks:

1 Ifu= =

1, 0

, then

Df= limh0

f(x0+a, y0)f(x0, y0)h

=fx(x0, y0).

2 Ifu= =0, 1, thenDf= lim

h

0

f(x0, y0+b)f(x0, y0)h

=fy(x0, y0).



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Theorem

Iff is a differentiable function ofx andy, thenf has a

directional derivative at any point(x0, y0) in the domain off,in the direction of any unit vector u=a, b and

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

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

g(0) = lim

h0

g(h) g(0)

h = lim

h0

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) = fx

dxdh

+ fy

dydh

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

Note that h= 0 x= x0, y=y0 so g(0) =fx(x0, y0)a+fy(x0, y0)b.

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



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Example

Example

EvaluateDu

f(1, 2) iff(x, y) =x

2

2xy y2

ifu=2

2 ,

2

2

.

Solution. First, we compute the partial derivatives

fx(x, y) = 2x2y fx(1, 2) =2

fy(x, y) =2x2y fy(1, 2) =6Hence,

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

2

2

+fy(1, 2)

2

2

= 22

2

6

22

= 2

2



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Example

Example

Determine the directional derivative off(x, y) =xy2

cos(xy)

at any point, in the direction of the vector 3 4.

Solution. The unit vector in the same direction as the givenvector is

u= 3,4 3,4 =

3,432 + (4)2 =

3

5,4

5

Hence,

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

35

+fy(x, y)

4

5

=

y2 +y sin(xy)

3

5

+ (2xy+x sin(xy))

4

5


Di i l D i i


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(x0, y0)

u

a

b

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

a = u cos = cos b = u sin = sin

and the formula in the previous theorem becomes

Duf(x, y) =fx(x, y)cos +fy(x, y)sin .


E l


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Example

Example

Find the directional derivative off(x, y) =ex2y

2x+y at the

point (1, 0) in the direction of the unit vector given by = 3

.

Solution: From the previous formula,

Duf(x, y) = fx(x, y)cos +fy(x, y)sin =

2xyex

2y 2

cos

3+x2ex

2y + 1

sin

3

Hence,

Duf(1, 0) = 2

1

2

+ 2

32

= 1 +

3


Th G di t V t


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The Gradient Vector

From the previous theorem,

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

=

fx(x, y), fy(x, y)

a, b

= fx(x, y), fy(x, y) u

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

This vector is called the gradient off, denoted grad f orf.


Th G di t V t


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The Gradient Vector

Definition

Iff is a function ofxand y, then the gradient off is the vectorfunctionfdefined as

f(x, y) =

fx

(x, y), fy

(x, y)

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

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


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Example

Example

Find the directional derivative off(x, y) =x2 ln y at the point

(2, 1) in the direction of 2 + .

Solution: We normalize the given vector

u= 2, 12

2

+ 12

= 25 , 1

5Next, compute the gradient

f(x, y) = fx(x, y), fy(x, y) =

2x ln y,x2

y

f(2, 1) = 0, 4Hence,

Duf(1, 2) =f(1, 2) u=0, 4

25,

15

=

45


Functions of Three Variables


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Functions of Three Variables

Definition

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

direction of the unit vector u=a,b,c islimh0

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

provided this limit exists.

Definition

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

f(x,y,z) =fx(x,y,z), fy(x,y,z), fz(x,y,z)

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

=

f(x,y,z)

u


The Gradient Vector


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The Gradient Vector

Theorem

Supposef is a differentiable function ofx andy. The

maximum value ofDuf isf and it occurs whenu is in thesame direction asf.

Proof. Let be the angle betweenf and u.Duf(x, y) = f u

= fu cos

= f cos

which is maximum when cos is 1 and is attained when = 0,i.e.,f and u are in the same direction.


Example


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Example

Example

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

h(t) = 1000 0.02x2 0.01y2. Standing at the point (50, 80), inwhich direction is the elevation changing fastest? What is themaximum rate of change of the elevation?

Solution:

The maximum rate of change occurs in thedirection off(50, 80).h(x, y) = 0.04x,0.02y

h(50, 80) = 0.04(50),0.02(80) =2,1.6

and the maximum rate of change ofh (Duf) is

f(1, 2)=

(2)2 + (1.6)2 = 2.5612.


The Gradient Vector


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The Gradient Vector


The Gradient Vector


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The Gradient Vector

Theorem

Iff andg are differentiable functions ofx andy, show that

(fg) =fg+gf andf

g

=

gffgg2

.

Example

FindF(x, y) ifF(x, y) =x2 sin(3xy).Solution: Let f(x, y) =x2 and g(x, y) = sin(3xy), thusF(x, y) =f(x, y)g(x, y). Hence,

F(x, y) = x2 3y cos(3xy), 3x cos(3xy) + sin(3xy) 2x, 0=

3x2y cos(3xy), 3x3 cos(3xy)

+2x sin(3xy), 0

=

3x2y cos(3xy) + 2x sin(3xy), 3x3 cos(3xy)


Exercises


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Exercises

1 Find the directional derivative of the given function at the givenpoint in the direction of the vector v.

a. f(r, ) =er sin ,

0, 3

, v= 32

b. g(x,y,z) = xy+z

, (4, 1, 1), v=1, 2, 32 Find the directional derivative off(x, y) = log7(x

2y) +xy in the

direction given by =

6, at the point (1, 2).

3 Iff(x, y) =xey, find the rate of change off at P(2, 0) in thedirection from P to Q

1

2, 2

. In what direction is fchangingfastest? What is the maximum rate of change off?

4 Find all points for which the direction of fastest change of

f(x, y) =x2

+y2

2x4y is + .5 Let fbe a function ofx and y and consider the points A(1, 3),

B(3, 3), C(1, 7) and D(6, 15). Given that D ABf(1, 3) = 3 andD ACf(1, 3) = 6, find the directional derivative off at A in the

direction of the vector AD.


References


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References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/


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01 Directional Derivatives and Gradient - Handout