Lec6[1]DerivadaDireccional y Gradiente

8/13/2019 Lec6[1]DerivadaDireccional y Gradiente

1/20

MATH 209

Calculus,III

Volker Runde

Directionalderivatives

The gradient

Tangentplanes to levelsurfaces

MATH 209Calculus, III

Volker Runde

University of Alberta

Edmonton, Fall 2011


2/20

MATH 209

Calculus,III

Volker Runde


The gradient


Directional derivatives, I

Recall...

Let z=f(x, y) be a function of two variables. Then:fx

: slope of the tangent line to the graph off in the

direction of the x-axis;fy

: slope of the tangent line to the graph off in thedirection of the y-axis.

QuestionWhat about the slope of a tangent line to the graph off in thedirection of an arbitrary line in the xy-plane?


3/20

MATH 209

Calculus,III

Volker Runde


The gradient


Directional derivatives, II

DefinitionThedirectional derivativeoff at (x0, y0) in the direction of theunit vector u=a, b is the limit

Duf(x0, y

0) = lim

h0

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

if it exists.

Example

What is Duf(1, 2), for f(x, y) =x3

3xy+ 4y2

and ugiven bythe angle = 6 , i.e.,

u= cos

6, sin

6 =

3

2 ,

1

2

?


4/20

MATH 209

Calculus,III

Volker Runde


The gradient


Directional derivatives, III

An application of the chain rule

Set g(t) :=f(x0+ta, y0+tb), so that

Duf(x0, y0) = limh0

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

= limh0

g(h)g(0)h

= dgdt

(0).


5/20

MATH 209

Calculus,III

Volker Runde


The gradient


Directional derivatives, IV

An application of the chain rule (continued)

With g(t) =f(x, y), x=x0+ta, and y=y0+tb, the chainrule yields

Duf(x0, y0) = dg

dt(0)

=f

x(x0, y0)

dx

dt(0) +

f

y(x0, y0)

dy

dt(0)

=f

x(x0, y0)a+

f

y(x0, y0)b.


6/20

MATH 209

Calculus,III

Volker Runde


The gradient


Directional derivatives, V

Theorem

If f is differentiable, then Duf exists for any unit vectoru=a, b and is computed as

Duf(x0, y0) =f

x(x0, y0)a+

f

y(x0, y0)b.


7/20

MATH 209

Calculus,III

Volker Runde


The gradient


Directional derivatives, V

Example (resumed)

For f(x, y) =x3 3xy+ 4y2 and u=

32 ,

12

, we have

f

x = 3x2 3y and f

y3x+ 8y,

so that

Duf(1, 2) = (36)32

+ (3 + 16) 12

=13332

.


8/20

MATH 209

Calculus,III

Volker Runde


The gradient


Definition of the gradient

Definition

Thegradientoff is the vector function defined by

f(x, y) =fx(x, y), fy(x, y)= fx(x, y)i+fy(x, y)j.

Important

Iff is differentiable, and u is a unit vector, then

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


9/20

MATH 209

Calculus,III

Volker Runde


The gradient


Examples, I

ExampleFind the directional derivative off(x, y) = ln x+e(x1)y at(1, 0) in the direction ofv=i+ 2j.As

fx= 1x

+ye(x1)y and fy= (x1)e(x1)y,

we havef(1, 0) =1, 0.Since|v|= 5, theunitvector u in the direction ofv isu= 15v= 15 i+ 25j.All in all:

Duf(1, 0) =1, 0

15,

25

= 1

5.


10/20

MATH 209

Calculus,III

Volker Runde


The gradient


Geometric meaning of the gradient, I

A geometric observation

Let fbe a differentiable function of two variables, let u be aunit vector, and let be the angle between u and

f.

We have:

Duf =fu=|f||u| cos =

|f|

cos .

This becomes maximal if cos = 1, i.e., = 0, which meansthat u has the same direction asf.


11/20

MATH 209

Calculus,III

Volker Runde


The gradient


Geometric meaning of the gradient, II

Theorem

Let f be a differentiable function of two variables, and letu be

a unit vector. Then Duf(x, y) is maximal when u has the same

direction asf(x, y). This maximal value of Duf(x, y) is|f(x, y)|.

Important

Everything we said about directional derivatives and gradientsof functions of two variables remains valid for functions of three(or more) variables.


12/20

MATH 209

Calculus,III

Volker Runde


The gradient


Examples, II

Example

The temparature in a room at (x, y, z) is given by

T = 80

1 +x2 + 2y2 + 3z2,

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 = Tx

i+T

y j+

T

zk,


13/20

MATH 209

Calculus,III

Volker Runde


The gradient


Examples, III

Example (continued)

...i.e.,

T =

160x

(1 +x2

+ 2y2

+ 3z2

)2i

320y

(1 +x2

+ 2y2

+ 3z2

)2j

480z(1 +x2 + 2y2 + 3z2)2

k

= 160

(1 +x2 + 2y2 + 3z2)2(xi2yj3zk).

It follows that

T(1, 1,2) =58

(i2j+ 6k).


14/20

MATH 209

Calculus,III

Volker Runde


The gradient


Examples, IV

Example (continued)

This vector has the same direction as1,2, 6.The unit vector in this direction is

u= 141

i 241

j+ 6

41k,

which gives the direction of the maximum increase.The maximum increase is

|T(1, 1,2)|= 58

414 C/m.


15/20

MATH 209

Calculus,III

Volker Runde


The gradient


More geometric meaning of the gradient, I

Another geometric observation

Let Sbe a level surface of a function F(x, y, z), i.e., given bythe equation F(x, y, z) =Cwith a constant C.Let P(x0, y0, z0) be a point on S, and let

r(t) =x(t), y(t), z(t) be a curve in Spassing through P.Then F(x(t), y(t), z(t)) =C, so that

0 = dC

dt

= Fx

dxdt

+ Fy

dydt

+ Fz

dzdt

=Fr(t).


16/20

MATH 209

Calculus,III

Volker Runde


The gradient


More geometric meaning of the gradient, II

Another geometric observation (continued)

Let t0 be such that r(t0) =x(t0), y(t0), z(t0).Then

0 =F(x0, y0, z0)r(t0).This means thatF(x0, y0, z0) is perpendicular to the tangent

vector r(t0) to any curve r(t) passing through P.


17/20

MATH 209

Calculus,III

Volker Runde


The gradient


Definition of tangent plane and normal line

DefinitionThetangent planeto the level surface Sgiven byF(x, y, z) =C at P(x0, y0, z0) is the plane passing through Pwith normal vectorF(x0, y0, z0), i.e., with the equation

Fx

(x0, y0, z0)(xx0)

+F

y(x0, y0, z0)(yy0) + F

z(x0, y0, z0)(zz0) = 0.

Thenormal lineto S through P is the line passing through Pperpendicularly to the tangent plane, i.e., with the symmetricequations

xx0Fx(x0, y0, z0)

= yy0

Fy(x0, y0, z0)

= zz0

Fz(x0, y0, z0)

.


18/20

MATH 209

Calculus,III

Volker Runde


The gradient


Examples, IV

ExampleLet Ebe the ellipsoid given by

x2

2 +

(y1)24

+ (z+ 2)2 = 4.

Find the tangent plane and the normal line to E atP(2, 3,1).Set

F(x, y, z) = x2

2

+(y1)2

4

+ (z+ 2)2.

Then E is the level surface F(x, y, z) = 4.We have

F

x

=x, F

y

=1

2

(y

1), and

F

z

= 2(z+ 2).


19/20

MATH 209

Calculus,III

Volker Runde


The gradient


Examples, V

Example (continued)

We obtain:

F

x(2, 3,1) =2,

F

y(2, 3,1) = 1, and F

z(2, 3,1) = 2.


20/20

MATH 209

Calculus,III

Volker Runde


The gradient


Examples, VI

Example (continued)

Hence:

the tangent plane has the equation

2(x+ 2) + (y3) + 2(z+ 1) = 0,i.e.,

2x+y+ 2z= 5;

the normal line has the symmetric equations

x+ 22

=y3 = z+ 12

.

Lec6[1]DerivadaDireccional y Gradiente

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