Directed Derivatives

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Directional Derivatives

Michael Daniel V. SamsonDigiPen Institute of Technology Singapore

18 July 2014

M. D. V. Samson (DigiPen SG) Directional Derivatives 18 July 2014 1 / 18


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Derivative in a direction

What derivatives are, in one dimension, graphically:

f(x) =x3

x2 + 3



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f(x) =x3

x2 + 3

P= (0.5, 2.625)



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f(x) =x3

x2 + 3

P= (0.5, 2.625)f (0.5) = 0.25

y= 0.25x+ 2.75



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f(x) =x3

x2 + 3

P= (0.5, 2.625)f (0.5) = 0.25

y= 0.25x+ 2.75

In one dimension (and whenever we look at traces on planes), thedirection is alwaysto the right, i.e. in the direction of positive x.



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Partial derivatives and directional derivatives

In two dimensions:

The surface and contour maps for f.

We consider the partial derivatives of f(x, y) =x2y4 4yat (0,2).



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In two dimensions, partial derivatives are:

The surface and contour maps for f with the plane for y = 2 and its trace.

We have thetraceof the function z=f(x, y) on the plane y=y0,yielding single variable function z=g(x) =f(x, y0). The partial derivativeat (x0, y0) is the value of the one-dimensional derivative.



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In two dimensions, partial derivatives are:

The surface and contour maps for f with the plane for x= 0 and its trace.

We have the trace of the function z=f(x, y) on the plane x=x0,yielding single variable function z=h(y) =f(x0, y). The partial derivativeat (x0, y0) is the value of the one-dimensional derivative.



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In two dimensions:

The surface and contour maps for f with the plane for y =x 2 and its trace.

The partial derivatives show different directions: on the plane y=y0, inthe direction of (positive) x; and on the plane x=x0, in the direction of(positive) y. Here, we choose the direction of1, 1.



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Vectors and directional derivatives

On the xy-plane, direction is indicated by a (unit) vector, u=a, b,a2 +b2 = 1 (or, alternatively, u() =cos , sin ).

Given unit vector u and differentiable function f(x, y), how do we find thederivative of f at (x0, y0) in the direction ofu, Duf(x0, y0)?



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Using the graph, as with partial derivatives:

Find the plane in the direction ofu passing through (x0, y0).



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Determine the trace of fon that plane.



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Evaluate the one-dimensional derivative.



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Evaluate the one-dimensional derivative.

Is there an easier way?



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Limits and derivatives

The one-dimensional derivative of f(x) at x0 is given by

f (x0) = limh0

f(x0+h)f(x0)h

= limxx0

f(x)f(x0)xx0 .



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f (x0) = limh0

f(x0+h)f(x0)h

= limxx0

f(x)f(x0)xx0 .

This is determines the slope of the tangent line as the limit of the slopesof approximating secant lines.

Source: Wikipedia article on derivative


Li i d d i i


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f (x0) = limh0

f(x0+h)f(x0)h

= limxx0

f(x)f(x0)xx0 .

This is determines the slope of the tangent line as the limit of the slopesof approximating secant lines.

Source: Wikipedia article on derivative

Important: these derivatives exist if and only ifthelimitsexist.M. D. V. Samson (DigiPen SG) Directional Derivatives 18 July 2014 5 / 18

Li i d di i l d i i


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Limits and directional derivatives

The partial derivative of f(x, y) at (x0, y0) with respect to x and y,

respectively, are given by

fx(x0, y0) =f

x(x0, y0) = lim

h0

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

,

fy(x0, y0) = f

y(x0, y0) = lim

h0

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

.


Li it d di ti l d i ti


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fx(x0, y0) =f

x(x0, y0) = lim

h0

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

,

fy(x0, y0) = f

y(x0, y0) = lim

h0

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

.

Likewise, the derivative of f(x, y) at (x0, y0) in the direction of unit vectoru=a, b is given by

Duf(x0, y0) =f

u (x0, y0) = lim

h0

f(x0

+ah, y0

+bh)

f(x0, y

0)

h .


Li it d di ti l d i ti


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fx(x0, y0) =f

x(x0, y0) = lim

h0

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

,

fy(x0, y0) = f

y(x0, y0) = lim

h0

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

.

Likewise, the derivative of f(x, y) at (x0, y0) in the direction of unit vectoru=a, b is given by

Duf(x0, y0) =f

u (x0, y0) = lim

h0

f(x0

+ah, y0

+bh)

f(x0, y

0)

h .

This means Dif(x0, y0) =fx(x0, y0), where i=1, 0, andDjf(x0, y0) =fy(x0, y0), where j=0, 1.


Di ectional de i ati es


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Directional derivatives

Theorem

The derivative of f (x, y) at(x0, y0) in the direction of unit vectoru=a, b is given by

Duf(x0, y0) = fu

(x0, y0) =afx(x0, y0) +bfy(x0, y0).




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Theorem


Duf(x0, y0) = fu

(x0, y0) =u fx(x0, y0), fy(x0, y0).

This follows from letting g(h) =f(x0+ah, y0+bh) =f(x, y) anddetermining Duf(x0, y0) =g

(0), which follows by the chain rule.




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Theorem


Duf(x0, y0) = fu

(x0, y0) =u fx(x0, y0), fy(x0, y0).

This follows from letting g(h) =f(x0+ah, y0+bh) =f(x, y) anddetermining Duf(x0, y0) =g

(0), which follows by the chain rule.

The directional derivative Duf(x, y) follows as usual.


Example


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Example

Find the directional derivative Duf(x, y) if

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

and u is the unit vector given by angle =/6. What is Duf(1, 2)?


Example


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Example


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

and u is the unit vector given by angle =/6. What is Duf(1, 2)?

fx(x, y) = 3x2 3y, fy(x, y) = 8y3x,


Example


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Example


f(x, y) =x3 3xy+ 4y2and u is the unit vector given by angle =/6. What is Duf(1, 2)?

fx(x, y) = 3x2 3y, fy(x, y) = 8y3x,u=cos(/6), sin(/6)=

3/2, 1/2,


Example


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Example




3/2, 1/2,

Duf(x, y) =1

2[

3fx(x, y) +fy(x, y)]

=12

[33x2 3x+ (833)y],


Example


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Example




3/2, 1/2,

Duf(x, y) =1

2[

3fx(x, y) +fy(x, y)]

=12

[33x2 3x+ (833)y],

Duf(1, 2) =1

2[3

33 + (166

3)] = (133

3)/2.


Gradient vector


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Gradient vector

A vector-valued function can be seen here:

Duf(x, y) =f

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

This vector-valued function is defined as thegradient of f:

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


Gradient vector


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Gradient vector

A vector-valued function can be seen here:

Duf(x, y) =f

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

This vector-valued function is defined as thegradient of f:

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

This gives a compact form (and alternative notation) for the derivative inthe direction of unit vector u:

Duf(x, y) = fu

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


Example


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xa p e

Find the directional derivative of the function f(x, y) =x2y3 4yat thepoint (2,1) in the direction of the vector v= 2i+ 5j.


Example


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p

Find the directional derivative of the function f(x, y) =x2y3 4yat thepoint (2,1) in the direction of the vector v= 2i+ 5j.First, determine the gradient at the point:

f(x, y) =2xy3, 3x2y2 4 f(2,1) =4, 8.


Example


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p


f(x, y) =2xy3, 3x2y2 4 f(2,1) =4, 8.

v is not a unit vector, so determine the unit vector u in the direction ofv:

u= v

|v| =2, 5

29.


Example


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p


f(x, y) =2xy3, 3x2y2 4 f(2,1) =4, 8.


u= v

|v| =2, 5

29.

Finally,

Duf(2,1) =2, 529

4, 8= 3229

.


Three-dimensional directional derivatives


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The derivative of a differentiable function f(x, y, z) at x= (x0, y0, z0) inthe direction of unit vector u=a, b, c, a2 +b2 +c2 = 1 is given by

Duf(x) =f

u(x) =u

f(x),

where the gradient of f is given by

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


Example


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If f(x, y, z) =xsin yz, (a) find the gradient of f and (b) find thedirectional derivative of the function f at (1, 3, 0) in the direction of

v=i+ 2jk.


Example


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v=i+ 2jk.(a) The gradient of f is

f(x, y, z) =sin yz, xzcos yz, xycos yz.


Example


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(b) This gives the gradient at the pointf(1, 3, 0) =0, 0, 3.


Example


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u=

v

|v| =1, 2,

1

6 .


Example


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u=

v

|v| =1, 2,

1

6 .Finally,

Duf(1, 3, 0) =1, 2,1

6 0, 0, 3= 3

6.


Direction of greatest ascent


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Theorem

Given differentiable function f ,

maxu

Duf(x) =|f(x)|, for all unit vectorsu,

where the maximum is attained when u is in the direction of f(x).




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Theorem


maxu



This follows from the fact that the dot product between a given vectorf(x) and a unit vector is maximum when the unit vector is in the samedirection as the given vector, i.e.

u= f(x)

|f(x)| ,and that uu=|u|2.




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Theorem


maxu



This implies that thegreatest ascent in the graph of f at x is in thedirection off(x)likewise,greatest descentis in the direction off(x).




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Theorem


maxu



Source: Wikipedia article on gradient


Example


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Suppose that the temperature is given at a point (x, y, z) in space is givenbyT(x, y, z) = 80/(1 +x2 + 2y2 + 3z2), where T is measured in degrees

Celsius and x, y, z in meters. In what direction does the temperatureincrease fastest at the point (1, 1,2)? What is the maximum rate ofincrease?


Example


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This is given byf(1, 1,2), where

f(x, y, z) = 160x, 320y, 480z(1 +x2 + 2y2 + 3z2)2

.

Thus,f(1, 1,2) =5,10, 30/8.


Example


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This is given byf(1, 1,2), where

f(x, y, z) = 160x, 320y, 480z(1 +x2 + 2y2 + 3z2)2

.

Thus,f(1, 1,2) =5,10, 30/8.

The maximum rate of increase is given by

|f(1, 1,2)|=58|1, 2,6|= 5

8

41.

This is slightly more than 4C / m.M. D. V. Samson (DigiPen SG) Directional Derivatives 18 July 2014 14 / 18

Normal vectors and tangent lines


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Given an implicitly-defined curve (such as the level set of a function)f(x, y) =k, the Implicit Function Theorem gives

dy

dx = fx(x, y)

fy(x, y),




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dy

dx = fx(x, y)

fy(x, y),

which means that the tangent line to the curve at (x0, y0) is parallel to the

vectorfy(x0, y0), fx(x0, y0).




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dy

dx = fx(x, y)

fy(x, y),


vectorfy(x0, y0), fx(x0, y0).It can be seen that

fy(x0, y0), fx(x0, y0) f(x0, y0) = 0;




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dy

dx = fx(x, y)

fy(x, y),



fy(x0, y0), fx(x0, y0) f(x0, y0) = 0;

in other words, the gradient vector of the function fat the point (x0, y0) isperpendicular to the tangent line to the function at that point.




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dy

dx = fx(x, y)

fy(x, y),



fy(x0, y0), fx(x0, y0) f(x0, y0) = 0;

in other words, the gradient vector of the function fat the point (x0, y0) isperpendicular to the tangent line to the function at that point.

It is said that the gradient vector is normalto thecurve.


Normal vectors and tangent planes


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The gradient vector is also normal to level surfaces in three dimensionsF(x, y, z) =k




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The gradient vector is also normal to level surfaces in three dimensionsF(x, y, z) =k, as shown in Stewarts Calculus, ch. 14.6.




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The gradient vector is also normal to level surfaces in three dimensionsF(x, y, z) =k, as shown in Stewarts Calculus, ch. 14.6. This leads to adefinition of thetangent planeto the surface at the point (x0, y0, z0):

F(x0, y0, z0) xx0, yy0, zz0= 0.




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The gradient vector is also normal to level surfaces in three dimensionsF(x, y, z) =k, as shown in Stewarts Calculus, ch. 14.6. This leads to adefinition of thetangent planeto the surface at the point (x0, y0, z0):

F(x0, y0, z0) xx0, yy0, zz0= 0.

That is, for any point on the tangent plane (x, y, z), the vector from(x0, y0, z0) to (x, y, z) is perpendicular to the gradient vector at that point.


Example


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Find the equations of the tangent plane and normal line at the point(

2, 1,

3) to the ellipsoid

x2

4 +y2 +

z2

9 = 3.


Example


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2, 1,

3) to the ellipsoid

x2

4 +y2 +

z2

9 = 3.

The gradient off(x, y, z) = 3 is given byf(x, y, z) =x/2, 2y, 2z/9,


Example


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2, 1,

3) to the ellipsoid

x2

4 +y2 +

z2

9 = 3.

The gradient off(x, y, z) = 3 is given byf(x, y, z) =x/2, 2y, 2z/9, sothe equation to the tangent plane to the point is given by

1, 2,2/3 x+ 2, y1, z+ 3=x+ 2y23

z6 = 0.


Example


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2, 1,

3) to the ellipsoid

x2

4 +y2 +

z2

9 = 3.

The gradient off(x, y, z) = 3 is given byf(x, y, z) =x/2, 2y, 2z/9, sothe equation to the tangent plane to the point is given by

1, 2,2/3 x+ 2, y1, z+ 3=x+ 2y23

z6 = 0.

The normal line is given by

(x+ 2) = y12

=3z+ 92

.


Summary


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The gradient of a differentiable multivariate function is a vector:

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

The derivative of a differentiable multivariate function in the direction of aunit vector u is given as

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

The gradient of a differentiable multivariate function at a point is a vectornormal to the graph of the function at that point, and is the direction ofsteepest ascent.

The length of the gradient is the maximum value of the directionalderivative over all unit vectors.


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Directed Derivatives