Notebook giving examples of

use of partial derivatives,

maximization and contour

plotting---useful for material

in Chapter 4 of Boas

Partial derivatives---using "D"

Partial deriv w.r.t. x

D@Sin@x + y^2D, xD

CosAx + y2E

and w.r.t. y

D@Sin@x + y^2D, yD

2 y CosAx + y2E

or both together

D@Sin@x + y^2D, 88x, y<, 1<D

9CosAx + y2E, 2 y CosAx + y

2E=

Second derivatives (partial w.r.t x first, then mixed partial, then mixed partial again, then second wr.t. y)

D@Sin@x + y^2D, 88x, y<, 2<D �� MatrixForm

-Sin@x + y2D -2 y Sin@x + y2D-2 y Sin@x + y2D 2 Cos@x + y2D - 4 y2 Sin@x + y2D

Taylor series with more than one variable

Expanding about the origin:

The O[y]^7 etc. terms tell us the size of the next terms in the expansion

Series@Cos@x + yD, 8x, 0, 5<, 8y, 0, 5<D

1 -y2

2

+y4

24

+ O@yD6+ -y +

y3

6

-y5

120

+ O@yD6x +

-1

2

+y2

4

-y4

48

+ O@yD6x2

+y

6

-y3

36

+y5

720

+ O@yD6x3

+

1

24

-y2

48

+y4

576

+ O@yD6x4

+ -y

120

+y3

720

-y5

14400

+ O@yD6x5

+ O@xD6

If you don’t want to see the missing terms you can truncate using “Normal”

Normal@%42D

1 -x2

2

+x4

24

+ -x +x3

6

-x5

120

y + -1

2

+x2

4

-x4

48

y2

+

x

6

-x3

36

+x5

720

y3

+1

24

-x2

48

+x4

576

y4

+ -x

120

+x3

720

-x5

14400

y5

Another example:

Normal@Series@x Log@1 + x y^2D, 8x, 0, 6<, 8y, 0, 6<DD

x2y2

-x3y4

2

+x4y6

3

Expanding about a different point (here (1,1)):

Normal@Series@x Log@1 + x y^2D, 8x, 1, 4<, 8y, 1, 4<DD3

8

H-1 + xL2-

1

12

H-1 + xL3+

5

192

H-1 + xL4+

1 +3

2

H-1 + xL +1

4

H-1 + xL2-1

8

H-1 + xL3+

1

16

H-1 + xL4 H-1 + yL +

1 - x

4

-1

4

H-1 + xL2+

1

16

H-1 + xL3 H-1 + yL2+

-1

6

+1 - x

6

+1

8

H-1 + xL2+

1

24

H-1 + xL3-

5

96

H-1 + xL4 H-1 + yL3+

1

8

+1

4

H-1 + xL +1

32

H-1 + xL2-

3

32

H-1 + xL3+

5

128

H-1 + xL4 H-1 + yL4+

Log@2D + H-1 + xL1

2

+ Log@2D

Implicit differentiation

Example of 4.6 #2

We are asked to find dy/dx given that

y Exp[x y]=Sin[x]

This seems to require 2 steps. First, differentiating the implicit definition of y[x]

2 Partial.nb

D@y@xD Exp@x y@xDD � Sin@xD, xDãx y@xD

y¢@xD + ã

x y@xDy@xD Hy@xD + x y

¢@xDL � Cos@xD

and then solving for y':

Solve@%, 8y'@xD<D

::y¢@xD ® -ã-x y@xD H-Cos@xD + ãx y@xD

y@xD2L1 + x y@xD

>>

Finding maxima and minima

Example 4.10 #9: extremizing a function within a region

We are asked to find the maximum of 2x^2-3y^2-2x within the circle of radius 1.

FindMaximum gives wrong answer if don't give a starting point:

FindMaximum@82 x^2 - 3 y^2 - 2 x, x^2 + y^2 £ 1<, 8x, y<D91.00688 ´ 10

-8, 9x ® 1., y ® -1.14485 ´ 10

-14==

To get the correct answer need to input a negative starting x:

FindMaximum@82 x^2 - 3 y^2 - 2 x, x^2 + y^2 £ 1<, 88x, -.5<, 8y, 0<<D84., 8x ® -1., y ® 0.<<

If don't give a starting guess then get only one of the possible minima:

FindMinimum@82 x^2 - 3 y^2 - 2 x, x^2 + y^2 £ 1<, 8x, y<D8-3.2, 8x ® 0.2, y ® 0.979796<<

Get other minimum by giving a starting point with negative y:

FindMinimum@82 x^2 - 3 y^2 - 2 x, x^2 + y^2 £ 1<, 88x, 0<, 8y, -1<<D8-3.2, 8x ® 0.2, y ® -0.979796<<

Clarify situation by making contour plot

contplot = ContourPlot@2 x^2 - 3 y^2 - 2 x, 8x, -1, 1<,8y, -1, 1<, FrameLabel ® Automatic, ContourLabels ® TrueD;

boundplot = ContourPlot@x^2 + y^2 � 1, 8x, -1, 1<,8y, -1, 1<, FrameLabel ® Automatic, ContourStyle ® 8Thick, Black<D;

Partial.nb 3

Show@contplot, boundplotD

-3

-3-2

-2

-1

-1

0

1

2

3

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0

x

y

This shows the dependence of the function along the boundary (either the top or the bottom half) show-

ing the minima at x=0.2

Plot@2 x^2 - 3 H1 - x^2L - 2 x, 8x, -1, 1<, PlotStyle ® 8Thick, Red<D

-1.0 -0.5 0.5 1.0

-3

-2

-1

1

2

3

4

Here’s another way of visualizing the problem, with a 3D contour plot.

The “RegionFunction” restricts the plot to the desired circle in x & y.

4 Partial.nb

ContourPlot3D@z == 2 x^2 - 3 y^2 - 2 x, 8x, -1, 1<, 8y, -1, 1<,8z, -4, 4<, RegionFunction ® Function@8x, y, z<, x^2 + y^2 £ 1DD

-1.0

-0.5

0.0

0.5

1.0

-1.0

-0.5

0.0

0.5

1.0

-4

-2

0

2

4

Example 4.13 #17 --- constrained minimization

Find position of minimum distance to the origin along the surface x=yz+10

Turns out there are two such positions:

FindMinimum with no starting point gives one of the two solutions:

FindMinimum@8x^2 + y^2 + z^2, x � y z + 10<, 8x, y, z<D819., 8x ® 1., y ® -3., z ® 3.<<

With a starting guess we can find the other minimum:

FindMinimum@8x^2 + y^2 + z^2, x � y z + 10<, 88x, 1<, 8y, 1<, z<D819., 8x ® 1., y ® 3., z ® -3.<<

Here we solve for x using the constraint and contour the distance vs. y and z. The two minima are clear

and also the saddle point at the origin.

Partial.nb 5

ContourPlot@Sqrt@Hy z + 10L^2 + y^2 + z^2D, 8y, -5, 5<,8z, -5, 5<, FrameLabel ® Automatic, ContourLabels ® TrueD

5

5

10

10

10

10

10

15

15

15

15

20

20

25

25

30

30

-4 -2 0 2 4

-4

-2

0

2

4

y

z

Another way of visualizing the function, though it’s hard to see exactly where the minumum distance

points are.

ContourPlot3D@x == y z + 10, 8x, -5, 5<, 8y, -5, 5<, 8z, -50, 50<, AxesLabel ® 8x, y, z<D

-5

0

5

x

-5 0 5

y

-50

0

50

z

6 Partial.nb