Function of Several Variables

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Function of Several Variables

• BET2533 Eng. Math. III
• R. M. Taufika R. Ismail
• FKEE, UMP

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Introduction

• Previously we have seen function of one variable,
  y f (x), where x is the independent variable
  and y is the dependent variable, that is y
  depends on x.
• There are also exist function of more than one
  variable such as z f (x,y), f f (x,y,z),
• v f (x,y,z,t), etc.
• For convenience, we are allow to express
• z z(x,y), f f(x,y,z), v v(x,y,z,t), etc.

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• For the function z f (x,y), x and y are
  independent variables and z is the dependent
  variable (z depends on x and y).
• However, notice that x and y are independent of
  each other, that is, y is NOT the function of x
  or vice versa.

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Function of two variables

• The graph of y f (x) is represented as a line
  or curve in 2-D, with respect to x-axis and
  y-axis.
• The graph of z f (x,y) is represented as a
  surface in 3-D, with respect to x-axis, y-axis
  and z-axis.

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• Some common surface

Paraboloid
Cylinder
Ellipsoid
Cone
Hyperboloid
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Example 1

• Sketch the graph of the following equations in
  three dimensions. Identify each of the surface.
• (i) (ii)
• (iii) (iv)
• (v)

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Solution

• (i)

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• (ii)

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• (iii)

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• (iii)

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• (iii)

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Partial derivatives

• For a function of one variable, the rate of
  change of the function is represented by its
  derivative
• For a function of more than one variable, we are
  interested in the rate of change of the function
  w.r.t. one of its variables while the other
  variables remain fixed
• This leads to the concept of partial derivatives
• In partial differentiation, operator is
  used instead of

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• For example, if z f (x,y), then the first
  partial derivatives of z are

and

• As can also be expressed as or
  simply , the following symbol can also be
  used to denote the partial derivatives of z
• or simply and (or and ).

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Example 2

• (a) If ,
  find

and . Then compute and .
(b) If , find
and .
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Solution

• (a) By regarding y as a constant, we
  differentiate z w.r.t. x to obtain

By regarding x as a constant, we differentiate z
w.r.t. y to obtain
Hence,
and
?
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• (b) By regarding y as a constant, we
  differentiate f w.r.t. x to obtain

By regarding x as a constant, we differentiate f
w.r.t. y to obtain
Hence,
and
?
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Higher order derivatives

• The second order derivative of a single variable
  function is denoted as

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• The second order derivatives of a multivariable
  function are denoted as

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Example 3

• Find all second derivatives of f where
• (i)
• (ii)

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Solution

• (i) The first derivatives of f are

The second derivatives of f are
?
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• (ii) The first derivatives of f are

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The second derivatives of f are
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?
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Critical points of a function

• Critical point (or stationary point) of graph z
  f(x,y) has three possibilities
• maximum point, minimum point or saddle point.
• The word extremum refer to either maximum or
  minimum values of a function.

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• (i) Maximum point

Point (a,b,f(a,b)) is a (local) maximum
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• (i) Minimum point

Point (a,b,f(a,b)) is a (local) minimum
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• (i) Saddle point

Point (a,b,f(a,b)) is a saddle point
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Maximum point
Saddle point
Saddle point
Minimum point
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Example 4

• Locate and identify the stationary points of the
  following functions
• (i)
• (ii)
• (iii)

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Solution
(i)

• Step 1 Find the 1st partial derivatives

Step 2 Solve fx 0 and fy 0 simultaneously
This yields (x,y) (0,1) and (0,-1).
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Step 3 Find the 2nd derivatives and G(x,y)
Step 4 Test the points (0,1) and (0,-1)
At (0,1)
Therefore, is a saddle point.
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At (0,-1)
Therefore, is a maximum point.
?
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(ii)

• Step 1 Find the 1st partial derivatives

Step 2 Solve fx 0 and fy 0 simultaneously
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From (1),
Substitute (3) into (2),
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This yields (x,y) (0,0), (2,-2) and (-2,2).
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Step 3 Find the 2nd derivatives and G(x,y)
Step 4 Test the points (0,0), (2,-2) and (-2,2).
At (0,0)
Therefore, is a saddle point.
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At (2,-2)
Therefore, is a minimum point.
At (- 2,2)
Therefore, is a minimum point.
?
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(iii)

• Step 1 Find the 1st partial derivatives

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Step 2 Solve fx 0 and fy 0 simultaneously
This yields (x,y) (1,1)
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Step 3 Find the 2nd derivatives and G(x,y)
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Step 4 Test the points (1,1)
Therefore, is a saddle point.
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At (0,-1)
Therefore, is a maximum point.
?
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DIY

• Locate and identify the critical points of the
  following functions.
• (i)
• (ii)
• (iii)

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• Answers
• (i) is a minimum point
• (ii) is a saddle point.
  is a
• minimum point.
• (iii) has no conclusion.
  and
• are saddle points.