Lecture 7: Partial...
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Lecture 7: Partial Differentiation
1. Key pointsDifferentiation of multivariable functionsPower series in two variables (Taylor series)Total differentiation vs partial differentiationChain rulesExtremum of multivariable functions and Hessian matrixLagrange multiplier
Maple commandsMultivariateCalculus packagediffTaylorApproximationLagrangeMultiplierplot3d
2. Partial Differentiation
Example 2.1
=
=
= 2
=
=
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(3)(3)
(4)(4)
(1)(1)
(2)(2)
Example 2.2 Ambiguity in expressionSuppose that a physical quantity is a function of coordinates. For example, using Cartesian coordinates in two-dimensional space
The same quantity can be expressed in polar coordinates:
We can also mix the Cartesian and polar coordinates like
or
All these expressions represent the same thing. Now, we want to calculate partial derivative .
Which expression should we use? That depends what quantity is fixed. If is fixed, we use Eq. (2).
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1. 1.
2. 2.
=
If is fixed, we use Eq. (3).
=
If is fixed, we use Eq. (4).
=
As you see, depends on what quantity is fixed. In order to avoid confusion, the fixed quantity
(second independent variable) is indicated by writing it as a subscript.
Exercises 1
For ,
(a) verify that
(b) verify that
If , , , find the following partial derivatives.
(a) (b) (c)
3. Taylor series in two variables
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(5)(5)
where all the derivatives are evaluated at and .
Example 3.1Expand in Maclaurin series.
=
The same result can be obtained from the expansion of sin and cos
Exercise 3.1Expand in a two-variable Maclaurin series.
Answer
=
4. Total differential
For , total differential of z is given by
Suppose we know the value of at a point . Now, we move to a new point . How much does the value of change?The change is defined as
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(7)(7)
(6)(6)
. Hence, we have
which is the meaning of (5).
Exercise 4.1Classical two-body problem can be mapped to an equivalent single-particle problem with an effective mass as a function of individual masses and . The reduced mass is
defined by . Find the change in when and are varied to and
, assuming and are small.
Answer
= = simplify
= = simplify
5. Chain rulesConsider a function of two variables . Now, if and are functions of , can be interpreted as a function as well: . Then, the regular derivative of with respect to can be obtained using the following chain rule:
Example 5.1
Consider a projectile motion and . The distance between
the initial position and the final position is given by Now, we calculate the time derivative of .
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(10)(10)
(11)(11)
(8)(8)
(9)(9)
Important formulae
ProofThere are three quantities, , , and and they are not independent. If and are independent variables, then . On the other hand, if and are independent variables, .Taking the total differential of and ,
Substituting (9) into (8), we obtain
Comparing the left and right hand sides,
and
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(13)(13)
(14)(14)
(15)(15)
(16)(16)
(17)(17)
(12)(12)
Multiplying to (12),
Here we used the reciprocal relation . Equations (11) and (13) are the
formulae.
Exercise 5.1
Show that for and ,
Answer
Substituting (15) to (14),
Compare it with
We find .
Exercise 5.2
Given , , , find
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(24)(24)
(20)(20)
(21)(21)
(19)(19)
(22)(22)
(18)(18)
(12)(12)
.
Answer
6. Minimum and maximum of two variable functions
Consider a function . Its total differential is .
At maximum or minimum positions, . Hence, and . However, this is not
a sufficient condition. The point satisfying this condition corresponds maximum, minimum or saddle.
Consider a Hessian matrix where , , . The differentials
are evaluated at the point obtained from and .
Since the Hessian matrix is symmetric, all eigenvalues are real. If both eigenvalues are positive, the point is a minimum point. If both eigenvalues are negative, it is a maximum point. If two eigenvalues have opposite sign, then it is a saddle point. If one or both eigenvalues are zero, we obtain no conclusion.
Example 6.1
Find the maximum and minimum points.
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(27)(27)
(28)(28)
(25)(25)
(26)(26)
(29)(29)
(12)(12)
(30)(30)
2
0
The Hessian matrix is . Its eigenvalues are 2 and -2. Hence, is a
saddle point.
Exercise 6.1
Find the maximum and mini points of .AnswerFirst, we find the candidates of the maximum and minimum points.
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(32)(32)
(31)(31)
(25)(25)
(12)(12)
There is only one point. We now try to find it is maximum or minimum by computing Hessianmatrix
= 2 = 2 = 0
The eigenvalues of the Hessian are both positive. Hence, is the minimum point.
7. Lagrange multipliers
We want to find maximum or minimum points of a function under a constraint given in a form of . We already learned how to find maximum or minimum points if there is no sucha constraint. We extend the previous approach by adding a new independent variable called Lagrange multiplier.Consider a new function
Maximum or minimum points must satisfy the following conditions:
Note that the third condition is exactly the constraint. Therefore, solving these equation simultaneously for , , and we find the maximum or minimum points under the constraint. Since our objective is to find and , it is not necessary to find we find first and used it in finding and .
Example 7.1
Find a minimum point of under a constraint .Without the constraint is the minimum point.
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(33)(33)
(25)(25)
(36)(36)
(34)(34)
(35)(35)
(12)(12)
x0 1 2
y 0
1
2
The blue line shows the constraint.
Equation (33) can be solved easily. The solution is either or .
For , we find from Eq. (35). This is the maximum point (as shown in the above plot). Note that we don't have to find in this case.
For , we find from Eq. (34) and from Eq. (35). This is the minimum
points.You can ask Maple to solve the simultaneous equation as follows:
Exercise 7.1Find the volume of the largest rectangular box, with edges parallel to the axes, inscribed on the
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(39)(39)
(44)(44)
(41)(41)
(25)(25)
(43)(43)
(40)(40)
(42)(42)
(38)(38)
(12)(12)
(37)(37)
ellipsoid: .
AnswerLet , , and be the corners of the box. The volume of the box is . We want to maximize this function under the constraint
Using Lagrange's multiplier, we define a new function
We solve Eqs. (38) for , , , and .
Simplifying (43) using (41) we find (38) we obtain
Solving (44) for , we find
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(47)(47)
(45)(45)
(25)(25)
(46)(46)
(12)(12)
(39) and (40), we obtain and