# SUMMARY SHEET PARTIAL DIFFERENTIATION Engineering Maths II (MAE111

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```							                           SUMMARY SHEET: PARTIAL DIFFERENTIATION
Engineering Maths II (MAE111), 2009
Fundamental deﬁnition: For function f = f (x, y) of two variables               Stationary Points: For f (x, y) a stationary point is deﬁned as where
x and y, the partial derivative wrt x is:                                       the rate of change is zero, so that

∂z       f (x + δx, y) − f (x, y)                                                             ∂f   ∂f
= lim                                                                                         =    = 0.
∂x  δx→0           δx                                                                         ∂x   ∂y
Note that this is effectively the rate of change of f in the x direction.       A stationary point may be minimum, maximum, or some other kind
Evaluation: Use the same rules as ordinary differentiation. When dif-           such as a saddle point.
ferentiating wrt one variable (say x), then treat other variables (say y)       For a MAXIMUM:
as constant.
2
Example: For                                                                          ∂2z                        ∂2z       ∂2z             ∂2z
<0       AND                            −                  >0
∂x2                        ∂x2       ∂y 2           ∂x∂y
2             2
z = f (x, y) = 4x + 3xy + 5y
For a MINIMUM:
the ﬁrst order partial derivatives are
2
∂2z                        ∂2z       ∂2z             ∂2z
∂z                                                             >0       AND                            −                  >0
=       8x + 3y                                      ∂x2                        ∂x2       ∂y 2           ∂x∂y
∂x
∂z
=       3x + 10y                               For a SADDLEPOINT:
∂y
2
∂2z        ∂2z           ∂2z
Higher Order Derivatives: For example, second order differential                                                       −                  <0
∂x2        ∂y 2         ∂x∂y
coefﬁcients:
∂2f ∂2f ∂2f ∂2f                                         (i.e., the function decreases in some directions, but increases in oth-
,    ,     ,      .
∂x2 ∂y 2 ∂x∂y ∂y∂x                                      ers).
Note: for most functions (actually continuous functions), the order of
Constrained Maxima and Minima: In order to ﬁnd a maximum or
partial differentiation is unimportant:
minimum of a function of several variables, which also satisﬁes a fur-
∂2f    ∂2f                                         ther constraint: use the constraint (or constraints) to eliminate one (or
=      .                                       more) of the variables. And then look for maxima/minima.
∂x∂y   ∂y∂x
Example: A cylindrical tin is to be made out of 600π cm2 of sheet
Small Increments: The approximate change in a function z(x, y) due              metal. What choice of radius r and height h will give the tin a maxi-
to small changes in x and y (denoted by δx, δy) is given by                     mum volume? Volume: V = πr2 h, surface area (including ends)
∂z      ∂z                                                                                          S
δz ≈       δx +    δy.                                                  S = 2πrh + 2πr2            ⇒h=         −r
∂x      ∂y                                                                                         2πr
This can be used for error estimation for a calculated quantity which           Substituting for h into equation for V :
depends on measurements with some fractional errors.
S       Sr
Total Derivative: For f (x, y) and if x = x(t) and y = y(t), where t                              V = πr2           −r =    − πr 3
is another variable, then                                                                                       2πr      2

df   ∂f dx   ∂f dy                                       Then ﬁnd value of r giving maximum V (remembering that S is ﬁxed.
=       +       .                                     And so on . . .
dt   ∂x dt   ∂y dt
Curve Fitting by Least Squares: For a set of n data points (xi , yi ),
Note mixture of partial and ordinary derivatives.
with i = 1, . . . n, choose a suitable curve to “ﬁt” the data, eg a straight
Change of Variables: Consider function z = z(x, y), where the vari-             line y = a + bx (but it could be some other function). For each data
ables x and y are themselves functions of another pair of variables u           point compute the y distance from the curve, and then square and then
and v, so that                                                                  add up for all the points:

x = x(u, v)     and        y = y(u, v).                                                 n

S=          (yi − a − bxi )2 = S(a, b)
If z is considered as a function of u and v then                                                        i=1

∂z          ∂z ∂x   ∂z ∂y
=           +                                       The best (“least squares”) ﬁt is when the function S(a, b) is a mini-
∂u          ∂x ∂u   ∂y ∂u                                 mum. Therefore the best ﬁt is given by the values of a and b for which
∂z          ∂z ∂x   ∂z ∂y                                 S is stationary:
=           +
∂v          ∂x ∂v   ∂y ∂v                                                         ∂S           ∂S
= 0,         = 0.
∂a           ∂b

Text Book References
Mathematics for Engineers         Engineering Mathematics
Croft & Davison                   Stroud (4th & 5th Editions)
Partial Differentiation        –                                 Programme 10, 11

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