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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
1
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