Section 11.3 Partial Derivatives
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Transcript Section 11.3 Partial Derivatives
Section 11.3
Partial Derivatives
Goals
Define partial derivatives
Learn notation and rules for calculating
partial derivatives
Interpret partial derivatives
Discuss higher derivatives
Apply to partial differential equations
Introduction
If f is a function of two variables x and y,
suppose we let only x vary while keeping
y fixed, say y = b, where b is a constant.
Then we are really considering a function
of a single variable x, namely, g(x) = f(x, b).
If g has a derivative at a, then we call it the
partial derivative of f with respect to x at (a, b)
and denote it by fx(a, b).
Introduction (cont’d)
Thus
The definition of a derivative gives
Introduction (cont’d)
Similarly, the partial derivative of f with
respect to y at (a, b), denoted by fy(a, b), is
obtained by holding x = a and finding the
ordinary derivative at b of the function
G(y) = f(a, y):
Definition
If we now let the point (a, b) vary, fx and fy
become functions of two variables:
Notations
There are many alternate notations for
partial derivatives:
Finding Partial Derivatives
The partial derivative with respect to x is
just the ordinary derivative of the function
g of a single variable that we get by
keeping y fixed:
Example
If f(x, y) = x3 + x2y3 – 2y2, find fx(2, 1) and
fy(2, 1).
Solution Holding y constant and
differentiating with respect to x, we get
fx(x, y) = 3x2 + 2xy3
and so
fx(2, 1) = 3 · 22 + 2 · 2 · 13 = 16
Solution (cont’d)
Holding x constant and differentiating
with respect to y, we get
fx(x, y) = 3x2y2 – 4y
and so
fx(2, 1) = 3 · 22 · 12 – 4 · 1 = 8
Interpretations
To give a geometric interpretation of
partial derivatives, we recall that the
equation z = f(x, y) represents a surface S.
If f(a, b) = c, then the point P(a, b, c) lies on
S. By fixing y = b, we are restricting our
attention to the curve C1 in which the
vertical plane y = b intersects S.
Interpretations (cont’d)
Likewise, the vertical plane x = a intersects
S in a curve C2.
Both of the curves C1 and C2 pass through
the point P.
This is illustrated on the next slide:
Interpretations (cont’d)
Interpretations (cont’d)
Notice that…
C1 is the graph of the function g(x) = f(x, b), so
the slope of its tangent T1 at P is g′(a) = fx(a, b);
C2 is the graph of the function G(y) = f(a, y), so
the slope of its tangent T2 at P is G′(b) = fy(a, b).
Thus fx and fy are the slopes of the tangent
lines at P(a, b, c) to C1 and C2.
Interpretations (cont’d)
Partial derivatives can also be interpreted
as rates of change.
If z = f(x, y), then…
∂z/∂x represents the rate of change of z with
respect to x when y is fixed.
Similarly, ∂z/∂y represents the rate of change
of z with respect to y when x is fixed.
Example
If f(x, y) = 4 – x2 – 2y2, find fx(1, 1) and
fy(1, 1).
Interpret these numbers as slopes.
Solution We have
Solution (cont’d)
The graph of f is the paraboloid
z = 4 – x2 – 2y2 and the vertical plane y = 1
intersects it in the parabola z = 2 – x2, y = 1.
The slope of the tangent line to this
parabola at the point (1, 1, 1) is fx(1, 1) = –2.
Similarly, the plane x = 1 intersects the
graph of f in the parabola z = 3 – 2y2, x = 1.
Solution (cont’d)
The slope of the tangent line to this
parabola at the point (1, 1, 1) is fy(1, 1) = –4:
Example
x
f
f
If f x , y sin
1 y , calculate x and y .
Solution Using the Chain Rule for
functions of one variable, we have
Example
Find ∂z/∂x and ∂z/∂y if z is defined
implicitly as a function of x and y by
x3 + y3 + z3 + 6xyz = 1
Solution To find ∂z/∂x, we differentiate
implicitly with respect to x, being careful
to treat y as a constant:
Solution (cont’d)
Solving this equation for ∂z/∂x, we obtain
Similarly, implicit differentiation with
respect to y gives
More Than Two Variables
Partial derivatives can also be defined for
functions of three or more variables, for
example
If w = f(x, y, z), then fx = ∂w/∂x is the rate of
change of w with respect to x when y and z
are held fixed.
More Than Two Variables (cont’d)
But we can’t interpret fx geometrically
because the graph of f lies in fourdimensional space.
In general, if u = f(x1, x2 ,…, xn), then
and we also write
Example
Find fx , fy , and fz if f(x, y, z) = exy ln z.
Solution Holding y and z constant and
differentiating with respect to x, we have
fx = yexy ln z
Similarly,
fy = xexy ln z
and
fz = exy/z
Higher Derivatives
If f is a function of two variables, then its
partial derivatives fx and fy are also
functions of two variables, so we can
consider their partial derivatives
(fx)x , (fx)y , (fy)x , and (fy)y ,
which are called the second partial
derivatives of f.
Higher Derivatives (cont’d)
If z = f(x, y), we use the following notation:
Example
Find the second partial derivatives of
f(x, y) = x3 + x2y3 – 2y2
Solution Earlier we found that
fx(x, y) = 3x2 + 2xy3
Therefore
fy(x, y) = 3x2y2 – 4y
Mixed Partial Derivatives
Note that fxy = fyx in the preceding
example, which is not just a coincidence.
It turns out that fxy = fyx for most functions
that one meets in practice:
Mixed Partial Derivatives (cont’d)
Partial derivatives of order 3 or higher can
also be defined. For instance,
and using Clairaut’s Theorem we can
show that fxyy = fyxy = fyyx if these functions
are continuous.
Example
Calculate fxxyz if f(x, y, z) = sin(3x + yz).
Solution
Partial Differential Equations
Partial derivatives occur in partial
differential equations that express certain
physical laws.
For instance, the partial differential
equation
u
2
x
2
u
2
y
2
0
is called Laplace’s equation.
Partial Differential Eqns. (cont’d)
Solutions of this equation are called
harmonic functions and play a role in
problems of heat conduction, fluid flow,
and electric potential.
For example, we can show that the
function u(x, y) = ex sin y is a solution of
Laplace’s equation:
Partial Differential Eqns. (cont’d)
Therefore, u satisfies Laplace’s equation.
Partial Differential Eqns. (cont’d)
The wave equation
u
2
t
2
u
2
a
2
x
2
describes the motion of a waveform,
which could be an ocean wave, sound
wave, light wave, or wave traveling along
a string.
Partial Differential Eqns. (cont’d)
For instance, if u(x, t) represents the
disaplacement of a violin string at time t
and at a distance x from one end of the
string, then u(x, t) satisfies the wave
equation. (See the next slide.)
Here the constant a depends on the
density of the string and on the tension in
the string.
Partial Differential Eqns. (cont’d)
Example
Verify that the function u(x, t) = sin(x – at)
satisfies the wave equation.
Solution Calculation gives
So u satisfies the wave equation.
Review
Definition of partial derivative
Notations for partial derivatives
Finding partial derivatives
Interpretations of partial derivatives
Function of more than two variables
Higher derivatives
Partial differential equations