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1
DERIVATIVES AND ITS APPLICATION
Unit Structure
1.0 Objective
1.1 Introduction
1.2 Review of Functions
1.3 Increasing and Decreasing Functions
1.4 Stationary Points
1.5 Maximum and Minimum Problems.
1.6 Graphing Polynomials
1.7 Newton’s Method
1.8 Summary
1.9 Exercise
1.10 References
1.0 OBJECTIVE: -
Derive the Newton -Raphson method formula,
Develop the algorithm of the Newton -Raphson method,
Use the Newton -Raphson method to solve a nonlinear equation.
Discuss the dra wbacks of the Newton -Raphson method.
Understanding of Mathematical concepts like limit, continuity,
derivative, integration of functions
1.1 INTRODUCTION: -
The course is designed to have a grasp of important concepts of Calculus
in a scientific way. It co vers topics from as basic as definition of functions
to partial derivatives of functions in a gradual and logical way. The learner
is expected to solve as many examples as possible to get a compete clarity
and understanding of the topics covered.
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2 1.2 RE VIEW OF FUNCTIONS: -
Definitions Let ƒ be a function with domain D. Then ƒ has an absolute
maximum value on D at a point c
If ƒ(x) ≤ ƒ(c) for all x in D and an absolute minimum value on D at c if
ƒ(x) ≥ ƒ(c) for all x in D.
Maximum and minimum values are c alled extreme values of the function
ƒ. Absolute maxima or minima are also referred to as global maxima or
minima.
Theorem 1 —The Extreme Value Theorem If ƒ is continuo us on a closed
interval [a, b]. T hen ƒ attains both an absolute maximum value M and an
absolute minimum value m in [a, b ] . That is, there are numbers
and
in [a, b] with ƒ(
) = m, ƒ(
) = M, and m ≤ ƒ(x) ≤ M for every
other x in [a, b] The proof of the Extreme Value Theorem requires a
detailed knowledge of the real number system (see Appendix 7) and we
will not give it here. Figure 4.3 illustrates possible locations for the
absolute ext rema of a continuous function on a closed interval [a, b] . As
we observed for the function y = cos x, it is possible that an absolute
minimum (or absolute maximum) may occur at two or more different
points of the interval
1.3 INCREASING AND DECREASING FUN CTIONS: -
Let f be some function defined on an interval.
Definition
The function f is increasing over this interval if, for all points x 1 and x 2 in
the interval,
This means that the value of the function at a larger number is greater than
or equal to the value of the function at a smaller number.
The graph on the left shows a differentiable function. The graph on the
right shows a piecewise -defined continuous function. Both these functions
are increasing.
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3 The function f is decreasing over this interval if, for all points x 1 and x 2 in
the interval,
The following graph shows an example of a decreasing function
Note that a function that is constant on the interval is both increasing and
decreasing over this interval. If we want to exclude such cases, t hen we
omit the equality component in our definition, and we add the word
strictly:
• A function is strictly increasing if x 1< x2 implies f (x 1) < f (x 2).
• A function is strictly decreasing if x 1< x 2 implies f (x 1) > f (x 2).
We will use the following re sults. These results refer to intervals where
the function is differentiable. Issues such as endpoints have to be treated
separately.
• If
for all x in the interval, then the function f is strictly
increasing.
• If
for all x in the interval, then the function f is strictly
decreasing.
• If
for all x in the interval, then the function f is constant.
1.4 STATIONARY POINTS: -
Definitions
Let f be a differentiable function.
• A stationary po int of f is a number x such that
.
• The point c is a maximum point of the func tion f if and only if
f (c) ≥ f(x), for all x in the domain of f . The value f (c) of the function at
c is called the maximum value of the function.
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4 • The point c is a minimum point of the function f if and only if
f (c) ≤ f (x), for all x in the domain of f . The value f (c) of the function at
c is called the minimum value of the function.
1.5 MAXIMUM AND MINIMUM PROBLEMS: -
• The point c is a local maximum point of the function f if there exists an
interval (a, b) with c ϵ (a,b) such that f (c) ≥ f (x), for all x ϵ (a,b).
• The point c is a local minimum point of the function f if there exists an
interval (a, b) with c ϵ (a,b) such that f (c) ≤ f (x), for all x ϵ (a,b).
These are sometimes called relative maximum and relative minimum
points. Local maxima and minima are often referred to as turning points
The followin g diagram shows the graph of y = f (x), where f is a
differentiable function. It appears from the diagram that the tangents to the
graph at the points which are local maxima or minima are horizontal. That
is, at a local maximum or minimum point c, we have f (c) = 0, and hence
each local maximum or minimum point is a stationary point
1.6 GRAPHING POLYNOMIALS: -
In the last module, we looked at the long -term behavior of polynomials.
After studying that module, you should be able to recognize that
polynomials like 43 2( ) 3 63 27 486fx x x x x and
43 2() 1 2 2 7 2 7 0 6 4 8gx x x x x have similar long -term behavior.
Since they are both 4th degree polynomials with a positive leading
coefficient, we know that their graphs must hav e arrows pointing up at the
extreme left - and right -sides (i.e., the outputs of both functions increases
without bound as the inputs increase without bound and as the inputs
decrease without bound). See Figure 1 below.
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5
Figure 1:()yf x and ()yg x
Although the functions 43 2( ) 3 63 27 486fx x x x x and
43 2() 1 2 2 7 2 7 0 6 4 8gx x x x x have similar long -run behavior,
they are not identical functions! Let’s study the short -run behavior of
their graphs to see how these functi ons differ. The short -run behavior of
the graph of a function concerns graphical features that occur when the
input values aren’t very large. (It’s hard to specify what “not large” means
since it will be different for each function, but we’ll for particu lar
graphical features rather than look within a particular interval, so we don’t
need to worry about being more specific.)
Clearly, 0x isn’t a large x-value, so the y-intercept will be part of the
short -run behavior of a polynomi al function’s graph. (Notice that the y-
coordinate of the y-intercept of a polynomial function is its the constant
term.)
The y-intercept of 43 2( ) 3 63 27 486fx x x x x is 0, 486 .
The y-intercept of 43 2() 1 2 2 7 2 7 0 6 4 8gx x x x x is 0, 648 .
Zeros (or roots ) are another important part of the short -run behavior of
the graph of a polynomial function. To find the roots of a polynomial
function, we can write it in factored form :
43 2( ) 3 63 27 486( 3)( 3)( 6)( 9)fx x x x x
xxxx
Each factor of f gives rise to a root, since when each factor equals zero,
the output for the function is zero. To find the roots, determine which
numbers make the factors equal to zero.
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Calculus
6 33336699xx
xx
xx
xx
FACTOR ROOT
In order to graph f, we can plot its roots and y-intercept; see Figure 2
below.
Figure 2
Now we can connect these points – making sure that our graph has the
correct long -run behavior – to complete the graph of
43 2( ) 3 63 27 486fx x x x x ; see Figure 3.
Figure 3: The graph of ()yf x.
Now let’s factor g to determine its roots:
43 2
2( ) 12 27 270 648( 3) ( 6)( 12)gx x x x x
xx x
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7 336612 12xx
xx
xx
FACTOR ROOT
Since the root 3x is associated with a squared (or “double”) factor, it
is often called a double root or a root of multiplicity two .
In order to graph g, we can plot its roots and y-intercept; see Figure 4.
(The orange point at 3x is a root of multiplicity two .)
Figure 4
Now we can connect these points – making sure that our graph has the
correct long -run behavior – to complete the graph of
43 2() 1 2 2 7 2 7 0 6 4 8gx x x x x (see Figure 5). Notice that the
only way to connect the points without using additional roots or incorrect
long-run behavior is to have the graph “bounce” of the double root at 3.x (Try it yourself!)
Figure 5: The graph of ()yg x.
EXAMPLE: Write an algebraic rule for the polynomial function p
graphed Figure 6. Note that the graph passes through the point ( 3, 18).
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8
Figure 6: The graph of ()yp x.
SOLUTION:
Since there are roots at 4x, 2x, 0x, and 3x, we know that
p has form () ( 3 ) ( 4 ) ( 2 )px k xx x x
where k is a constant. To find k we can use the point ( 3, 18).
( 3, 18) ( 3) 18 ( 3)( 3 3)( 3 4)( 3 2)18 ( 3)( 6)(1)( 1)
18 18
1pk
k
k
k
Therefore, () ( 3 ) ( 4 ) ( 2 ) .px xx x x
EXAMPLE: Write an algebraic rule for the polynomial function w
graphed in Figure 7. Note that the y-intercept of w
is (0, 12).
Figure 7: The graph of ()yw x.
SOLUTION:
Since there are roots at 3x, 1x, 1x, and 3x, and since the
graph bounces off the x-axis at 1x, this is a root of multiplicity two
(i.e., a double -root). Thus, we know that w has form
2( ) ( 3)( 1) ( 1)( 3)wx k x x x x munotes.in
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9 where k is a constant. To find k we can use the fact that the y-interc ept
of w is (0, 12):
2
2
12 4
93(0, 12) (0) 12 (0 3)(0 1) (0 1)(0 3)12 (3)(1) ( 1)( 3)
12 9fk
k
k
k
Thus, 2 4
3( ) ( 3)( 1) ( 1)( 3)wx x x x x .
EXAMPLE: Write an algebraic rule for the polynomial function h
graphed in Figure 8. Note that the y-intercept of h is (0, 13).
Figure 8: The graph of ()yh x.
SOLUTION:
Notice that the graph of ()yh x does not have roots at integer values.
But there are some “nice” points on the graph of ()yh x along the
horizontal line 5y. If we treat this line as the x-axis (i.e., imagine that
the graph of h is shifted down 5 units) then the graph would have roots 4x, 1x, and 4x. So to find a rule for h we can create a
function that has these three roots and then shift it up 5 units: () ( 4 ) ( 1 ) ( 4 ) 5hx k x x x
To find k we can use the y-intercept (0, 13):
1
2(0) 13 (0 4)(0 1)(0 4) 513 ( 4)( 1)(4) 5(0, 13)
13 16 5
81 6hk
k
k
k
k
Therefore, an algebraic rule for h is 12() ( 4 ) ( 1 ) ( 4 ) 5hx x x x . munotes.in
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Calculus
10 Properties of Polynomial Functions
■ The graph of a polynomial is a smooth unbroken curve. (By
“smooth” we mean that the graph does not have any sharp corners
as turning points.)
■ The graph of a polynomial always exhibits the characteristic that as x gets very lar ge, y gets very large.
■ If p is a polynomial of degree n, then the polynomial equation () 0px has at mostn distinct solutions; that is, p has at most n
zeros. This is equivalent to saying that the graph of ()yp x
crosses the x-axis at most n times. Thus a polynomial of degree
five can have at most five x-intercepts.
■ The graph of a polynomial function of degree n can have at
most1n turning points (see Key Point below). For example, the
graph of a polynomial of degree five can have at most four turning
points. In particular, the graph of a quadratic (2nd degree)
polynomial function always has exactly one turning point – its
vertex.
EXAMPLE: What is th e minimum possible degree of the polynomial
function in Figure 9?
Figure 9
SOLUTION:
The polynomial function graphed in Figure 9 has four zeros and five
turning points. The properties of polynomials tell us that a polynomial
function with four zeros mus t have a degree of at least four. These
properties also tell us that if a polynomial has degree n then it can have at
most 1n turning points. In other words the degree of a polynomial must
be at least one more than the number of turning points. Since this graph
has four turning points, the degree of the polynomial must be at least six.
Keep in mind that although a 6th degree polynomial may have as many as
six real zeros, it need not have that many. The graph in Figure 9 only ha s
four real zeros.
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11 Concavity
There are four graphs in Figure 10 (below). The first two graphs ( a and b)
have different shapes, but they are both increasing on the interval ( 2, 1).
The second two graphs ( c and d) also have different shapes, but they are
both decreasing on the interval ( 2, 1).
a
b
c
d
Figure 10
As the first graph ( a) rises, it bends or curves upward; but as the second
graph ( b) rises, it bends or curves downward. Similarly, the thir d graph
(c) is falling and curved upward, whereas the fourth graph ( d) is falling
and curved downward. A graph the curves upward (like the graphs a and
c) is called concave up ; a graph that curves downward (like the graphs b
and d) is called concave down . It might help to remember thata parabola
that opens upward is concave up and a parabola that opens downward is
concave down.
The graph in Figure 11 is always increasing, concave down on the interval (, 1 ), and concave up on the inter val (1, ). At the point (1, 1) the
concavity changes from concave up to concave down. Such a point is
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12
Figure 11: An inflection point occurs at (1, 1).
The polyn omial 322yx x x is graphed in Figure 12; the graph
appears to have an inflection point at 13x.
Figure 12: An inflection point occurs at 13x.
EXAMPLE: Use the graph in Figure 13 to find …
a. the approximate intervals where the graph of f is increasing.
b. the approximate intervals where the graph of f is decreasing.
c. the approximate intervals where the graph of f is concave up.
d. the approximate intervals where the graph of f is co ncave down.
e. the number of inflection points on the graph of f.
Figure 13: The graph of ()yf x. munotes.in
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13 SOLUTION:
a. The approximate intervals where the graph of f is increasing are ( , 4.5) and (0, 4.5).
b. The approximate intervals where the graph of f is decreasing are ( 4.5, 0) and (4.5, ).
1.7 NEWTON’S METHOD: -
Methods such as the bisection method and the false position method of
finding roots of a nonlinear equation 0)(xf require bracketing of the
root by two guesses. Such methods are called bracketing methods . These
methods are always convergent since they are based on reducing the
interval between the two guesses so as to zero in on the root of the
equation.
In the Newton -Raphson method, the root is not bracketed. In fact, only
one initial guess of the root is needed to get the iterative process started to
find the root of an equation. The method hence falls in the category of
open methods . Convergence in open methods is not guaranteed but if the
method does converge, it does so much faster than the bracketing
methods.
The Newton -Raphson method is based on the principle that if the initial
guess of the root of 0)(xf is at ix, then if one draws the tangent to the
curve at )(ixf , the point 1ix where the tangent crosses the x-axis is an
improved estimate of the root (Figure 1).
Using the definition of the slope of a function, at ixx θ = xfi tan 10
i ii
xxxf = ,
which gives ii
i ixfxf = xx1 (1)
Equation (1) is called the Newton -Raphson formula for solving nonline ar
equations of the form 0xf . So starting with an initial guess, ix, one
can find the next guess, 1ix, by using Equation (1). One can repeat this
process until one finds the root withi n a desirable tolerance.
Algorithm
The steps of the Newton -Raphson method to find the root of an equation 0xf are munotes.in
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14 1. Evaluate xf symbolically
2. Use an initial guess of the root, ix, to esti mate the new value of the
root, 1ix, as ii
i ixfxf = xx1
3. Find the absolute relative approximate error a as
010
11
ii i
ax x x =
4. Compare the absolute relative approximate error with the pre -specified
relative error tolerance, s. If a>s, then go to Step 2, else stop the
algorithm. Also, check if the number of iterations has exceeded the
maximum number of iterations allow ed. If so, one needs to terminate
the algorithm and notify the user.
Figure 1 Geometrical illustration of the Newton -Raphson method.
Example 1
You are working for ‘DOWN THE TOILET COMPANY’ that makes
floats for ABC co mmodes. The floating ball has a specific gravity of 0.6
and has a radius of 5.5 cm. You are asked to find the depth to which the
ball is submerged when floating in water.
f (x)
f (x) f (xi+1)
xi+2 xi+1 xi x θ [xi, f(xi)]
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15
Figure 2 Floating ball problem.
The equation that gives the depth x in meters to which the ball is
submerged under water is given by
0 10 993.3 165.04 2 3 x x
Use the Newton -Raphson method of finding roots of equations to find
a) the depth x to which the ball is submerged under water. Conduct three
iterations to estimate the root of the above equation.
b) the absolute relative approximate error at the end of each iteration, and
c) the number of significant digits at least correct at the end of e ach
iteration.
Solution 4 2 310 993.3 1650 x. xxf x. x xf 330 32
Let us assume the initial guess of the root of 0xf is . . x m 0500
This is a reasonable guess (discuss why 0x and m 11.0x are not
good choices) as the extreme values of the depth x would be 0 and the
diameter (0.11 m) of the ball.
Iteration 1
The estimate of the root is 00
0 1xfxfx x 050330 050310 993.3 0501650 05005024 2 3
. . .. . ..
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16 3410910 118.1050. 01242.0 050.
062420.
The absolute relative approximate error a at the end of Iteration 1 is
100
10 1xxx
a 19.90% 100062420050 062420
.. .
The number of significant digits at least correct is 0, as you need an
absolute relative approximate error of 5% or less for at least one
significant digit to be correct in your result.
Iteration 2
The estimate of the root is 11
1 2xfxfx x 062420330 062420310 993.3 0624201650 06242006242024 2 3
. . .. . ..
371090973.810 977813062420 .. 510 4646.4 062420 .
062380.
The absolute relative approximate error a at the end of Iteration 2 is
100
21 2xxx
a
100062380062420 062380.. .
% 07160.
The maximum value of m for which m
a2105.0 is 2.844. Hence,
the number of significant digits at least correct in the answer is 2. munotes.in
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17 Iteration 3
The estimate of the root is 22
2 3xfxfx x 062380330 062380310 993.3 0623801650 06238006238024 2 3
. . .. . ..
3111091171.81044.4062380 .
910 9822.4 062380 .
062380.
The absolute relative approximate error a at the end of Iteration 3 is
100062380062380 062380.. .
a
0
The number of significant digits at least correct is 4, as only 4 significant
digits are carried through in all the calculations.
Drawbacks of the Newton -Raphson Method
1. Divergence at inflection points
If the selection of the initial guess or an iterated value of the root turns out
to be close to the inflection point (see the definition in the appendix of this
chapter) of the function xf in the equation 0xf , Newton -Raphson
method may start diverging away from the root . It may then start
converging back to the root. For example, to find the root of the equation 0 512.0 13x xf
the Newton -Raphson method reduces to
23 3
1)1 (3512.0)1 (
ii
i ixx = xx
Starting with an initial guess of 0.50x , Table 1 shows the iterated
values of the root of the equation. As you can observe, the root starts to
diverge at Iteration 6 because the previous estimate of 0.92589 is close to
the inflection point of 1x (the value of xf' is zero at the inflection
point). Eventually, after 12 more iterations the root converges to the exact
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18 Table 1 Divergence near inflection point.
Iteration
Number ix
0 5.0000
1 3.656 0
2 2.7465
3 2.1084
4 1.6000
5 0.92589
6 –30.119
7 –19.746
8 –12.831
9 –8.2217
10 –5.1498
11 –3.1044
12 –1.7464
13 –0.85356
14 –0.28538
15 0.039784
16 0.17475
17 0.19924
18 0.2 munotes.in
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19
Figure 3 Divergence at inflection point for 0 13x xf .
2. Division by zero
For the equation 0 1042 0306 2 3. x. x xf
the Newton -Raphson method reduces to
i ii i
i ix x. x. x = xx
06.0 31042 030
26 2 3
1
For 00x or 02.00x , division by zero occurs (Figure 4). For an init ial
guess close to 0.02 such as 01999.00x , one may avoid division by zero,
but then the denominator in the formula is a small number. For this case,
as given in Table 2, even after 9 iterations, the Newton -Raphson method
does not converge .
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Calculus
20 Table 2 Division by near zero in Newton -Raphson method.
Iteration
Number ix )(ixf %a
0
1
2
3
4
5
6
7
8
9 0.019990
–2.6480
–1.7620
–1.1714
–0.77765
–0.51518
–0.34025
–0.223 69
–0.14608
–0.094490 -610 1.60000
18.778
–5.5638
–1.6485
–0.48842
–0.14470
–0.042862
–0.012692
–0.0037553
–0.0011091 100.75
50.282
50.422
50.632
50.946
51.413
52.107
53.127
54.602
-1 .00E-05-7.50E-06-5.00E-06-2.50E-060.00E+ 002.50E-065.0 0 E-0 67.50E-061.0 0 E - 0 5
-0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04xf(x)
0.02 Figure 4 Pitfall of division by zero or a near zero number.
3. Oscillations near local maximum and minimum
Results obtained from the Newton -Raphson method may oscillate about
the local maximum or minimum without converging on a root but
converging on the local maximum or minimum. E ventually, it may lead to
division by a number close to zero and may diverge.
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21 For example, for 022xxf
the equation has no real roots (Figure 5 and Table 3).
-10123456
-2 -1 0 1 2 3f(x)
x 3
4 2
1 -1.75 -0.30400.53.142 Figure 5 Oscillations around local minima for 22xxf .
Table 3 Oscillations near local maxima and minima in Newton -Raphson
method.
Iteration Number ix )(ixf %a
0
1
2
3
4
5
6
7
8
9 –1.0000
0.5
–1.75
–0.30357 3.1423
1.2529
–0.17166 5.7395
2.6955
0.97678 3.00
2.25
5.063
2.092
11.874 3.570
2.029
34.942 9.266
2.954 300.00
128.571 476.47
109.66
150.80
829.88
102.99
112.93
175.96
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22 4. Root jumping
In some case where the function )(xf is oscillating and has a number of
roots, one may choose an initial guess close to a root. However, the
guesses may jump and converge to some other root. For example for
solving the equation 0 sinx if you choose 539822.7 4.20 x as an
initial guess , it converges to the root of 0x as shown in Table 4 and
Figure 6. However, one may have chosen this as an initial guess to
converge to 28318536 2 . x.
Table 4 Root jumping in Newton -Raphson method.
Iteration Number ix )(ixf %a
0
1
2
3
4
5 7.539822
4.462
0.5499
–0.06307
410 376.8
1310 95861.1 0.951
–0.969
0.5226
–0.06303
510 375.8
1310 95861.1 68.973
711.44
971.91
41054.7
101028.4
-1 .5-1-0.500.511. 5
-2 0 2 4 6 8 10xf(x)
-0.06307 0.5499 4.461 7.539822 Figure 6 Root jumping from intended location of root for 0 sinx xf .
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23 Appendix A. What is an inflection point?
For a function xf, the point where the concavity changes from up -to-
down or down -to-up is called its inflection point. For example, for the
function 31x xf , the concavity changes at 1x (see Figure 3),
and hence (1,0) is an inflection point.
An inflection points MAY exist at a point where 0)(xf and where
)(''xf does not exist. The reason we say that it MAY exist is because if
0)(xf , it only makes it a pos sible inflection point. For example, for
16 )(4x xf , 0)0(f , but the concavity does not change at 0x .
Hence the point (0, –16) is not an inflection point of 16 )(4x xf .
For 31x xf , )(xfchanges sign at 1x ( 0)(xf for 1x, and
0)(xf for 1x), and thus brings up the Inflection Point T heorem for a
function )(xf that states the following.
“If )('cf exists and )(cf changes sign at cx, then the point ))(,( cfc
is an inflection poin t of the graph of f.”
Newton -Raphson method can also be derived from Taylor series. For a
general function xf, the Taylor series is i i i i i x xxf xf xf 1 1 + 2
1!2i iix xxf"
As an approximation , taking only the first two terms of the right hand side, i i i i i x xxf xf xf 1 1
and we are seeking a point where , xf 0 that is, if we assume , xfi 01 i i i i x xxf xf 1 0
which gives ii
i ixf'xfx x 1
This is the same Newton -Raphson method formula series as derived
previously using the geometric method.
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Calculus
24 a. It appears that the graph of f is concave up on the interval (2 , 2 ).
b. It appears that the graph of f is concave down on the intervals (, 2 ) and (2, ).
c. As we move from left to right, the graph changes from concave down
to concave up and then back to concave down. Each change occurs at
an inflection point, so the graph of f has two inflection points.
EXAMPLE: Suppose p is a polynomial function that satisfies the
following conditions: The graph of p has exactly three turning points : (2 . 5 , 5 ), (0, 2), (2.5, 5) and the graph of p has exactly two
inflection points: (1 , 0 ) and (1, 0). Sketch a graph of p based upon
this information. How many real zeros does p have?
SOLUTION:
Let’s start by plotting the given points; see Figu re 14.
Figure 14
There is only one way to connect the points to create a polynomial
function without adding turning points or inflection points; see Figure 15.
You should verify this for yourself.
Figure 15: The graph of ()yp x.
Since it has four x-intercepts, p has four real zeros.
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25 1.8 SUMMARY: -
This chapter is mainly focusing on basic concepts in derivatives.
1.9 EXERCISE: -
The accompanying figure shows some level curves of an unspecified
function f(x, y). Which of the th ree vectors shown in the figure is most
likely to be ∇f ? Explain
1.10 REFERENCES: -
Calculus: Early transcendental (10th Edition): Howard Anton, Irl
Bivens, Stephen Davis, John Wiley & sons, 2012.
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26 2
INTEGRATION AND ITS APPLICATIONS
Unit Structure
2.0 Objective
2.1 Introduction
2.2 An overview of the area problem
2.3 The indefinite integral ant i derivatives
2.4 The indefinite integral
2.5 Area between two curves
2.6 Length of A plane curve
2.7 Simpson’s rule
2.8 Summary
2.9 Exercise
2.10 References
2.0 OBJECTIVE
Derive the Simpson’s method formula,
Develop the algorithm of the Simpson’s method,
2.1 INTRODUCTION
In this chapter we will begin with an overview of the problem of finding
areas —we will discuss what the term “area” means, and we will outline
two approaches to defining and calculating areas. Following this
overview, we will discuss the Fundamental Theorem of Calculus, which is
the theore m that relates the problems of finding tangent lines and areas,
and we will discuss techniques for calculating areas. We will then use the
ideas in this chapter to define the average value of a function, to continue
our study of rectilinear motion, and to examine some consequences of the
chain rule in integral calculus. We conclude the chapter by studying
functions defined by integrals, with a focus on the natural logarithm
function.
2.2 AN OVERVIEW OF THE AREA PROBLEM
Formulas for the areas of polygons, such as squares, rectangles,
triangles, and trapezoids, were well known in many early civilizations.
However, the problem of finding formulas for regions with curved
boundaries (a circle being the simplest example) caused difficulties
for early mathematici ans. munotes.in
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27 The first real progress in dealing with the general area problem was
made by the Greek Mathematician Archimedes, who obtained areas of
regions bounded by circular arcs, parabolas, spirals, and various other
curves using an ingenious procedure that was later called the method
of exhaustion. The method, when applied to a circle, consists of
inscribing a succession of regular polygons in the circle and allowing
the number of sides to increase indefinitely (F igure 5.1.1). As the
number of S ides increases, the polygons tend to “exhaust” the region
inside the circle, and the areas of the polygons become better and
better approximations of the exact area of the circle.
To see how this works numerically, let
denote the area of a regular
n-sided polygon inscri bed in a circle of radius 1. Table 5.1.1 shows the
values of A(n) for various choices of n. Note that for large values of n
the area A(n) appears to be close to π (square units),
as one would expect. This suggests that for a circle of radius 1, the metho d
of exhaustion is equivalent to an equation of the form
Since Greek mathematicians were suspicious of the concept of “infinity,”
they avoided its use in mathematical arguments. As a result, computation
of area using the method of exhaustion was a very cumbersome procedure.
It remained for Newton and Leibniz to obtain a general method for finding
areas that explicitly used the notion of a limit. We will discuss their
method in the context of the following problem
THE RECTANGLE METHOD FOR FINDING AREAS
One approach to the area problem is to use Archimedes’ method of
exhaustion in the following way:
• Divide the interval [a, b]inton equal subintervals, and over each
subinterval construct a rectangle that extends from the x -axis to any point
on the curve y = f(x) that is above the subinterval; the particular point does
not matter —it can be above the center, above an endpoint, or above any
other point in the subinterval. In Figure 5.1.3y = f(x)it is above the center.
• For each n, the total area of the rectang les can be viewed as an
approximation to the exact area under the curve over the interval [a, b].
Moreover, it is evident intuitively that as n increases these approximations
will get better and better and will approach the exact area as a limit munotes.in
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Calculus
28 (Figure 5. 1.4). That is, if A denotes the exact area under the curve and An
denotes the approximation to A using n rectangles, then
We will call this the rectangle method for computing A
to illustrate this idea, we will use the rectangle method to approximate the
area under the curve y = x2 over the interval [0 , 1] (Figure 5.1.5). We will
begin by dividing the interval [0 , 1] into nequal subintervals, from which it
follows that each subinterval has length 1 /n; the endpoints of the
subintervals occur at,
0 ,
, ,
(Figure 5.1.6). We want to construct a rectangle over each of these
subintervals whose height is the value of the function f(x) = x2 at some
point in the subinterval. To be specific, let us use the right endpoints, in
which case the heights of our rectangles will be
and since each rectangle has a base of width 1 /n, the total area Anof the n
rectangles will be
For example, if n = 4, then the total area of the four approximating
rectangles would be
Table 5.1.2 shows the result of evaluating (1) on a computer for some
increasingly large values of n. These computations suggest that the exact
area is close to 1 3. Later in this chapter we will prove that this area is
exactly 1 3 by showing that
2.3 THE INDEFI NITE INTEGRAL ANTIDERI VATIVES:
THE INDEFINITE INTEGRALANTIDERIVATIVES
Definition: A function F is called an ant derivative of a function f on a
given open interval if
for all x in the interval.
For example, the function F(x) =
is an ant derivative of f(x) = x2 on the
interval (-∞, +∞) because for each x in this interval. munotes.in
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29
However, F(x) =
is not the only ant derivative of f on this interval. If
we add any constant C to
, then the function G(x) =
+ C is also an
ant derivative of f on ( −∞, +∞), since
In general, once any single antiderivative is known, other antiderivatives
can be obtained by adding constants to the known antiderivative. Thus,
are all ant derivatives of f(x) = x2.
2.4 THE INDEFINITE INTEGRAL: -
The process of finding antiderivatives is called ant differentiation or
integration. Thus, if
then integrating (or ant differentiating) the function f(x) produces an ant
derivative of the form F (x) + C. To emphasize this process, Equation (1)
is recast using integral notatio n,
then integrating (or ant differentiating) the function f(x) produces an ant
derivative of the form F (x) + C. To emphasize this process, Equation (1)
is recast using integral notation,
Note that if we differentiate an ant derivative of f(x), we obta in f(x) back
again. Thus,
The expression ∫ f(x) dx is called an indefinite integral. The adjective
“indefinite” emphasizes that the result of ant differentiation is a “generic”
function, described only up to a constant term. The “elongated s” that
appear s on the left side of (2) is called an integral sign, ∗ the function f(x) munotes.in
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30 is called the integrand, and the constant C is called the constant of
integration. Equation (2) should be read as: The integral of f(x) with respect
to x is equal to F (x) plus a const ant.
THE DEFINITION OF AREA AS A LIMIT; SIGMA NOTATION
SIGMA NOTATION
To simplify our computations, we will begin by discussing a useful
notation for expressing lengthy sums in a compact form. This notation is
called sigma notation or summation notation be cause it uses the uppercase
Greek letter Ʃ (sigma) to denote various kinds of sums. To illustrate how
this notation works, consider the sum
12+22+32+42+52
in which each term is of the form k2 , where k is one of the integers from 1 to 5. In
sigma notation this sum can be written as,
Which is re ad “the summation of k2, where k runs from 1 to 5.” The notation tells us
to form the sum of the terms that result when we substitute successive integers for
k in the expression k2, starting with k = 1 and ending with k = 5.
More generally, if f(k) is a fu nction of k, and if m and n are integers such that m ≤ n,
then
Denotes the sum of the terms that result when we substitute successive
integers for k, starting with k = m and ending with k = n
THE DEFINITE INTEGRAL
A function f is said to be integrable on a finite closed interval [a, b] if the
limit
exists and does not depend on the choice of partitions or on the choice of
the points
in the subintervals. When this is the case we denote the limit
by the symbol
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31 Which is called the definite integral of f from a to b. The numbers a and b
are called the lower limit of integration and the upper limit of integration ,
respectively, and f(x) is called the integrand
2.5 AREA BETWEEN TWO CURVES: -
Suppose that f and g are continuous functions on an interval [a, b] and
f(x) ≥ g(x) for a ≤ x ≤ b [This means that the curve y = f(x) lies above the
curve y = g(x) and that the two can touch but not cross.] Find the area A of
the region bounded above by y = f(x), below by y = g(x), and on the sides
by the lines x = a and x = b
To solve this problem, we divide the interval [a, b] into n subintervals,
which has the effect of subdividing the region into n strips (Figure 6.1.3b).
If we assume that the width of the kth strip is ∆xk, then the area of the
strip can be approximated by the area of a rectangle of width ∆xk and
height f(
) − g(
), where
is a point in the kth subinterval. Adding
these approximations yields the following Riemann sum that approximates
the area A:
Taking the limit as n increases and the widths of all the subintervals
approach zero yields the following definite integral for the area A between
the curves:
Area F ormula : If f and g are continuous functions on the interval [a, b],
and if f(x) ≥ g(x) for all x in [a, b], then the area of the region bounded
above by y = f(x), below by y = g(x), on the left by the line x = a, and on
the right by the line x = b is
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32 Example 1 Find the area of the region bounded above by y = x + 6,
bounded below by y = x2, and bounded on the sides by the lines x = 0 and
x = 2.
Solution. The region and a cross section are shown in Figure 6.1.4. The
cross section extends from g(x) = x2 on the bottom to f(x) = x + 6 on the
top. If the cross section is moved through the region, then its leftmost
position will be x = 0 and its rightmost position will be x = 2. Thus, from
(1)
It is possible that the upper and lower boundaries of a region may intersect
at one or both endpoints, in which case the sides of the region will be
points, rather than vertical line segments (Figure 6.1.5). When that occurs
you will have to determine the points of intersection to obtain the limits of
integration.
2.6 LENGTH OF A PLANE CURVE: -
Our first objective is to define what we mean by the length (also called the
arc length) of a plane curve y = f(x) over an interval [a, b] (Figure 6.4.1).
Once that is done we will be able to focus on the problem of computing
arc lengths. To avoid some complications that would otherwise occur, we
will impos e the requirement that f’ be continuous on [a, b], in which case
we will say that y = f(x) is a smooth curve on [a, b] or that f is a smooth
function on [a, b]. Thus, we will be concerned with the following problem.
arc length problem Suppose that y = f(x) is a smooth curve on the interval
[a, b]. Define and find a formula for the arc length L of the curve y = f(x)
over the interval [a, b].
To define the arc length of a curve we start by breaking the curve into
small segments. Then we approximate the curve segments by line
segments and add the lengths of the line segments to form a Riemann sum.
Figure 6.4.2 illustrates how such line segments tend to become better and
better approximations to a curve as the number of segments increases. As
the number of segme nts increases, the corresponding Riemann sums
approach a definite integral whose value we will take to be the arc length
L of the curve. To implement our idea for solving Problem 6.4.1, divide munotes.in
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33 the interval [a, b] into n subintervals by inserting points x 1, x2, . . . , x n−1
between a = x 0 and b = x n. As shown in Figure 6.4.3a, let P 0, P1, . . . ,P n be
the points on the curve with x -coordinates a = x 0,
x1, x2, . . . , x n−1, b = x n and join these points with straight line segments.
These line segments form a polygonal path that we can regard as an
approximation to the curve y = f(x).
As indicated in Figure 6.4.3b, the length Lk of the kth line segment in the
polygonal path is
If we now add the lengths of these line segments, we obtain the following
approxim ation to the length L of the curve
To put this in the form of a Riemann sum we will apply the Mean -Value
Theorem (4.8.2). This theorem implies that there is a point
between xk-1
and xksuch that munotes.in
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34
and hence we can rewrite (2) as
Thus, takin g the limit as n increases and the widths of all the subintervals
approach zero yields the following integral that defines the arc length L:
In summary, we have the following definition.
definition If y = f(x) is a smooth curve on the interval [a, b], th en the arc
length L of this curve over [a, b] is defined as
This result provides both a definition and a formula for computing arc lengths.
Where convenient, (3) can also be expressed as
Moreover, for a curve expressed in the form x = g(y), where gis
continuous on [c, d], the arc length L from y = c to y = d can be expressed
as
2.7 SIMPSON’s RULE: -
SIMPSON’s RULE
MODELING WITH DIFFERENTIAL EQUATIONS
A function y = y(x) is a solution of a differential equation on an open
interval if the equation i s satisfied identically on the interval when y and
its derivatives are substituted into the equation. For example, y = e2x is a
solution of the differential equation munotes.in
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35
on the interval ( −∞, +∞), since substituting y and its derivative into the left
side of this equation yields
for all real values of x. However, this is not the only solution on ( −∞, +∞),
for example, the function
is also a solution for every real value of the constant C, since
After developing some techniques for solving equations such as (1), we
will be able to show that all solutions of (1) on ( −∞, +∞), can be obtained
by substituting values for the constant C in (2). On a given interval, a
solution of a differential equation from which all solutions on that interval
can be derived by substituting values for arbitrary constants is called a
general solution of the equation on the interval. Thus (2) is a general
solution of (1) on the interval ( −∞, +∞) ,
The graph of a solution of a differential equation is called an integral
curve for the equation, so the general solution of a differential equation
produces a family of integral curves corresponding to the different
possible choices for the arbitrary con stants. For example, Figure 8.1.1
shows some integral curves for (1), which were obtained by assigning
values to the arbitrary constant in (2)
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36 Separation of Variables
Step 1. Separate the variables in (1) by rewriting the equation in the
differential for m
h(y) dy = g(x) dx
Step 2. Integrate both sides of the equation in Step 1 (the left side with
respect to y and the right side with respect to x):
∫ h(y) dy = ∫ g(x) dx
Step 3. If H(y) is any antiderivative of h(y) and G(x) is any antiderivative
of g(x) ,then the equation
H(y) = G(x) + C
will generally define a family of solutions implicitly. In some cases it may
bepossible to solve this equation explicitly for y.
SLOPE FIELDS
In Section 5.2 we introduced the concept of a slope field in the context of
differential equations of the form y = f(x); the same principles apply to
differential equations of the form
y’ = f(x, y)
To see why this is so, let us review the basic idea. If we interpret y as the
slope of a tangent line, then the differential equation sta tes that at each
point (x, y) on an integral curve, the slope of the tangent line is equal to
the value of f at that point (Figure 8.3.1). For example, y’= y – x
A geometric description of the set of integral curves can be obtained by
choosing a rectangula r grid of points in the xy -plane, calculating the
slopes of the tangent lines to the integral curves at the gridpoints, and
drawing small segments of the tangent lines through those points. The
resulting picture is called a slope field or a direction field for the
differential equation because it shows the “slope” or “direction” of the
integral curves at the gridpoints.
The more grid points that are used, the better the description of the integral
curves. For example, Figure 8.3.2 shows two slope fields for (1)—the first
was obtained by hand calculation using t he 49 gridpoints shown in the
companying table, and the second, which gives a clearer picture of the
integral curves, was obtained using 625 gridpoints and a CAS
Euler’s Method
To approximate the solu tion of the initial -value problem
, y(x 0) = y 0
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37 Proceed as follows:
Step 1. Choose a nonzero number x to serve
as an increment or step size along thex -axis, and let
x1 = x 0 + ∆ x, x 2 = x 1 + ∆ x, x 3 = x 2 + ∆x, . . .
Step 2. Compute successively
y1 = y 0 + f(x0, y0)∆x
y2 = y 1 + f(x 1, y1)∆x
y3 = y 2 + f(x 2, y2)∆x
yn+1 = y n + f(x n, yn)∆x
The numbers y 1, y2, y3, . . . in these equations are the approximations of
y(x 1),y(x 2), y(x 3),
The Method of Integrating Factors
Step 1. Calculate the integrating factor
μ =
Since any μ will suffice, we can take the constant of integration to be zero
in this step.
Step 2. Multiply both sides of (3) by μ and express the result as
(μy) = μq(x)
Step 3. Integrate both sides of the equation obtained in Step 2 and then
solve for y. Be sure to include a constant of integration in this step.
2.8 SUMMARY: -
This chapter basically focuses on intergration and application
2.9 EXERCISES: -
The area A(x) under the graph of f and over the interval[a, x] is give n.
Find the function f and the value of a
2.10 REFERENCES -
Calculus: Early transcendental (10th Edition): Howard Anton, Irl
Bivens, Stephen Davis, John Wiley & sons, 2012.
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38 3
PARTIAL DERIVATIVES AND ITS
APPLICATIONS
Unit Structure
3.0 Objective
3.1 Introduction
3.2 Functions
3.3 Limits and Continuity
3.4 Partial Derivatives
3.5 Differentiability, Differentials and Local Linearity
3.6 Chain Rules
3.7 Directional Derivatives and t he Gradient
3.8 Maxima and Minima
3.9 Summary
3.10 Exercise
3.11 References
3.0 OBJECTIVE
Understanding of Mathematical concepts like limit, continuity,
derivative, integration of functions.
Ability to a ppreciate real world applications which uses these
concepts.
Skill to formulate a problem through Mathematical modeling and
simulation
3.1 INTRODUCTION
In this chapter we will extend many of the basic concepts of calculus to
functions of two or more variab les, commonly called functions of several
variables. We will begin by discussing limits and continuity for functions
of two and three variables, then we will define derivatives of such
functions, and then we will use these derivatives to study tangent plan es,
rates of change, slopes of surfaces, and maximization and minimization
problems. Although many of the basic ideas that we developed for
functions of one variable will carry over in a natural way, functions of
several variables are intrinsically more co mplicated than functions of one
variable, so we will need to develop new tools and new ideas to deal with
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39 3.2 FUNCTIONS
Definition A function f of two variables, x and y, is a rule that assigns a
unique real number f(x, y) to each point (x, y) in some set D in the xy -
plane.
13.1.2 definition A function f of three variables, x, y, and z, is a rule that
assigns a unique real number f(x, y, z) to each point (x, y, z) in some set D
in three dimensional space.
Example 1 Let f(x, y) =
+ ln(x2 − y). Find f(e, 0) and sketch the
natural domain of f
3.3 LIMITS AND CONTINUITY: -
Continuity can fail in the following ways:
The limit fails to exist. In some texts, this is called an essential
discontinuity. Any of the examples in the section on limits apply here.
The limit exists, but the function isn't defined at the point. sinxyx at
x = 0 is an example.
The limit exists and the function is defined at the point, but the
function output is different from the limit. The function
sin for 02 for 0xxfx x
x
is an example.
The latter two cases (where the limit exists as x approaches the point in
question) are called removable discontinuities.
To understand these last three points we need to start taking a look at the
concept of limit more precisely. What does it really mean when we say
that a function f is continuous at x = c if the values of f(x) approach f(c) as
x approaches c? What does it mean to approach c? How close to c does x
to get?
The concept of limit is the underpinning o f calculus.
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40
The informal definition or notation is L lim
cx
if the values of f(x) approach
L as x approaches c.
We will look for trends in the values of f(x) as x gets closer to c but xc.
Example 1:
sinlim
0 (Use radians.)
It appears from the graph that as approaches 0 from either side that the
value of sin appears to approach __________. The actual value of
sin when = 0 is __________.
Therefore the limits exists but the function is not continuous at =
0.While it appears that approaches 0 from either side that the value of
1
0
sinlim we are still very vague about what we mean by words like
“approach” and “close”.
Here is the formal definition of limit:
We define xflim
cxto be the number L (if one exists) such that for every
positive number (epsilon) > 0 (as small as we want), there is a positive
number (delta) > 0 (sufficiently small) such that if cx and cx then L)x(f .
The following f igure will help us with what this definition means:
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41 When we say “ f(x)” is close to L” we measure closeness by the distance
between f(x) and L. L)x(f = Distance between f(x) and L.
When we say “as close to L as we want,” we use th e (the Greek letter
epsilon) to specify how close.
We write L)x(f to indicate that we want the distance between f(x)
and L to be less than .
Similarly, we in terpret “ x is sufficiently close to c” as specifying a
distance between x and c: cx , where (the Greek letter delta) tells
us how close x should be to c.
If L)x(flim
cx
, then we know that no matter how narrow the horizontal
band determined by , there is always a which makes the graph stay
within that band for cx c .
Basically what we are trying to do is can we guarantee that the inputs
(sufficiently close to the value we are approaching but not equal to the
value) will make the outputs as close to L as we want.
We will use a graphic illustration to help make sense of this so lets go
back to sin)x(f .
How close should be to 0 ) ? ( 0 in order to make sin within
0.01 of 1? 0010.
First, set the y -range to go from ymin = 0.99 to ymax = 1.01. (0.99 < y <
1.01)
Making sure that the graph does not leave the window through the top or
bottom (meaning it goes below 0.99 or above 1.01), change the range
symmetrically.
Example 2 Use the definition of limit to show that the 6 2
3
x lim
x
We must show how, given any > 0, that we can find a >0 such that
If 3x and 3x , then 6 2x .
Since 3 26 2 x x the to get 6 2x would require that
3 2x or 23x .
Since dcx then 2. munotes.in
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42 One- and Two -Sided Limits
When we write )x(flim
x2 we mean that the number f(x) approaches as x
approaches 2 from both sides. This is a Two-Sided Limit .
If we want x to approach 2 only through values greater than 2 (like 2.1,
2.01, 2.003), we write )x(flim
x2. This i s called a right -hand limit .
(Similar to the concept of right difference quotient)
If we want x to approach 2 only through values less than 2 (like 1.9, 1.99,
1.994), we write )x(flim
x2. This is called a left-hand limit . (Similar to
the concept of left difference quotient)
Right -hand limits and left-hand limits are examples of One-Sided Limits .
If both the left -hand and right -hand limits are equal, then it can be proved
that )x(flim
x2 exists.
Whenever there is no number L that L)x(flim
cx
, we say )x(flim
cxdoes
not exist.
*Limits have to be a number and it has to be unique for that function.
Examples of Limits That Do Not Exist
1) Right – Hand Limit and Left -Hand Limit are different.
The one -sided limits exist but are different. At any integer, for example,
the greatest integer function doesn't have a limit. Functions with split
definitions can fall in this category at the point where the split occurs.
For example, with
22
2xxlim
x,
21 for 1 for 1xxfx
xx, 1lim
xfx doesn't exist.
Each of the following functions fails to have a limit at x = 0:
xgxx
and 1arctan hxx.
2) The function does not approach any finite number L a s cx.
The outputs grow without bound as the inputs approaches the point
from either one side munotes.in
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43 or the other, or both. For example,
201
xlim
x,
2lim tanxx,
and
0lim lnxx
don't exist because in each case, the outputs from the function
grow without bound.
3) The function does not settle down on a single value but oscillates
madly.
The function outputs fail to settle down on a single value, instead
oscillating madly.
A typical example is
xsinlim
x1
0.
3.4 PARTIAL DERIVATIVES
Given a certain multidimensional function, ),,,( tzyxA , a partial
derivative at a specific point defines the local rate of change of that
function in a particular direction. F or the4 -dimensional variable,
),,,( tzyxA , the partial derivatives are expressed as xtzyxAtzyx xA
xA
tzyx
tzy
,,, ,,,lim
constant ,,0
,,= slope of A in the x
direction ytzyxAtzy yxA
yA
tzxy
tzx
,,, ,, ,lim
constant ,,0
,,= slope of A in the y
direction ztzyxAtz zyxA
zA
tyxz
tyx
,,, , ,,lim
constant ,,0
,,= slope of A in the z
direction ttzyxAt tzyxA
tA
zyxt
zyx
,,, ,,,lim
constant ,,0
,,= the local time rate of
change of A
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44 The subscripts on the brackets indicate that those dimensions are held
constant.
Notice that the definition of a partial derivative of a multi -variable
function is the same as derivatives of functions of a single variable, but
with the other variables of the function being held constant. Whenever
you see the “backward -six” notation for the derivative, you should think
about what variable you are operating on, as indicated in the denominator
of the expression, while holding the other variables constant.
It is common convention that the directions being held constant are
implied and not explicitely written with subscripts.
2. Higher order partial derivatives
We can apply th e partial derivative multiple times on a scalar
function or vector. For example, given a multivariable function, yxf,,
there are four possible second order partial derivatives:
xyf
xf
y yxf
yf
x yf
yf
y xf
xf
x
2 2
22
22
; ; ;
The last two partial derivat ives, yxf
2
and xyf
2
are called “mixed
derivatives.” An important theorem of multi -variable calculus is the
mixed derivative theorem . The proof is beyond the scope of this course
and only the results are stated.
Mixed derivative Theorem: If a functionyxf, is continous and
smooth to second order, then the order of operation of the partial
derivatives does not matter. In other words: xyf
yxf
2 2
for a continous
and smooth (to second order) function yxf,
Example: For the function yx xyyxf2 2exp , , show xyf
yxf
2 2
Answer Provided:
yx yx xy yx x xyxyx xyyx yf
x yxf2 2 2 2 2 22
exp 1 2 exp 2 exp
yx yx xy yx xy yyyx xyxy xf
y xyf2 2 2 2 2 22
exp 1 2 exp2 exp
We can see that the order of operation of the p artial derivative on a
continous and smooth scalar function does not matter.
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Partial Derivatives and
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45 3. Del operator:
The del operator is a linear combination of spatial partial derivatives. In
rectangular coordinates, it is expressed as
zkyjxi^ ^ ^
(1)
Notice the second equality above is missing the vector arrow. is
always a vector operator and thus it is common convention to just leave
off the vector arrow.
The analysis of the del operator on various objects such as scalar funct ions
or vectors can be rather complex. In rectangular coordinates, however, the
rules we learned about in chapter 2 on “multiplying” vectors apply to the
del operator as well. It is important to notice however that the order is
extremely important in th e use of equation (1). The del operator acts on
all objects to the right of it. It is cruicial to note that the del operator is
not commutative when applied to scalars or vectors! You only apply
del operators on what is to the right in the term and neve r on the
objects to the left.
4. Gradient Operator
Applying the gradient operator, , on a scalar function zyx,,,
simply requires scalar multiplication. The gradient of yields the
following:
kzjyix zkyjxi ˆ ˆ ˆ ˆ ˆ ˆ
(2)
Notice that equation (2) is a linear combination of vector components and
basis vectors. In other words the gradient of a scalar yields a vector. You
will be tested on the application leading to equation (2) as w ell as the
fact that the result of is a vector. Since the gradient of a scalar
function is a vector, it obeys all the rules that we learned about in
chapter 2.
Example: For scalar function xyz show that x x
Example – Given a velocity vector ^ ^ ^
kwjviuu and the gradient of a
scalar function, as defined in equation (2), expand out u in
rectangular c oordinates:
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Calculu s
46 Answer provided: Using equation (10) from chapter 2,
zwyvxu kzjyixkwjviu u
ˆ ˆ ˆ^ ^ ^
We took the dot product of the vector u with the vector . We could
have just as well taken the dot product of the vector u with the operator
and then applied that on the scalar function : zwyvxuzwyvxu kzjyixkwjviu u
ˆ ˆ ˆ^ ^ ^
.In other words, u u . This equality is only rel avent when we
are operating on a scalar .
In this course, we will only take gradients of scalar functions . It is
possible to take gradients of vectors but you obtain a 9 element matrix
called the Dyadic product of the vector field, xu. For example, given the
vector ^ ^ ^
kwjviuu , the gradient of u is
zw
zv
zuyw
yv
yuxw
xv
xu
kzujyuixuu^ ^ ^
You can see why we want to avoid operations like this.
Gradient properties: magnitude
Equation (2) is a vector since it has a magnitude and direction. For a
function zyxff ,,, the magnitude of f is simply found using the
rules of chapter 2.
2 2 2
zf
yf
xff f f (3)
Gradient properties: direction
The direction of f is a bit more complicated. From the previous
chapter we can see that the direction of f can be expressed by the unit
vector,
ff
, but we also can interpret the direction of f in a more
geometric or physical way. First we need to use the differential of f,
which is labeled df. A differential is an infintesimal (meaning really munotes.in
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Partial Derivatives and
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47 small) change in the value of the multivariable function f and has
components:
dzzfdyyfdxxfdf
If we define the vector line element, ^ ^ ^
kdzjdyidx d , then we can
see by inspection that the differential takes the simple form
df df
Now let us apply the geometric definition of the dot product df :
cos df df df where is the coplanar angle between
the vector f and d.
If d is perpendicular f then o90 and 0df . In other words, d
is along lines of constant fwhen it is perpendicular to f. Alternatively,
we find that dfis a maximum when d is parallel to f. This means
that df is maximum when d is in the same direction as f(and also
perpendiculal to contours of constant f). This also means that f must
always be in the direction that leads to the greatest df. The direction of
f is also called the asecenda ntof f. Figure 1, on page 6, shows you a
picture relating the direction of fto lines of constant f.
5: The change of a quantity in the direction of the velocity field
(Advection)
We can find the change of a scalar, zyxf ,, , in an arbitrary direction,
^
3^
2^
1^
kujuiuu where 12
32
22
1 u u u , by taking the dot product
of ^
u with f. The results is:
fudtdf^
(4)
To derive equation (4), parametrize the spatial curve,
^ ^ ^
ktzjtyitx with respect to the variable t:
1 1 utxtuxtxo
2 2 utytuytyo munotes.in
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48 3 3 utztuztzo
Then, using the chain rule , we obtain equation (4):
^
uftz
zf
ty
yf
tx
xf
dtdf
Finding variations in a specific direction often occurs when we try
to find that variation of a physical quantity in the direction of the flow
field, u. We usually discuss the rate of change of the scalar quantity, zyxf ,, due to variations in f along the flow field, u. This is represented
mathematically as:
fudtdf
The term on the right side of the equality is called the advective term and
is one of two con tributions to the total or material derivative that we will
learn more about later in the semester. Often we are interested in
determining if there is any variation in the direction of flow. If one
obtain’s the result:
0fu
We say t hat the function, f, is spatially constant along the flow field, u.
For example, if our scalar quantity is a time -independent pressure field, zyxp ,, , then the equation 0pu , tells us that isobars are constant
along the flow field which also means that isobar contours are everywhere
parallel to the velocity vector field.
I. The gradient product rule of two scalar functions: gffg fg
II. The divergence product rule with a v ector and a scalar: u uu
III. The divergence of the gradient of a scalar – The Laplacian:
2
IV. The curl of the gradient of a scalar: 0,0,0 0
Notice the solution is the vector zero, 0, of which each component is
zero. It is common notation to imply the vector symbol of the vector zero
since the curl is always a vector result. munotes.in
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Partial Derivatives and
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49 V. The divergence of the curl of a vector: 0 u
Notice that this is just the scalar number 0 since the divergence always
results in a scalar function or number.
VI. The cross -product product rule with a vector and a scalar: uu u
VII. The divergence of the cross product: b aa bba
VIII. The curl of the cross product of a vector: a bb ab aa bba
IX. the gradient of the dot product of two vectors: a bb aa bb aba
X: The curl of the curl of a vector: aa a2
3.5 DIFFERENTIABILITY, DIFFERENTIALS AND
LOCAL LINEARITY: -
Definition. We say that the function y = f(x) is locally linear, or
differentiable, at the point x = a if the limit
exists.We simply say “f is locally linear” (or “differentiab le”) if it’s locally
linear at all points in a specified domain
Continuous functions We say that a function f is continuous at a point x =
a if • it is defined at the point, and • we can achieve changes in the output
that are arbitrarily small by restricti ng changes in the input to be
sufficiently small. This second condition can also be expressed in the
following form (due, in essence, to Augustin Cauchy in the early 1800’s):
Given any positive number ɛ (the proposed limit on the change in the
output is tr aditionally designated by the Greek letter ɛ, pronounced
‘epsilon’), there is always a positive number (the Greek letter ‘delta’),
such that whenever the change in the input is less than , then the
corresponding change in the output will be less than ɛ. A function is said munotes.in
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Calculu s
50 to be continuous on a set of real numbers if it is continuous at each point
of the set.
3.6 CHAIN RULES
In this lesson, we will need to use the Power Rule for rational exponents.
We will prove the Power Rule for rational exponents in Les son 11. Recall
that we have proved the Power Rule for positive integers in Lesson 8 and
for negative integers in Lesson 9.
Recall: nm n ma a/
Examples Differentiate the following functions.
1. 3 2x y
3/2xy3/13/13232xx y
Answer : 3/132xy
2. xxf1)(
2/1)( x xf2/32/32121)(xx xf
Answer:2/321)(xxf
3.
45 7 35)(
uuu uh
4/15/7 35)(uuu uh =
4/3 5/73 5 u u4/1 5/2497)( u u uh munotes.in
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Partial Derivatives and
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51 =20/5 20/8497u u = )9 28(4120/13 20/5u u =
4/120/1349 28uu
Answer :4/120/1349 28)(uuuh
4. 103 )(3 4 3 t tt ts
4/3 4/7 4/15 3 4/310 3 )103 ( )( t t t t ttts
4/1 4/3 4/11430421415)( t t t ts =
)10 7 5(433 4/1 t t t =
4/134)10 7 5(3tt t
Answer: 4/134)10 7 5(3)(tt tts
Theorem (The Chain Rule) If f and g are two differentiable functions
and ))(( )() ()( xgf xgf xk, then
)())(( )( xgxgf xk.
Proof By definition, hxk hxkxk
h)( ) (lim)(
0
.
COMMENT: The Chain Rule tells us how to differentiate the
composition of two functions f and g. In this form of the Chain Rule,
you would have to identify both functions.
Example Differentiate10 2 3)3 4 2( x x y using this form of the
Chain Rule.
Let 10)( x xf and let 3 4 2)(2 3 x x xg . Then
))((xgf = munotes.in
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52 )3 4 2(2 3 x xf = 10 2 3)3 4 2( x x . Thus,
))((xgfy.
10)( x xf 910)( x xf ))((xgf =
)3 4 2(2 3 x xf =
9 2 3)3 4 2(10 x x
3 4 2)(2 3x x xg x x xg 8 6)(2
Thus, )())(( xgxgf y = )8 6()3 4 2(102 9 2 3x x x x
Answer: )8 6()3 4 2(102 9 2 3x x x x y
Clearly, we need a better way than this!
Another way to state the Chain Rule: If )(ufy, where
)(xgu, then dxdududydxdy .
COMMENT: Since in the statement above, ))((xgfdudy and
)(xgdxdu , then this is the same statement of the Chain Rule given
earlier. However, in this form you only have to identify the function g,
which is being called u. In the first statement of the Chain Rule given
above, you had to identify bot h the functions of f and g.
Example Differentiate10 2 3)3 4 2( x x y using this form of the
Chain Rule.
Let 3 4 22 3 x x u . Then 10uy . Thus, 910ududy =
9 2 3)3 4 2(10 x x .
NOTE: When you write the answer for the derivative dudy
, you say
910u to munotes.in
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Partial Derivatives and
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53 yourself (silently), but you write 9 2 3)3 4 2(10 x x . Since dxdududydxdy , then
)8 6()3 4 2(102 9 2 3x x x x y since x xdxdu8 62 . With
this form of the Chain Rule, we are back to writing down the answer for
the derivative of a function.
Answer: )8 6()3 4 2(102 9 2 3x x x x y
COMMENT: The fastest way to confuse a calculus student about the
Chain Rule is to give them a fun ction of u. So, let’s address this problem.
First, we need to understand that the symbols t and T are not the same.
Because of this, t can be used to represent one expression and T can be
used to represent another expression. In physics, it is very comm on for t
to represent time and for T to represent temperature. Thus, for function
10 2 3)3 4 2( x x y , that was differentiated above, we could
have used X for the substitution variable instead of u. That is, let
3 4 22 3 x x X . We wil l call X “big X” instead of capital X.
So, if the function above was 10 2 3)3 4 2( t t y , then we would
have used a “big T” for the substitution variable instead of u. That is, let
3 4 22 3 t t T . If the function above was
10 2 3)3 4 2( u u y , then we would have used a “big U” for the
substitution variable. That is, let 3 4 22 3 u u U .
Examples Differentiate the following functions.
1. 3 2 3)3 2 4()( x x x xf
Let big X = 3 2 42 3 x x x . Thus, 3) ()( Xbig xf . By
the Power Rule and the Chain Rule,
) ( ) (3 ) (2 3XbigD Xbig XbigDx x . In general, we have
that ) ( ) ( ) (1XbigD Xbign XbigDxn n
x . Thus,
3 2 3)3 2 4()( x x x xf
)1 4 12()3 2 4(3)(2 2 2 3 x x x x x xf
Answer: )1 4 12()3 2 4(3)(2 2 2 3 x x x x x xf munotes.in
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54 2. 5)17 8(3 x y
Let big X = 17 8x . Thus, 5) (3 Xbig y . Then
) ( ) (156XbigD Xbig yx . Thus,
5)7 8(3x y 8)7 8(156 x y =
6)7 8(120 x
Answer:6)7 8(120 x y or6)7 8(120
xy
3. 5/3 2)2 3()( w w wg
Let big W = w w 2 32 . Thus, 5/3) ()( Wbig wg . Then
) ( ) (53)(5/2WbigD Wbig wgw . Thus,
5/3 2)2 3()( w w wg
)2 6( )2 3(53)(5/2 2 w w w wg
Answer: )2 6( )2 3(53)(5/2 2 w w w wg
or )1 3( )2 3(56)(5/2 2 w w w wg
or 5/2 2)2 3(5)1 3(6)(w wwwg
4. 4 3 5) 3 4(75)(t t tts
First, we may write 4 3 5) 3 4(75)( t t t ts .
Let big T = t t t 3 53 4 . Thus, 4) (75)( Tbig ts .
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55 Then
) ( ) (720)(5TbigD Tbig tst . Thus,
4 3 5) 3 4(75)( t t t ts
)1 9 20() 3 4(720)(2 4 5 3 5 t t t t t ts
Answer: )1 9 20() 3 4(720)(2 4 5 3 5 t t t t t ts
or 5 3 52 4
) 3 4(7)1 9 20(20)(t t tt tts
5. 153)(xxf
We differentiate this function in Lesson 9 using the Quotient Rule.
Now, we will differentiate it using the Chain Rule.
First, we may write 1)15 (3 )( x xf .
Let big X = 15x . Thus, 1) (3 )( Xbig xf . Then
) ( ) (3)(2XbigD Xbig xfx . Thus,
1)15 (3 )( x xf
)1()15 (3)(2 x xf = 2)15 (3x
Answer: 2)15 (3)( x xf or2)15 (3)(
xxf
6. 17 921)(xxg
First, we may write 1)17 9(21)( x xg .
Let big X = 17 9x . Thus, 1) (21)( Xbig xg . Then
) ( ) (21 )(2XbigD Xbig xgx . Thus, munotes.in
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Calculu s
56 1)17 9(21)( x xg
9)17 9(21 )(2 x xg =
2)17 9(189 x
Answer:
2)17 9(189 )( x xg or 2)17 9(189)(
xxg
7. 22 3 46
2 t ty
First, we may write 1 2)22 3 4(6 t t y .
Let big T = 22 3 42 t t . Thus, 1) (6 Tbig y . Then
) ( ) (62TbigD Tbigdtdy
t . Thus,
1 2)22 3 4(6 t t y
)3 8()22 3 4(62 2 t t tdtdy
Answer: )3 8()22 3 4(62 2 t t tdtdyor
2 2)22 3 4()3 8(6
t tt
dtdy
8. 3 2 2)18 3 ( 10)( z z zh
3/2 2)18 3 (10)( z z zh
Let big Z = 18 32 z z . Thus, 3/2) (10)( Zbig zh .
Then
) ( ) (320)(3/1ZbigD Zbig zhz . Thus,
3/2 2)18 3 (10)( z z zh munotes.in
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Partial Derivatives and
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57 )3 2( )18 3 (320)(3/1 2 z z z zh
Answer: )3 2( )18 3 (320)(3/1 2 z z z zh
or 3/1 2)18 3 (3)3 2(20)(
z zzzh
Since )3 ()6 ( 18 32 z z z z ,then
3/1])3 ()6 ([3)3 2(20)(
z zzzh
Sign of )(zh: + +
6 23 3
NOTE: In Lesson 14, we conclude that the function h is
increasing on the interval ),3(23,6
and is
decreasing on the interval
3,23)6, ( . There is
a local maximum occurring when 23x and since
3 2 2)3 ()6 ( 10)( z z zh , the local maximum is
32 2
26
23
212
231023
h =
32 2
29
2910
=
32 2
29
2910
= 34
2910
= 3
2945 = 3
83645 = 236 453
. There is a local minimum occurring when 6x
and since 3 2 2)3 ()6 ( 10)( z z zh , the local minimum munotes.in
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58 is 0)6(h . There is a local minimum occurri ng when
3x and the local maximum is 0)3(h .
9. 85 9 )( w wg
This function is from Lessons 6 and 7. In Lesson 6, we found the
slope of the tangent line to the graph of )(wgy at the point
)7,4())4(,4(g . In Lesson 7, we use the definition of
derivative to find the derivative of this function. Now, we will use
the Chain Rule to find the derivative of this function.
2/1)85 9()( w wg
Let big W = 85 9w . Thus, 2/1) ()( Wbig wg . Then
) ( ) (21)(2/1WbigD Wbig wgw
. Thus,
2/1)85 9()( w wg
9 )85 9(21)(2/1 w wg = 2/1)85 9(29w
Answer:2/1)85 9(29)( w wg or
2/1)85 9(29)(
wwg
10. 5 2 3)9 8()67( x x y
NOTE: )6()67(3 )67(2 3 x x Dx
x x x Dx 16)9 8(5 )9 8(4 2 5 2
Using the Product Rule, we obtain
x x x x x y 16)9 8(5)67( )9 8()6()67(34 2 3 5 2 2 =
4 2 3 5 2 2)9 8()67(80 )9 8()67(18 x x x x x = munotes.in
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Partial Derivatives and
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59 ])67(40)9 8(9[)9 8()67(22 4 2 2x x x x x =
) 240 280 81 72()9 8()67(22 2 4 2 2x x x x x =
)81 280 312()9 8()67(22 4 2 2 x x x x =
)81 280 312()9 8()67(22 4 2 2 x x x x
Answer: )81 280 312()9 8()67(22 4 2 2 x x x x y
or )81 280 312()9 8()67(22 4 2 2 x x x x y
11. 5 4) 2(3/2
zzy
12. 4
32
85 3)(
xxxh
13. 9
356)(
tt tg
14. 2 2 2)3 5 ()4 3( x xx x y
15. ) 16()9 ()(4 8 2w w wf
3.7 DIRECTIONAL DERIVATIVES AND THE
GRADIENT: -
f(x,y) be a real -valued function with domain D in R2 , and let (a,b) be a
point in D . Let v be a unit vector in R2 . Then the directional derivative
of f at (a,b) in the direction of v , deno ted by D vf(a,b) , is defined as
Dv f(a,b) =
For a real -valued function f(x,y) , the gradient of f , denoted by ∇f , is
the vector
∇f=(∂f/∂x,∂f/∂y)
In R2 . For a real -valued function f(x,y,z) , the gradient is the vector
∇f=(∂f/∂x,∂f/∂y,∂f/∂z)
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60 3.8 MAXIMA AND MINIMA
Let f(x,y)f(x,y) be a real -valued function, and let (a,b)(a,b) be a point in
the domain of ff. We s ay that ff has a local maximum at (a,b)
(a,b) if f(x,y)≤f(a,b)f(x,y) ≤f(a,b) for all (x,y)(x,y) inside some disk of
positive radius centered at (a,b) (a,b), i.e. there is some sufficiently
small r>0r>0 such that f(x,y)≤f(a,b) for all (x,y) f(x,y) ≤f (a,b) for all (x,y)
for which (x−a)2+(y−b)2Likewise, we say that ff has a local minimum at (a,b) if f(x,y)>f(a,b) for
all (x,y)(a,b) if f(x,y)>f(a,b) for all (x,y) inside some disk of positive
radius centered at (a,b)(a,b).
If f(x,y)≤f(a,b) for all (x,y)f(x,y) ≤f(a,b) for all (x,y) in the domain of ff,
then ff has a global maximum at (a,b)(a,b). If f(x,y)≥f(a,b) for all (x,y) in
the domain of f, then ff(x,y) ≥f(a,b) for all (x,y) in the domain of f,
then f has a global minimum at (a,b)(a,b).
3.9 SUMMARY
This chapter mainly covers the basic concepts of partial derivatives.
3.10 EXERCISES
Find (a) the equation of the tangent line to the graph of the function
4 2)3 8 4( x x y at the point )81,2( and (b) the point(s)
on the graph at which the tangent line is horizontal.
3.11 REFERENCES
Calculus: Early transcendental (10th Edition): Howard Anton, Irl
Bivens, Stephen Davis, John Wiley & sons, 2012.
munotes.in