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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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Calculus

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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Derivatives and Its

Application

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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Calculus

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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Derivatives and Its

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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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Derivatives and Its

Application

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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Calculus

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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Derivatives and Its

Application

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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Derivatives and Its

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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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Calculus

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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Derivatives and Its

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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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Calculus

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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Derivatives and Its

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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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Calculus

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,

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Derivatives and Its

Application

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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Calculus

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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Derivatives and Its

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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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Derivatives and Its

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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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Calculus

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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Derivatives and Its

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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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Derivatives and Its

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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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Integration and Its

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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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Integration and Its

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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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Calculus

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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Integration and Its

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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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Integration and Its

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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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Calculus

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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Integration and Its

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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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Integration and Its

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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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Partial Derivatives and

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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:

munotes.in

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Partial Derivatives and

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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

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Partial Derivatives and

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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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## Page 44

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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.

munotes.in

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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:

munotes.in

## Page 46

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

## Page 47

Partial Derivatives and

Its Applications

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

## Page 51

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

## Page 52

Calculu s

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

## Page 53

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

## Page 54

Calculu s

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 .

munotes.in

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Partial Derivatives and

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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

## Page 56

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

## Page 57

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

## Page 58

Calculu s

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