chapter 3 application of differentiation

8/12/2019 Chapter 3 Application of Differentiation

http://slidepdf.com/reader/full/chapter-3-application-of-differentiation 1/33

BA201 ENGINEERING MATHEMATICS 2012

57

CHAPTER 3 APPLICATION OF DIFFERENTIATION

3.1 MAXIMUM, MINIMUM AND INFLECTION POINT & SKETCHING THE GRAPH

Introduction to Applications of Differentiation

In Isaac Newton's day, one of the biggest problems was poor navigation at sea .

Before calculus was developed, the stars werevital for navigation.Shipwrecks occured because the ship was notwhere the captain thought it should be. Therewas not a good enough understanding of how theEarth, stars and planets moved with respect toeach other.Calculus (differentiation and integration) wasdeveloped to improve this understanding.Differentiation and integration can help us solvemany types of real-world problems .

We use the derivative to determine the maximum and minimum values of particular

functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.).Derivatives are met in many engineering and science problems , especially whenmodelling the behaviour of moving objects.Our discussion begins with some general applications which we can then apply tospecific problems.NOTES:

a. There are now many tools for sketching functions (Mathcad, LiveMath, ScientificNotebook, graphics calculators, etc). It is important in this section to learn thebasic shapes of each curve that you meet. An understanding of the nature ofeach function is important for your future learning. Most mathematical

modelling starts with a sketch.b. You need to be able to sketch the curve, showing important features. Avoid

drawing x-y boxes and just joining the dots.c. We will be using calculus to help find important points on the curve.




58

3 types of turning points:

a. Maximumpoint

Use2

2d ydx

<0 sign: −ve

b. Minimumpoint

Use2

2

d ydx

>0 sign: +ve

Inflection point

Use

2

2

d ydx = 0




59

The kinds of things we will be searching for in this section are:

x-intercepts Use y = 0NOTE: In many cases, finding x -intercepts is not so easy.

If so, delete this step.

y-intercepts Use x = 0

Guide to solve the problem:

a. Determine stationary point, 0dydx

Find the value of x .

b. Find the value of y , when ? x

c. Determine the nature of stationary point,2

2 ?d ydx




60

Example 1:

Find the stationary point of the curve 2 6 y x x and sketch the graph.

Solution:

Determine stationary point,

dydx

=

0dydx

Find the value of x ,

Find the value of y , when12 x

2 6 y x x




61

So, stationary point is

Determine the nature of the point,

So, the point is a minimum point.

y - intercept, when 0 x

( ) ( )

x-intercept, when 0 y

62 x x y

062 x x

0)3)(2( x x

and




62

(1/2,-25/4)

0-2

-6

3

y

x

Example 2:

Find stationary point for the curve and determine the nature of the point.

Solution:

Determine stationary point,




63

( )

and

When when

( ) ( ) ( ) ( )

The stationary points are ( ) and ( )

Nature :




64

At point ( ) when x=3 At point ( ) when x=

( ) ( )

( ) ( )

Stationary point ( ) stationary point ( )

is a minimum point is a maximum point

Sketch graph

x-intercept when y=0

( )

and




65

-5.2

(3,54)

5.2

(-3,54)

0

y-intercept when x=0

( ) ( )

x

y




66

Example 3:

Find stationary point for the curve 32 4 y x and determine the nature of the point.

Solution:

Determine stationary point, 0dydx

dydx

=

Find the value of x ,

Find the value of y , when 0 x

32 4 y x




67

So, stationary point is ( , ).

Determine the nature of the point,

2

2

d ydx

So, the point ( , ) is a




68

Example 4:

Find the stationary point on the curve 22 4 3 y x x and determine their nature.




69

Example 5:

Determine whether the equation 24 5 6 f x x x is a maximum, minimum or

inflection and sketch the graph of ( ).




70

Exercise 3.1

a. Find the coordinates of stationary points of the curve 3 23 9 y x x and determine

their nature.

b. Find the coordinates of stationary points of the curve 24 y x and determine

their nature.

c. Find the coordinates of stationary points of the curve 3 48 2 y x x determine their

nature and sketch the graph.

d. Find turning point for the curve 3 26 9 5 y x x x and determine maximum point

and minimum point for the curve.




71

3.2 RECTILINEAR MOTION

ONE OF THE most important applications of calculus is to motion in a straightline, which is called rectilinear motion.

In this matter, we must assume that the object is moving along acoordinate line. The object that moves along a straight line with position s = f (t ),

has corresponding velocity dsv

dt , and its acceleration

2

2

dv d sa

dt dt .

If t is measured in seconds and s in meters, then the units of velocity aremeters per second , which we abbreviate as m/sec. The units of acceleration arethen meters per second per second , which we abbreviate as m/sec².

s=0 -The particle at the beginning

- The particle returns back to O again

v=0 -the particle is instantaneously at rest

-maximum displacement

a=0 -constant velocity

-the particle is begin

t=0 -initial velocity

-initial acceleration




72

Example 1:

An object P moving on a straight line has position 2 38 s t t in meter and theinterval time of t seconds. Find

a. The velocity of P at the time t .

b. The acceleration of P at the time t .

c. The velocity of P when 3t s .

d. The displacement of P during the 6 th seconds of moving.

Solution

a. Given that 2 38 s t t ,

So dsv

dt

v 2316 t t

b. We know that2

2

dv d sa

dt dt

,

a t 616

c. When 3t s ,

v 2316 t t

2)3(3)3(16

2748 121 ms




73

d. During 6 s means 6t s ,

m s st 72216288)6()6(8 326

m s st 75125200)5()5(8 325

56

t t s s s

7572

m3

Example 2:

A car moves along a straight line so that the displacement, s in meter and t inseconds passes through O is given by 3 26 5 s t t t . Find

a. The displacement of the car when 3t s .

b. The velocity of the car when its acceleration is 212ms .

c. The time when the car returns back to O.

Solution:

a. When 3t s .

3 26 5 s t t t

)3(5)3(63 23

155427

m12




74

b. Value of t when its acceleration is 212ms .

Given,

212a ms

6 12a t

So, 6 12 12t

12126 t

246 t

st t

4

6

24

c. The time when the car returns back to O, 0 s m ,

3 2

3 2

6 5

6 5 0

s t t t

t t t

0)56( 2

t t t

,0t 0)5)(1(

0562

t t

t t

st

t

1

01

st

t 5

05




75

Example 3:

A car is moving at a straight line with position 319

3 s t t , where its displacement,

s in meter and the time, t in seconds. Find the:

a. Displacement, s of the car after 3 seconds.

b. Displacement, s of the car in fourth second.

c. Velocity, v of the car when the acceleration, 210 /a m s

d. Acceleration of the car at the time t = 2 seconds.




76

Exercise:

1. A car moves from a static condition in a straight line. Its displacement s meter

from a fixed point after t second is given by 2

2 s t t

, find

a. Acceleration of the car when 4t s .

b. The time when the velocity is 139ms .

c. Displacement in the 4 th second.

2. A ball is rolling along a straight ground from the fixed point O and its

displacement, s meter and time t second is given by3 2

2 20 s t t t .

a. Calculate the acceleration of the ball when it begins to move.

b. Calculate the acceleration when 3t s .

c. Find the displacement when the ball is instantaneously at rest.

d. Find displacement of the ball during 4th seconds.

3. A particle move in linear and the distance, S from fixed point 0, t seconds after

passing point 0, is given by 3 29 24 1 s t t t . Calculate

a. Distance in the 5 th second.

b. Values of t when the particles stop for a while.

c. Velocity when acceleration is zero.

d. Acceleration when 4t s .




77

3.3 RATE OF CHANGE

If 2 variables both vary with respect to time and have a relation between them, we can

express the rate of change of one in terms of the other. We need to differentiate bothsides w.r.t. (with respect to) time d

dt .

If y f x , then dy dy dxdt dx dt

where dydt

and dxdt

are the rates of change of y and x

respectively.

Example 1:

If the radius of a circle increase at the rate of 115

cms , find the rate of area at the instant

when the radius is 10cm .

Solution:

Given, Find,

dr

dt 5

1 ?

dA

dt

r 10cm

We know that the area of a circle is,

2 A r

Differentiate A with respect to r,

r dr dA

2




78

Using the formula,

dA dA dr

dt dr dt

,51

2 r

when

cmr 10

)10(52

124 scm




79

Example 2:

An open cylinder has a radius of 20 cm . Water is poured into the cylinder at the rate of3 1

40 cm s . Find the rate of increase of the height of the water level. hr v 2

r= 20cm

1340 scmdt dv

Find, ?dt dh




80

Example 3:

The radius of a circle is increasing at a rate of 5 /cm m . Find the rate of increase of the

area of the circle at the instant when its radius is 12cm . (Ans:2

120 /dA

cm mdt )




82

Exercise:

1. The formulae for the volume and surface area of a sphere of radius r are 34

3

V r

and 24 A r respectively. When 5r cm , V is increasing at the rate of 3 110cm s .

Find the rate of increase of A at that instant. (Ans: 0.032 /dj

cm sdt

,

21.28 /dA

cm sdt

)

2. Given that the height of a cylinder is 3cm and the radius increase at the rate of0.2cm/s. Find the rate of change of its volume when the volume is




83

3.4 OPTIMIZATION PROBLEM

Important!

The process of finding maximum or minimum values is called optimisation . We are trying to dothings like maximise the profit in a company, or minimise the costs, or find the least amount ofmaterial to make a particular object.

These are very important in the world of industry.

Example 1:

If the sum of height, h cm and radius, r cm of a cone is 15 cm . What is the maximum volume ofthe cone?

Solution:

( 1)

(2)

Substitute (1) into (2)

( )




84




85

Example 2:

A wire with length 2 meters is bent to form a rectangular with a maximum area. Find themeasurement of its sides.

Solution:




86

Example 3:

The rectangular above has perimeter of 40 meters. Given the perimeter equation is2 2 40 x y . Find

a. Find the equation of the area in term x .

b. Find the maximum area of the rectangle.

y

x




87

POLITEKNIK KOTA BHARU

JABATAN MATEMATIK, SAINS DAN KOMPUTER

BA 201 ENGINEERING MATHEMATICS 2

PAST YEAR FINAL EXAMINATION QUESTIONS

1. A particle is moving along a straight line where s is the distance travelled by theparticle in t seconds. Find the velocity and acceleration of the particle by usingthe following equations.

a. 2 36 2 s t t

b. 3 28 48 72 s t t t

c. 2 464 16 s t t

2. For the curve of 3 26 2 y x x , find

a. The coordinates of all the turning points.

b. The maximum and minimum points.

c. Sketch the graph for the above curve.

3. A particles moves along a straight line such that its distance, s meter from a fixed

point O is given by 2 39 6 s t t .

a. Find the velocity of the particles after 2 seconds and the acceleration

after 4 seconds.b. Find the acceleration when the velocity is 9 /m s .




89

9. i. Find the stationary point and sketch the graph, if any of the function3 26 9 5 y x x x .

ii. The curve3 2

6 11 6 y x x x has 2 stationary points. Find the

points and state the condition of the points.

10. i) A cone shape container that placed upside down having a base radius

12cm and height 20cm . Then one boy pouring some water to the

cone which is x cm height, the volume of that cone is 3V cm , prove

that 3325

V x . (Given volume of cone, 213

V r h )

ii) The water is flow out of the cone through the hole at the cone peak. Find

the nearest alteration for V when x decrease from 5 cm to 4.98cm .

chapter 3 application of differentiation

Documents