3.1 MAXIMUM, MINIMUM AND INFLECTION POINT & SKETCHING THE GRAPH. In Isaac Newton's day, one of the biggest problems was poor navigation at sea.

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 were vital for navigation.

Shipwrecks occured because the ship was not where the captain thought it should be. There was not a good enough understanding of how the Earth, stars and planets moved with respect to each other.

Calculus (differentiation and integration) was developed to improve this understanding.

Differentiation and integration can help us solve many 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 when modelling the behaviour of moving objects.

Our discussion begins with some general applications which we can then apply to specific problems.

NOTES:

a. There are now many tools for sketching functions (Mathcad, LiveMath, Scientific Notebook, graphics calculators, etc). It is important in this section to learn the basic shapes of each curve that you meet. An understanding of the nature of each 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. Maximum

point

Use

2 2

d y

dx <0 sign: −ve

b. Minimum

point Use

2 2

d y

dx >0 sign: +ve

c. Inflection point

Use

2 2

d y dx = 0

59

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

x-intercepts Use y = 0

NOTE: 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,dy 0 dx 

Find the value of x.

b. Find the value of y, when x?

c. Determine the nature of stationary point,

2

2 ?

d y dx 

60 Example 1:

Find the stationary point of the curve yx² x 6and sketch the graph.

Solution:

Determine stationary point, dy

dx=

dy 0 dx 

Find the value of x ,

Find the value of y, when ¹ x2

2 6

yx  x

61 So, stationary point is

(

)

Determine the nature of the point,

So, the point

(

)

y - intercept, when x0

( ) ( )

x-intercept, when y0

2  6

x x y

0

2 x6 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 y2x³4 and determine the nature of the point.

Solution:

Determine stationary point, dy 0 dx 

dy dx=

Find the value of x ,

Find the value of y, when x0 2 3 4

y x 

67 So, stationary point is ( , ).

Determine the nature of the point,

2 2

d y dx 

So, the point ( , ) is a

68 Example 4:

Find the stationary point on the curve y2x²4x3 and determine their nature.

69 Example 5:

Determine whether the equation ^{f x}

 

70 Exercise 3.1

a. Find the coordinates of stationary points of the curve y3x³9x² and determine their nature.

b. Find the coordinates of stationary points of the curve ^y





c. Find the coordinates of stationary points of the curve y8x³2x⁴ determine their nature and sketch the graph.

d. Find turning point for the curve yx³6x²9x5 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 straight line, which is called rectilinear motion.

In this matter, we must assume that the object is moving along a coordinate line. The object that moves along a straight line with position s = f(t), has corresponding velocity v ^ds

 dt , and its acceleration

2 2

dv d s a dt  dt .

If t is measured in seconds and s in meters, then the units of velocity are meters per second, which we abbreviate as m/sec. The units of acceleration are then 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 s8t²t³ in meter and the interval 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 t 3s.

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

Solution

a. Given that s8t² t³, So v ^ds

 dt

v16t3t²

b. We know that

2 2

dv d s a dt  dt , a166t

c. When t3s, v16t3t² 16(3)3(3)²

4827 21ms^1

73 d. During 6s means t6s,

m s_t6s 8(6)² (6)³ 28821672

m s_t5s 8(5)² (5)³ 20012575

5

6 

 

s_t s_t s

7275 m

3



Example 2:

A car moves along a straight line so that the displacement, s in meter and t in seconds passes through O is given by s t³ 6t²5t. Find

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

b. The velocity of the car when its acceleration is 12ms^2. c. The time when the car returns back to O.

Solution:

a. When t3s.

3 2

6 5

s t t  t

^{336(3)25(3)} 275415

12m

74 b. Value of t when its acceleration is 12ms^2.

Given, 12 2

a ms^ 6 12 a t

So, 6t12 12 6t1212 6t 24

s t t

4 6 24





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

3 2

3 2

6 5

6 5 0

s t t t

t t t

  

  

0 ) 5 6

(t²  t  t

,

0

t ( 1)( 5) 0

0 5

2 6











 t t

t t

s t t

1 0 1







s t t

5 0 5







75 Example 3:

A car is moving at a straight line with position ^{1 3} 9

s3t  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, a10 /m s² 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 ^st2





a. Acceleration of the car when t4s. b. The time when the velocity is 39ms^1. c. Displacement in the 4^thsecond.

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 by s t³ 2t²20t.

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

b. Calculate the acceleration when t3s.

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 s t³ 9t²24t1. 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 t4s.

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 both sides w.r.t. (with respect to) time ^d

dt. If ^{y f x}

 

dt  dxdt where ^dy

dt and ^dx

dt are the rates of change of y and x respectively.

Example 1:

If the radius of a circle increase at the rate of ^{1 1}

5cms^ , 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, Ar2

Differentiate A with respect to r, dr r

dA 2

78 Using the formula,

dA dA dr dt  dr dt

^5, 2 1

 r

when

r 10cm

^{5 (10)} 2



1

4 2 ^

 cm s

79 Example 2:

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

3 1

40 cm s^ . Find the rate of increase of the height of the water level.





r= 20cm

1

40 3 ^

 cm s dt

dv

Find, ? dt dh

80 Example 3:

The radius of a circle is increasing at a rate of 5cm m/ . Find the rate of increase of the area of the circle at the instant when its radius is12cm . (Ans: dA 120 ²/

cm m

dt   )

81 Example 4:

The volume of a spherical balloon is decreasing at a rate of 4 cm s ^{3 1}. Find the rate of decrease of its radius when its radius is 5cm . (Ans:dj 0.04 /

dt   cm s)

82 Exercise:

1. The formulae for the volume and surface area of a sphere of radius r are ^{4 3} V  3r and A4r² respectively. When r5cm, V is increasing at the rate of 10cm s^{3 1}. Find the rate of increase of A at that instant. (Ans: dj 0.032 /

dt  cm s, 1.28 2/

dA cm s

dt   )

2. Given that the height of a cylinder is 3cm and the radius increase at the rate of 0.2cm/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 do things like maximise the profit in a company, or minimise the costs, or find the least amount of material 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 15cm. What is the maximum volume of the 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 the measurement of its sides.

Solution:

86 Example 3:

The rectangular above has perimeter of 40 meters. Given the perimeter equation is 2x2y40. 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 the particle in t seconds. Find the velocity and acceleration of the particle by using the following equations.

a. s6t²2t³ b. s8t³48t²72t c. s64t²16t⁴

2. For the curve of yx³6x²2, 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 s 9 6t²t³.

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.

88 4. The curve is given as y3x³9x². Find

a. The coordinates of the stationary points.

b. The nature of the points.

5. A particle moves in a straight line. Its displacement, s meter, from the fixed point O after t second is given by s t³ 3t²8t.

a. Find the velocity when time is 3 and 5 seconds.

b. Find the time when the velocity is 8ms^1.

6. A balloon is filled with air at the rate of 20cm³per minute until a sphere is formed. Find the rate of change of the radius when the radius is 4cm ( ^{4 3}

v 3r ) 7. A car moves from a static condition from point O with displacement s meter on a

straight road in time t seconds is given by s75tt³.

a. Find the total displacement when the car is instantaneously at rest.

b. Find the velocity when the acceleration is 3ms². c. Find the initial acceleration.

d. Find the acceleration when it goes back to point O.

8. i. Find the stationary points for yx³x²and determine the maximum points.

ii. Given f x( )x³6x²9x. Determine the nature of the stationary point and hence sketch the graph of f x( ).

89

9. i. Find the stationary point and sketch the graph, if any of the function

3 2

6 9 5

yx  x  x .

ii. The curve yx³6x²11x6 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 cmheight, the volume of that cone is V cm³, prove that ^{3 3}

V 25x . (Given volume of cone, ^{1 2} V 3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 5cm to 4.98cm.