BA201 Engineering Mathematic Unit 03 - Appofdiff

TOPIC 3

APPLICATION OF DIFFERENTIATION

1

PREPARED BY NIK AKHTAR NASUHA /WONG TUCK SUNG

3.1 UNDERSTAND THE APPLICATION OF DIFFERENTIATION

The differentiation is the subfield of Calculus and there is various Application of

Differentiation applied in real world. The differentiation is very important part of

Mathematics as it is used in many scientific fields. Differentiation can be defined as the

process of finding the Derivatives of the Functions. The examples of what can we find using

differentiation are the gradient of a curve, the stationary point of a curve, the maximum or

minimum value of an area and etc.

3.1.1 USE DIFFERENTIATION TO FIND THE GRADIENT OF A CURVE

Basically, gradient is another word for "slope". The higher the gradient of a graph at a point,

the steeper the line is at that point. A negative gradient means that the line slopes

downwards.

Figure 3.1

To find the gradient of a curve, we must draw an accurate sketch of the curve. At the point

where we need to know the gradient, draw a tangent to the curve. A tangent is a straight

line which touches the curve at one point only. We then find the gradient of this tangent.

TOPIC 3 : APPLICATION OF THE

DIFFERENTIATION

http://www.tutorcircle.com/calculus-cE9t.html

http://www.tutorcircle.com/math-c27t.html

http://www.tutorcircle.com/derivatives-t1bXp.html

http://www.tutorcircle.com/functions-s8IUi.html

TOPIC 3


2


Example 3.1 :

a) Find the equation of the tangent of y = 4x2 - 12x + 10 at the point where x = 4.

Solution:

Given that y = 4x2 – 12 x + 10.

Differentiating y with respect to x,

dx

dy = 8x – 12.

Substituting x = 4 into dx

dy ,

We get dx

dy = 20 .

Figure 3.2

y= 4x2 – 12 x + 10

10

x

y

4

26

Tangent

b) Find the gradient of each of the following curves at the point given. (Let’s try this question in class) i) at the point (5, 4) [Ans : 10] ii) ( ) at the point (5,0) [Ans : 45]

TOPIC 3


3


3.1.2 Find turning point/stationary point

For a function: y = f(x), a stationary point is a point on the function graph where the gradient

of the function is zero.If the gradient of the function changes sign at the stationary point,

then it is called a turning point, which can be a local maximum or local minimum:

Figure 3.3

c) Given that the curve with equation has gradient 2 at the point (3, -1). a and b are constants. Find the value of a and b. (Let’s try this question in class)

[Ans :

]

(turning point)

(turning point)

TOPIC 3


4


Example 3.2

1) Given that 732 xxy , find the stationary point/ turning point of this curve.

Solution :

32

732

xdx

dy

xxy

at turning stationary/turning point, gradient is zero. Means :

2

3

32

032

x

x

x

Substitute x=3/2 into equation y :

4

19

72

9

4

9

72

33

2

32

y

y

y

so, turning point occured at

4

19,

2

3

2)Find the stationary point on the curve . (Let’s try this question in class) [Ans : (0,0) and (-2,-8)]

TOPIC 3


5


3.1.3 Determine maximum, minimum and point of inflection

1) We use the first derivative to know the coordinate of the stationary / turning points.

2) Then, we use the second derivative,

to determine whether the turning points is

maximum, minimum or point of inflection. The way to ensure that a stationary point

is the point of inflexion is that dx

dy = 0 and

2

2

dx

yd = 0.

Figure 3.4

a) Maximum point

b) Minimum point

c) Points of inflexion

Figure 3.5

f’(x) = 0

f’(x) = 0

f’(x) > 0 f’(x) < 0 f’(x) > 0

f’(x) < 0

Point of inflection

f” (x) = 0

f” (x) < 0

f” (x) > 0

f’(x) > 0

f’(x) > 0

f’(x) < 0

f’(x) < 0

TOPIC 3


6


Example 3.3:

Find the coordinates of the stationary points on each of the following curves and state their

nature (i.e. whether they are maxima, minima or points of inflection). Then, sketch the

curve.

a)

Solution :

1) Find the first derivative :

2) The turning point happen when the gradient is zero, so find the coordinate x when

.

3) Substitute the value of x into equation y to find the coordinate y.

( )

So, the coordinate of turning point : ( )

4) Use the second derivative,

to determine the nature of the point.

5) Find the Intersection point at x-axis (y=0) [only for quadatric equation]

( )( )

( )( )

6) Intersection point at y-axis (x=0) [only for quadatric equation]

( ) ( )

( )

7) Sketch the graph

y

-8

-2 4

(1, -9)

x

TOPIC 3


7


b)

(Let’s try this question in class)

c) ( )


TOPIC 3


8


d)


e) ( )


TOPIC 3


9


Example 3.4

a) Find the stationary point on the curve y = 2x3 - 5 and determine whether it is a

minimum, maximum or point of inflexion.

Solution:

y = 2x3 - 5

Therefore, dx

dy = 6x

2.

For a stationary point, dx

dy=0

Therefore, 6x2

= 0

x = 0.

When x = 0 , y = 2 ( 0 ) 3

- 5 = -5.

Therefore, the stationary point is ( 0 , -5 ).

Differentiating dx

dy with respect to x,

2

2

dx

yd = 12x

When x = 0 , 2

2

dx

yd = 0

Therefore, the point ( 0 , -5 ) is a point of inflexion.

b) Find the point of inflection on the curve of y = f(x) = 2x3 − 6x2 + 6x – 5.


TOPIC 3


10


3.2 Rate of change

A lot of physical quantities that exist in our daily lives change with respect to time. For

example, an electrical engineer may need to know the change of current in an electrical

circuit with respect to time. A building contractor may need to know the change of

expansion of concrete mix with respect to time in the building of a bridge. We call these

changes with respect to time, the rate of change. For the electrical engineer, we say he is

interested in the rate of change of current. For the building contractor, he is interested in

the rate of change of expansion of concrete mix.

In differentiation, we write dt

dI or )(tI to the rate of change of current, where I is the

current and t the time. Similarly, dt

dE or )(' tE represents the rate of change of expansion

of concrete mix, where E is the expansion of concrete mix.

Example 3.5

a) If the radius of a circle increases at a rate of 5

1 cms-1, find the rate of change of area

of the circle when its radius is 10 cm.

Solution:

Let r = radius,

A = area of circle

Therefore, A = r2

dr

dA= 2 r

r

TOPIC 3


11


Given that dt

dr=

5

1 cms-1

The rate of change of area, dt

dA =

dr

dA

dt

dr

= 2r 5

1

= 5

2 r

When r = 10, dt

dA =

5

)10(2

dt

dA= 4 cm2 s-1

b) If f ( t ) = t 3 – 4t + 8, what is the rate of change of f(t) when t = 3 and t = 5 ?

Solution:

Given that f(t) = t 3 – 4t + 8. The rate of change of f(t) is f (t). Then,

f (t) = 3t2 – 4

When t = 3 ,

f (t) = 3 (3)2 – 4 = 23.

When t = 5,

f (5) = 3 ( 5 )2 – = 71.

c) Air is being pumped into a spherical balloon such that its radius increases at a rate of

0 .75 cm/s. Find the rate of change of its volume when the radius is 5 cm.

Solution :

The volume ( V) of a sphere with radius r is

Differentiating with respect to t, you find that

TOPIC 3


12


The rate of change of the radius dr/dt =0 .75 cm/s because the radius is increasing with

respect to time.

At r = 5 cm, you find that

scm

scmcmdt

dV

/75

)/75.0()5(4

3

2

hence, the volume is increasing at a rate of 75π cm3/s .

d) A metallic cube expands in such a way that all its sides change at a rate of 2 cms-1.

Find the rate of change of its surface area when its volume is 125cm3.

[120 cm2 s

-1]


e) The radius of a circle increases at a rate of 0.5 ms-1. Find the rate of change of its

area when its circumference is 12 m long.[ 6 m2

s-1

]


TOPIC 3


13


f) Sand is poured on a flat floor at a rate of 4 cm3 per second. The accumulating sand

forms a circular cone with its height always at 4

3 of its radius. Find the rate of

change of radius when its radius is 4 cm. [3

1cm s

-1]


TOPIC 3


14


3.3 Solve Optimization Problems

Many application problems in calculus involve functions for which you want to find

maximum or minimum values. The restrictions stated or implied for such functions will

determine the domain from which you must work. The function, together with its domain,

will suggest which technique is appropriate to use in determining a maximum or minimum

value.

GUIDELINES FOR SOLVING MAX./MIN. PROBLEMS

Steps in solving problems on maxima and minima :

1) Express the quantity that has to be maximized or minimized in terms of only one

variable based on the given information.

Let V (volune) be the quantity that has to be maximized or minimized. Thus, express V in terms of only one variable, let it be x.

2) Find

and solve the equation

to determine the value of x.

3) Find

. Substitute the value of x that is found into

. If the value negative, value

V is maximum. If the value positive, value V is minimum.

4) Calculate the maximum or minimum value of V by substituting the value of x into V.

Example 3.6

a) A rectangle has a perimeter of 100m. Find the length and the width of the rectangle

so that ractangle has a maximum area.[x=25, y=25]


TOPIC 3


15


b) The variables x and y are related by the equation Find the maximum

value of L for which .


c) A box has the shape of a cuboid. Its height is y m. It has a square base of side x m. If the volume of the box is 64 , find the value of x and y such that the total surface area of the box is a minimum.


TOPIC 3


16


d) An open rectangular box with square base is to be made from 48 cm2 of material.

What dimensions will result in a box with the largest possible volume ?

Solution :

Let variable x be the length of one edge of the square base and variable y the height of the

box.

The total surface area of the box is given to be

48 = (area of base) + 4 (area of one side) = x2 + 4 (xy) ,

so that

4xy = 48 - x2

or

.

We wish to MAXIMIZE the total VOLUME of the box

TOPIC 3


17


V = (length) (width) (height) = (x) (x) (y) = x2 y .

However, before we differentiate the right-hand side, we will write it as a function of x only.

Substitute for y getting

V = x2 y

= 12x - (1/4)x3 .

Now differentiate this equation, getting

V' = 12 - (1/4)3x2

= 12 - (3/4)x2

= (3/4)(16 - x2 )

= (3/4)(4 - x)(4 + x)

= 0

for

x=4 or x=-4 .

But since variable x measures a distance and x > 0 .

If

x=4 cm. and y=2 cm

then V = 32 cm3 is the largest possible volume of the box.

e) A container in the shape of a right circular cylinder with no top has surface area 3

cm2 What height h and base radius r will maximize the volume of the cylinder ?

Solution :

Let variable r be the radius of the circular base and variable h the height of the cylinder.

TOPIC 3


18


The total surface area of the cylinder is given to be

(area of base) + (area of the curved side)

,

so that

or

.

We wish to MAXIMIZE the total VOLUME of the cylinder

V = (area of base) (height) .

However, before we differentiate the right-hand side, we will write it as a function of r only.

Substitute for h getting

.

Now differentiate this equation, getting

TOPIC 3


19


= 0

for

r=1 or r=-1 .

But since variable r measures a distance and r > 0 .

If

r=1 cm and h=1cm

then cm3 is the largest possible volume of the cylinder.

3.4 DISPLACEMENT, VELOCITY AND ACCELERATION

Let’s consider an object that moves along a straight line from a fixed point O.

After t seconds, let s represents its displacement from O. As s is dependent on t, we say

that s is a function of t. Hence,

Similarly, if v represents its velocity and a represents its acceleration,

Acceleration, dt

dva .

Therefore,

Displacement, s = f (t)

Velocity, v = dt

ds ,

dt

dsv and a =

dt

dv .

a = dt

dv =

dt

d

dt

ds =

dt

sd 2

.

TOPIC 3


20


Example 3.7

1) An object P moves along a straight line such that its displacement, s meter, from the

fixed point O after t seconds is given as s = 8t2 – t3. Find

a. the distance traveled by P in the third second.

b. the velocity of P after 1 second.

c. the acceleration of P after 2 seconds.

Solution:

a. s = 8t2 – t3

at t = 3 , S3 = 8 ( 3 )2 - ( 3 )3 = 8 (9) - 27 = 45 m

at t = 2 , S2 = 8 ( 2 )2 - ( 2 )3 = 8 (4) - 8 = 24 m

Therefore, the distance traveled by P in the third second = S3 – S2

= 45 – 24

= 21 m .

b. The velocity, v = dt

ds

= 16 t – 3t2 .

When t = 1,

v = 16 (1) – 3 (1)2 = 16 –3 = 13ms-1

c. The acceleration, a = dt

dv = 16 –6t .

When t = 2,

a = 16 – 6(2) = 16 – 12 = 4ms-2

2) A police patrol car is chasing a robber along the straight highway in Kuantan. The

displacement s meter, t seconds after passing through Kuantan is given as s = t3 - 6t

+ 5t. Find

a. the velocity of the car when it returns to Kuantan.

b. the velocity of the car when its acceleration is zero.

Solution:

TOPIC 3


21


a. When the car is back in Kuantan, s = 0. Hence,

t3 – 6t2 + 5 t = 0

t ( t2 – 6t + 5 ) = 0

t (t – 1 ) ( t – 5 ) = 0

Therefore, t = 0 , 1 or 5 s.

The velocity, v = dt

ds= 3 t2 – 12t + 5

When t = 0, v = 3 (0)2 – 12 (0) + 5 = 5 ms-1

When t = 1, v = 3 (1)2 – 12 (1) + 5 = -4 ms-1

When t = 5, v = 3 (5)2 – 12 (5) + 5 = 20 ms-1

When the car first passes through Kuantan, v = 5 ms-1 .

The velocity of the car when it returns to Kuantan -4 ms-1 and 20 ms-1.

b. The acceleration, a =dt

dv = 6 t – 12.

When a = 0 ,

6 t – 12 = 0

6t =12

t =6

12 = 2s

Hence, the velocity then is

v = 3 (2)2 – 12(2) + 5

= 12 – 24 + 5

= -7 ms-1.

3) A particle moves in a straight line from fixed point O such that its displacement,

s meter, t seconds after passing O is given by

a. Find the initial velocity.

b. Find the acceleration when the velocity is 6 m/s.

c. What is the displacement and velocity of the particle at 2 seconds.

a. The initial velocity occurs when

TOPIC 3


22


( ) + ( )

= 30 m/s

b.

⁄

24

( )( )

When, ( )

When, ( )

c. When

( ) ( ) ( )

( ) ( )

4) A particle moves along a straight line. Its displacement, meter from point

second after passing through point is given by . Calculate :

a. the distance of the particle at the fourth seconds,

b. the displacement of the particle when its stop at moment,

c. the acceleration of the particle at . Solution : a.

( )

TOPIC 3


23


( )

b.

( )

and

when t = 0s , ( ) ( )

when t = 6s , ( ) ( )

c.

When , ( )

5) A particle moves on a straight line according to in meter and time, t

in seconds after passing through the fixed point O.

a. Find the total distance of the particle in the first 4 seconds.

b. Find the distance at the time t=2s.

c. Find the displacement travelled by the particle in the third seconds.

d. Find the velocity of the particle when t=3s.

e. Find the acceleration when t=2s.


TOPIC 3


24


TOPIC 3


25


Activity 3(a)

1. Find the stationary points on the following curves and determine whether these

points are minimum, maximum or point of inflexion.

a. y = 2x - x2

b. y = 4x3 - 3x4

c. y = 2x ( x - 1 )2

d. x

xy

2)2(

e. y = x3 - 3x2 + 2

f. y = x3 - 9x2 + 24x - 6

2. Find the minimum and maximum point of these functions and sketch their graphs.

a. y = x2 – x – 6

b. y = -x2 + 3

TOPIC 3


26


)4

25,

2

1(

-x

y

(0 , 3)

x

y

Feedback for Activity 3(a)

1. a. ( 1, 1 ) maximum point.

b. ( 1 , 1 ) maximum point, ( 0 , 0 ) point of inflexion.

c. ( 1 , 0 ) minimum, (1/3 , - 4/9 ) maximum.

d. ( 2 , 0 ) minimum, ( -2 , -8 ) maximum.

e. ( 0 , 2 ) maximum, ( 2 , -2 ) minimum.

f. ( 2 , 14 ) maximum, ( 4 , 10 ) minimum.

2. a. minimum point at )4

25,

2

1( b. maximum point at ( 0 , 3 )

.

TOPIC 3


27


Activity 3(b)

1. An object moving along a straight line travels s meter, t seconds after passing

point O. Find its velocity and acceleration after 2 seconds.

a. s = t3

b. s = 3t2 + t + 1

c. s = 3t4 – 4t3

2. An object moving along a straight line travels s meter, t seconds after passing

point O is given as s = t 3 –3 t2. find its velocity when its acceleration is

.

3. A car starts from a stationary position and moves s meter along a straight line

after t seconds. Given that s = t2 ( t + 2 ), find

a. its acceleration, a, after 4 seconds.

b. the time, t , when its velocity is 39 ms-1.

c. the distance traveled during the fourth second.

4. A car is traveling along a straight highway in Ipoh. The displacement s meter,

t seconds after passing through Ipoh is given as s = 27 t – t3. Find

a. the value of s when the car stops momentarily.

b. its velocity, v, when the car is passing through Ipoh again.

TOPIC 3


28


Feedback for Activity 3(b)

1. a. 12ms-1, 12 ms-2

b. 13ms-1, 6 ms-2

c. 48 ms-1

2. 24ms-1

3. a. 28 ms-2

b. 3 s

c. 51 m.

4. a. 54

b. -54ms-1

TOPIC 3


29


Activity 3(c)

1) The demand equation of a good is , where Q is the level of output. Given that the total revenue is T=PQ. Find the level of output that maximizes total revenue. Calculate the maximum total revenue.

2) The demand equation of a good is √( ) , where Q is the level of output.

Given that the total revenue is T=PQ. Find the level of output that maximizes total revenue. Calculate the maximum total revenue.

3) A firm’s output function is given by , where L denotes the number

of workers. Find the size of the workforce that maximizes output. Calculte the maximum output.

4) A firm’s output function is given by , where L denotes the number of

workers. Find the size of the workforce that maximizes output. Calculte the maximum output.

5) The sum of two positive numbers is 50. Find the numbers if the sum of their squares

is a minimum.

6) An open water tank with square base has a volume of . Given that the square base has sides of length x metres. Show that the surface area of the tank, A, is given

by

. Find the minimum surface area of the tank and the corresponding

value of x.

7) A closed container is in the shape of a cuboid with a base length twice its width, is to

be made with 48m2 of thin metal sheet. Show that the volume of the cuboid is given

by

where x is the width of the base. Hence, find the maximum

volume of the cuboid.

TOPIC 3


30


Feedback for Activity 3(c)

1. Q = 10 ; T = 200

2. Q = 166.7 ; T = 3042.9

3. L = 20 ; Q = 800

4. L = 12.25 ; Q = 22500

5. 25 ; 25

6. 108m2 ; 6

7.

TOPIC 3


31


SELF ASSESSMENT 3 (a)

1. For each of the following functions, find the stationary point and state its type

a. f(x) = x2 - 4x + 2

b. b. f(x) = x3 – 12x

c. f(x) = x

x )16( 3

d. d. f(x) = ( x – 1)(x + 2 )2

2. A curve is given as y = 2

33

x

x. Show that this curve has only one stationary point.

Subsequently, state whether this point is the minimum or maximum point.

3. The curve y = x3 + ax

2 + bx + c passes the point (1 , 1) and its stationary points are at x

= -1 and x = 3 . Find the values of a, b and c.

4. Find the minimum and maximum point of these functions and sketch their graphs

a. y = 2x2 – 5x – 3

b. y = 4x2 + 3x – 1

TOPIC 3


32


SOLUTION : SELF ASSESSMENT 3(a)

1. a. (2 , -2) minimum .

b. (2 , - 16) minimum, (-2 , 16) maximum.

c. (2 , 12) minimum .

d. (0 , -4) minimum, (-2 , 0) maximum.

2. minimum

3. a = -3, b = -9 and c = 12.

4. minimum point at ( 8

49,

4

5 ) b. minimum point at )

16

25,

8

3(

-x

y

-x

y

TOPIC 3


33


SELF ASSESSMENT 3 (b)

1. Find the maximum and/or minimum point for the following functions

a. y = 2x2

-5x + 4

b. y = 4 -3x2

2. Find the minimum velocity if 33

32

t

tv

.

3. Find the minimum value of x2 + 2y

2 if x and y are related by the equation

x +2y = 1.

4. Given that y = 2x 3 + 2x

2 - 7x - 8. Determine the gradient of the tangent at the

point (2, 2). Subsequently, find the coordinates of another point that has a

tangent with the same gradient.

5. An object is moving along a straight line such that its displacement, s meter

from a fixed point O, after t seconds, is given as s = t2 - 5t + 6.

a. How far is its displacement from O initially?

b. When will the object be at rest momentarily?

c. Find the time when the object is back to its starting position.

6. Tenaga National estimates that the demand for electricity t months from

January is 10205.0 2 ttx , where x is in millions of units per month. The

profit U (in million RM) is estimated as 25.01.0001.0 2 xxU .

Find the rate of change of profit 6 months from January.

8. A cuboid with a square base is to be made with of thin metal sheet.

Show that the volumes, V is given by

, where x is the length of

the base. Find the maximum value of the cuboid.

9.

X cm

r cm

TOPIC 3


34


The above diagram consists of rectangle of length x cm and two semicircles,

each radius r cm. The perimeter of the diagram is 400cm. Given that the area

of the diagram is A , show that ( ). Hence, find

a) the value of r such that A is maximum.

b) the maximum value of A.

SOLUTIONS: SELF ASSESSMENT 3(b)

1. a. minimum point is )8

7,

4

5(

b. maximum point is (0, 4)

2. minimum value is 2.25.

3. 0.333

4. 25, ( 2.67 ,-13.04 )

5. a. 6 meter

b. t =2

5

c. at t = 2 and t = 3 seconds.

6. 0.596.

7.

8. a )

)

Documents

BA201 Engineering Mathematic Unit 03 - Appofdiff