APPLICATION OF DIFFERENTIATION AND

INTEGRATION

CHAPTER 4

APPLICATION OF DIFFERENTIATION

Derivatives are used to analyze the motion of an object in a straight line.

If t of x is the displacement of moving object from a fixed plane and t is time, then

Velocity: A rate of change of s with respect to t.

Acceleration: The rate of change of velocity (or speed) with respect to time.

DISPLACEMENT, VELOCITY AND ACCELERATION

'ds

v sdt

2

2' ''

d sa v s

dt

Example 1:A particle is moving in straight line and its distance, s, in metres from a fixed point in the line after t seconds is given by the equation, . Find velocity and acceleration of particle after 3 seconds.Solution:

2 312 15 4s t t t

2 3

2

2

2

2

2

12 15 4

12 30 12

3, 12 30(3) 12(9) 30 / sec

12 30 12

30 24

3, 30 24(3) 42 / sec

s t t t

dsv t t

dtt v m

dst t

dt

d sa t t

dt

t a m

velocity

acceleration

displacement

1. Find the velocity of an object at any time t, given that its acceleration is with t = 2, given the initial velocity is 0. Ans : 40m/s

2. The height of a pebble dropped off a building at time t = 0 is at time t. The pebble strikes the ground at t = 3 seconds. What is the acceleration and velocity when its strikes the ground. Ans : v = -132.3m/s,

3. Suppose that denotes the position of a bus at time t. Find the velocity and acceleration.

Ans :

EXERCISE

( ) 18 2a t t

3( ) 44.1 4.9 metresh t t

244.1 /a m s

212

4s t t

1 11,

2 2v t a

Related rates means related rates of change, and since rates of change are derivatives, related rates really means related derivatives.

Process of finding a rate which a quantity changes by relating that quantity to other quantities.

Rate changes usually respect to time, t.Procedure for Solving Related Rates Problems:1. Draw a figure (if appropriate) and assign variables.2. Find a formula relating the variables.3. Use implicit diff erentiation to find how the rates are

related.4. Substitute any given numerical information into the

equation in step 3 to find the desired rate of change.

RELATED RATES

Example (The Falling Ladder) :A ladder is sliding down along a vertical wall. If the ladder is 10 metres long and the top is slipping at the constant rate of 10m/s, how fast is the bottom of the ladder moving along the ground when the bottom is 6 metres from the wall?

2 2 2

2 2 2

2

2

Purpose : Find

6

?

10

Use phytagorean Theorem: x

Value of the height:

6 10

100 36

64

8

Implicitly differentiation:

2 2 0

2(6) 2(8)( 10) 0

12 ( 160)

dx

dtx base

y height

dy

dx

y z

y

y

y

y

dx dyx ydt dxdx

dtdx

dt

1

0

160

12

dxms

dt

x

yz

Example (Shrinking Balloon):Gas is escaping from a spherical balloon at the rate of 2 cubes feet per minute. How fast is the surface area shrinking when the radius of the balloon is 12 feet?

3 24Volume of Spherical Volume of Surface Area 4

3r r

3 2

2

2

2

4 4

34

3 2 4 83

4

Finding ,where 2

4 12 2

2 1

576 288

Now, already know va

V r sA r

dV dr dsA dsA dr drr r r

dt dt dt dr dt dtdr

rdt

dr dV

dt dtdr

dtdr

dt

lue of , so ?

1 18 (12) / min

288 3

dr dsA

dt dtdsA

ftdt

shrinking

Example (Circular Cone) :A water tank has the shape of an inverted circular cone with base radius 2m and height 4m. If water is being pumped into the tank at the rate of , find the rate at which the water is rising when the water is 3m deep.Given : rate at which volume of water is being increased in the tankUnknown : Rate at which the height of water changing.

32 / minm

2

2

3

?

1

3Eliminate r as we dont know r, or dr/dt.

dV

dth

dh

dt

V r h

2

2

3

2

2

2

2

1

3Cone ratio:

4

2

1,

2 3 2

1

12

1 = (3)

121

= 4

4 =

4 8 = (2) / min

(3 ) 9

V r h

h

r

h hr V h

h

dV dV dh

dt dh dtdh

hdt

dhhdt

dh dV

dt h dtdh

mdt

Example (Cylinder):Suppose a right circular cylinder’s radius is expanding at the rate of 0.5cm/sec while its height is shrinking at the rate of 0.8cm/sec. How fast is the cylinder’s volume changing when its radius is 3cm and its height is 2 cm?

2

2

2

?, 0.8, 0.5, 3, 2

2

= (3 )( 0.8) 2 3 2 0.5 1.2

dV dh drr h

dt dt dt

V r h

dV dV dh dV dr dh drr rh

dt dh dt dr dt dt dt

1. A basin has the shape of inverted cone with altitude 100 cm and radius at the top of 50 cm. Water is poured into the basin at the constant rate of 40 cubic cm/minute. At the instant when the volume of water in the basin is cubic centimetres, find the rate at which the level of water is rising. Ans :0.1572cm/min

EXERCISE

486

31

3V r h

2. The radius of a right circular cylinder is increasing at a rate of 2 in/min and the height is decreasing at a rate of 3 inch/min. At what rate is the volume changing when the radius is 8 inches and the height is 12 inches? Is the volume increase or decrease? Ans :

3. Gas escaping from a spherical balloon at the rate ofHow fast is the surface area changing when the radius is 14ft?Ans :

3192 / mininches

34 / minft

20.5714 / minft

Example :Sketch the curve and find the points on the graph function

1. Find the stationary points,2. Do fi rst derivative test

3. Inflexion point,

SKETCHING THE GRAPH

( 1)(1 )y x x x

0dy

dx

Interval x Sign Conlcusion

2

20

d y

dx

4. Second derivative test :

5. Finding maximum and minimum points*Equation using the inflexion, points of x take from stationary points.

Interval x Sign Conclusion

1. Find the stationary point on the graph of the function below: Hence sketch the graph.

EXERCISE

2

4 3

2

i. 3

ii. 4

iii. 4

y x

y x x

y x

Area of a Region Bounded by the Curve and the Axis

1. Area under a curve – region bounded by the given function, vertical lines and the x- axis.

AREA OF REGION

Area = ( )b

a

f x dx

Steps in calculating the area under the curve f(x) :

1. Sketch the area.2. Determine the boundaries a and b.

3. Set up the definite integral.4. Integrate.

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Example :

Calculate the area bounded by and the x-axis.

Solutions :

2( ) 4f x x x

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2. Determine a and b

Therefore the boundaries are a = 0 and b = 4.3. Set up integral

2

2

( ) 4

0

4 0

(4 ) 0

0, 4

f x x x

y

x x

x x

x x

42

0

44 2 32

0 0

2 3 2 3

2

( ) (4 )

4(4 )

2 3

1 1 = 2(4) (4) 2(0) (0)

3 3

32 = 0

3

32 =

3

b

a

A f x dx x x dx

x xx x dx

units

2. Area under a curve – given function, region bounded by the horizontal lines and the y – axis.

Example :Find the area of region enclosed bySolutions :Transform f (x) to g (y)

Area = ( )d

c

g y dy

1y x

2

2

1

1

1

y x

y x

x y

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Finding a and b

Integral

11 32

1 1

3 3

2

13

1 ( 1) = - 1 ( 1)

3 3

4 =

3

yy dy y

unit

2

0,

1 0

1

1, 1

x

y

y

y y

3. Area between two curvesUniversal formula for finding the area between two curves:Using the vertical elements:

Using the horizontal elements:

2 1Area = b

a

y y dx

2 1Area = d

c

x x dy

Case 1

Case 2

( ) ( ) or b b b b

a a a a

upper lowerA f x dx g x dx dx dx

function function

( ) ( ) or d d d d

c c c c

right leftA f y dy g y dy dy dy

function function

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Example :

Find the area of region enclosed by the following curves :

Solutions :

1. Sketch the graph

2. Intersection points (Finding a and b)

2 , 6, 0, 5y x y x x x

2

2

6

6 0

( 3)( 2) 0

3, 2

x x

x x

x x

x x

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3 3 5 52 2

0 0 3 3

3 52 2

0 3

3 52 3 3 2

0 3

6 6

6 6

1 1 1 1= 6 6

2 3 3 2

x dx x dx x dx x dx

x x dx x x dx

x x x x x x

2 3 3 2

3 2

2

1 1 1 1= (3) 6(3) (3) 0 (5) (5) 6(5)

2 3 3 2

1 1 (3) (3) 6(3)

3 2

157=

6unit

1. Find the area of region bounded by the given curves,

Ans :

2. Calculate the area of segment from curve

Ans :

EXERCISES

23 , 0 and 3y x x x

(3 ) by line y x x y x

227unit

24

3unit

3. Find the area bounded by following curves :Ans :

1sin by the line

2y x y 24

2 2unit

It is important whether you can visualize the whole solid. You just need to know what happen to a typical slice when you rotate it.

When rotate the typical slice, then will get two shapes :

- disk (volume, where r is the radius- Washer (volume, where R is the outer radius

and r is the inner radius.

Volume Solid Generated by Region Bounded by the Curve and the Axis

VOLUME

2

2

{ ( )}

{ ( )}

b

a

d

c

x axis V f x dx

y axis V g y dy

2r2 2R r

Example :Find the volume of solid formed by the area bounded by the curves from x = 0 to x = 2 about the x – axis.

Solution :

2 1y x

2 22 22 2

0 0

24 2

0

25 3

0

5

Volume 1 1

= 2 1

2 =

5 3

2 =

5

x dx x dx

x x dx

x xx

3 5 3

3

2(2) 0 2(0)2 0

3 5 3

206 =

15unit

Example :Find the volume of solid formed by the area bounded by curves

Solution :

2 8 , the straight line 2y x x

22

0

23

0

3

1

8

1 =

8 3

1 8 = 0

8 31

= 3

V y dy

y

unit

Volume of Solid Generated by Region Bounded by the Two Curve

Revolving R about the x – axis

Revolving R about the y - axis

2 2

2 2

2 2

( ) ( )

( ) ( )

= outer radius inner radius

b b

a a

b

a

V f x dx g x dx

f x g x dx

2 2

2 2

2 2

( ) ( )

( ) ( )

= outer radius inner radius

b b

a a

b

a

V f y dx g y dy

f y g y dy

Example :Calculate the volume of solid formed when the area bounded by the curvesSolution :1. Sketch the graph.2. Intersection point

and y x y x

2 2

2

2 0

(1 ) 0

0, 1

y x y x

x x

x x

x x

x x

x x

x x

3. Integrate

2 2

21 1 112 22

0 0 0

12 3

0

2 3

3

( ) ( )

2 3

(1) (1)0

2 3

1

6

b b b

a a a

Adr f x dx g x dx

x dx x dx x x dx

x x

unit

1. Calculate the volume of solid formed when the area bounded by the curves which is revolving at x-axis.

Ans :

2. Calculate the volume of solid formed when the area bounded by the curves which is revolving at y – axis. Ans :

EXERCISE

2 22 and y x y x

364

15unit

23 ,x y x y

345

2unit