CHAPTER 4
APPLICATION OF
DIFFERENTIATION AND
INTEGRATION
APPLICATION OF
DIFFERENTIATION
DISPLACEMENT, VELOCIT Y AND
ACCELERATION
 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.
ds
v  s' 
dt
 Acceleration: The rate of change of velocity (or speed) with
respect to time.
d 2s
a  v '  s ''  2
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
2
3
the equation, s  12t  15t  4t . Find velocity and acceleration of
particle after 3 seconds.
Solution:
s  12t  15t 2  4t 3
displacement
ds
v

 12  30t  12t 2
velocity
dt
t  3, v  12  30(3)  12(9)  30m / sec
ds
 12  30t  12t 2
dt
d 2s
a  2  30t  24t
acceleration
dt
t  3, a  30  24(3)  42m / sec 2
EXERCISE
1. Find the velocity of an object at any time t, given that its
acceleration is a (t )  18t  2 with t = 2, given the initial
velocity is 0. Ans : 40m/s
2. The height of a pebble dropped of f a building at time t = 0
is h(t )  44.1  4.9t 3 metres 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,
a  44.1m / s 2
1
3. Suppose that s  t 2  t  2 denotes the position of a bus at
4
time t. Find the velocity
and acceleration.
Ans : v  1 t  1, a  1
2
2
RELATED RATES
 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 dif ferentiation 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.
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?
Purpose : Find
z
y
dx
dt
x  base  6
y  height  ?
dy
 10
dx
Use phytagorean Theorem: x 2  y 2  z 2
Value of the height:
62  y 2  102
y 2  100  36
x
y 2  64
y 8
Implicitly differentiation:
dx
dy
2x  2 y
0
dt
dx
dx
2(6)  2(8)(10)  0
dt
dx
12  (160)  0
dt
dx 160 1

ms
dt 12
 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 ?
4
Volume of Spherical   r 3
3
Volume of Surface Area  4 r 2
4
V   r3
sA  4 r 2
3
dV
4
dr
dsA dsA dr
dr
  3  r 2

   2  4 r  8 r
dt
3
dt
dt
dr dt
dt
dr
 4 r 2
dt
dr
dV
Finding
, where
 2
shrinking
dt
dt
2 dr
4 12 
 2
dt
dr
2
1


dt 576
288
dr
dsA
Now, already know value of
, so
?
dt
dt
dsA
1 
1

 8 (12)  


ft / min

dt
3
 288 
 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 2m3 / min , 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
tank
Unknown : Rate at which the height of water changing.
1
V   r 2h
3
Cone ratio:
h 4

r 2
dV
2
dt
h3
dh
?
dt
1
V   r 2h
3
Eliminate r as we dont know r, or dr/dt.
2
h
1 h
r  ,V     h
2
3 2
1
  h3
12
dV dV dh


dt
dh dt
1
dh
= (3)  h 2 
12
dt
1
dh
=  h2
4
dt
dh
4 dV
=
dt
 h 2 dt
dh
4
8
=
(2)

m / min
2
dt
 (3 )
9
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?
dV
dh
dr
 ?,  0.8,  0.5, r  3, h  2
dt
dt
dt
V   r 2h
dV  dV dh   dV dr   2 dh  
dr 

 
    r

2

rh
 

dt  dh dt   dr dt  
dt  
dt 
=  (32 )(0.8)  2  3 2  0.5   1.2
EXERCISE
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 486 cubic centimetres, find the
rate at which the level of water is rising. Ans :0.1572cm/min
1 3 

V

r h

3


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 : 192 inches 3 / min
3
3. Gas escaping from a spherical balloon at the rate of 4 ft / min
How fast is the surface area changing when the radius is 14ft?
Ans : 0.5714 ft 2 / min
SKETCHING THE GRAPH
 Example :Sketch the curve and find the points on the graph
function y  x( x  1)(1  x)
dy
1. Find the stationary points, dx  0
2. Do first derivative test
Interval
x
d2y
0
3. Inflexion point,
2
dx
Sign
Conlcusion
4. Second derivative test :
Interval
x
Sign
Conclusion
5. Finding maximum and minimum points
*Equation using the inflexion, points of x take from stationary
points.
EXERCISE
1. Find the stationary point on the graph of the function below:
Hence sketch the graph.
i. y  3  x 2
ii.y  x 4  4 x 3
iii.y  4  x 2
AREA OF REGION
 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.
b
Area =
 f ( x)dx
a
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.
Example :
2
Calculate the area bounded by f ( x)  4 x  x and
the x-axis.
Solutions :
2. Determine a and
b
2
f ( x)  4 x  x
y0
4x  x2  0
x(4  x)  0
x  0, x  4
Therefore the boundaries are a = 0 and b = 4.
3. Set up integral
b
4
a
0
A   f ( x)dx   (4 x  x 2 )dx
4
 4x
x 
2
(4
x

x
)
dx


 2
0
3  0

1
1

 

=  2(4) 2  (4)3    2(0) 2  (0)3 
3
3

 

4
2
3
 32 
=  0
3
32
=
units 2
3
2. Area under a curve – given function, region bounded by the
horizontal lines and the y – axis.
d
Area =  g ( y )dy
c
Example :
Find the area of region enclosed by
Solutions :
Transform f (x) to g (y)
y  x 1
y2  x 1
x  y2 1
y  x 1
Finding a and b
x  0,
y2 1  0
y  1
y  1, y  1
Integral
1
y

  y 2  1dy     y 
3
 1
1
1
3
13   (1)3

= -   1  
 (1) 
3   3

4
= unit 2
3
3. Area between two curves
Universal formula for finding the area between two curves:
 Using the vertical elements:
b
Area =
y
2
 y1 dx
a
 Using the horizontal elements:
d
Area =
x
2
c
 x1 dy
 Case 1
b
b
a
a
A   f ( x)dx   g ( x)dx
 upper 
or  
dx 

function 
a
 lower 
a  function  dx
 right

or  
dy 

function 
c
 left

c  function  dy
b
b
 Case 2
d
d
c
c
A   f ( y )dy   g ( y )dy
d
d
Example :
Find the area of region enclosed by the
following curves :
y  x 2 , y  x  6, x  0, x  5
Solutions :
1. Sketch the graph
2. Intersection points (Finding a and b)
x2  x  6
x2  x  6  0
( x  3)( x  2)  0
x  3, x  2
3
5
5
3



2
2
  x  6dx   x dx     x dx   x  6dx 
0
3
0
 3

3
5
  x  6  x dx   x 2  x  6dx
2
0
3
3
5
1  1
1
1

=  x 2  6 x  x3    x3  x 2  6 x 
3 0  3
2
2
3
1
1
1

1

=  (3) 2  6(3)  (3)3   0   (5)3  (5) 2  6(5)  
3
2
2

3

1 3 1 2

(3)

(3)

6(3)
 3

2
157
=
unit 2
6
EXERCISES
1. Find the area of region bounded by the given curves,
y  3 x 2 , x  0 and x  3
Ans : 27unit 2
2. Calculate the area of segment from curve y  x(3  x) by line y  x
4
2
Ans : unit
3
3. Find the area bounded by following curves :
1
y  sin x by the line y 
Ans : 4   unit 2
2
2 2
VOLUME
 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,  r 2 where r is the radius
2
2
- Washer (volume,  R   r where R is the outer radius and r is
the inner radius.
 Volume Solid Generated by Region Bounded by the Curve and
the Axis
b
x  axis
V    { f ( x)}2 dx
a
d
y  axis V    {g ( y )}2 dy
c
Example :
Find the volume of solid formed by the area bounded by the
curves y  x 2  1 from x = 0 to x = 2 about the x – axis.
2
2
2
Volume


x

1
dx


x



Solution :

  1 dx
2
0
2
2
0
2
=    x 4  2 x 2  1 dx
0
2
 x 2x

= 
 x
3
 5
0
5
3
 25 2(2)3
  05 2(0)3

= 
 2   
 0
3
3
 5
  5

206
=
 unit 3
15
Example :
Find the volume of solid formed by the area bounded by curves
y 2  8 x, the straight line x  2
Solution :
2
1 2
V   y dy
80
2
1 y 
 
8  3 0
1 8
=  0
8 3
1
=  unit 3
3
3
=
 Volume of Solid Generated by Region Bounded by the Two
Curve
 Revolving
R aboutb the x – axis
b
2
2
V     f ( x) dx     g ( x) dx
a
a
b
2
2
    f ( x)   g ( x)  dx


a
=  outer radius    inner radius 
 Revolving
R aboutb the y - axis
b
2
2
V     f ( y ) dx     g ( y ) dy
2
2
a
a
b
2
2
    f ( y )   g ( y )  dy


a
=  outer radius    inner radius 
2
2
Example :
Calculate the volume of solid formed when the area bounded by
the curves y  x and y  x
Solution :
1. Sketch the graph.
2. Intersection point
y xyx
xx
 
2
x
  x
x  x2
x  x2  0
x(1  x)  0
x  0, x  1
2
3. Integrate
b
b
b
 Adr     f ( x) dx    g ( x) dx
2
a
a
2
a
2
1
1
 
2
2

x
dx


x
dx


x

x



 dx


0  
0
0
1
1
2
1
 x 2 x3 
   
 2 3 0
 (1) 2 (1)3 
 

0

3 
 2
1
  unit 3
6
EXERCISE
1. Calculate the volume of solid formed when the area
bounded by the curves y  2  x 2 and y  x 2 which is revolving
at x-axis.
Ans : 64
 unit 3
15
2. Calculate the volume of solid formed when the area bounded
2
by the curves x  3 y , x  y which is revolving at y – axis.
Ans : 45  unit 3
2