The document discusses differential calculus, particularly focusing on calculating average velocity over specific time intervals and estimating instantaneous velocity at t = 2 seconds. It presents a series of calculations that show that the instantaneous velocity ranges between 19.551 m/s and 19.649 m/s, derived from various approaches to limit time intervals. Additionally, it explains the concept of derivatives and their application in estimating the rate of change of distance with respect to time.

2. DIFFERENTIAL CALCULUS:
One way to approach this problem is to find the average velocity for
various intervals of time ending at t = 2 seconds.
Average velocity between t = t1 and t = t2 equals distance travelled
between t = t1 and t = t2 seconds divided by (t2– t1).
Hence the average velocity in the first two seconds:
t (sec) S (m)
0 0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4

3. DIFFERENTIAL CALCULUS:
Similarly we can compute the average velocities between t = t1 and t = 2
for various t1 as follows:
t = t1 0 1 1.5 1.8 1.9 1.95 1.99
v 9.8 14.7 17.15 18.62 19.11 19.355 19.551
From the above table, it can be observed that the average velocity is
gradually increasing. As we make the time intervals ending at t = 2
smaller, we see that we get a better idea of the velocity at t = 2. Hoping
that nothing really dramatic happens between 1.99 seconds and 2
seconds, we conclude that the average velocity at t = 2 seconds is just
above 19.551m/s.
t (sec) S (m)
0 0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4

4. DIFFERENTIAL CALCULUS:
Second way to approach this problem is to find the average velocity
for various intervals of time starting at t = 2 seconds.
Average velocity between t = 2 seconds and t = t2 equals distance
travelled between t = 2 sec and t = t2seconds divided by (2– t2).
t (sec) S (m)
0 0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4

5. DIFFERENTIAL CALCULUS:
Average velocities between t = 2 and t = t2 for various t2 as follows:
t = t2 4 3 2.5 2.2 2.1 2.05 2.01
v 29.4 24.5 22.05 20.58 20.09 19.854 19.649
Here again we note that if we take smaller time intervals starting at t = 2,
we get better idea of the velocity at t = 2.
 In both set of computations we found the average velocities for
increasing time interval ending at t = 2 sec and for the decreasing
time interval ending at t = 2 sec.
 Hoping that nothing dramatic happen just before t = 2 sec and just
after t = 2 sec both these sequences of average velocities must
approach a common limit.
OBSERVATIONS:
t (sec) S (m)
0 0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4

6. DIFFERENTIAL CALCULUS:
 Velocity of the body at t = 2 sec is between 19.551m/s and 19.649 m/s.
 Technically, we say that the instantaneous velocity at t = 2 sec is
between 19.551 m/s and 19.649 m/s.
COCLUSIONS:
DISCUSSION:
 From the given data of distance covered at various time instants we
have estimated the rate of change of the distance at a given instant of
time.
 We say that the derivative of the distance function s = 4.9t2 at t = 2 is
between 19.551 and 19.649.
t (sec) S (m)
0 0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4

7. DIFFERENTIAL CALCULUS:
0
10
20
30
40
50
60
70
80
90
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
Time, t (sec)
Distance,
S
(m)
An alternate way of viewing this limiting process is by drawing a
plot of distance ‘S’ versus the time ‘t’ elapsed t (sec) S (m)
0 0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4

8. DIFFERENTIAL CALCULUS:
Time (t)
Distance
(S)
t Δt
S
ΔS
Let t be changed by a small amount Δt, and the corresponding change in
S be ΔS.
t (sec) S (m)
0 0
1 4.9
1.5 11.025
1.8 15.876
1.9 17.689
1.95 18.63225
2 19.6
2.05 20.59225
2.1 21.609
2.2 23.716
2.5 30.625
3 44.1
4 78.4

9. DIFFERENTIAL CALCULUS:
When t changes by Δt, S changes by ΔS so that the rate of
change seems to be equal to
ΔS
Δt
. If A be the point (t, S) and B be
the point (t + Δt, S + ΔS), the rate
ΔS
Δt
equals the slope of the
line AB. We have
Time (t)
Distance
(S)
t Δt
S
ΔS
θ
A
B
C
ΔS
Δt
=
BC
AC
=
However, this cannot be the precise definition of the rate.
Because the rate also varies between the points A and B. The
curve is steeper at B than at A.
Thus, to know the rate of change of S at a particular value of t, say at A, we have to take Δt very small.
So, if we go on drawing the point B closer to A and every time calculate
ΔS
Δt
= , we shall see that as Δt is made
smaller and smaller the slope of the line AB approaches the slope of the tangent at A.
The “rate of change” of S with respect to t can be defined as:

10. DIFFERENTIAL CALCULUS:
This slope of the tangent at A thus gives the rate of change of S
with respect to t at A. This rate is denoted by
dS
dt
Time (t)
Distance
(S)
t Δt
S
ΔS
θ
A
B
C
t
S
dt
dS
t 



 0
lim
For small changes Δt we can approximately write
t
dt
dS
S 



11. DIFFERENTIAL CALCULUS:
x
y
x Δx
y
Δy
θ
A
B
C
Now, for any function y(x) if y increases with an
increase in x at a point,
dy
d
is positive there, because
both Δy and Δx are positive. If the function y
decreases with an increase in x, Δy is negative when
Δx is positive. Then
y
and hence
dy
d
is negative.
 If we are given the graph of y versus x, we can find
dy
d
at any point of the curve by drawing the tangent at that
point and finding its slope.
 Even if the graph is not drawn and the algebraic relation between y and x is given in the form of an equation,
we can find
dy
d
algebraically
Finding
dy
d
algebraically:

12. DIFFERENTIAL CALCULUS:
L ΔL
L
ΔL
A
ΔA
2
L
A
, 
uare
Area of sq
2
ΔL)
L
(
ΔA
A 


L
ΔL
2
ΔL)
(
L
ΔA
A 2
2






L
ΔL
2
ΔL)
(
ΔA 2




L
2
ΔL
ΔL
ΔA



?
L
A

d
d
Now if ΔL is made smaller and smaller i.e. ΔL → 0, 2L + ΔL
will approach 2L.
L
2
ΔL
ΔA
lim
L
A
,
0
ΔL



d
d
Thus

13. DIFFERENTIAL CALCULUS:
2
A
, r
rcle
Area of ci 

2
)
Δ
(
ΔA
A r
r 

 
]
Δ
2
)
Δ
(
[
ΔA
A 2
2
r
r
r
r 




 
r
r
r 


 Δ
2
)
Δ
(
ΔA 2


r
r
r

 2
Δ
Δ
ΔA



?
A

dr
d
Now if Δr is made smaller and smaller i.e. Δr → 0, 2πr + πΔr
will approach 2πr.
r
r
dr
d
Thus
r

2
Δ
ΔA
lim
A
,
0
Δ



Δr
r
A
ΔA

14. for some common functions
1
)
( 
 n
n
nx
x
dx
d

15. for some common functions
x
x
dx
d
cos
)
(sin 

16. for some common functions
x
x
dx
d
sin
)
(cos 


17. for some common functions
x
x
dx
d 2
sec
)
(tan 

18. for some common functions
x
x
dx
d 2
cosec
)
(cot 


19. for some common functions
x
x
x
dx
d
tan
sec
)
(sec 


20. for some common functions
x
x
x
dx
d
cot
cosec
)
(cosec 



21. for some common functions
x
x
dx
d 1
)
(ln 

22. for some common functions
x
x
e
e
dx
d

)
(

23. Rules of finding for some composite functions
)
constant
a
is
(where
)
(
c
dx
dy
c
cy
dx
d


24. Rules of finding for some composite functions
dx
dv
dx
du
v
u
dx
d


 )
(

25. Rules of finding for some composite functions
dx
du
v
dx
dv
u
uv
dx
d


)
(

26. Rules of finding for some composite functions
2
v
dx
dv
u
dx
du
v
v
u
dx
d









27. Rules of finding for some composite functions
dx
du
du
dy
dx
dy



28. Maxima & Minima
x1 x2
y
x
The function y = f(x) becomes maximum at x1 and
minimum at x2
At at x1 and x2 the tangent to the curve is parallel
to the X-axis and hence its slope tanq = 0.
0

dx
dy
So, at maxima or minima
0

dx
dy
Condition for Maxima & Minima

29. Maxima & Minima
x1 x2
y
x
ve
dx
dy

 ve
dx
dy


0

dx
dy
The rate of change of is negative at a maximum
maximum
at
dx
y
d
dx
dy
dx
d
0
2
2








0
0
2
2


dx
y
d
dx
dy
Condition for Maximum
Condition for Maxima

30. Maxima & Minima
x1 x2
y
x
ve
dx
dy


ve
dx
dy


0

dx
dy
The rate of change of is positive at a minimum
minimum
at
dx
y
d
dx
dy
dx
d
0
2
2








0
0
2
2


dx
y
d
dx
dy
Condition for Minimum
Condition for Minima