Denition approach of learning new topics

26 4 DERIVATIVES
seconds will travel 2t metres. So the function relating the time taken t to
travel x distance is given by 5 x = 2t
or x(t) = 2t
So what is the rate of change of displacement with respect to time, between
t = 1 and t = 2.
We know that rate of change of displacement is ∆x
∆t = x 2 − x 1
= 4 − 2
t 2 − t 1 2 − 1 =
2. In physics, rate of change of diplacement is termed as velocity.
Hence the average velocity is 2.
What if we want to nd the instateneous velocity at t = 1. For that we
need to get the secant at t = 1 which is a tangent. And we know that the
tangent is the derivative. So the velocity at t = 1 is dx
dt = d dt (2t) = 2
2. A particle starts moving from a starting point A such that velocity at
every point is given by v = 2t 3 where t is the time taken. Then how is
acceleration related to velocity and time taken.
v = 2t 3
The rate of change of velocity is termed in physics as acceleration i.e.
a = dv . Dierentiating we get,
dt
a = dv
dt = d dt (2t3 ) = 2 d dt (t3 ) = 2 · 3t 2 = 6t 2
4.5 Instantaneous rate of change[3]
Now we know that average rate of change in the function y = f(x) is represented
as ∆y and to get the instantaneous rate we use the limit of average
∆x
∆y
lim
∆x→0 ∆x
5 Since x can be expressed in terms of variable t. Second notation describes that. We have
see if a function f is expresses in terms of x then we write it as y = f(x).