# AVERAGE RATE OF CHANGE AND INSTANTANEOUS RATE OF CHANGE

The average rate of change in an interval [a,b] is

[f(b) - f(a)] / (b-a)

whereas,

the instantaneous rate of change at a point x is

f'(x)

for the given function.

Problem 1 :

A point moves along a straight line in such a way that after t seconds its distance from the origin is

s  =  2t2 + 3t meters.

(i) Find the average velocity of the points between t = 3 and t = 6 seconds.

(ii) Find the instantaneous velocities at t = 3 and t = 6 seconds.

Solution :

(i)  Average rate of change between t = 3 and t = 6

=  [f(6) - f(3)] / (6-3)  ---(1)

f(x)  =  2t2 + 3t meters.

f(6)  =  2(6)2+3(6)  ==>  72+18  ==>  90

f(3)  =  2(3)2+3(3)  ==>  18+9  ==>  27

Applying in (1), we get

=  (90-27)/(6-3)

=  63/3

=  21 meter/.second

(ii)  f(x)  =  2t2 + 3t

f'(x)  =  4t+3

 at t  =  3f'(3)  =  4(3)+3f'(3)  =  12 + 3f'(3)  =  15 m/sec ar t  =  6f'(6)  =  4(6)+3f'(6)  =  24 + 3f'(6)  =  27 m/sec

Problem 2 :

A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of

s  =  16t2 in t seconds.

(i) How long does the camera fall before it hits the ground?

(ii) What is the average velocity with which the camera falls during the last 2 seconds?

(iii) What is the instantaneous velocity of the camera when it hits the ground?

Solution :

(i)  s  =  16t2

To reach the ground it has to cover 400 meter.

16t =  400

t =  400/16  =  25

t  =  5 sec

So, it takes 5 seconds to reach the ground.

(ii) average velocity with which the camera falls during the last 2 seconds.

Last 2 seconds means  t  =  3 to t  =  5

f(t)  =  16t2

f(3)  =  16(3) ==>  144

f(5)  =  16(5) ==>  400

Average velocity  =  [f(5) - f(3)] / (5-3)

=  (400-144)/2

=  256/2

=  128 ft/sec

(iii)  Instantaneous rate of change

=  Instantaneous velocity

f(t)  =  16t2

f'(t)  =  32t

f'(5)  =  32(5)

f'(5)  =  160 ft/sec Kindly mail your feedback to v4formath@gmail.com

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