The derivative can also be used to
determine the rate of change of one variable with respect to another. A few examples are population
growth rates, production rates, water flow rates, velocity, and acceleration.
Rates of change with respect to some other quantity in our daily life are given below :
1) Slope is the rate of change in vertical length with respect to horizontal length.
2) Velocity is the rate of displacement with respect to time.
3) Acceleration is the rate of change in velocity with respect to time.
4) The steepness of a hillside is the rate of change in its elevation with respect to linear distance
The average rate of change in an interval [a, b] is
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 = 2t^{2 }+ 3t meters.
(i) Find the average velocity of the points between t = 3 and t = 6 seconds.
(ii) Find the instantaneous velocity at t = 3 and t = 6 seconds.
Solution :
Average velocity = [f(b) - f(a)]/(b - a)
Substitute a = 3 and b = 6.
f(b) = f(6) = 2(6)^{2} + 3(6) = 2(36) + 18 = 72 + 18 = 90 |
f(a) = f(3) = 2(3)^{2} + 3(3) = 2(9) + 9 = 18 + 9 = 27 |
Average velocity = (90 - 27)/(6 - 3)
= 63/3
= 21
Hence the average velocity is 21 meter/sec.
(ii) The instantaneous velocity is given by s'(t) = 4t + 3.
Instantaneous velocity at t = 3 sec is
s'(3) = 4(3) + 3
= 15 meter/sec
Instantaneous velocity at t = 6 sec is
s'(6) = 4(6) + 3
= 27 meter/sec
Problem 2 :
A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of s = 16t^{2} 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) Given s = 16t^{2 }----(1).
The camera has to fall 400 ft to hit the ground.
To calculate the time taken by the camera before it hits the ground, substitute s = 400 in (1).
400 = 16t^{2}
t^{2 }= 25, then t = 5 sec.
(ii) Since the total time is 5 seconds, the average velocity with which the camera falls during the last 2 seconds is given by
S(5) = 16(5)^{2} = 400
S(3) = 16(3)^{2} = 144
Average velocity = [S(5) - S(3)]/(5 - 3)
= (400 - 144) / 2
= 256/2
The average velocity is 128.
(iii) When the camera hits the ground time t = 5. The instantaneous velocity is given by
s'(t) = 32t
= 32(5)
= 160
The instantaneous velocity is 160 feet/sec.
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