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 = 2t^{2} + 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) = 2t^{2} + 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) = 2t^{2} + 3t
f'(x) = 4t+3
at t = 3 f'(3) = 4(3)+3 f'(3) = 12 + 3 f'(3) = 15 m/sec |
ar t = 6 f'(6) = 4(6)+3 f'(6) = 24 + 3 f'(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 = 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) s = 16t^{2}
To reach the ground it has to cover 400 meter.
16t^{2 } = 400
t^{2 } = 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) = 16t^{2}
f(3) = 16(3)^{2 } ==> 144
f(5) = 16(5)^{2 } ==> 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) = 16t^{2}
f'(t) = 32t
f'(5) = 32(5)
f'(5) = 160 ft/sec
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