Differentiation-rate of chang

用戶3618

2009-01-09 10:00 am

1. An object moves along a horizontal line. Its displacement s from a fixed point at time t is given by
s = t^3 - 9t^2 + 24t
(a) What is the velocity of the object at time t?
(b) When is s increasing and when decreasing?
(c) What is the total distance travelled from t = 0 to t = 7?

2. When a colony of bees was studied, it was found that its population P after t days can be approximated by the modal P = 80 e^(0.1t).
(a) What was the population of the bee colony when the study began?
(b) What was the population of the bee colony 10 days after the study began?
(c) Find the average rate of increase in the population in the first 10 days of the study.
(d) Find the instantaneous rate of increase in the population just after 10 days.

回答 (1)

六呎將軍

2009-01-09 5:21 pm

✔ 最佳答案

(1a) Taking differentiation of s w.r.t. t:
ds/dt = 3t² - 18t + 24
= 3(t² - 6t + 8)

(b) ds/dt = 3(t - 4)(t - 2)
So when 0 < t < 2, it is increasing
When 2 < t < 4, it is decreasing
When t > 4, in it increasing

(c) If we are talking about the distance, direction should be neglected and then:
When t = 0, s = 0 m
When t = 2, s = 20 m
When t = 4, s = 16 m
When t = 7, s = 70 m
So from t = 0 to t = 2, distance travelled = 20 m
From t = 2 to t = 4, distance travelled = 4 m
From t = 4 to t = 7, distance travelled = 54 m
So total distance travelled = 20 + 4 + 54 = 78 m

(2a) Sub t = 0, we have P = 80 which is the initial population.

(b) Sub t = 10, we have P = 80e = 217 (corrected to the nearest integer)

(c) Average rate of increase = (217 - 80)/10 = 13.7 bees per day

(d) Taking differentiation of P w.r.t. t:
dP/dt = 80 e^(0.1t) d(0.1t)/dt (Chain Rule)
= 8 e^(0.1t)

So sub t = 10, the instantaneous rate of increase is:
8e = 21.7 bees per day (Corrected to 1 d.p.)

參考: Myself

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