Question

一道微积分题目。。

The high (in metres) of a model rocket t seconds after launching is given by
h(t)=3t^2
until the fuel runs out after 20s.

(1). find the average velocity of the rocket during each of the first four seconds.
(2). Find (in simplified form) the average velocity during an interval of time △t starting at t=2, t=6, and t=18.
(3). Using the results of part (2), , find the velocity at times t=2, t=6, and t=18

求详细解释过程 谢谢T T~~~

Answer 1

(1)
h(0) = 0 m

h(1) = 3 m

h(2) = 12 m
h(3) = 27 m
h(4) = 48 m
During the 1st second, the average velocity is (h(1) - h(0))/(1-0) = (3-0)/(1-0) = 3 m/s
During the 2nd second, the average velocity is (h(2) - h(1))/(2-1) = (12-3)/2-1) = 9 m/s
During the 3rd second, the average velocity is (h(3) - h(2))/(3-2) = (27-12)/(3-2) = 15 m/s
During the 4th second, the average velocity is (h(4) - h(3))/(4-3) = (48-27)/(4-3) = 21 m/s

(2)
When t = 2, h(t) = h(2) = 12 m
At the moment of t+△t = 2+△t, h(2+△t) = 3(2+△t)² = 12 + 12△t+ 3△t²
The average velocity between t = 2 and t = 2+△t second is:
(h(2+△t) - h(2))/(2+△t - 2) = (12 + 12△t+ 3△t² - 12)/△t = 12 + 3△t m/s
Similarly, the average velocity between t = 6 and t = 6+△t second is:
(h(6+△t) - h(6))/(6+△t - 6) = (108 + 36△t+ 3△t² - 108)/△t =36 + 3△t m/s
the average velocity between t = 6 and t = 6+△t second is:
(h(18+△t) - h(186))/(18+△t - 18) = (972 + 108△t+ 3△t² - 972)/△t = 108 + 3△t m/s

(3)
When △t ->0, 12 + 3△t -> 12 m/s, this is the velocity at t = 2
Similarly, when t = 6, the velocity is 36 m/s;
when t = 18, the velocity is 108 m/s