The depth of water in a leaking storage tank is d cm at time t hours after midnight on Sunday. The value of 𝑑 is
given by 𝒅 = 𝟏𝟎 − 𝟐√𝟑𝒄𝒐𝒔𝟒𝒕 − 𝟐𝒔𝒊𝒏𝟒𝒕
. Find
(i) The least and greatest depth of water possible with this model (6 marks)
(ii) The time at which the depth first reaches these values (3 marks)
(i)
d = 10 - 2√3cos4t - 2sin4t
d' = 8√3sin4t - 8cos4t
0 = 8√3sin4t - 8cos4t
8cos4t = 8√3sin4t
cos4t = √3sin4t
1 = √3sin4t/cos4t
1 = √3tan4t
1/√3 = tan4t
√3/3 = tan4t
π/6 + πn = 4t
π/24 + πn/4 = t
Least depth when n=0:
π/24 = t
d = 10 - 2√3cos4(π/24) - 2sin4(π/24)
d = 6
Greatest depth when n=1:
7π/24 = t
d = 10 - 2√3cos4(7π/24) - 2sin4(7π/24)
d = 14
(ii)
Again, t=π/24 will be the first minimum, and t=7π/24 will be the first maximum. Feel free to verify this by looking at the graph.