2.5cos(0.675(t β 3.5)) + 4.3
D' = -1.6875sin(0.675(t β 3.5))
10am is 10 hours past midnight, so just find D'(10)
please do not blindly regurgitate specific instructions on how to proceed. I am not responsible for following your teacher's directions, and I certainly have no idea about the "terminology and formulae learned in this unit."
The depth of water, π, in metres, varies with the tides throughout the day and can be modelled by the equation π·(π‘) = 2.5πππ (0.675(π‘ β 3.5)) + 4.3, where π‘ is the number of hours past 12ππ.
Abithan claims that at exactly 10ππ, the depth is decreasing by 0.5π per hour. Is Abithanβs claim, correct? Describe and explain the steps you would take to check Abithanβs claim. Use terminology and formulae learned in this unit. Provide calculations as needed.
2 answers
t = 10 hours past midnight
dD/dt = -2.5 (0.675) sin (0.675(π‘ β 3.5))
dD/dt = -2.5 (0.675) sin (0.675(π‘ β 3.5))