Solving Trig Applications: Docking Safety in Seaports

[tsaitea]

By using the averages of high and low tide levels. The depth of the water in a seaport can be approximated by the sinusoid d=3.2sin0.166pi(t-2.5)+14.1 where d is the depth and t is the hours after midnight. if a ship needs at least 12 m of water in a seaport to dock safely, how long could the ship dock safely for?

What I have tried so far... My attempt at this problem is to find the two points where the wave intersects y=12 and subtract the two points.
12=3.2sin0.166pi(t-2.5)+14.1
t=2.5+(asin((12-14.1)/3.2)/(0.166pi))=1.12
second solution is pi-1.12=2.02
subtracting the two gives me 0.90 hours which is incorrect the answer is 8.769 hours.

Any ideas what I am doing wrong here?

 

[gopher_p]

You're finding your multiple answers in the wrong manner. There are two solutions in ##[-\frac{\pi}{2},\frac{3\pi}{2})## to the equation ##\frac{12-14.1}{3.2}=\sin\theta##; call them ##\theta_1## and ##\theta_2##. You'd then solve ##0.166\pi(t-2.5)=\theta_1## and ##0.166\pi(t-2.5)=\theta_2## to get the first two times greater than ##0## where the tide was at ##12##.

Now you should find some way to justify to yourself (a) why these are the first two solutions and (b) why their difference gives you the answer to the question that was asked.

 

[tsaitea]

Okay, so I solved for theta1 and theta2 and got the positive angles to be 3.8 rads and 5.6 rads. I got there by plotting down the corresponding angle for -0.716 rads and the other solution for this angle which would be in quadrant 3. I then added pi to that 0.716 to get 3.8 and then I subtracted 2pi from 0.716 to get 5.6. Am I on the right track?

 

[gopher_p]

You correctly found the first solution, ##-0.716## (by plugging and chugging with a calculator, I'm guessing), but you failed to recognize it as one of the solutions that I suggested you look for. You also correctly found the second solution, ##3.8##; it looks like by using a reference angle in the third quadrant.

##5.6## is a solution, but it's not one of the ones that I suggested you look for. I'd also note that you subtracted your reference angle from ##2\pi## to get this solution, not the other way around as you wrote it. I'm guessing that was maybe just a typo, but just in case ...