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March 13th, 2014, 02:21 PM  #1 
Senior Member Joined: Apr 2008 Posts: 194 Thanks: 3  What did I do wrong?
By using the averages of high and low tide levels, the depth of water can by approximated by d(t) = 2.5 sin(0.164*pi*(t1.5)) + 13.4. A cruise ship needs a water depth of least 12 meters to dock safely. For how many hours per tide cycle, can the ship dock safely? My attempt 12 = 2.5 sin(0.164*pi*(t1.5)+13.4 reference angle t = 0.594 so, I put this reference angle in quadrants 3 and 4 where the ratio of sine is negative. Then, the standard position angles in 3 and 4 are 8.75 and 12.54 respectively. Hence, 12.54  8.75 = 3.79. But, this is wrong. Could someone please explain it? Thanks. 
March 13th, 2014, 05:38 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  Re: What did I do wrong?
Are you told what depth of water the ship needs to dock safely?

March 14th, 2014, 03:34 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,834 Thanks: 733  Re: What did I do wrong?
The ship can certainly dock safely when the sine is in the first two quadrants. Therefore you should have more than half a day. You need to exclude the range of t where sin < .56.


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