Question #2d74e

1 Answer
Dec 16, 2016

drawn

The equation #d(t) = 3.5 cos ((pit)/6)+4.5# models the water at the beach. The variation of water level with time as per the given equation has been shown in the above figure.graphically.
Graph has been plotted taking 12 O clock night as zero hour i,e, t = 0

The condition to dive off safely is that the water must be at least #3m# deep.
So inserting #d(t)=3# in the given equation we get

#3.5 cos ((pit)/6)+4.5=3#

#=>3.5 cos ((pit)/6)=3-4.5=-1.5#

#=>(pit)/6=2npi+-cos^-1(-1.5/3.5)#

#=>t=12n+-6/picos^-1(-1.5/3.5)" where " n in NN#

#=>t=12n+-3.84#

when n =0

#=>t=12n+3.84=3.84 hr#

when n =1

#=>t=12*1-3.84=8.16 hr->color(red)(8hr" " 9min " "36 sec)#

and

#=>t=12*1+3.84=8.16 hr->color(red)(15hr" " 50min " "24sec)#

Hence from above calculation as well as from graphical presentation of variation of depth of water with time it is obvious that during daytime the water level remains at least 3m deep for the period

#color(red)(8hr" " 9min " "36 sec) to color(red)(15hr" " 50min " "24sec) #
and this period is safe to dive off .