The height (h) of the water in metres at a certain point at the wave pool over a period of seconds is modelled by the equation #h(s)=sin^2s+1/2sins+3/2#?

How high is the water after 2 seconds?
During the first 10 seconds, how many times does the wave height reach 3 metres?
During the first 5 seconds, at what 3 points is the water level at 2 metres?

1 Answer
Apr 14, 2018

Simply find the value of #h# when #s = 2# for the first part.

#h(2) = sin^2(2) + 1/2sin(2) + 3/2#
#h(2) = 2.78# meters

For the second part, we must solve for when #h(s) = 3#.

#3 = sin^2s + 1/2sins + 3/2#

#0 = sin^2s + 1/2sins - 3/2#

#0 = 2sin^2s + sins - 3#

#0 = 2sin^2s - 2sins + 3sins - 3#

#0 = 2sins(sins - 1) + 3(sins - 1)#

#0 = (2sins + 3)(sins - 1)#

#sins = -3/2 or sins = 1#

Since #-3/2 > 1#, #sins = -3/2# has no solution. Thus #s = pi/2 + 2pin# seconds. The only time this happens in #0 ≤ s≤ 10# is when #s = pi/2~~ 1.57 s# and #s = (5pi)/2~~7.85 s#.

For the final part, we must set #h(s) = 2#.

#2 = sin^2s + 1/2sins + 3/2#

#0 = sin^2s + 1/2sins - 1/2#

#0 = 2sin^2s + sins - 1#

#0 = 2sin^2s + 2sins - sins - 1#

#0 = 2sins(sins + 1) - (sins + 1)#

#0 = (2sins - 1)(sins + 1)#

#sins = 1/2 and sins = -1#

Since #2pi ~~6.3# and #(3pi)/2 ~~4.7# and because sine is only positive in the first and second quadrants, we can consider the interval #0 ≤ s ≤ (3pi)/2#

#s = pi/6, (5pi)/6, (3pi)/2#

Thus

#s ~~0.52 s, 2.62 s, 4.71 s#

Hopefully this helps!