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An oceanographer measured a set of sea waves during a storm and modelled the vertical displacement of waves in meters using the...Asked by Farah
An oceanographer measured a set of sea waves during a storm and modelled the vertical displacement of waves in meters using the equation h(t)=0.6cos2t+0.8sint, where t is the time in seconds.
a) Determine the vertical displacement of the wave when the velocity is 0.8m/s
Ans: -1.2sin2t+0.8cost = 0.8
-2.4(sint)(cost)+0.8cost = 0.8
cost(-2.4sint+0.8) = 0.8
cost = 0.8
t = cos-1(0.8) OR -2.4sint +0.8 = 0.8
=0.6 t = 0
b) Determine the maximum velocity of the wave and when it occurs.
Ans: cost(-2.4sint+0.8)=0
therefore t= 1.5 and 0.3 and Vmax occurs at t=1.5s
c) When does the wave first change from a hill to a trough? Explain.
Please check the above answers and help if they are incorrect, and need guidance with part c, is it asking for the height?
a) Determine the vertical displacement of the wave when the velocity is 0.8m/s
Ans: -1.2sin2t+0.8cost = 0.8
-2.4(sint)(cost)+0.8cost = 0.8
cost(-2.4sint+0.8) = 0.8
cost = 0.8
t = cos-1(0.8) OR -2.4sint +0.8 = 0.8
=0.6 t = 0
b) Determine the maximum velocity of the wave and when it occurs.
Ans: cost(-2.4sint+0.8)=0
therefore t= 1.5 and 0.3 and Vmax occurs at t=1.5s
c) When does the wave first change from a hill to a trough? Explain.
Please check the above answers and help if they are incorrect, and need guidance with part c, is it asking for the height?
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a) To find the vertical displacement when the velocity is 0.8m/s, we need to find the first derivative of the given equation:
h(t) = 0.6cos(2t) + 0.8sin(t)
h'(t) = -1.2sin(2t) + 0.8cos(t)
Now, set h'(t) = 0.8:
-1.2sin(2t) + 0.8cos(t) = 0.8
-3(sin(2t) - cos(t)) = 2
sin(2t) - cos(t) = 2/3
Now, use the double angle identity, sin(2t) = 2sin(t)cos(t):
2sin(t)cos(t) - cos(t) = 2/3
cos(t)(2sin(t) - 1) = 2/3
For the wave's vertical displacement when the velocity is 0.8m/s, plug in the values of t that satisfy the above equation into the original equation h(t).
b) To find the maximum velocity, we need the second derivative of h(t):
h''(t) = -2.4cos(2t) - 0.8sin(t)
To find the critical points, set h''(t) = 0:
-2.4cos(2t) - 0.8sin(t) = 0
Now, find the values of t that make h''(t) = 0, and then plug them into the first derivative equation, h'(t), to find the maximum velocity. I made an error in part a, Vmax occurs when h''(t) = 0, not when h'(t) = 0.
c) The wave changes from a hill to a trough when the vertical displacement is 0, so we need to find when h(t) = 0:
0.6cos(2t) + 0.8sin(t) = 0
Divide by 0.6:
cos(2t) + (4/3)sin(t) = 0
Now, find the values of t that make h(t) = 0. The first value of t for which this happens is when the wave switches from a hill to a trough.
h(t) = 0.6cos(2t) + 0.8sin(t)
h'(t) = -1.2sin(2t) + 0.8cos(t)
Now, set h'(t) = 0.8:
-1.2sin(2t) + 0.8cos(t) = 0.8
-3(sin(2t) - cos(t)) = 2
sin(2t) - cos(t) = 2/3
Now, use the double angle identity, sin(2t) = 2sin(t)cos(t):
2sin(t)cos(t) - cos(t) = 2/3
cos(t)(2sin(t) - 1) = 2/3
For the wave's vertical displacement when the velocity is 0.8m/s, plug in the values of t that satisfy the above equation into the original equation h(t).
b) To find the maximum velocity, we need the second derivative of h(t):
h''(t) = -2.4cos(2t) - 0.8sin(t)
To find the critical points, set h''(t) = 0:
-2.4cos(2t) - 0.8sin(t) = 0
Now, find the values of t that make h''(t) = 0, and then plug them into the first derivative equation, h'(t), to find the maximum velocity. I made an error in part a, Vmax occurs when h''(t) = 0, not when h'(t) = 0.
c) The wave changes from a hill to a trough when the vertical displacement is 0, so we need to find when h(t) = 0:
0.6cos(2t) + 0.8sin(t) = 0
Divide by 0.6:
cos(2t) + (4/3)sin(t) = 0
Now, find the values of t that make h(t) = 0. The first value of t for which this happens is when the wave switches from a hill to a trough.
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