To find the times when the buoy is exactly 6 meters above sea level, we set the equation f(t) equal to 6:
5 cos(pi/6 t) + 4 = 6
5 cos(pi/6 t) = 2
cos(pi/6 t) = 2/5
To find the first two times the buoy is exactly 6 meters above sea level, we need to find the first two solutions to the equation cos(pi/6 t) = 2/5.
The first solution to cos(pi/6 t) = 2/5 is when pi/6 t = arccos(2/5). Using a calculator, we can find that arccos(2/5) is approximately 1.107 radian. So, pi/6 t is approximately 1.107. Solving for t, we get t ≈ 6.64 hours after midnight.
The second solution to cos(pi/6 t) = 2/5 is when pi/6 t = -arccos(2/5) + 2π. Using a calculator, we can find that -arccos(2/5) + 2π is approximately 5.176 radians. So, pi/6 t is approximately 5.176. Solving for t, we get t ≈ 31.056 hours after midnight.
Therefore, approximately 24.4 hours elapse between the first two times the buoy is exactly 6 meters above sea level.
A buoy floats on the surface of the water. The height in meters of the buoy t hours after midnight, relative to sea level, is given by the equation f(x)=5 cos pi/6 t+4. Estimate how many hours elapse between the first two times the buoy is exactly 6 m above sea level
1 answer
1 answer