The height in metres of a tsunami is modelled by the equation `h(t)=12sin(pi/6-t)` where time is in minutes. How long after the tsunami will the wave reach a height of 6 metres? Ok I know this!! 6/12=sin((pi/6)-t) 1/2=sin((pi/6)-t) when sin=1/2 = pi/6 so, pi/6=sin((pi/6)-t) sin^-1(pi/6)=pi/6 - t 0.0274= -t Look I am not sure about this!! help, help

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We are given that `h(t)=12sin(pi/6-t)` and we are asked to find t so that h(t)=6.

`6=12sin(pi/6-t)`

`sin(pi/6-t)=1/2`

`sintheta=1/2 ==> theta=pi/6 +-2kpi` or `theta=(5pi)/6 +- 2kpi`

So `pi/6-t=(pi/6+-2kpi)` ==> `t=pi/6-pi/6+-2kpi=2kpi`

`pi/6-t=((5pi)/6+-2kpi)`  ==> `t=pi/6-(5pi)/6+-2kpi=(-2pi)/3+-2kpi`

The first instance after t=0 that the height is 6 is at `t=-(2pi)/3+2pi=(4pi)/3`

So `t~~4.19"min"`

The graph:

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We are given that `h(t)=12sin(pi/6-t)` and we are asked to find t so that h(t)=6.

`6=12sin(pi/6-t)`

`sin(pi/6-t)=1/2`

`sintheta=1/2 ==> theta=pi/6 +-2kpi` or `theta=(5pi)/6 +- 2kpi`

So `pi/6-t=(pi/6+-2kpi)` ==> `t=pi/6-pi/6+-2kpi=2kpi`

`pi/6-t=((5pi)/6+-2kpi)`  ==> `t=pi/6-(5pi)/6+-2kpi=(-2pi)/3+-2kpi`

The first instance after t=0 that the height is 6 is at `t=-(2pi)/3+2pi=(4pi)/3`

So `t~~4.19"min"`

The graph:

Note that `y=12sin(pi/6-t)=12sin(-(t-pi/6))`

The 12 is the amplitude. The negative 1 inthe function reflects the sin wave across teh y-axis, and the `pi/6` is a phase shift of `pi/6` units to the right.

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