The height in metres of a tsunami is modelled by the equation h(t)=10sin(pi/8 - t)
where time is in minutes, after de occurrence of an earthquake. Approximately How long after the tsunami will the wave reach a height of 10m?
I know this:
10=10sin(pi/8 - t)
-10 sin (t-pi/8)
-1 = sin (t - pi/8)
sin (3pi/2) = -1 = sin(t- pi/8)
3pi/2 =t - pi/8
t= pi/8+ 3pi/2 = 13 pi/8
3 Answers | Add Yours
Since the problem provides the information that the wave reaches the height of 10 m, you need to substitute 10 for `h(t) ` in the given equation `h(t) = 10sin(pi/8 - t)` such that:
`10 = 10sin(pi/8 - t)`
You need to divide by 10 such that:
`1 = sin(pi/8 - t) => pi/8 - t = (-1)^n*sin^(-1) (1) + n*pi`
`pi/8 - t = (-1)^n*(pi/2) + n*pi`
`-t = (-1)^n*(pi/2) + n*pi - pi/8`
`t = (-1)^(n+1)*(pi/2)- n*pi+ pi/8`
You need to consider n having an odd value, such that:
`t = pi/2 - (2k+1)*pi + pi/8 => t = 5pi/8 - pi - 2k*pi`
`t = (5pi - 8pi)/8 - 2k*pi => t = -(3pi)/8 - 2k*pi`
You need to consider n having an even value, such that:
`t = -pi/2 - 2k*pi + pi/8 => t = -(3pi)/8 - 2k*pi`
Hence, evaluating the general solution to the given equation yields `t = -(3pi)/8 - 2k*pi` .
You got it right. You have equation
`10sin(pi/8-t)=10` Now devide whole equation by 10.
Now ask youself where is sine equal to 1? ` ` Sine is equal to 1 in `pi/2+2k pi` ,`k inZZ`. So we have:
` ` `-t=-pi/8+pi/2+2kpi`
Since time can only be positive we are interested only in positive values of `t.` So for `k=1` we get our solution:
`2 pi`minutes after that wave would again reach the height of 10 meters and again `2pi`minutes after that etc.
From this graph you can see that sine is equal to 1 in `pi/2`, if we were looking where sine si equal to -1 we would have `-pi/2` etc. You can also look at that on trigonometric circle (unit circle).
We’ve answered 319,858 questions. We can answer yours, too.Ask a question