Trig general solutions Question (1 Viewer)

coyazayo · Jul 16, 2016

You are at Risser's beach to search for interesting shells. At 12pm on 19th June, the tide is in (i.e the water is at its deepest.) At that time you find that the depth of the water at the end of the breakwater is 15m. At 8pm the same day when the tide is out, you find that the depth of the water is 11m. Assume that the depth of the water varies sinusodially with time.

1. Derive an equation expressing depth in terms of the number of hours that have elapsed since 12 noon on 19th June

2. Using the general solutions formula, find out what the earliest time on 20th June that the water will be at 12.7 m deep

InteGrand · Jul 16, 2016

coyazayo said:
You are at Risser's beach to search for interesting shells. At 12pm on 19th June, the tide is in (i.e the water is at its deepest.) At that time you find that the depth of the water at the end of the breakwater is 15m. At 8pm the same day when the tide is out, you find that the depth of the water is 11m. Assume that the depth of the water varies sinusodially with time.

1. Derive an equation expressing depth in terms of the number of hours that have elapsed since 12 noon on 19th June

2. Using the general solutions formula, find out what the earliest time on 20th June that the water will be at 12.7 m deep

$\noindent 1) Let $x$ be the depth in metres at time $t$ hours after 12 noon on $19^{\text{th}}$ June. The extreme positions are $x =11$ and $x=15$, so the amplitude is clearly $2$. The centre of motion is at $x = 13$. It takes 8 hours to go from the peak to the trough, which is a half-period. So the period is $16$, so the angular frequency is $\omega = \frac{2\pi}{16} = \frac{\pi}{8}$. Since we start at the extreme maximal position, the function is thus $x(t) = 13 + 2\cos \left(\frac{\pi}{8}t\right)$.$

$\noindent 2) When $x=12.7$, we have $12.7 = 13 + 2\cos \left(\frac{\pi}{8}t\right) \Rightarrow \cos \left(\frac{\pi}{8}t\right) = -0.15$. Now use the general solution for cosine and select the one that gives the first one on 20 June (so the smallest solution after $t=12$).$

Trig general solutions Question (1 Viewer)

coyazayo

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InteGrand

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