Apply trigonometric methods in solving problems

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The discussion focuses on applying trigonometric methods to determine safe crossing times for an inlet affected by tidal changes. Given the maximum and minimum water depths of 3.9 meters and 0.7 meters, the amplitude and midline of the sine or cosine function are calculated to model the tide's behavior. The equations derived indicate that the water depth can be modeled using trigonometric functions, with specific values for amplitude and phase shift. The safe crossing depth is identified as 0.8 meters, and the timing for this depth is calculated based on the trigonometric model. The Department of Conservation's recommendations align with the findings, emphasizing safety during tidal fluctuations.
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Summary: Hey, I'm getting confused with this question and don't think I'm doing it right, I was wondering if anyone could help me

Tides vary so the high tide and low tide height of the water is different every day. At certain times of the year, such as a Spring tide, the water can be very deep and it may not be safe to cross the inlet. During one big tide, the water was 3.9 metres deep at high tide and 0.7 metres deep at low tide.
The Department of Conservation (DOC) recommends that walkers only cross the inlet within one and a half hours before low tide and two hours after low tide.
Find a safe time to cross the inlet during this big tide and discuss DOC’s recommendation in relation to your findings for both tidal situations.

Water is 3.9 meters deep at high tide, Maximum value = 3.9m
Water is 0.7 meters deep at low tide, Minimum value 0.7m
The time period is 12.5 hours between high tides

d(t)=a sin b (x-c) + d or d(t)=a cos b (x-c) + d
where a is the amplitude = (max - min)/2
d=(max + min)/2

a=(3.9-0.7)/2=1.6
b=4pi/25
c= Points from graph (6.05 and 6.45 (2dp))
d= (3.9+0.7)/2=2.3

Equations of model
1.6 cos (4pi/25) (x-6.05) + 2.3
1.6 sin (4pi/25) (x-6.45) + 2.3

It is safe to cross the inlet at 0.8m, time for which the depth is at 0.8 meters
1.6 cos (4pi/25) t + 2.3=0.8
 

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Here's video that teaches you how to do this kind of problem:

 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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