Would love for someone to confirm if my Trig answer is correct

[JimmyBarns2]

Info::::: There is a lake in the Karuu Wildlife park in South Africa.

A trig function can model the depth of the water.

The water is 5.5 meters deep at the highest tide and only 1.5 meters deep at the lowest tide. There are also 14 hours between each high tide and the lake is able to be walked through when the depth of the lake is less than 2 meters.

Questions

1: Model the depth of the water at the lake and state the period when its safe to walk in the lake.


Answer:

Period is 14 hours as this is the time between each high tide. This means 2Pi/B =14 so B = 2Pi/14.

Amplitude is 5.5-1.5/2 so it's 2 metres.

Distance to mid point is 5.5 metres + 1.5 / 2 so that's 3.5 metres

and when it's safe to walk should be 14/2 so 7 hours between the low tides.

The equation would then be D = 2 cos 2pi/14 + 3.5

Would this be correct?

Many thanks.

 

[Otis]

… D = 2 cos 2pi/14 + 3.5

Would this be correct? …

Hi Jimmy. You forgot the time variable. (Also, the ratio 2/14 can be simplified.) Otherwise, your function for water depth in terms of elapsed time is correct.

By the way, have you learned about 'function notation' yet?

… when it's safe to walk should be … 7 hours between the low tides …

I don't understand your conclusion above, but I can say that it's not safe to walk in the lake for seven hours between low tides (if that's what you're thinking). Or, maybe you meant to say that 7 hours after t=0 is halfway between the first two high tides. That would be the first low tide (D=1.5 when t=7), but low tide is not the only time when it's safe to walk in the lake.

Your model is based on high tide starting at t=0. If you're familiar with the graph of y=cos(x) over one period, starting at x=0, then you know the curve starts and ends at its high points and drops down to its lowest point halfway through the period. Your model does the same thing over one period; it starts and ends at high tide and drops down to low tide halfway through the period. The question asks about the time interval where the curve appears below the line y=2 (because people can walk when "the depth of the lake is less than 2 meters"). In other words, for what values of t is D<2.

Can you find the first two values of t where D=2? It's safe to walk in the lake during the interval in between those two points in time.

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[Otis]

… Amplitude is 5.5-1.5/2 …

Here's a helpful note about typing math. When a numerator (or a denominator) in a ratio contains more than a single number (or symbol), we need to enclose it within grouping symbols, in order to show the correct order of operations. Like this:

(5.5 - 1.5)/2

Without those grouping symbols, the expression 5.5-1.5/2 means \(5.5 - \frac{1.5}{2}\) because we do division before subtraction.

Cheers

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