Question

On February 10, 1990, high tide in Boston was at midnight. The water level at high tide was 9.9 feet; later, at low tide, it was 0.1 feet. Assuming the next high tide is exactly 12 hours later and that the height of the water is given by a sine or cosine curve, find a formula for water level in Boston as a function of time. On Feb 10, when is the water level 6.5 feet high? On Feb 10, what is the water level at 10 am? 4pm?

Answer 1

Answer:

a) Model

$H(t)=4.9cos(\frac{\pi}{6}t)+5$

b) The tide will be at 6.5 ft high at 2:24 am (going down).

As the tide will rise again, the tide will be again at 6.5 ft high at 9:36 am (going up).

c) H(10 am) = 7.45

H(4 pm) = 2.55

Step-by-step explanation:

We can model this as a sine or cosine wave, for which we will calculate its parameters by the data given.

The model we will use is:

$H(t)=Acos(\omega t+\phi)+B$

t will be in hours, starting from midnight (t=0)

1) We know that at midnight happens the high tide. That means that

$H(0)=Acos(\omega* 0+\phi)+B=A+B=9.9\\\\ \phi=0\\\\A+B=9.9$

2) Six hours later (t=6) happens the low tide (0.1 ft)

$H(0)=Acos(\omega*6)+B=-A+B=0.1\\\\\\ cos(\omega*6)=-1,\, \omega=\pi/6\\\\\\B=0.1+A=9.9-A\\\\2A=9.9-0.1\\\\A=9.8/2=4.9\\\\\\B=0.1+A=0.1+4.9=5$

Now that we have calculated all the parameters, the model for the height of the tide is:

$H(t)=4.9cos(\frac{\pi}{6}t)+5$

On Feb 10, when is the water level 6.5 feet high?

$H(t)=4.9cos(\frac{\pi}{6}t)+5=6.5\\\\cos(\frac{\pi}{6}t)=(6.5-5)/4.9=0.306\\\\t=\frac{6}{\pi}arcos(0.306)=2.4$

The tide will be at 6.5 ft high at 2:24 am (going down).

As the tide will rise again, the tide will be again at 6.5 ft high at 9:36 am (going up).

On Feb 10, what is the water level at 10 am? 4pm?

At 10 am is t=10:

$H(10)=4.9cos(\frac{\pi}{6}10)+5=4.9*0.5+5=2.45+5=7.45$

At 4 pm, is t=12+4=16

$H(16)=4.9cos(\frac{\pi}{6}16)+5=4.9*(-0.5)+5=-2.45+5=2.55$