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

The formula for the water level in Boston as a function of time is W(t) = 4.9 * cos((π / 6) * (t - 0)) + 5. The water level is 6.5 feet high around 1:16 AM on Feb 10. The water level at 10 AM is approximately 6.761 feet and at 4 PM is approximately 0.959 feet.

Step-by-step explanation:

To find a formula for the water level in Boston as a function of time, we can use a sinusoidal function since the water level is given by a sine or cosine curve. Since the high tide occurred at midnight, which is the starting point, we can use a cosine function. The general formula for a cosine function is A * cos(B(x - C)) + D, where A is the amplitude, B is the frequency, C is the phase shift, and D is the vertical shift. In this case, the amplitude (A) is half the difference between the high and low tide (9.9 - 0.1) / 2 = 4.9. The frequency (B) is 2π divided by the period, which is 12 hours, so B = 2π / 12 = π / 6. The phase shift (C) is 0 for the high tide location and the vertical shift (D) is the average of the high and low tide, (9.9 + 0.1) / 2 = 5.

Therefore, the formula for the water level in Boston as a function of time is:

W(t) = 4.9 * cos((π / 6) * (t - 0)) + 5

To find when the water level is 6.5 feet high on Feb 10, we set the formula equal to 6.5 and solve for t:

6.5 = 4.9 * cos((π / 6) * (t - 0)) + 5

Subtracting 5 from both sides gives:

1.5 = 4.9 * cos((π / 6) * (t - 0))

Dividing by 4.9 gives:

0.306 = cos((π / 6) * (t - 0))

Taking the inverse cosine gives:

t - 0 = cos-1(0.306)

Solving for t gives:

t = 0 + cos-1(0.306)

Using a calculator, we find that cos-1(0.306) is approximately 1.266 radians.

Therefore, on Feb 10, when the water level is 6.5 feet high, the time is approximately 1.266 hours after midnight, which is around 1:16 AM.

To find the water level at 10 AM on Feb 10, we substitute t = 10 into the formula:

W(10) = 4.9 * cos((π / 6) * (10 - 0)) + 5

Calculating, we get:

W(10) ≈ 4.9 * cos((π / 6) * 10) + 5 ≈ 1.761 + 5 ≈ 6.761

Therefore, the water level at 10 AM on Feb 10 is approximately 6.761 feet.

To find the water level at 4 PM on Feb 10, we substitute t = 16 into the formula:

W(16) = 4.9 * cos((π / 6) * (16 - 0)) + 5

Calculating, we get:

W(16) ≈ 4.9 * cos((π / 6) * 16) + 5 ≈ -4.041 + 5 ≈ 0.959

Therefore, the water level at 4 PM on Feb 10 is approximately 0.959 feet.

Answer 2

a) Model

`H(t)=4.9cos((\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((\pi)/(6)t)+5`

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

`H(t)=4.9cos((\pi)/(6)t)+5=6.5\\\\cos((\pi)/(6)t)=(6.5-5)/4.9=0.306\\\\t=(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((\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((\pi)/(6)16)+5=4.9*(-0.5)+5=-2.45+5=2.55`