Question

Ocean tides can be modeled by a sinusoidal function. Suppose that there is a low and high tide every 12 hours, and that high tide in Seattle occurs at 1:00 a.m. and 1:00 p.m. with the low tides 6 hours after high tides. Also suppose that the water level at high tide is 10 ft above the water level at low tide. a. Find a formula for the function that computes the height of the tide above low tide at time . (In other words, corresponds to low tide.) b. What is the height of the tide at 11:00 a.m.?

Answer 1

Answer:

(a)$y=5sin(\frac{\pi}{6}(x+2))+5$

(b) 7.5ft above the low tide.

Step-by-step explanation:

(a) To find the function that computes the height of the tide, you need to select the form of the sinusoidal function. For example, use the form:

$y=Asin(B(x-C))+D$

Where A is the amplitude, B the frequency, C the phase shift and D the vertical shift.

The  amplitude is half the distance between the highest and the lowest tide:

$A=10/2=5ft$

The frequency is related to the period T by:

$B=\frac{2\pi}{T}$

The period is 12 hours, then

$B=\frac{2\pi}{12}=\frac{\pi}{6}$

The high tide is at 1:00 a.m. and 1:00 p.m. , this is the moment when $sin(B(x-C))=1$, if $sin(\frac{\pi}{2})=1$ then $B(x-C)$ must be equal to $\frac{\pi}{2}$ when $x=1$:

$B(x-C)=\frac{\pi}{2}\\\frac{\pi}{6}(1-C)=\frac{\pi}{2}\\\frac{1}{6}(1-C)=\frac{1}{2}\\(1-C)=\frac{6}{2}\\-C=3-1\\C=-2$

The vertical shift is the sum of the lowest value, the height of the low tide ($lt$) and the amplitude:

$D=5+lt$

The function is:

$y=5sin(\frac{\pi}{6} (x+2))+5+lt$

Because the function must be the height above low tide height, subtract this heigh from the function:

$y=5sin(\frac{\pi}{6} (x+2))+5+lt-lt$

$y=5sin(\frac{\pi}{6} (x+2))+5$

(b) Use x=11 in the function

$y=5sin(\frac{\pi}{6} (11+2))+5=2.5+5=7.5ft$ above the low tide.