Question

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cycle in 14 hours. What is the amplitude, period, and midline of a function that would model this periodic phenomenon? Amplitude = 3 feet; period = 14 hours; midline: y = 6 Amplitude = 3 feet; period = 7 hours; midline: y = 3 Amplitude = 6 feet; period = 14 hours; midline: y = 6 Amplitude = 6 feet; period = 7 hours; midline: y = 3

Answer 1

Answer:

                $y(t) = 6 -3cos(\frac{2\pi }{14} )t$

                $y(t) = 6 -3cos(\frac{2\pi }{7} )t$

                $y(t) = 6 - 6cos(\frac{2\pi }{14} ) t$

                $y(t) = 3- 6cos(\frac{2\pi }{7} )t$

Step-by-step explanation:

Given that,

Hudson Bay tides vary between $3 ft$ and $9 ft$.

Tide is at its lowest when $t=0$

Completes a full cycle in 14 hours.

To find:- What is the amplitude, period, and midline of a function that would model this periodic phenomenon?

So, The periodic function of this model is

                                         $y(t) = y^{'} + Acos(\omega\ t)$                ...................(1)

                                                                 where, $A- Amplitude of cycle$

                                                                              $\omega = Angular speed (in Radian.)$

Then putting the value in given Equation(1) we get,

        Amplitude = $\frac{9-3}{2} ft = 3ft$

                      $y^{'} = (3+ 3 )ft = 6ft$

Now, At $t=0 sec$ it complete full cycle in $14 hours.$ $-cos(\omega t)$ because it is at lowest at t=0sec.

∵                    $\omega t= 2\pi$

                     $\omega (t+14) = 2\pi$

∴                           $\omega = \frac{2\pi }{14}$            

Hence     $y(t) = 6 -3cos(\frac{2\pi }{14} )t$

                $y(t) = 6 -3cos(\frac{2\pi }{7} )t$

                $y(t) = 6 - 6cos(\frac{2\pi }{14} ) t$

                $y(t) = 3- 6cos(\frac{2\pi }{7} )t$