Question

Graphing population edfintity homework help asap

Answer 1

To find the population at specific times, you can plug in values of t into the formula. At
`\( t = 0 \)` (January 1),
`\( P(0) = 600 \sin\left((\pi)/(6)(-(3)/(2))\right) + 1000 \)`. At
`\( t = 6 \)` (July 1),
`\( P(6) = 600 \sin\left((\pi)/(6)(6 - (3)/(2))\right) + 1000 \)`

To model a population that oscillates sinusoidally, we can use a sinusoidal function of the form:

`\[ P(t) = A \sin(B(t - C)) + D \]`

where:

-
`\( A \)` is the amplitude (half the difference between the maximum and minimum values),

-
`\( B \)` is the frequency (related to the period of the function),

-
`\( C \)` is the phase shift (horizontal shift of the graph),

-
`\( D \)` is the vertical shift (midline of the graph).

In this case, we are given that the population oscillates between a low of 400 and a high of 1600. The amplitude is half the difference between these values:

`\[ A = (1600 - 400)/(2) = 600 \]`

The period of the sinusoidal function is one year, which means the time it takes to complete one full cycle is 12 months. Therefore, the frequency
`\( B \)` is given by:

`\[ B = \frac{2\pi}{\text{period}} = (2\pi)/(12) = (\pi)/(6) \]`

Now, we need to determine the phase shift
`\( C \)` and the vertical shift
`\( D \)`. The function oscillates between a low of 400 and a high of 1600, so the midline is the average of these values:

`\[ D = (1600 + 400)/(2) = 1000 \]`

Now, the function oscillates between the low and high points, so we need to find when the function reaches its maximum (which is on July 1). Since the function is sinusoidal, the maximum occurs when the argument of the sine function is at its maximum, and that corresponds to a phase shift:

`\[ B(t - C) = (\pi)/(2) \]`

Solving for
`\( C \)`, we get:

`\[ C = t - (\pi)/(2B) \]`

Substitute
`\( B = (\pi)/(6) \)` into the equation:

`\[ C = t - (\pi)/(2\left((\pi)/(6)\right)) \]`

Simplify:

`\[ C = t - (3)/(2) \]`

Now, we have all the information to write the formula for the population
`\( P(t) \)`

`\[ P(t) = 600 \sin\left((\pi)/(6)(t - (3)/(2))\right) + 1000 \]`

This formula models the population
`\( P \)` as a function of time
`\( t \)` in months since the start of the year.

Below is the attached graph.