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The population sizes of many animal species rise and fall over time. Suppose that the population size of a certain species can be modeled by the following:

[tex]\[ p(t) = 4249 - 1180 \cos \left(\frac{2 \pi}{7} t \right) \][/tex]

In this equation, \( p(t) \) represents the total population size and \( t \) is the time (in years). Suppose we start at \( t = 0 \) years.

During the first 7 years, when will the population size be 4800?

Do not round any intermediate computations, and round your answer(s) to the nearest hundredth of a year. (If there is more than one answer, enter additional answers with the "or" button.)

[tex]\[ t = \][/tex] [tex]$\square$[/tex] years

or

[tex]\[ t = \][/tex] [tex]$\square$[/tex] years

Sagot :

GinnyAnswer
To find when the population size will be 4800 during the first 7 years, we need to solve the equation given by:

[tex]\[ p(t) = 4249 - 1180 \cos\left(\frac{2 \pi}{7} t\right) \][/tex]

We set this equal to 4800 to find the time \( t \):

[tex]\[ 4249 - 1180 \cos\left(\frac{2 \pi}{7} t\right) = 4800 \][/tex]

First, solve for \( \cos\left(\frac{2 \pi}{7} t\right) \):

[tex]\[ 4249 - 1180 \cos\left(\frac{2 \pi}{7} t\right) = 4800 \][/tex]

Subtract 4249 from both sides of the equation:

[tex]\[ - 1180 \cos\left(\frac{2 \pi}{7} t\right) = 4800 - 4249 \][/tex]

Simplify the right side:

[tex]\[ - 1180 \cos\left(\frac{2 \pi}{7} t\right) = 551 \][/tex]

Now, divide both sides by -1180 to isolate the cosine term:

[tex]\[ \cos\left(\frac{2 \pi}{7} t\right) = \frac{551}{-1180} \][/tex]

[tex]\[ \cos\left(\frac{2 \pi}{7} t\right) = -\frac{551}{1180} \][/tex]

This further simplifies to:

[tex]\[ \cos\left(\frac{2 \pi}{7} t\right) = -0.4678... \][/tex]

Next, find the inverse cosine of -0.4678 to solve for \( \frac{2 \pi}{7} t \):

[tex]\[ \frac{2 \pi}{7} t = \cos^{-1}(-0.4678) \][/tex]

Using a calculator to find \( \cos^{-1}(-0.4678) \):

[tex]\[ \frac{2 \pi}{7} t \approx 2.2913 \text{ radians} \][/tex]

Now, solve for \( t \):

First instance:
[tex]\[ t = \frac{7}{2 \pi} \cdot 2.2913 \][/tex]

Calculate \( t \):

[tex]\[ t \approx 2.29 \text{ years} \][/tex]

There is another solution because cosine is periodic with period \( 2 \pi \). The other solution within one period (0 to 7 years) would be found as follows:

[tex]\[ \frac{2 \pi}{7} t = 2 \pi - 2.2913 \][/tex]

[tex]\[ \frac{2 \pi}{7} t \approx 3.9919 \text{ radians} \][/tex]

Solve for \( t \):

[tex]\[ t = \frac{7}{2 \pi} \cdot 3.9919 \][/tex]

Calculate \( t \):

[tex]\[ t \approx 4.71 \text{ years} \][/tex]

So, the solutions to the given problem are \( t = 2.29 \) years and \( t = 4.71 \) years.

Thus, the population will be 4800 at:

[tex]\[ t = 2.29 \text{ years} \][/tex]
or
[tex]\[ t = 4.71 \text{ years} \][/tex]
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