Question

1. The rate at which people enter a movie theater on a given day is modeled by the function S defined by S(t) = 80-12cos The rate at which people leave the same movie theater is modeled by the function R 7 5 defined by R(t) = 12e/o +20. Both S(1) and R(1) are measured in people per hour and these functions are valid for 10(Solved A, but need the explanations for B to D)
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Answer 1

The given rates indicates the number of people that enter or leave the

theater in a unit time.

Responses (approximate values):

a. 899 people

b. 139 people

c. Approximately 20 people are leaving per unit time at t = 20

d. The number of people in the theater are is a maximum at t ≈ 17.776

How is a function integrated, and what is rate of change?

$S(t) = \mathbf{80 - 12 \cdot cos \left(\dfrac{t}{5} \right)}$

$R(t) = \mathbf{12 \cdot e^{t/10} + 20}$

a. The number of people that have entered the movie theater at t = 20

seconds given to the nearest whole number (by rounding down) are as

follows;

$\displaystyle \mathbf{\int\limits^{20}_{10} 80 - 12 \cdot cos \left(\dfrac{t}{5} \right) \, dt }= 899\pi$

• 899 people have entered at t = 20

b. The number of people in the theater, at time, t, P(t) given to the nearest whole number at t = 20 are found as follows;

$Number \ in \ the \ theater, \ P(t) = \displaystyle \mathbf{\int\limits^{t}_{10} \left( S(t) - R(t) \right)\, dt}$

Which gives;

$\displaystyle P(20) = \mathbf{\int\limits^{20}_{10} \left(80 - 12 \cdot cos \left(\dfrac{t}{5} \right) \right) - \left(12 \cdot e^{t/10} + 20 \right) dt} \approx 139$

• There are approximately 139 people in the theater at t = 20

c. The value of P'(t), is found as follows;

$P(t) = \displaystyle \mathbf{\int\limits^{t}_{10} \left( S(t) - R(t) \right)\, dt}$

Therefore;

$\displaystyle P(t) = \mathbf{ \int\limits^{t}_{10} \left(80 - 12 \cdot cos \left(\dfrac{t}{5} \right) \right) - \left(12 \cdot e^{t/10} + 20 \right) dt}$

Which gives;

$P(20) = \left(60 \cdot sin(2) + 120 \cdot e + 60 \cdot t - 600\right) - 120 \cdot e^{\dfrac{t}{10} } + 60 \cdot sin \left(\dfrac{t}{5} \right)$

By differentiation, we have;

$P'(t) = \mathbf{ -\left(120 \cdot e^{\dfrac{t}{10} } \cdot \dfrac{10}{100} +60 \cdot cos \left(\dfrac{t}{5} \right) \cdot \dfrac{5}{25} -60 \right)}$

Which gives;

$P'(20) = -\left(120 \cdot e^{\dfrac{20}{10} } \cdot \dfrac{10}{100} +60 \cdot cos \left(\dfrac{20}{5} \right) \cdot \dfrac{5}{25} -60 \right) \approx -20.825$

• P'(20) indicates that at t = 20, approximately 20 people are leaving the movie theater per unit time

d. The time at which the maximum number of people are in the theater is given as follows;

At the maximum number in the theater, P'(t) = 0

Therefore;

$P'(t) =0 = \mathbf{-\left(120 \cdot e^{\dfrac{t}{10} } \cdot \dfrac{10}{100} +60 \cdot cos \left(\dfrac{t}{5} \right) \cdot \dfrac{5}{25} -60 \right)}$

Which gives;

$\mathbf{120 \cdot e^{\dfrac{t}{10} } \cdot \dfrac{10}{100} +60 \cdot cos \left(\dfrac{t}{5} \right) \cdot \dfrac{5}{25}} = 60$

By using a graphing calculator, we get;

• t ≈ 17.776

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