CALC OK: The rate people enter a stadium for a baseball game per hour is given byE(t) = -500 cos(0.9t + 1) + 400for 0

Name: CALC OK: The rate people enter a stadium for a baseball game per hour
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Description: CALC OK: The rate people enter a stadium for a baseball game per hour is given byE(t) = -500 cos(0.9t + 1) + 400for 0

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Algebra & Trigonometry with Analytic Geometry

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ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.2: Trigonometric Functions Of Angles

Problem 32E

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CALC OK: The rate people enter a stadium for a baseball game per hour is given by
E(t) = -500 cos(0.9t + 1) + 400
for 0 <t< 5 hours where t = 0 corresponds to 5:00PM. At t = 0, there are already 350 people in the
stadium. Starting at t = 2, the rate people begin to leave the stadium per hour is given by
L(t) = 380e0.6t-2 – 150 for 2 < t < 6 hours.
a) How many people enter the stadium from 5:00PM to 10:00PM? Show your work.
b) At t=4, is the number of people in the stadium increasing or decreasing? Provide numerical
evidence of your answer.
c) Write, but do not solve, an integral equation that gives the number of people, P(t), in the
stadium at t=5.
d) At what time, t, is the number of people in the stadium at a maximum during the interval
2sts5 hours? Justify your answer.
Posted Mon Feb 22, 2021 at 2:05 pm

Expert Solution

Step 1

$(a)$

The rate at which people enter in the stadium for a baseball game per hour is

$E (t) = - 500 \cos (0.9 t + 1) + 400, 0 \leq t \leq 5$

Number of people in the stadium from 5:00 PM to 10:00 PM is given by integrating both sides

$\int E (t) = \int_{0}^{5} [- 500 \cos (0.9 t + 1) + 400] d t E_{t} = \frac{- 500}{0.9} \times \sin {(0.9 t + 1)}_{0}^{5} + {[400 t]}_{0}^{5} = \frac{- 5000}{9} [\sin 5.5 - \sin 1] + 2000 = \frac{- 5000}{9} [- 0.7055 - 0.8414] + 2000 = 860 + 2000 = 2860$

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