
Sophie L.
asked 10/19/20Is there a time t , for 1<t<8 , at which the rate of change of the number of people at the venue changes from negative to positive? Give a reason for your answer.
"For time 0 ≤ t ≤ 8, people arrive at a venue for an outdoor concert at a rate modeled by the function A defined by A(t) = 0.3sin(1.9t) + 0.3cos(0.6t) + 1.3. For time 0 ≤ t ≤ 1, no one leaves the venue, and for time 1 ≤ t ≤ 8, people leave the venue at a rate modeled by the function L defined by L(t) = 0.2cos(1.9t) + 0.2sin (t) + 0.8. Both A(t) and L(t) are measured in hundreds of people per hour, and t is measured in hours. The number of people at the venue, in hundreds, at time t hours is given by P(t)."
1 Expert Answer
Janelle S. answered 10/19/20
Penn State Grad for ME, Math & Test Prep Tutoring (10+ yrs experience)
P(t) = A(t) - L(t)
= (0.3sin(1.9t) + 0.3cos(0.6t) + 1.3) - (0.2cos(1.9t) + 0.2sin(t) + 0.8)
= 0.3sin(1.9t) + 0.3cos(0.6t) - 0.2cos(1.9t) - 0.2sin(t) + 0.5
P'(t) = .57cos(1.9t) - .18sin(0.6t) + .38sin(1.9t) - 0.2cos(t)
Set the derivative equal to 0 to solve for inflection points which indicate a minimum or maximum:
P'(t) = .57cos(1.9t) - .18sin(0.6t) + .38sin(1.9t) - 0.2cos(t) = 0
Solving for when P'(t) = 0 on the interval of 1 < t < 8:
t = 2.783, 4.422, 6.176, 7.897
Plug a value between each inflection point into P'(t) to see if it is positive or negative. You'll find that P'(t) goes from negative to positive at:
t = 2.783, 6.176
At both those points, P'(t) is going from negative to positive which means that P(t) is going from decreasing to increasing and has a local minimum.
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Paul M.
10/19/20