Please help with calculus problem

Question

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) + 0.2sint + 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).

a) At time t=2 , there are 6 hundred people at the venue. Write an equation for the locally linear approximation of P at t=2 , and use it to approximate the number of people at the venue at time t=2.5 .

b) Find P′′(5). Using correct units, interpret the meaning of P′′(5) in the context of the problem.

c) Is 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.

d) A rectangular “standing-only” section at the venue changes size as t increases in order to manage the flow of people. Let x represent the length, in feet, of the section, and let y represent the width, in feet, of the section. The length of the section is increasing at a rate of 6 feet per hour, and the width of the section is decreasing at a rate of 3 feet per hour. What is the rate of change of the area of the section with respect to time when x=16 and y=10 ? Indicate units of measure.

Answer 1

Part A

A locally linear approximation is calculated using a the value of P(t) at a given point and the derivative of P(t) at that point.

Near any given time t = a, the locally linear approximation gives that

P(t) is approximately equal to P(a) + P'(a)(t - a).

So near t = 2, P(t) is approximately P(2) + P'(2)(t - 2)

We know that P(2) = 6, but we still have to find P'(2)

From the problem, we know that P'(t) = A(t) - L(t), because the rate at which the number of people in the venue is changing (P'(t)) is the rate at which people are coming in (A(t)) minus the rate at which people are leaving (L(t)).

We are given A(t) and L(t), so all we have to do to find P'(2) is to evaluate A(2) - L(2), which yields (and you can check yourself) that P'(2) = 0.4015 (hundreds of people / hour)

So the equation for our linear approximation is P(t) ~= P(2) + P'(2)(t - 2)

so P(t) ~= 6 + (0.4015)(t - 2)

So P(2.5) ~= 6 + (0.4015)(0.5) = 6.20

Which means that P(2.5) is approximately 620 people.

Part B

For this part, the question asks us for P''(5). We first need to find P''(t) in general. We know from part a that P'(t) = A(t) - L(t)

Differentiating both sides, we get that

P''(t) = A'(t) - L'(t)

Evaluating this on the given functions A(t) and L(t), you should get (again you should check for yourself) that

P''(t) = 0.57cos(1.9t) - 0.18sin(0.6t) + 0.38sin(1.9t) - 0.2cos(t)

So P''(5) = 0.57cos(9.5) - 0.18sin(3) + 0.38sin(9.5) - 0.2cos(5) = -0.679 (hundreds of people / hour²)

In words this means that at t = 5 hours, the rate at which the number of people in the venue is changing is decreasing by 67.9 people per hour.

Part C

Assuming that you can use a calculator, the easiest way to do this part is by graphing. The "rate of change of the number of people at the venue" is P'(t). So what you would want to do is graph P'(t) and find where it changes from positive to negative.

Part D

This part of the question is actually a related rates problem.

It's useful to draw a rectangle with one side labeled x and one side labeled y.

We're eventually going to want to find the "rate of change of the area of the section with respect to time", which in math terms is dA/dt, which suggests we're going to need an expression for the area.

Because the section is a rectangle, we know that A = xy.

Differentiating with respect to time, we need to use the chain rule and the product rule, and we get (check for yourself) that

dA/dt = (dx/dt)y + (dy/dt)x

The problem tells us that the length x is increasing by 6 ft / hour, so (dx/dt) = 6 ft/hour

We are also given that the width y is decreasing by 3 ft / hour, so (dy/dt) = -3 ft/hour

Finally, the problem tells us to calculate dA/dt for when x = 16ft and y = 10ft. Putting this all together with our equation, we get that

dA/dt = (6 ft/hour)(10 ft) + (-3 ft/hour)(16 ft)

= 60 (ft²/hr) - 48 (ft²/hr)

= 12 (ft²/hr)

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Hopefully this was helpful! I tried to make a video response, but it didn't save. If any parts of the question were confusing, feel free to contact me. I'd also be happy to help with future calculus assignments and problems if you'd like!

-- Alex