Alex S. answered • 11/28/20

Stanford Undergrad tutoring Math, Physics, and Languages

__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 / hour**^{2}**)**

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 (ft^{2}/hr) - 48 (ft^{2}/hr)

= **12 (ft**^{2}**/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