Suppose that the population (in thousands) of a certain kind of insect after t months is given by the following formula.
P(t) = 3t + sin(4t) + 100
Determine the minimum and maximum population in the first 4 months.

Solution:

The question that we’re really asking is to find the absolute extrema of P(t) on the interval [0,4].


Step 1: Find the derivative
P'(t) = 3 + 4cos(4t)

Step 2: Isolate t
3 + 4cos(4t) = 0
4cos(4t) = -3
cos(4t) = -3/4
4t = arccos(-3/4)

4t = 2.4189 + 2πn, n = 0, ±1, ±2, ...  -> I get why it's 2.4189 and the degree is 138.59, but I don't get why it had + 2πn, n - 0, ±1, ±2, ... 
4t = 3.8643 + 2πn, n= 0, ±1, ±2, ... -> I don't know where 3.8643 came from and I want to know it's degree too.

t = 2.4189/4  + 2πn, n= 0, ±1, ±2, ...
t = 3.8643/4  + 2πn, n= 0, ±1, ±2, ...

t = 0.6047 + πn/2, n= 0, ±1, ±2, ...
t = 0.9661 + πn/2, n= 0, ±1, ±2, ...

 

Step 3: Find the critical points

at n = 0 -> t = 0.6047, t = 0.9661 -> i don't get why we had to use positive n=0,1,2 

at n = 1 -> t = 0.6047 + π/2 = 2.1755, t = 0.9661 + π/2 = 2.5369

at n = 2 -> t =  0.6047 + π = 3.7463, t = 0.9661 + π = 4.1077 -> because [0,4] we don't need this

 

Step 4: Plug critical points into the original function

P(t) = 3t + sin(4t) + 100
 

P(0) = 100
P(0.6047) = 102.4756
P(0.9661) = 102.2368
P(2.1755) = 107.1880
P(2.5369) = 106.9492
P(3.7463) = 111.9004
P(4) = 111.7121
* remember that P is in thousands

Answer:
Absolute Minimum Population is 100,000 which occurs at t = 0 or (0, 100)
Absolute Maximum Population is 111,900 which occurs at t = 3.7463 or (3.7463, 111.9004)


 

 



 

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A graph may help you. Remember that we are dealing in radians, not degrees for t. Also, it’s not clear whether we are only looking at integer values for t. You can see that there are maxima and minima in between months. The approx values shown on the graph will tell you which values of t are meaningful. So when using calculus you need to keep in mind what the graph looks like so that you know which value of t applies. You can also draw up a table of values for t and P(t). The value of t itself is not the issue: sin(4t) is what matters because it’s that value which is a component of P(t). You already know that t ranges between 0 and 4 (months).

The absolute minimum is at the start (t=0, P(0)=100). The absolute maximum is at t=3.746, P(3.746)=111.9. 3.746 is the value of t without having to add 2πn, in other words, n=0 for the whole range of t. You get a better idea of how sin(4t) contributes to P, if you plot Q(t)=3t+100 on the same graph, you will see how the sine wave is superimposed on what is otherwise a linear function. See below:

 

The sine function has a cycle of 2π radians (about 6.263), which corresponds to 4t, so the interval t=π/2 (1.571) indicates how often P(t) will cycle on the line Q(t)=3t+100. sinθ=0 when θ=0, π, 2π, 3π, etc., so 4t=0, π, 2π, 3π, etc., making intervals of π/4=0.785 between values of t where P(t) intersects with Q(t). And when θ=π/2, 5π/2, 9π/2, ... (t=π/8=0.393, 5π/8=1.963, 9π/8=3.534, ...) we have maxima, and when θ=3π/2, 7π/2, ... (t=1.178, 2.749, ...) we have minima.

Now, bring in the calculus: cos(4t)=-¾, 4t=2.419, t=0.605 for an extremum. To find out if it’s max or min we take the next derivative: -16sin(4t), which is negative when t=0.605, implying a maximum. The next maximum is ¼arccos(-¾)+π/2=2.1755, the next at ¼arccos(-¾)+π=3.7463, and so on. The P values for these t values can then be calculated.

Note that at t=0 we are in the middle of a cycle, and this is the absolute minimum on the interval t in [0,4], because P(0)=100. We would need t<0 in order to drop below 100. When is -16sin(4t)>0? sin(4t) must be negative. So we look for a solution of cos(4t)=-¾ such that sin(4t)<0. Since cos(4t)=cos(-4t)=-¾, t=-0.605 is also a solution, but we need t>0, so we move to the next minimum by adding π/2=0.966. The next minimum is 2.536, and so on.

This should explain in detail how to think this one out.

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