Math in the Real World

Problem 1:
From the growth and decay of bacteria to the population of squirrels in your neighbor's backyard, functions can model all sorts of things. But what about when you want to know what happens a long time from now? How many squirrels can that giant oak tree really hold before the squirrels start looking elsewhere? How do we get there?
While we don't have all the answers, we know limits can help.
First, we need to know or be refreshed about one our favorite equations: the equation for logistic growth.

We like it for what it can do, not necessarily how it looks. All those letters stand for:
y = the population of whatever we might be analyzing

L = the carrying capacity

k = a constant
t = time
Okay, we admit that's a lot to take in and a little bit confusing to boot. Let's give you some context to help things along.
Suppose we had a population of Emperor Penguins in Antarctica, which we learn, after careful study, can be modeled by the equation:

The question here isn't so much how many penguins are hanging around right now or even how many there will be in 10 years. After all, that just becomes a simple substitution problem. The real question we'd like to answer is will the penguin population grow without bound or level off?
This sounds an awful lot like end behavior to us. So lets take the limit as t approaches infinity.

This is kind of a tricky limit to evaluate. Lucky for us, we learned some sweet properties of limits that allow us to think about this thing as:

Since e raised to a giant negative number definitely approaches zero, we know this limit approaches 400,000 Emperor Penguins. Good thing too, because the carrying capacity is actually defined as the maximum population that the model can sustain.https://www.shmoop.com/precalculus-limits/real-world.html
This problem uses the equation for logistical growth and uses it to model the population of Emperor Penguins. Then it uses the limit of the problem to find out the number of Emperor Penguins Antarctica can support. I chose this problem because we just learned about limits in class and I was curious to see how using limits is applied to the real world. This is useful if you are looking to repopulate a species since so many species are becoming endangered.

Problem 2:
Basketball is one sport that can easily be related to the different concepts of integral calculus. One of the most important parts of the game of basketball, shooting the ball, can be described through arc length equations.
 The path taken by a basketball when shot can be split into two components, the horizontal (x) direction and the vertical (y) direction. These two components can be represented by the parametric equations:   t vxtx o o  cos
    2 2 1 sin gt tvyty oo   

The variables are considered to be;
xo is the initial horizontal position of the basketball.
yo is the initial vertical position of the basketball. vo is the initial velocity of the basketball.  is the angle the ball is projected with respect to the x-axis. g is the acceleration due to gravity, -9.81 m/s2.
t is the time traveled.
 In order to fully understand the path of a basketball, one must consider the situation of a person shooting a basketball. Imagine a basketball court as a coordinate plane system where the shooters feet rest at the origin, (0,0), and the basketball hoop is located a distance, d, away from the shooter. A regulation hoop stands 10 feet high or 3.05 meters high, therefore the final destination of the basketball is the point (d,3.05). Using my height of 6 feet 4 inches and an average jump when shooting of 8 inches the release point of the ball would be 7 feet or 2.13 meters in the air, point (0,2.13). Inputting the data into the two parametric equations the equations change to:   t vtx o  cos    2 905.4sin13.2 t tvty o  

 With the new revised equations, the distance the basketball travels can be found using the arc length equation,                 tdt dt dy dt dx L , 22 . The derivatives of x(t) and y(t) with respect to time t are:    cosov dt dx    tv dt dy o 81 .9sin  

Therefore
           22 81.9sincos t vvL o o =                      dt tvtvv ooo       22222 24.96sin**62.19sincos =   dt tvtv oo      22 24.96sin**62.19
 Using the average velocity of a basketball shot, 2.24 m/s, the shot angle that would produce maximum efficiency, 45 degrees, and the time it would take the ball to travel from the free throw line, about 2 s, the arc length can be calculated.
L=   dt tt  2 0 22 24.9645sin24.2(**62.1924.2 = 17.34 https://math.la.asu.edu/~nbrewer/spring2010/honorsprojects/jordan_walker_mat266.pdf
This problem uses distance, velocity, and arc length equations to find the best and most accurate way to shoot a basketball. First the problem considers all the factors of shooting a basketball like the height of the basket, the height of the person, and the average jump height of the person taking the shot. Then the arc length equation is used to find the distance the ball will travel. To find this answer you use the average velocity of the shot, shot angle, and the time the ball would take to reach the hoop. I chose this problem because I play basketball and found it interesting how calculus is applied to something I actually do.