Math 181 Miniproject 2: Population and Dosage.md
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Math 181 Miniproject 2: Population and Dosage
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**Overview:** In this miniproject you will use technological tools to turn data and into models of real-world quantitative phenomena, then apply the principles of the derivative to them to extract information about how the quantitative relationship changes.

**Prerequisites:** Sections 1.1--1.6 in *Active Calculus*, specifically the concept of the derivative and how to construct estimates of the derivative using forward, backward and central differences. Also basic knowledge of how to use Desmos.

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1\. A settlement starts out with a population of 1000. Each year the population increases by $10\%$. Let $P(t)$ be the function that gives the population in the settlement after $t$ years.
(a) Find the missing values in the table below.
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(a)
| $t$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|--------|------|---|---|---|---|---|---|---|
| $P(t)$ |1000|1100| 1210|1331|1464|1611|1772|1949

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(b) Find a formula for $P(t)$. You can reason it out directly or you can have Desmos find it for you by creating the table of values above (using $x_1$ and $y_1$ as the column labels) and noting that the exponential growth of the data should be modeled using an exponential model of the form
\\[
y_1\sim a\cdot b^{x_1}+c
\\]
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(b) Using desmos we can find this and then we just input the values that are given in the table above this is following the formula:
P(t) = 994.072 + $1.10044^{t}$ + 6.08172

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(c\) What will the population be after 100 years under this model?
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(c\) So with this we can put in 100 into that equation that is above:
$P\left(t\right)=\ 994.072\ +\ 1.10044^{100}+\ 6.08172\ =15343$ people.

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(d) Use a central difference to estimate the values of $P'(t)$ in the table below. What is the interpretation of the value $P'(5)$?
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(d) We now use this formula and using the data from the table above.
$P'\left(t\right)=\frac{f\left(t+1\right)=f\left(t-1\right)}{2}$
Now again we input those values and solve it:
$P'\left(t\right)=\frac{1210-1000}{2}=105$
| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ | 105 |115.5|127.05|139.7|153.7|169.1

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(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
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(e) Now we use this formula for these:
$P''\left(x\right)=\frac{\ f\left(x+h\right)-2f\left(x\right)+f\left(x-h\right)}{h^{2}}$
x=3 is the varible we know and plug in the known values to get the P''(3)
$P''\ \left(3\right)=\frac{f\left(3+1\right)-2f\left(3\right)+3\left(3-1\right)}{1^{2}}=12.1$

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(f) **Cool Fact:** There is a constant $k$ such that $P'(t)=k\cdot P(t)$. In other words, $P$ and $P'$ are multiples of each other.

What is the value of $k$? (You could try creating a slider and playing with the graphs or you can try an algebraic approach.)
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(f) The value is 1.1

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2\. The dosage recommendations for a certain drug are based on weight.
| Weight (lbs)| 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 |
|--- |--- |--- |--- |--- |--- |--- |--- |--- |--- |
| Dosage (mg) | 10 | 30 | 70 | 130 | 210 | 310 | 430 | 570 | 730 |
(a) Find a function D(x) that approximates the dosage when you input the weight of the individual. (Make a table in Desmos using $x_1$ and $y_1$ as the column labels and you will see that the points seem to form a parabola. Use Desmos to find a model of the form
\\[
y_1\sim ax_1^2+bx_1+c
\\]
and define $D(x)=ax^2+bx+c$.)
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(a) I put a table into desmos and put those values in as well and it got a, b, c and it was set up in this equation:
$D\left(x\right)=0.025x^{2}+\left(0.5\right)x+10$

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(b) Find the proper dosage for a 128 lb individual.
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(b) To find this 128 lb we would have to put in 128 for x using that last equation:
$f\left(128\right)=0.025\left(128\right)^{2}+\left(0.5\right)\left(128\right)+10\ =\ 355.6\ mg$

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(c\) What is the interpretation of the value $D'(128)$.
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(c\) This would be for the interpretation:
D'(128) = 2.7 mg

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(d) Estimate the value of $D'(128)$ using viable techniques from our calculus class. Be sure to explain how you came up with your estimate.
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(d) We can use linear approximation:
$y=f\left(a\right)+f'\left(a\right)\left(x-a\right)$
Now with plugging everything in we can rewrite it into the correct way with the formula:
$y=\ D\left(a\right)+D'\left(a\right)\left(x-a\right)$
$310=D\left(128\right)+D'\left(128\right)\left(120-128\right)$
$D'\left(128\right)=\frac{310-356}{-8}=5.75$

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(e) Given the value $D'(130)=6$, find an equation of the tangent line to the curve $y=D(x)$ at the point where $x=130$ lbs.
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(e) Now what we can do is we can plug in 130 for that x
$D\left(130\right)=0.025\left(130\right)^{2}+\left(-0.5\right)\left(130\right)+10=367.5$
now following this: $y-y_{0}=m\left(x-x_{0}\right)$
we can use 367.5 aand 130 and plug it into that new formula:
$y-367.5=6x-780$
$y=6x-412$ and this is our final answer

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(f) Find the point on the tangent line in the previous part that has $x$-coordinate $x=128$. Does the output value on the tangent line for $x=128$ lbs give a good estimate for the dosage for a 128 lb individual?
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(f) We need to use this equation that waas used in the question previous and use 128:
y = 6 (128) - 412.5= 355.5 mg
Then this being said its a good enoguh estimate for thaat 128 lb we found just in a different way than paart b

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