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) After every year, our value increases by 10%
| $t$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|--------|------|---|---|---|---|---|---|---|
| $P(t)$ | 1000 |1100 | 1210 | 1331 | 1464.1 | 1610.5 | 1771.6 | 1948.7 |

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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) After inputing these values in Desmos, the formula generated is\\[
P(t)= 1000.11\cdot 1.09999^{x_1}+(-0.109203)
\\]

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(c\) What will the population be after 100 years under this model?
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(c\)Since we found a formula for P(t), x=100yrs and we plug it into the formula.\\[
P(t)= 1000.11\cdot 1.09999^{x_1}+(-0.109203)
\\]\\[
P(100)= 1000.11\cdot 1.09999^{100}+(-0.109203)
\\]\\[
P(100)= 13769604.53 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)The formula for central difference for f'(x) is\\[
f'\left(x\right)=\ \frac{f\left(x+h\right)-f\left(x-h\right)}{2h}
\\]\\[
f'\left(1\right)=\ \frac{f\left(1+1\right)-f\left(1-1\right)}{2(1)}
\\]\\[
f'\left(1\right)=\ \frac{f\left(2\right)-f\left(0\right)}{2(1)}
\\]\\[
f'\left(1\right)=\ \frac{1210-1000}{2}
\\]\\[
f'\left(1\right)=105 people
\\]\\[
f'\left(2\right)=115.5 people
\\]\\[
f'\left(3\right)=127.1 people
\\]\\[
f'\left(4\right)=139.8 people
\\]\\[
f'\left(5\right)=153.8 people
\\]\\[
f'\left(6\right)=169.1 people
\\]
P'(5) indicates the rate at which the population is increasing at the end of the fifth year is 153.8 people.

| $t$ | 1 | 2 | 3 | 4 | 5 | 6 |
|--- |---|---|---|---|---|---|
| $P'(t)$ |105 | 115.5 | 127.1 | 139.8 | 153.8 | 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) The formula for central difference for f''(x) is\\[
f''\left(x\right)=\ \frac{f\left(x+h\right)-2f(x)+f\left(x-h\right)}{h^2}
\\]\\[
f''\left(3\right)=\ \frac{f\left(3+1\right)-2f(3)+f\left(3-1\right)}{1^2}
\\]\\[
f''\left(3\right)=\ f\left(4\right)-2f(3)+f\left(2\right)
\\]\\[
f''\left(3\right)=\ 12.1 people
\\]

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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) K=1.1 This can be confirmed with the constant change of 1.1 in P(t) values when multiplied that in turn makes P'(t) and P(t) multiples of each other.

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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) W=The values Desmos generated according to the formula is \\[
y_1\sim 0.025x_1^2+-0.50x_1+10
\\]\\[D(x)=0.025x^2+-0.50x+10\\]

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(b) Find the proper dosage for a 128 lb individual.
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(b) Use the formula we found from Desmos and replace x= 128lb.\\[D(x)=0.025x^2+-0.50x+10\\]\\[D(128)=0.025(128)^2+-0.50(128)+10\\]\\[D(128)=355.6mg\\]

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(c\) What is the interpretation of the value $D'(128)$.
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(c\)D'(128) indicates the rate at which the dosage is increasing is 5.9mg for someone who is 128lb.

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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)I decided to use central difference to estimate my values just as I did with the problems before. Central difference formula is\\[
f'\left(x\right)=\ \frac{f\left(x+h\right)-f\left(x-h\right)}{2h}
\\]\\[
f'\left(128\right)=\ \frac{f\left(128+1\right)-f\left(128-1\right)}{2(1)}
\\]\\[
f'\left(128\right)=\ \frac{f\left(129\right)-f\left(127\right)}{2}
\\]\\[
f'\left(128\right)=\ \frac{361.5-349.7}{2}
\\]\\[
f'\left(128\right)=5.9mg\\]

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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)The function we have found with Desmos was \\[D(x)=0.025x^2+-0.50x+10\\] We also have to find the derivative of the function which is \\[D'(x)=0.05x+-0.5\\] We are asked to find the eqaution of the tangent line where x=130lbs. So we plug our values into D(x).\\[D(130)=0.025(130)^2+-0.50(130)+10\\]\\[D(130)=417.1mg\] So, the point is at (130,417.1) These values we use to plug into the derivative to find the slope. \[D'(x)=0.05x+-0.5\\]\\[D'(130)=0.05(130)+-0.5\\]\\[D'(130)=6.5mg\\] The slope is 6.5. Now, we use the point slope form to find the equation. \\[y-y_0=m(x-x_0)\\]\\[y-417.1=6.5(x-130)\\]\\[y=6.5x-427.9\\]

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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 use the same point slope form eqaution\\[y=6.5x-427.9\\] So we plug 128 into our x-value\\[y=6.5(128)-427.9\\]\\[y=404.1mg\\] The output value gives a fair estimate to the actual value of x=128. The value is off by 48.5mg from the actual of 355.6mg.

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