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title: 'Math 181 Miniproject 2: Population and Dosage'

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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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:::info
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.
:::

(a)
| $t$    | 0    | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|--------|------|---|---|---|---|---|---|---|
| $P(t)$ | 1000 |  1100 |1210   | 1331  | 1464.1 	|  1610.5	|1771.561	|   1958.7171|





:::info
(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
\\]
:::

(b)The formula for P(t) will be 

P(t)=$1000(1+10/100)^t+0$

P(t)=$1000(1x1)^t+0$

P(t)=$1000(1x1)^t$




:::info
(c\) What will the population be after 100 years under this model?
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(c\) The population after 100 years would be 

P(100)=$1000(1x1)^100$

=$13780612.34$




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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)
| $t$     | 1 | 2 | 3 | 4 | 5 | 6 |
|---      |---|---|---|---|---|---|
| $P'(t)$ |  100 | 110  |121   | 133  | 146.6  |161 |

The interpretation of P'(5) is that in the 5th year the increase of population is at 146.6






:::info
(e) Use a central difference to estimate the values of $P''(3)$. What is the interpretation of this value?
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(e) we must differentiate P'(t) on both sides we get

P''(t)=$(1002.29)(1.09975)^t(log(1.09975))^2 -2.6115$

at t=3 the value is 

P''(3)=$(1002.29)(1.09975)^3 In(1.09975)^ -2.6115=9.79606$

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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)




:::success
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)





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(b) Find the proper dosage for a 128 lb individual.
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(b) The proper dosage for 128lb is

D(x)=$0.025(128)^2+-0.5(128)+10$

=$356$ lb of dosage





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(c\) What is the interpretation of the value $D'(128)$. 
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(c\) The interpretation of the value D'(128) is

D'(x)=$0.05x-0.5$

D'(128)=$0.05x128-0.5$

D'(128)=5.9

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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.
:::


(d)To estimate the value of D'(128) you have to hse the linear approximation formula

Formula: y=f(a)+f'(a)(x-a)

From the table we use values x=120 then corresponding value of y=310 and a=128 since we have to find D'(128)

We can write the formula as 

y=D(a)+D'(a)(x-a)

$310=D(128)+D'(128)(120-128)$

$310=356+D'(128)(-8)$

$D'(128)=310-356/(-8)=5.75$


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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)Given, D'(130)=6

We have to find the equation of tangent where x=130lbs

$y130=D(130)=0.025(130)^2-0.5(130)+10$

$y130=D(130)=367.5$

evaluation of tangent line can be given as: y-y0=m(x-x0)

$y-367.5=G(x-130)$

$y-367.5=6x-780$

$Gx-y=412.5$

That's the equation of a tangent line


:::success
(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) From the previous equation we have tangent line equation

$Gx-y=412.5$

at x=128lbs

$Y=Gx-412.5$

$y=6x128-412.5$

$y=355.5$

This is an estimate as dosage 128.








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