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Exponential models

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Bacteria multiplying

Suppose we have a population of bacteria that doubles every day. For example, each divided in half everyday.

Let $y$ be the number of bacteria, $x$ is time in days. Then $$ y = f(x). $$ Re-write the above $$\underbrace{f(x + 1)}_{\text{population: at time } x+1} = \underbrace{2f(x)}_{\text{ at time } x}$$ To know $f(x)$ for all $x$, we need to know $f(0)$. But this isn't good for calculus since incremental functions aren't even continuous. So we create a continuous model: the rate of growth is proportional to the size of population. What does this mean?

$$\begin{alignat}{3} & \text{time, } x &\quad &\text{time, } x+1 & & \\ y & = 10 & \quad y &= 20 &\quad \Delta y &= 10 \\ y & =100 & \quad y&= 200 &\quad \Delta y &= 100 \end{alignat}$$ If triples: $$\begin{alignat}{3} & \text{time, } x &\quad &\text{time, } x+1 & & \\ y & = 10 & \quad y &= 30 &\quad \Delta y &= 20 \\ y & =100 & \quad y&= 300 &\quad \Delta y &= 200 \end{alignat}$$

We can restate this in calculus terms: the derivative of $y$ is proportional to $y$. Or: $$\frac{dy}{dx} = ky.$$ This is called a differential equation. Here $k$ is the growth rate.

This is the solution: $$y = Ce^{kx}$$ for any $C$.

Verify: $$\begin{aligned} \frac{dy}{dx} &= (Ce^{kx})^{\prime} \\ \text{CMR } &= C(e^{kx})^{\prime} \end{aligned}$$

Right Hand Side: $$\begin{aligned} \text{CR } & = C \cdot e^{kx} \cdot (kx)^{\prime} \\ & = Ce^{kx}\cdot k \end{aligned}$$

Left Hand Side: $$ky = k \cdot e^{kx}$$

What is $C$? Given $$ y = Ce^{kx} $$ subsitute $x=0$. Then $$y(0) = Ce^{k\cdot 0} = ce^{0} = C $$ So, $C$ is the initial population. Re-phrase our solution: $$y(x) = y(0)e^{kx} $$

This is what we deal with mathematically:

Growth 
Know if $k > 0$, then this function is increasing and $\lim\limits_{x \to \infty} y(x) = \infty $.
Decay 
if $k < 0$, the this function is decreasing and $\lim\limits_{x \to \infty} y(x) = 0$.
Graphs showing exponential growth and decay rates.

Until this point, this is all algebra, not calculus.

Population Growth

Population grows by 10% a year. How long will it take to double? $$ y = Ce^{kx}, \quad x \text{ time }, \quad k=? $$ look at the algebra: assume $$y(0) = 1$$ then $$y(1) = 1.1$$ Substitute these into the model: $$\begin{aligned} Ce^{k\cdot 0} & = 1 \to C = 1 \\ Ce^{k\cdot 1} & = 1.1 \to e^{k} = 1\\ \to k &= \ln 1.1 \end{aligned} $$

To answer the question, we are looking for $x$, time, satisfying $$\begin{aligned} y(x) & = 2, \quad \text{ substitute} \\ e^{kx} & = 2, \quad \text{ solve } \\ kx & = \ln 2 \\ x &= \frac{\ln 2}{k} = \frac{\ln 2}{\ln 1.1} \end{aligned} $$

Radioactive decay and radiocarbon dating

Once the tree is but carbon start to decay, loses half of its mass over a certain period of time. The loss follows the exponential decay model: $$Ce^{k\cdot x}.$$

Percentage of this elements, $^{14}\text{C}$

Idea:

  1. Know the element's decay constant, $k$.
  2. Measure % of the element present vs the amount normally present,
  3. $\Rightarrow$ Calculate the time when tree was cut.

Example. Half life is 5730 years (i.e., time takes to go from 100% to 50%). Parchment has 74% of $^{14}\text{C}$ left. How old is it?

Estimate assumes that decay is linear.

Let's estimate first by assuming that decay is linear.

The period is close to $$\frac{1}{4}\text{-life} = \frac{1}{2}\frac{1}{2}\text{-life} = 2865.$$ (Actual age is younger.)

Now the exponential model. Given $y = Ce^{kx}$, find $k$.

Set $$\begin{aligned} y(0) = 1, y(5730) & = \frac{1}{2}, \text{ substitute } \\ e^{k\cdot 5730} & = \frac{1}{2}, \text{ solve for } k \\ k \cdot 5730 & = \ln \frac{1}{2} \\ k &= \frac{\ln \frac{1}{2}}{5730} \end{aligned} $$

Find $x$ such that, $y(x) = 0.74$, or $$\begin{aligned} e^{kx} & = 0.74 \\ kx &= \ln 0.74 \\ x & = \frac{\ln 0.74}{k} \\ &= \frac{\ln 0.74}{\ln 0.5}\cdot 0.5730 = 2455 \end{aligned}$$

Newton's Law of Cooling

The rate of cooling of an object is proportional to the difference between its temperature and the temperature of the atmosphere.

Let $T$ be the temperature and we assume that $T > T_{0}, where $T_{0}$ is the ambient temperature.

Previous discussion suggests the differential equation: $$ \frac{dT}{dx} = k( T - T_{0}).$$ Indeed, this is simply:

  • rate of change of $T$ is proportional to the difference between $T$ and $T_0$.

What about warming? Does this still apply? Yes. But first,

Question: What is the sign of $k$?

Answer: $k$ is negative.

Verify:

  1. Warmer object: $k< 0, T - t_{0} > 0$, so $\frac{dT}{dx} < 0$, so $T \searrow$: cooling.
  2. Cooler object: $k< 0, T - t_{0} < 0$, so $\frac{dT}{dx} > 0$, so $T \nearrow$: warming.

Next, what is $T$? The solution to this new differential equation?

Try this. What if $T_{0} = 0$, then $$ T = Ce^{kx}.$$ We know from before as the standard exponential model.

What if $T \neq 0$, then $$T = T_{0} + Ce^{kx} $$ We just shift the graph up by $T_{0}$.

Shift the graph up by $T_{0}$.

Exercise: Substitute to confirm.

Fact: Horizontal asymptote is $y = T_{0}$.

Review Exercise.

Composition of three functions.png

Differentiate $ y = \sin(\cos(\tan x ))$.

Flowchart: $$\begin{alignat}{4} y &= \sin (v) & \gets v & = \cos(u) & \gets u &= \tan(x) &\gets x \\ \frac{dy}{dv} & = \cos(v) & \frac{dv}{du} &= -\sin(u) & \frac{du}{dx} &= \sec^{2}x & \\ \end{alignat}$$ Multiply these (Chain Rule: substitute $v = \cos u$ and $u = \tan x$) $$\begin{aligned} \frac{dy}{dx} & = \cos(v)\left(-\sin(u)\right)\sec^{2}(x) \\ &= \cos(\cos(u)\cdot(-\sin(u)\cdot\sec^{2}x \\ &= -\cos\cos\tan x \cdot \sin\tan x \cdot \sec{2}x. \end{aligned}$$

Compounded interest

Suppose we have money in the bank at APR 10% compounded annually. Then after a year, given $\$1,000$ initial deposit, you have $$\begin{aligned} 1000 + 1000\cdot 0.10 &= 1000(1 + 0.1) \\ &= 1000 \cdot 1.1 \end{aligned}$$ Same every year. After $t$ years, it's $1000\cdot1.1^{t}$. What if it is compounded semi-annually, with the same APR? After $\frac{1}{2}$ year, $1000\cdot 0.05$, or total $$ 1000 + 1000\cdot 0.05 = 1000\cdot 1.05$$ after another $\frac{1}{2}$ year $$\left(1000\cdot 1.05\right)\cdot 1.05 = 1000 \cdot 1.05^{2} $$ After $t$ years $$ 1000\cdot (1.05^{2})^{t} = 1000\cdot 1.05^{2t} $$ Note: $1.05^{2} = 1.1025 > 1.1 $

Try compound quarterly, $$ 1000\cdot 1.025^{4t} $$ If compounded $n$ times, then $$ 1000 \cdot \left(1 + \frac{1}{n}\right)^{nt} $$ where $\frac{1}{n}$ is the interest in one period.

Generally, for APR $r$ (given as a decimal) and for the initial deposit $A_{0}$, after $t$ years, the current amount is $$A(t) = A_{0}\left(1 + \frac{r}{n} \right)^{nt},$$ if compounded $n$ times per year.

What if we compounded more and more, $n \to \infty$?

$$ \begin{aligned} \lim_{n \to \infty} A(t) & = \lim_{n \to \infty} A_{0} \left( 1 + \frac{r}{n} \right)^{nt} = \infty ? \text{ NO.}\\ \text{CMR} &= A_{0} \lim_{n \to \infty} \left( 1 + \frac{r}{n} \right)^{nt} \\ & = A_{0} \left( \lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^{n}\right)^{t} \\ & = A_{0} (e^{r})^{t} \end{aligned}$$ Note: $\lim\limits_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n} = e$.

With $\text{APR} = r$, initial deposit $A_{0}$, after $t$ years you have $$A(t) = A_{0} e^{rt}, $$ if interest is compounded continuously.

Example: APR 10%, $A_{0} = 1000$, $t = 1$. $$A(1)=1000\cdot e^{1.1} = 1000\cdot e^{0.1} \approx \$1,105 $$ interest $ \$1,105 > \$100 $ (annual).

Example: APR 100% $$ A(1) = 1000\cdot e = \$2,718 $$ annual: $ \$1000 < \$1,718 $.

Example: How long does it take to triple your money with APR=5%, compounded continuously? Plug in $A_{0} = 1$. Solve for $t$. $$\begin{aligned} 3 & = 1\cdot e^{0.05t} \\ \ln 3 &= 0.05t \\ t &= \frac{\ln e}{0.05} = 22 \text{ years.} \end{aligned}$$