Exponentials [1.5]
Ahhh... remember!
Multiplication of exponentials: $$10^2*10^3=100*1,000=100,000 = 10^{(2+3)}=10^5.$$
Meaning of negative exponentials: $$2^{-3}=\frac{1}{2^3}=\frac18=0.125 .$$
Division of exponentials: (as long as they have the same base!) $$\frac{10^2}{ 10^3}= 100/1000=1/10=0.1=10^{2-3}=10^{-1}=\frac{1}{10}.$$
Exponential raised to a power: $$(3^2)^3=9^3=9\times9\times9=729 = 3^{2*3}=3^6.$$ and in particular... $$e^{kx}= (e^k)^x = (e^x)^k.$$
Useful in problem solving... The second raising "reverses" the effect of the first raising: $$(17.8^3)^{1/3}=17.8^{(3*1/3)}=17.8^1= 17.8.$$ so to solve this for $x$... $$\begineq x^4 &=13\\ (x^4)^{1/4}&=13^{1/4}\\ x &= 13^{0.25}=1.8988 \text{ using a calculator} \endeq$$
[any number]${}^{\color{red}0} =1; \ \ e^0=1; \ \ 17.8^0=1$.
But first...HW [1.1]
The most common issue was on Problem 5
The population of Washington DC grew from 1900 to 1950, stayed approx constant during the 1950's and decreased from about 1960 to 2005. Graph the population as a function of years since 1900.
So, the $x$-axis should run from 0 (=1900) to 105 (=2005)
Exponential functions [1.5]
Exponential compared to linear function
$$\nonumber \color{red}{f(x)=mx+b;}\ \ \ \color{blue}{g(x)=b\cdot m^x}$$
Linear function: $\color{red}{f(x)=mx+b}$
|
Exponential function: $\color{blue}{g(x)=b\cdot m^x}$
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Problem 6
A product costs \$80 today. How much will the product cost in $t$ days if the price is decreased by
- \$4 a day
- 5% per day
a) $c=80-4t$
b) $c=80(1-0.05)^t=80(0.95)^t$
Problem 20
The number of passengers using a railway fell from 190,205 to 174,989 during a 5-year period. Find the annual percentage decrease over this period.
$N(5)=N(0)m^5$, so
$$S(t)=1000(0.82)^t=1000(1-0.18)^t.$$
So, the multiplier, 0.82, is equal to $1-r$, with $r=-0.18$. This means that sales are g. decreasing by 18% per year.
...A problem like this has been on previous exams.
Determine whether each of the following tables of values could correspond to a linear function, an exponential function, or neither. Find formulas for the linear and exponential functions.
(a) Let's see... $\frac{y(1)}{y(0)} = \frac{12.7}{10.5} = 1.21 $, $\frac{y(2)}{y(1)}=\frac{18.9}{12.7}=1.49$. Oh, already the first two ratios are not the same. This is not an exponential function.
(b) Let's see... $\frac{s(0)}{s(-1)} = \frac{50.2}{30.12} = 0.60$, $\frac{s(1)}{s(0)}=\frac{18.072}{30.12}=0.60$
. Keep checking the remaining ratio...
$\frac{s(2))}{s(1)}=\frac{10.8432}{18.072}=0.60$. They are all the same, so (b) is an exponential function and the mulitiplier is $m=0.6$. An equation is:
$$s(t)=30.12\cdot m^t$$
(c) Each time $u$ increases by 2, $g(u)$ changes by -3, therefore (c) is a linear function, a line with a slope, $m=\frac{\Delta y}{\Delta x}=\frac{-3}{2}=-1.5.$ An equation is:
$$g(u)=27 -1.5u$$
1000 dollars, compounded annually (once a year) at 12% interest
$$A(t)=1000(1+0.12)^t=1000*1.12^t$$
1000 dollars, compounded quarterly (4 times a year) at 12% interest
$$A(t)=1000(1+0.12/4)^{4t}=1000[(1.03)^{4}]^t=1000[1.1255]^t$$
12.55% is the Annual Percentage Rate (APR) of this scheme
In general, compounding $n$ times a year with an interest rate of $r$
$$A(t)=A_0(1+r/n)^{nt}.$$
Continuous compounding:
$$A(t)=\lim_{n\to \infty} A_0(1+r/n)^{nt}=A_0e^{rt}$$
Where $e$ is Euler's number, an irrational number, approximately equal to 2.71828.
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$$\begineq \frac{N(5)}{N(0)}=\frac{174989}{190205}
=0.92&= m^5\\
0.92^{0.2}&=(m^5)^{0.2}\\
0.983461&=m\\
1-0.01654&=1+r
\endeq
$$
So, $r=-0.01654\approx-0.017\Rightarrow$ an average decrease of 1.7% per year.
ConcepTest problem
Sales at a company are changing according to the formula $$S(t) = 1000(0.82)^t,$$ where $S$ is sales in thousands of dollars and $t$ is measured in years. Sales at this company are:
$S(0)=1000$ means that in the starting year ($t=0$) sales were \$1000.
Problem 25
Compounding