1.1 Self-Introductions

Who are you?

Speed round: (4 questions < 3min)

• Your name and how you would like to be called
• Where are you from?
• What brings you to Chiang Mai?
• Where and what do you want to be in 5 years time?
• How enjoyable has your acquisition of math skills been?

1.1.1 Attitudes towards Calculus

1.2 Your Instructor

linkedin

email: robert_b@payap.ac.th
Office: PC314 (Office hrs by appointment

1.3 Signing Up for the Class Resources

• Canvas LMS: https://canvas.instructure.com/enroll/RWWR4A

• Google Classroom: https://classroom.google.com/c/NTM3NzI0MDgxMjUx?cjc=tcitkp5

• Slides: https://rpubs.com/rbatzing/

1.4 Course description

Definition and properties of the limit, rate-of-change and the derivative, differentiation laws, development of the definite integral as area under the curve, the indefinite integral and the Fundamental Theorem of Calculus, integrals of standard functions, techniques of integration, applications of integration, simple differential equations, introduction to partial differentiation and multi-variable integration.

1.5 Learning Objectives of this Course

By the end of the course, students should be able to:

2.1 Case 1: Autonomous Quadracopters

• Agile flying robots TED 2014
• Making athletics quadracopters

Flight state: x,y,z,yaw,spin1,spin2,spin3,spin4

Flight path: Jerk, Snap, Acceleration, Velocity, Distance and Time

\(\begin{eqnarray} J_{t} &=& J_{(t-1)} + S_{t} dt\cr A_{t} &=& A_{(t-1)} + J_{t} dt\cr V_{t} &=& V_{(t-1)} + A_{t} dt\cr D_{t} &=& D_{(t-1)} + V_{t} dt\cr \end{eqnarray}\)

2.2 Case 2: Modeling snowfall in Dec 2010

Snowfall in Syracuse, Dec 3-10

Snowfall in Syracuse, Dec 10-18

2.2.1 Mathematical modeling of snowfall

\[y = \frac{1}{a}\left(\left(x-3\right)\left(x-11\right)\left(x-18\right)\right)^4\]

2.2.2 Graphing and Modelling

2.2.3 Minimizing Error

2.3 Case 3: Chirping crickets as thermometers

• Chirping crickets

• Data:

2.3.1 Modeling

3.1 Origins of Algebra

3.2 Properties of the equality \(a=b\)

3.3 Some useful properties

3.4 Strategy to solving equations

1. Using properties of equality, migrate variables to one side and constants to the other
2. The Simplify left and right sides of the equation
3. Rearrange the terms
4. Apply operations to both sides
5. Reduce the terms to simple terms

3.4.1 Example of an Algebratic expression

\[\frac{x+2}{4} = \frac{x-1}{3} +2\]

3.4.2 Solution: Part 1

\[\eqalign{\frac{x+2}{4} \color{red}{{- \frac{x-1}{3}}} &=& \color{yellow}{\frac{x-1}{3}} + 2 \color{red}{- \frac{x-1}{3}}\cr \color{red}{\left(\frac{3}{3}\right)}\left(\frac{x+2}{\color{yellow}{4}}\right) - \color{red}{\left(\frac{4}{4}\right)}\left(\frac{x-1}{\color{yellow}{3}}\right) &=& 2\cr \color{red}{12}\left(\frac{3x +6}{\color{yellow}{12}} - \frac{4x-4}{\color{yellow}{12}}\right) &=& 2 \color{red}{\times 12}}\]

3.4.3 Solution: Part 2

3.4.4 Challenges: Solve for \(x\)

\[2(x-3) - 17 = 13 - 3(x+2)\]

\[\frac{3}{x+6} + \frac{1}{x-2} = \frac{4}{x^2 + 4x -12}\] \[\frac{1}{x} + \frac{1}{q} = \frac{1}{f}\]

\[(x^2 -25)(x + 5) = 11(x-5)\]

3.5 Multiplication of polynomials

\(\eqalign{(ax+b)(cx+d) &=& (acx^2 + adx) + bcx + bd)\\ &=& acx^2 + (ad+bc)x + bd}\)

\[(x+3)(x-2) = (x^x -2x) + (3x - 6)= x^2 + x -6\]

\[(x+3)(x+3) = (x^x +3x) + (3x + 9)= x^2 + 6x + 9\] \[(x+3)(x-3) = (x^x -3x) + (3x - 9)=x^2 - 9\]

3.6 Graphically Factoring polynomials

Given this graph of an unknown relationship what can we say about its root factors and its behavior?

3.6.1 Effects of order of the function

3.6.2 Functions that share root factors

3.7 Solving systems of equations

3.8 Quadratic formula

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

* For \(x^2-25 = 0\) : \((x+5)(x-5)\)

3.8.1 Maximum/Minimum values of a Quadratic

\(\frac{-b}{2a} = \cases{x_{(y_{max})}, & if a < 0\cr x_{(y_{min})}, & if a > 0\cr}\)

3.9 Multiple variables in a linear system

\(\begin{eqnarray} 2x & + y & + z &=& 5\cr 4x &- 6y & &=& -2\cr -2x &+ 7y & + 2z &=& 9 \cr \end{eqnarray}\)

• Matrix representation

\(\left[\begin{matrix} 2 & 1M& 1\cr 4&-6& 0\cr -2& 7& 2&\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} 5\cr -2\cr 9 \cr \end{matrix}\right]\)

3.9.1 Solution by matrix rearrangments (2)

• Gaussian elimination of first variable

\(\left[ \begin{matrix} 1 & \frac{1}{2} & \frac{1}{2}\cr 4-2\times2 &-6-2\times1 & 0-2\times1\cr -2+2& 7+1& 2+1\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} \frac{5}{2}\cr -2-2\times5\cr 9 + 5\cr \end{matrix}\right]\)

• Rearrange

\(\left[ \begin{matrix} 1 & \frac{1}{2} & \frac{1}{2}\cr 0 &-8 & -2\cr 0& 8& 3\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} \frac{5}{2}\cr -12\cr 14\cr \end{matrix}\right]\)

3.9.2 Solution by matrix rearrangments (3)

• Gaussian elimination of second variable

\(\left[ \begin{matrix} 1 & \frac{1}{2} & \frac{1}{2}\cr 0 &\frac{-8}{-8} & \frac{-2}{-8}\cr 0& 8-8& 3-2\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} \frac{5}{2}\cr \frac{-12}{-8}\cr 14-12\cr \end{matrix}\right]\)

• Rearrange

\(\left[ \begin{matrix} 1 & \frac{1}{2} & \frac{1}{2}\cr 0 & 1 & \frac{1}{4}\cr 0& 0& 1\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} \frac{5}{2}\cr \frac{3}{2}\cr 2\cr \end{matrix}\right]\)

3.9.3 Solution by matrix rearrangments (4)

• Back substitute the third variable \(\left[ \begin{matrix} 1 & \frac{1}{2} & \frac{1}{2}-\frac{1}{2}\cr 0 & 1 & \frac{1}{4}-\frac{1}{4}\cr 0& 0& 1\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} \frac{5}{2} -\frac{2}{2}\cr \frac{3}{2}-\frac{2}{4}\cr 2\cr \end{matrix}\right]\)

• Rearrange

\(\left[ \begin{matrix} 1 & \frac{1}{2} & 0 \cr 0 & 1 & 0 \cr 0& 0& 1\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} \frac{3}{2}\cr 1 \cr 2\cr \end{matrix}\right]\)

3.9.4 Solution by matrix rearrangments (5)

• Back substitution of 2nd variable

\(\left[ \begin{matrix} 1 & \frac{1}{2}-\frac{1}{2} & 0 \cr 0 & 1 & 0 \cr 0& 0& 1\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right]= \left[\begin{matrix} \frac{3}{2} - \frac{1}{2} \cr 1 \cr 2\cr \end{matrix}\right]\)

• Rearrange

\(\left[ \begin{matrix} 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0& 0& 1\cr\end{matrix}\right] \left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} 1 \cr 1 \cr 2\cr \end{matrix}\right]\)

• Solution

\(\left[ \begin{matrix} x \cr y \cr z \cr\end{matrix}\right] = \left[\begin{matrix} 1 \cr 1 \cr 2\cr \end{matrix}\right]\)

3.9.5 Solution by matrix rearrangments (6)

• Validation

\[\left[\begin{matrix} 2 & 1 & 1\cr 4&-6& 0\cr -2& 7& 2&\cr\end{matrix}\right] \left[\begin{matrix} 1 \cr 1 \cr 2\cr \end{matrix}\right] = \left[\begin{matrix} 2\times 1 + 1\times 1 +1\times 2\cr 4 \times 1 + -6\times 1 + 0\times 2\cr -2\times 1+ 7\times 1 + 2\times 2 \cr\end{matrix}\right] =\]

\[\left[\begin{matrix} 2 + 1 + 2\cr 4 + -6 \cr -2 + 7 + 4\cr\end{matrix}\right] = \left[\begin{matrix} 5\cr -2\cr 9 \cr \end{matrix}\right]\]

3.10 Chemical Stoichiometry

\[C_6H_{12}O_6 + O_2 \longrightarrow CO_2 + H_2O\]

Setup the Stoichiometry Table

3.10.1 Matrix version

\[\left[\begin{matrix} 1 & 0 & 0 & 0\cr 6 & 2 & -2 & -1\cr12 & 0 & 0 & -2\cr 6 & 0 & -2 & 0 \cr\end{matrix}\right]\left[\begin{matrix}x\cr y\cr z\cr w\cr\end{matrix}\right] = \left[\begin{matrix}1\cr0\cr0\cr0\cr\end{matrix}\right]\]

• Gaussian elimination of the first variable

\[\left[\begin{matrix} 1 & 0 & 0 & 0\cr 0 & 2 & -2 & -1\cr 0 & 0 & 0 & -2\cr 0 & 0 & -2 & 0 \cr\end{matrix}\right]\left[\begin{matrix}x\cr y\cr z\cr w\cr\end{matrix}\right] = \left[\begin{matrix}1\cr -6\cr -12\cr -6\cr\end{matrix}\right]\]

3.10.2 Stoichiometry

• Gaussian elimination of the second variable

\[\left[\begin{matrix} 1 & 0 & 0 & 0\cr 0 & 1 & -1 & -\frac{1}{2}\cr 0 & 0 & 0 & -2\cr 0 & 0 & -2 & 0 \cr\end{matrix}\right]\left[\begin{matrix}x\cr y\cr z\cr w\cr\end{matrix}\right] = \left[\begin{matrix}1\cr -3\cr -12\cr -6\cr\end{matrix}\right]\]

• Gaussian elimination of the 3rd and 4th variable

\[\left[\begin{matrix} 1 & 0 & 0 & 0\cr 0 & 1 & -1 & -\frac{1}{2}\cr0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1\cr \end{matrix}\right]\left[\begin{matrix}x\cr y\cr z\cr w\cr\end{matrix}\right] = \left[\begin{matrix}1\cr -3\cr 3\cr 6\cr\end{matrix}\right]\]

3.10.3 Stoichiometry

• Back substitution

\[\left[\begin{matrix} 1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0 \cr0 & 0 & 1 & 0 \cr 0 & 0 & 0 & 1\cr \end{matrix}\right]\left[\begin{matrix}x\cr y\cr z\cr w\cr\end{matrix}\right] = \left[\begin{matrix}1\cr 3\cr 3\cr 6\cr\end{matrix}\right]\]

• Solution

\[C_6H_{12}O_6 + 3 O_2 \longrightarrow 3 CO_2 + 6 H_2O\]

3.10.4 Stoichiometry

• Validation

\[\left[\begin{matrix} 1 & 0 & 0 & 0\cr 6 & 2 & -2 & -1\cr12 & 0 & 0 & -2\cr 6 & 0 & -2 & 0 \cr\end{matrix}\right]\left[\begin{matrix}1\cr 3\cr 3\cr 6\cr\end{matrix}\right] = \left[\begin{matrix}1\cr6+6-6-6\cr12-12\cr6-6\cr\end{matrix}\right]=\left[\begin{matrix}1\cr0\cr0\cr0\cr\end{matrix}\right]\]

3.10.5 Exercise:

\[IO_3^- + I^- + H^+ \longrightarrow I_3^- + H_2O\]

4.0.1 Identify the functions and non-functions

4.1 Simple Polynomials

4.1.1 Polynomials with Offsets

4.1.2 Polynomials with Offsets

5.1 Slide Rule

• Pickett Slide Rule simulator

• Make your own slide rule

The slide rule was a mechanical analog computer based on logrithms. It was invented in the 1600’2 to multiply and divide numbers and provide values for logarithmic, and trigonometric functions. Output was 3 significant decimal digits in scientific notation.

5.1.1 Multiplication on a slide rule

Multiplying with a slide rule

• Adding distance of \(a\) to that of \(b\) determine the product \(ab\)
• Place the 1 value of the slide C scale over the value of \(a\)
• Use the cursor to find the value of \(b\) on the same scale.
• Read the corresponding value on the fixed C scale.

5.1.2 Multiplication

5.1.3 Other operations

5.1.4 Converting between log and numbers

\[x = 10^{\log(x)}\]

displayLog <- function(x) {
  cat(
    sprintf("%8.5f %7.5f %8.5f\n",
            x,log10(x),10**log10(x)))
}
for (x in 1:10) { displayLog(x) }

 1.00000 0.00000  1.00000
 2.00000 0.30103  2.00000
 3.00000 0.47712  3.00000
 4.00000 0.60206  4.00000
 5.00000 0.69897  5.00000
 6.00000 0.77815  6.00000
 7.00000 0.84510  7.00000
 8.00000 0.90309  8.00000
 9.00000 0.95424  9.00000
10.00000 1.00000 10.00000

5.1.5 Converting between \(\log_{10}(x)\) and \(\log_n(x)\)

\[log_n(x) = \frac{\log_{10}(x)}{\log_{10}{(n)}}\]

logn <- function(n,x) {
  logx10 = log10(x)
  logn10 = log10(n)
  logxn = logx10 / logn10
  cat(
    sprintf(
      "%8.5f %7.5f %7.5f %7.5f %8.5f\n",
      x,logx10,logn10,logxn,n**logxn))
}
cat("x log10(x) log10(n) logN(x) n^(logN(x)\n")
for (x in 2:10) { logn(2,x) }

5.1.6 Output

 x log10(x) log10(n) log2(x) 2^(log2(x)

   2  0.3010  0.3010  1.0000  2.0000
   3  0.4771  0.3010  1.5850  3.0000
   4  0.6021  0.3010  2.0000  4.0000
   5  0.6990  0.3010  2.3219  5.0000
   6  0.7782  0.3010  2.5850  6.0000
   7  0.8451  0.3010  2.8074  7.0000
   8  0.9031  0.3010  3.0000  8.0000
   9  0.9542  0.3010  3.1699  9.0000
  10  1.0000  0.3010  3.3219 10.0000

6.1 Estimates of Pi

• Archimedes: \(\frac{22}{7}\)

• Zu Chongzhi: \(\frac{355}{113}\)

• Gregory-Leibniz Series:
  \(\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots\)

• Nilakantha Series:
  \(\pi= 3+ \frac{4}{2\times 3 \times 4}-\frac{4}{4\times 5 \times 6}+\frac{4}{6\times 7 \times 8}-\frac{4}{8\times 9 \times 10}+\ldots\)

7.0.1 Equivalences

$$\[\begin{eqnarray} 1 &=& sin(x)^2 + cos(x)^2\cr sin(x) &=&\sqrt{1-cos(x)^2}\cr \end{eqnarray}\]$$

7.0.2 Law of Sines

\(\frac{A}{sin(a)} = \frac{B}{sin(b)} = \frac{C}{sin(c)}\)