$\newcommand{\dis}{\displaystyle} \newcommand{\m}{\hspace{1em}} \newcommand{\mm}{\hspace{2em}} \newcommand{\x}{\vspace*{1ex}} \newcommand{\xx}{\vspace*{2ex}} \let\limm\lim \renewcommand{\lim}{\dis\limm} \let\fracc\frac \renewcommand{\frac}{\dis\fracc} \let\summ\sum \renewcommand{\sum}{\dis\summ} \let\intt\int \renewcommand{\int}{\dis\intt} \newcommand{\sech}{\text{sech}} \newcommand{\csch}{\text{csch}} \newcommand{\Ln}{\text{Ln}} \newcommand{\p}{\partial} $

Lecture 14, 9/23/2022. This page is for Section 1 only.
ACMS 20550: Applied Mathematics Method I
Instructor: Bei Hu, b1hu@nd.edu, Hurley 174A

  1. Total Differentials
    1. Suppose $y = f(x)$, $\fcolorbox{white}{yellow}{$dy = f'(x) dx $}$ and $\fcolorbox{white}{yellow}{$\Delta y = f(x+\Delta x)- f(x) \approx f'(x) \Delta x $}$, $ \m dy \approx \Delta y$.
    2. Suppose $z = f(x,y)$, then
      $d z = \frac{\p z}{\p x} dx + \frac{\p z}{\p y} dy, \m dz \approx \Delta z $
  2. Examples: Approximation by differentials.
    Estimating $\sqrt{(2.98)^2+(4.03)^2}$ using differential.
    Sol. Step 1: Choose $z = f(x,y) = \sqrt{x^2+y^2}$
    $\m$ Step 2: $ \m x=3, \m y=4, \m dx = -0.02, \m dy = 0.03$.
    $\m $ Step 3: $\m dz = \frac{\p z}{\p x} dx + \frac{\p z}{\p y} dy = \frac{x}{\sqrt{x^2+y^2}} dx + \frac{y}{\sqrt{x^2+y^2}} dy = \frac35(-0.02)+\frac45(0.03) = 0.012$
    $\m $ Step 4: $\m \sqrt{(2.98)^2+(4.03)^2} = f(2.98, 4.03) = f(3,4)+\Delta z \approx f(3, 4) + dz = 5 + 0.012 = 5.012$.
  3. Chain rules
    If $z = f(x,y), \m x=x(t), \m y=y(t),\m$ then $\m \frac{dz}{dt} = \frac{\p z}{\p x} \frac{ dx}{dt} + \frac{\p z}{\p y}\frac{ dy}{dt}$
    . Compare this with 3(b)!
  4. Implicit differentiation: Differentiate using chain rule without solving the equation.
    Eg. Finding a tagent line of a curve with its equation given implicitly.