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Rates of Change.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=1cm]{geometry} \begin{document} {\large Rates of Change} \begin{align*} \text{Interdependent variables }&\text{on time:}\\ &\text{If $x=f(t)$ and $y=g(t)$ , then }\frac{dy}{dx}=\frac{g'(t)}{f'(t)}\\ &\text{Likewise, if $\frac{dy}{dx}$ is known, }g'(t)=\frac{dy}{dx}\cdot f'(t)\:,\:\:\text{and }f'(t)=\frac{g'(t)}{\frac{dy}{dx}}\:,\:\:\text{or in chain-rule form}\\ &\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\:,\quad\text{and}\quad\frac{dx}{dt}=\frac{dy}{dt}\Big/\frac{dy}{dx}\\ \\ \text{Initial Conditions:}\quad&\text{To determine a measurement at time $t$, you need the rate}\\ &\text{of change and a boundary condition, which is usually the}\\ &\text{\it initial condition \rm \--- the measurement when $t=t_0$.}\\ &\text{Given }\int f'(t)\:dt=f(t)+C\\ &\int_{t_0}^{t_1}f'(t)\:dt=\Big[f(t)\Big]^{t_1}_{t_0}=f(t_1)-f(t_0)\:,\quad C=-f(t_0)\text{ is the initial condition.}\\ &\boxed{f(T)=f(t_0)+\int_{t_0}^{T}f'(t)\:dt}\quad\text{gives the measurement at time $T$}\\ \\ &\text{In practice, for measurement $U(t)$ with rate of change $u(t)$, first}\\ &\text{integrate $u(t)$ and get the integral $I(t)$, so }U(t)=I(t)+C\:,\\ &\text{then determine $C$ by substituting the boundary condition $U(t_0)$ .}\\ &C=U(t_0)-I(t_0)\:,\quad\text{then finally derive the measurement:}\\ &U(t)=I(t)-I(t_0)+U(t_0)\\ \\ &\text{The boundary condition may be at $+\infty$ (e.g. when a portion of}\\ &\text{the measurement is a constant over time). In this case,}\\ &\int_{t_1}^{+\infty}f'(t)\:dt=\Big[f(t)\Big]^{+\infty}_{t_1}=f(+\infty)-f(t_1)\\ &f(T)=f(+\infty)-\int_{T}^{+\infty}f'(t)\:dt\qquad\text{gives the measurement at time $T$}\\ % % % %\text{Generally},\quad&f(x)=e^{k(x-x_1)}+y_1\qquad\left(f(x)=e^{-kx_1}\cdot e^{kx}+y_1,\text{ so }c=e^{-kx_1}\text{ and }C=y_1\right)\\ %\text{i.e.}\quad&\frac{d}{dx}e^x=e^x\quad\text{ is invariant under scalings, translations and reflections.}\\ \end{align*} \begin{align*} \text{Exponential growth }&\text{and decay:}\quad\boxed{\frac{d}{dt}f(t)=kf(t)\quad\Leftrightarrow\quad f(t)=f(0)\:e^{kt}\:,\quad\text{where }k,f(t)\in\mathbb{R}}\\ &\qquad\qquad\text{Note: When }t=0\:,\quad f(0)=f(0)\:e^0\:\:.\\ \\ f(t)\text{ is in }&\text{exponential growth when }k>0\quad\text{(as }\lim_{t\to+\infty}f(t)=\text{$+\infty$ when $c>0$, or $-\infty$ when $c<0$)}\\ \text{or }&\text{exponential decay when }k<0\quad\text{(as }\lim_{t\to+\infty}f(t)=0\text{)}\\ &\text{(When $k=0, f(t)=f(0)$ . i.e. the function is a constant \--- independent of time.)}\\ \\ &\text{This property means the rate of change is proportional to the function value, so}\\ &\boxed{y=y_0 e^{kt}}\quad\text{where $y=f(t)$ and $y_0=f(0)$}\\ \\ &\text{If $k$ is restricted to }\mathbb{+R}\text{, i.e. $k>0$, then exponential decay can be presented as}\\ &y=y_0 e^{-kt}\\ \\ \text{Natural Growth }&\text{and GPs:}\quad\text{For }t=0,1,2,\ldots\quad y_0,y_1,y_2,\ldots\text{ become }V_0,V_0 e^k,V_0 e^{2k},\ldots\\ &\text{So natural growth is a GP with }a=V_0\quad\text{and}\quad r=e^k\:\:.\\ \\ &\text{Also, the difference between any two consecutive terms (the gain or loss) is}\\ &V_0 e^{(n+1)k}-V_0 e^{nk}=V_0\:(e^k-1)\:e^{nk}\:,\\ &\text{which is also a GP, with }a=V_0\:(e^k-1)\quad\text{and}\quad r=e^k\:\:.\\ \\ &\text{The above partially explains why the derivative of an}\\ &\text{exponential function is proportional to itself.}\\ \end{align*} \end{document}