The increment \(\Delta x\) represents the change of a variable \(x\) as it increases or decreases from one value \(x = x_0\) to value \(x = x_1\). The increment of variable \(x\) is calcualted as: $$\Delta x = x_1 - x_0. $$ If the variable \(x\) is given increment \(\Delta x\) and the variable \(x\) is the argument of the function \(y=f(x)\) then the increment of a function can be written in the following form $$ \Delta y = f(x_0 + \Delta x) - f(x_0).$$ The average rate of change of the function on the interval between \(x = x_0\) and \(x = x_0 + \Delta x\) is defined as the quotient $$ \frac{\Delta y}{\Delta x} = \frac{f(x_0 + \Delta x) - f(x_0)}{x_1 - x_0}.$$
The Derivative of a function \(y = f(x)\) with respect to \(x\) at the point \(x=x_0\) is limes of quotient \(\frac{\Delta y}{\Delta x}\) where \(\Delta x\) tends to zero. The derivative can be written as
$$ \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x_0 + \Delta x) - f(x_0)}{\Delta x}$$ This limit is also called instantaneous rate of change or simply rate of change of \(y\) with respect ot \(x\) at \(x=x_0.\)
- Example 1 - Calculate the increment of a function \(y = x^2\) which corresponds to increase from:
- \(x = 1\) to \(x_1 = 2\)
- \(x = 1\) to \(x_1 = 1.1\)
- \(x = 1\) to \(x_1 = 1 + h\)
- $$ \Delta x = 2-1 = 1$$ $$ \Delta y = f(x_1) - f(x) = 2^2 -1^2 = 4 - 1 = 3 $$
- $$ \Delta x = 1.1 - 1 = 0.1$$ $$\Delta y = f(x_1) - f(x) = 1.1^2 -1^2 =1.21 - 1 = 0.21 $$
- $$\Delta x = 1 + h - 1 = h$$ $$\Delta y = (1+h)^2 - 1^2 = 1+2h +h^2 - 1 = h(h+2) $$
- Example 2 - Calculate the increment of a function \(y = \sqrt[3]{x}\) which corresponds to increase from:
- \(x = 0\) with increment \(\Delta x = 0.001\)
- \(x = 8\) with increment \(\Delta x = -9\)
- \(x = a\) with increment \(\Delta x = h\)
Solution- $$\Delta x = x_1 - x \Rightarrow x_1 = \Delta x + x = 0.001 + 0 = 0.001 $$ $$\Delta y = f(x_1) - f(x) = \sqrt[3]{0.001} - \sqrt[3]{0} = 0.1 $$
- $$\Delta x = x_1 - x \Rightarrow x_1 = \Delta x + x = -9 + 8 = -1$$ $$ \Delta y = f(x_1) - f(x) = \sqrt[3]{-1} - \sqrt[3]{8} = -1 -2 = -3$$
- $$\Delta x = x_1 - x \Rightarrow x_1 = \Delta x + x = h + a$$ $$\Delta y = f(x_1) - f(x) = \sqrt[3]{h+a} - \sqrt[3]{a}$$
- Example 3 - Calculate the increment \(\Delta y\) and the quotient \frac{\Delta y}{\Delta x} for functions
- \(y = \frac{1}{(x^2 - 2)^2}\) for \(x = 1\) and \(\Delta x = 0.4\)
- \(y = \sqrt{x}\) for \(x = 0\) and \(\Delta x = 0.0001\)
- \(y = \log x\) for \(x = 100000\) and \(\Delta x = -90000\)
Solution- \begin{eqnarray} x_1 &=& \Delta x + x = 0.4 +1 = 1.4 \\\nonumber \Delta y &=& f(x_1) - f(x) = \frac{1}{(1.4^2-2)^2} - \frac{1}{(1^2 - 2)^2} \\\nonumber \Delta y &=& 624 \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{624}{0.4} = 1560\end{eqnarray}
- \begin{eqnarray} x_1 &=& \Delta x + x = 0.0001 + 0 = 0.0001 \\\nonumber \Delta y &=& f(x_1) - f(x) = \sqrt{0.0001} - \sqrt{0} = 0.01 \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{0.01}{0.0001} = \frac{10^{-2}}{10^{-4}} = 100\end{eqnarray}
- \begin{eqnarray}x_1&=& \Delta x + x = -90000 + 100000 = 10000\\\nonumber \Delta y &=& f(x_1) - f(x) = \log 10^5 - \log 10^4 = 5-4 = 1 \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{1}{10000} = 0.0001\end{eqnarray}
- Example 4 Calculate the increment \(\Delta y\) and quotient \(\frac{\Delta y}{\Delta x}\) that corresponds the change of argument from \(x\) to \(x+\Delta x\) for functions
- \(y= ax + b\)
- \(y = x^3\)
- \(y = \frac{1}{x^2}\)
- \(y = \sqrt{x}\)
- \(y = 2^x\)
- \(y = \ln x\)
Solution- \begin{eqnarray}\Delta y &=& f(x+\Delta x) - f(x) = ax + a\Delta x + b - ax - b = a\Delta x\\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{a\Delta x}{\Delta x} = a \end{eqnarray}
- \begin{eqnarray}\Delta y &=& f(x+\Delta x) - f(x) \\\nonumber \Delta y &=& x^3 + 3x^2 \Delta x + 3x \Delta x^2 + \Delta x^3 - x^3 \\\nonumber \Delta y &=& 3x^2\Delta x + 3x\Delta x^2 + \Delta x^3\\\nonumber \frac{\Delta y }{\Delta x} &=& 3x^2 + 3x\Delta x + \Delta x^2 \end{eqnarray}
- \begin{eqnarray}\Delta y &=& f(x+\Delta x) - f(x) = \frac{1}{x^2 +\Delta x^2 + 2x\Delta x} - \frac{1}{x^2} \\\nonumber \Delta y &=& \frac{x^2 - x^2 - 2x\Delta x - \Delta x^2}{x^2 (x + \Delta x)^2} \\\nonumber \Delta y &=& - \frac{2x\Delta x + \Delta x^2}{x^2 (x+\Delta x)^2}\end{eqnarray}
- \begin{eqnarray}\Delta y &=& f(x+\Delta x) - f(x) = \sqrt{x+\Delta x} - \sqrt{x}\\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{\sqrt{x+\Delta x} - \sqrt{x}}{\Delta x} \cdot \frac{\sqrt{x+\Delta x} + \sqrt{x}}{\sqrt{x+\Delta x} + \sqrt{x}}\\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{1}{\sqrt{x+\Delta x} + \sqrt{x}}\end{eqnarray}
- \begin{eqnarray} \Delta y &=& f(x+\Delta x) - f(x) = 2^{x+\Delta x} - 2^x \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{2^{x+\Delta x} - 2^x}{\Delta x}\end{eqnarray}
- \begin{eqnarray}\Delta y &=& f(x+\Delta x) - f(x) = \ln(x+\Delta x) - \ln x \\\nonumber \Delta y &=& \frac{x+\Delta x}{x} \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{\ln \frac{x+\Delta x}{x}}{\Delta x}\end{eqnarray}
- Example Calculate the quotient \(\frac{\Delta y}{\Delta x}\) for parabola wich is defined by function \(y = 2x- x^2 \) for argument increase from
- \(x_1 = 1\) to \(x_2 = 2\)
- \(x_1 = 1\) to \(x_2 = 0.9\)
- \(x_1 = 1\) to \(x_2 = 1+h\)
Solution- \begin{eqnarray}\Delta x &=& x_2 - x_1 = 2- 1 = 1 \\\nonumber \Delta y &=& f(x_2) - f(x_1) = (2\cdot 2 - 2^2) - (2 \cdot 1 - 1^2 ) = -1 \\\nonumber \frac{\Delta y}{\Delta x} = -1 \end{eqnarray}
- \begin{eqnarray}\Delta x &=& x_2 - x_1 = 0.9 -1 = -0.1 \\\nonumber \Delta y &=& f(x_2) - f(x_1) = 2\cdot \frac{9}{10} - \frac{81}{100} - 2 \cdot 1 + 1^2 \\\nonumber \Delta y &=& \frac{180-81}{100} - 1 = -0.01\\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{0.01}{0.1} = 0.1\end{eqnarray}
- \begin{eqnarray}\Delta x &=& x_2 - x_1 = 1+h -h = h\\\nonumber \Delta y &=& f(x_2) - f(x_1) = 2(1+h) - (1+h)^2 - 2\cdot 1 + 1^2 \\\nonumber \Delta y &=& 2 + 2 h - 1 -2h -h^2 - 2 + 1 = -h^2 \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{63}{3} = 21\end{eqnarray}
- Example 6 Determine the average rate of change of function \(y=x^3\) if \(x\) is in range from 1 to 4.
Solution\begin{eqnarray}\Delta x = x_2 - x_1 = 4- 1 = 3 \\\nonumber \Delta y &=& f(x_2) - f(x_1) = 4^3 - 1^3 = 63 \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{63}{3} = 21.\end{eqnarray} - Example 7 The law of moving point is defined as \(s = 2t^2 + 3t + 5\) where distance s is defined in centimeters and time t in seconds. Determine the avrage speed of point (average rate of change) in time from 1 to 5 s?
Solution\begin{eqnarray}\Delta s &=& s(t_2) - s(t_1) = 2\cdot 5^2 +3 \cdot 5 + 5 - 2 - 3 - 5 \\\nonumber \Delta s &=& 50 + 15 + 5 - 2- 3- 5 = 60 \\\nonumber \frac{\Delta s}{\Delta t} &=& \frac{60}{4} = \frac{30}{2} = 15\end{eqnarray} - Example 8 DEtermine the average rate of change for function \(y=2^x\) on interval \( 1\leq x \leq 5\).
Solution\begin{eqnarray}\Delta y &=& f(x_2) - f(x_1) = 2^5 - 2 = 30 \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{30}{2} = \frac{15}{2} = 7.5 \end{eqnarray} - Example 9 Determine the average rate of change for function y = f(x) on interval \([x, x+\Delta x]\)
Solution\begin{eqnarray}\Delta y&=& f(x_2) - f(x_1) = f(x+\Delta x) - f(x) \\\nonumber \frac{\Delta y}{\Delta x} &=& \frac{f(x+\Delta x)- f(x)}{\Delta x}\end{eqnarray} - Example 10 Find the derivative of function \(y=f(x)\) in given point \(x\).
Solution\begin{eqnarray}\frac{\Delta y}{\Delta x} &=& \frac{f(x+\Delta x)- f(x)}{\Delta x}\\\nonumber y' = f'(x) &=& \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}\end{eqnarray}
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