The Rate of Change of a function describes how fast the function is changing. We have seen how to calculate the average and instantaneous rate of change for polynomial functions and rational functions. We can also calculate these quantities for trigonometric functions.
The Average Rate of Change represents the slope of a secant line between two points on the function. It is expressed algebraically as:
Where:
The Instantaneous Rate of Change represents the change in the rate at a particular point representing the slope of the tangent at the point. We can estimate the slope of a tangent by calculating the slope of a secant using two points that are very close. It is expressed algebraically as:
Where:
When evaluating \(m_t\), its good to make \(h\) as small as possible. For the following examples, we will set \(h = 0.01\).
Determine the Average Rate of Change for \(f(x) = 4\cos(x) - 6\) within the interval \(0 \leq x \leq \pi\).
First, we can evaluate \(f(x_2)\) by substituting \(\pi\) for \(x\) in the original function:
\(f(\pi) = 4\cos(\pi) - 6\)
\(f(\pi) = 4(-1) - 6\)
\(f(\pi) = -4 - 6\)
\(\textcolor{blue}{f(\pi) = -10}\)
Next, we can evaluate \(f(x_1)\) by substituting \(0\) for \(x\) in the original function:
\(f(0) = 4\cos(0) - 6\)
\(f(0) = 4(1) - 6\)
\(f(0) = 4 - 6\)
\(\textcolor{red}{f(0) = -2}\)
Finally, we can substitute all pertinent values into the secant formula to determine its slope at these points:
Therefore, we can determine that the Average Rate of Change for \(f(x)\) within the interval \(0 \le x \le \pi\) is \(\boldsymbol{\approx -2.55}\).
Determine the Instantaneous Rate of Change for \(g(x) = 2.5\sin(5x)\) at \(x = 8\).
Next, we can evaluate \(g(x + h)\) by substituting \(8.01\) for \(x\):
\(g(8.01) = 2.5\sin[5(8.01)]\)
\(g(8.01) = 2.5\sin(40.05)\)
\(g(8.01) = 2.5(0.710848952)\)
\(\textcolor{blue}{g(8.01) = 1.777122382}\)
Next, we can evaluate \(g(x)\) by substituting \(8\) for \(x\):
\(g(8) = 2.5\sin[5(8)]\)
\(g(8) = 2.5\sin(40)\)
\(g(8) = 2.5(0.74511316)\)
\(\textcolor{red}{g(8) = 1.862782901}\)
Finally, we can substitute all pertinent values into the tangent formula to determine its slope at these points:
Therefore, we can determine that the Instantaneous Rate of Change for \(g(x)\) at \(x = 8\) is \(\boldsymbol{\approx -8.6}\).
First, we can evaluate \(h(x_2)\) by substituting \(5\) for \(x\) in the original function:
\(h(5) = -3.2\tan(5) + 7\)
\(h(5) = -3.2(-3.380515006) + 7)\)
\(h(5) = 10.81764802 + 7\)
\(\textcolor{blue}{h(5) = 17.81764802}\)
Next, we can evaluate \(h(x_1)\) by substituting \(2\) for \(x\) in the original function:
\(h(2) = -3.2\tan(2) + 7\)
\(h(2) = -3.2(-2.185039863) + 7\)
\(h(2) = 6.992127562 + 7\)
\(\textcolor{red}{h(2) = 13.99212756}\)
Finally, we can substitute all pertinent values into the secant formula to determine its slope at these points:
Therefore, we can determine that the Average Rate of Change for \(h(x)\) within the interval \(2 \le x \le 5\) is \(\boldsymbol{\approx 1.28}\).
First, we can evaluate \(i(x + h)\) by substituting \(3.01\) for \(x\):
\(i(3.01) = 10\cos\left[\cfrac{\pi}{0.2}(3.01)\right]\)
\(i(3.01) = 10\cos(47.28096944)\)
\(i(3.01) = 10(-0.98768834)\)
\(\textcolor{blue}{i(3.01) = -9.876883406}\)
Next, we can evaluate \(i(x)\) by substituting \(3\) for \(x\):
\(i(3) = 10\cos\left[\cfrac{\pi}{0.2}(3)\right]\)
\(i(3) = 10\cos(47.1238898)\)
\(i(3) = 10(-1)\)
\(\textcolor{red}{i(3) = -10}\)
Finally, we can substitute all pertinent values into the tangent formula to determine its slope at these points:
Therefore, we can determine that the Instantaneous Rate of Change for \(i(x)\) at \(x = 3\) is \(\boldsymbol{\approx 12.3}\).