Explore: Modeling Applications Solutions

Modeling Applications Solutions

Mathematical models of real-life scenarios can be created from tables, graphs, and/or equations. 
The models do not always match the scenario exactly, but they are close representations . 
They can be used to analyze and predict using interpolation and extrapolation .

Example 4

After making a cup of tea at 8 a.m., Rheema decided to measure how quickly it was cooling. She noted two points on an exponential decay graph.

Write an equation to represent the graph of $y = a b^{x} + 55 .$

At what approximate time will the temperature of the tea reach $66 ° F$ ?

Find the rate of change between the given points.

$(12, 178.2), (34, 134) y = a b^{x} + 55 134 = a b^{34} + 55 79 = a b^{34} a = \frac{79}{b^{34}}$

$\begin{array}{rcl} y & = & a b^{x} + 55 \\ y & = & \frac{79}{b^{34}} b^{x} + 55 \\ 178.2 & = & \frac{79}{b^{34}} (b^{12}) + 55 \\ 123.2 & = & \frac{79}{b^{22}} \\ 123.2 b^{22} & = & 79 \\ \sqrt[22]{b^{22}} & = & \sqrt[22]{\frac{79}{123.2}} \\ b & = & 0.98 \\ a = \frac{79}{0 . 98^{34}} = 157.01 \approx 157 \end{array}$

$y = 157 {(0.98)}^{x} + 55$

$\begin{array}{rcl} y & = & 157 {(0.98)}^{x} + 55 \\ 66 & = & 157 {(0.98)}^{x} + 55 \\ 11 & = & 157 {(0.98)}^{x} \\ \frac{11}{157} & = & 0 . 98^{x} \\ \log \frac{11}{157} & = & \log 0 . 98^{x} \\ \log \frac{11}{157} & = & x \log 0.98 \\ x & = & \frac{\log \frac{11}{157}}{\log 0.98} = 131.58 \end{array}$

$131.58 \approx 132 minutes 60 minutes = 1 hour \frac{132}{60} = 2.2 hr$

At approximately 10:12 a.m., the tea reaches $66 ° F$ .

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{178.2 - 134}{12 - 34} = - \frac{44.2}{22} \approx - 2.009$

Note

Q: Does the rate of change in part C represent the changing temperature for the entire graph?

A: No, because the rate of change is linear. It represents the section of the graph between 12 and 34 minutes, but not the entire graph.