91Ó°ÊÓ

The model $$t=16.625 \ln \left(\frac{x}{x-750}\right), \quad x>750$$ approximates the length of a home mortgage of \(\$ 150,000\) at \(6 \%\) in terms of the monthly payment. In the model, \(t\) is the length of the mortgage in years and \(x\) is the monthly payment in dollars. (a) Use the model to approximate the lengths of a \(\$ 150,000\) mortgage at \(6 \%\) when the monthly payment is \(\$ 897.72\) and when the monthly payment is \(\$ 1659.24\) (b) Approximate the total amounts paid over the term of the mortgage with a monthly payment of \(\$ 897.72\) and with a monthly payment of \(\$ 1659.24 .\) (c) Approximate the total interest charges for a monthly payment of \(\$ 897.72\) and for a monthly payment of \(\$ 1659.24\) (d) What is the vertical asymptote for the model? Interpret its meaning in the context of the problem.

Four-legged animals run with two different types of motion: trotting and galloping. An animal that is trotting has at least one foot on the ground at all times, whereas an animal that is galloping has all four feet off the ground at some point in its stride. The number of strides per minute at which an animal breaks from a trot to a gallop depends on the weight of the animal. Use the table to find a logarithmic equation that relates an animal's weight \(x\) (in pounds) and its lowest galloping speed \(y\) (in strides per minute). $$ \begin{array}{|c|c|} \hline \text { Weight, } x & \text { Galloping speed, } y \\ \hline 25 & 191.5 \\ 35 & 182.7 \\ 50 & 173.8 \\ 75 & 164.2 \\ 500 & 125.9 \\ 1000 & 114.2 \\ \hline \end{array} $$

A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C}\). The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). \(\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)\) \(\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right)\) (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\). (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use a graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form \(\ln (T-21)=a t+b\). Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points \(\left(t, \frac{1}{T-21}\right)\) Use a graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of a graphing utility to fit a line to these data. The resulting line has the form \(\frac{1}{T-21}=a t+b\) Solve for \(T,\) and use a graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

The values \(y\) (in billions of dollars) of U.S. currency in circulation in the years \(\begin{array}{lllll}2000 & \text { through } 2007 & \text { can be } & \text { modeled } & \text { by }\end{array}\) \(y=-451+444 \ln t, 10 \leq t \leq 17,\) where \(t\) represents the year, with \(t=10\) corresponding to 2000 . During which year did the value of U.S. currency in circulation exceed \$690 billion? (Source: Board of Governors of the Federal Reserve System)

Automobiles are designed with crumple zones that help protect their occupants in crashes. The crumple zones allow the occupants to move short distances when the automobiles come to abrupt stops. The greater the distance moved, the fewer g's the crash victims experience. (One \(g\) is equal to the acceleration due to gravity. For very short periods of time, humans have withstood as much as 40 g's.) In crash tests with vehicles moving at 90 kilometers per hour, analysts measured the numbers of g's experienced during deceleration by crash dummies that were permitted to move \(x\) meters during impact. The data are shown in the table. A model for the data is given by \(y=-3.00+11.88 \ln x+(36.94 / x),\) where \(y\) is the number of g's. $$ \begin{array}{|c|c|} \hline x & \text { g's } \\ \hline 0.2 & 158 \\ 0.4 & 80 \\ 0.6 & 53 \\ 0.8 & 40 \\ 1.0 & 32 \\ \hline \end{array} $$ (a) Complete the table using the model. $$ \begin{array}{|l|l|l|l|l|l|} \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline y & & & & & \\ \hline \end{array} $$ (b) Use a graphing utility to graph the data points and the model in the same viewing window. How do they compare? (c) Use the model to estimate the distance traveled during impact if the passenger deceleration must not exceed \(30 \mathrm{~g}\) 's. (d) Do you think it is practical to lower the number of g's experienced during impact to fewer than \(23 ?\) Explain your reasoning.